Draft of solutions to exercises in chapter of An introduction to game theory by Martin J. Osborne

[email protected]; www.chass.utoronto.ca/~osborne/index.html Version: 00/11/6. c 1995–2000 by Martin J. Osborne. All rights reserved. No part of this book may be reCopyright produced in any form by any electronic or mechanical means (including photocopying, recording, or information storage and retrieval) without permission in writing from Martin J. Osborne. On request, permission to make one copy for each student will be granted to instructors who wish to use the book in a course, on condition that copies be sold at a price not more than the cost of duplication. 1 Introduction 5.3 Altruistic preferences Person 1 is indifferent between (1, 4) and (3, 0), and prefers both of these to (2, 1). Any function that assigns the same number to (1, 4) and to (3, 0), and a lower number to (2, 1) is a payoff function that represents her preferences. 6.1 Alternative representations of preferences The function v represents the same preferences as does u (since u(a) < u(b) < u(c) and v(a) < v(b) < v(c)), but the function w does not represent the same preferences, since w(a) = w(b) while u(a) < u(b). 1 Draft of solutions to exercises in chapter of An introduction to game theory by Martin J. Osborne

[email protected]; www.chass.utoronto.ca/~osborne/index.html Version: 00/11/6. c 1995–2000 by Martin J. Osborne. All rights reserved. No part of this book may be reCopyright produced in any form by any electronic or mechanical means (including photocopying, recording, or information storage and retrieval) without permission in writing from Martin J. Osborne. On request, permission to make one copy for each student will be granted to instructors who wish to use the book in a course, on condition that copies be sold at a price not more than the cost of duplication. 2 Nash Equilibrium 14.1 Working on a joint project The game in Figure 3.1 models this situation (as does any other game with the same players and actions in which the ordering of the payoffs is the same as the ordering in Figure 3.1). Work hard Goof off Work hard 3, 3 2, 0 Goof off 0, 2 1, 1 Figure 3.1 Working on a joint project (alternative version). 16.1 Hermaphroditic fish A strategic game that models the situation is shown in Figure 3.2. Either role Preferred role 1 2 (H Either role + L), 12 (H + L) H, L Preferred role L, H S, S Figure 3.2 A model of encounters between pairs of hermaphroditic fish whose preferred roles differ. In order for this game to differ from the Prisoner’s Dilemma only in the names of the players’ actions, there must be a way to associate each action with an action in the Prisoner’s Dilemma so that each player’s preferences over the four outcomes are the same as they are in the Prisoner’s Dilemma. Thus we need L < S < 12 (H + L). That is, the probability of a fish’s encountering a potential partner must be large enough that S > L, but small enough that S < 12 (H + L). 17.2 Games without conflict Any two-player game in which each player has two actions and the players have the same preferences may be represented by a table of the form given in Figure 4.1, where a, b, c, and d are any numbers. 3 4 Chapter 2. Nash Equilibrium T B L a, a c, c R b, b d, d Figure 4.1 A strategic game in which conflict is absent. 25.1 Altruistic players in the Prisoner’s Dilemma a. A game that model the situation is given in Figure 4.2. Quiet Fink Quiet 4, 4 3, 3 Fink 3, 3 2, 2 Figure 4.2 The payoffs in a variant of the Prisoner’s Dilemma in which the players are altruistic. This game is not the Prisoner’s Dilemma because one (in fact both) of the players’ preferences are not the same as they are in the Prisoner’s Dilemma. Specifically, player 1 prefers (Quiet, Quiet) to (Fink, Quiet), while in the Prisoner’s Dilemma she prefers (Fink, Quiet) to (Quiet, Quiet). (Alternatively, you may note that player 1 prefers (Quiet, Fink) to (Fink, Fink), while in the Prisoner’s Dilemma she prefers (Fink, Fink) to (Quiet, Fink), or that player 2’s preferences are similarly not the same as they are in the Prisoner’s Dilemma.) b. For an arbitrary value of α the payoffs are given in Figure 4.3. In order that the game be the Prisoner’s Dilemma we need 3 > 2(1 + α) (each player prefers Fink to Quiet when the other player chooses Quiet), 1 + α > 3α (each player prefers Fink to Quiet when the other player choose Fink), and 2(1 + α) > 1 + α (each player prefers (Quiet, Quiet) to (Fink, Fink)). The last condition is satisfied for all nonnegative values of α. The first two conditions are both equivalent to α < 12 . Thus the game is the Prisoner’s Dilemma if and only if α < 12 . If α = 12 then all four outcomes (Quiet, Quiet), (Quiet, Fink), (Fink, Quiet), and (Fink, Fink) are Nash equilibria; if α > 12 then only (Quiet, Quiet) is a Nash equilibrium. Quiet Fink Quiet 2(1 + α), 2(1 + α) 3, 3α Fink 3α, 3 1 + α, 1 + α Figure 4.3 The payoffs in a variant of the Prisoner’s Dilemma in which the players are altruistic. Chapter 2. Nash Equilibrium 5 25.2 Selfish and altruistic social behavior a. A game that model the situation is shown in Figure 5.1. Sit Stand Sit 1, 1 0, 2 Stand 2, 0 0, 0 Figure 5.1 Behavior on a bus when the players’ preferences are selfish (Exercise 25.2). This game is not the Prisoner’s Dilemma. If we identify Sit with Quiet and Stand with Fink then, for example, (Stand, Sit) is worse for player 1 than (Sit, Sit), rather than better. If we identify Sit with Fink and Stand with Quiet then, for example, (Stand, Stand) is worse for player 1 than (Sit, Sit), rather than better. The game has a unique Nash equilibrium, (Sit, Sit). b. A game that models the situation is shown in Figure 5.2, where α is some positive number. Sit Stand Sit 1, 1 2, 0 Stand 0, 2 α, α Figure 5.2 Behavior on a bus when the players’ preferences are selfish (Exercise 25.2). If α < 1 then this game is the Prisoner’s Dilemma. It has a unique Nash equilibrium, (Stand, Stand) (regardless of the value of α). c. Both people are more comfortable in the equilibrium that results when they act according to their selfish preferences. 28.1 Variants of the Stag Hunt a. The equilibria of the game are the same as those of the original game: (Stag, . . . , Stag) and (Hare, . . . , Hare). Any player that deviates from the first profile obtains a hare rather than the fraction 1/n of the stag. Any player that deviates from the second profile obtains nothing, rather than a hare. An action profile in which at least 1 and at most m − 1 hunters pursue the stag is not a Nash equilibrium, since any one of them is better off catching a hare. An action profile in which at least m and at most n − 1 hunters pursue the stag is not a Nash equilibrium, since any one of the remaining hunters is better off joining the pursuit of the stag (thereby earning herself the right to a share of the stag). 6 Chapter 2. Nash Equilibrium b. The set of Nash equilibria consists of the action profile (Hare, . . . , Hare) in which all hunters catch hares, and any action profile in which exactly k hunters pursue the stag and the remaining hunters catch hares. Any player that deviates from the first profile obtains nothing, rather than a hare. A player who switches from the pursuit of the stag to catching a hare in the second type of profile is worse off, since she obtains a hare rather than the fraction 1/k of the stag; a player who switches from catching a hare to pursuing the stag is also worse off since she obtains the fraction 1/(k + 1) of the stag rather than a hare, and 1/(k + 1) < 1/k. No other action profile is a Nash equilibrium, by the following argument. • If some hunters, but fewer than m, pursue the stag then each of them obtains nothing, and is better off catching a hare. • If at least m and fewer than k hunters pursue the stag then each one that pursues a hare is better off switching to the pursuit of the stag. • If more than k hunters pursue the stag then the fraction of the stag that each of them obtains is less than 1/k, so each of them is better off catching a hare. 28.2 Extension of the Stag Hunt Every profile (e, . . . , e), where e is an integer from 0 to K, is a Nash equilibrium. In the equilibrium (e, . . . , e), each player’s payoff is e. The profile (e, . . . , e) is a Nash equilibrium since if player i chooses ei < e then her payoff is 2ei − ei = ei < e, and if she chooses ei > e then her payoff is 2e − ei < e. Consider an action profile (e1 , . . . , en ) in which not all effort levels are the same. Suppose that ei is the minimum. Consider some player j whose effort level exceeds ei . Her payoff is 2ei − e j < ei , while if she deviates to the effort level ei her payoff is 2ei − ei = ei . Thus she can increase her payoff by deviating, so that (e1 , . . . , en ) is not a Nash equilibrium. (This game is studied experimentally by van Huyck, Battalio, and Beil (1990). See also Ochs (1995, 209–233).) 29.1 Hawk–Dove A strategic game that models the situation is shown in Figure 6.1. The game has two Nash equilibria, (Aggressive, Passive) and (Passive, Aggressive). Aggressive Passive Aggressive 0, 0 1, 3 Passive 3, 1 2, 2 Figure 6.1 Hawk–Dove. Chapter 2. Nash Equilibrium 7 31.1 Contributing to a public good The following game models the situation. Players The n people. Actions Each person’s set of actions is {Contribute, Don’t contribute}. Preferences Each person’s preferences are those given in the problem. An action profile in which more than k people contribute is not a Nash equilibrium: any contributor can induce an outcome she prefers by deviating to not contributing. An action profile in which k people contribute is a Nash equilibrium: if any contributor stops contributing then the good is not provided; if any noncontributor switches to contributing then she is worse off. An action profile in which fewer than k people contribute is a Nash equilibrium only if no one contributes: if someone contributes, she can increase her payoff by switching to noncontribution. In summary, the set of Nash equilibria is the set of action profiles in which k people contribute together with the action profile in which no one contributes. 32.1 Guessing two-thirds of the average If all three players announce the same integer k ≥ 2 then any one of them can deviate to k − 1 and obtain $1 (since her number is now closer to 23 of the average than the other two) rather than $ 13 . Thus no such action profile is a Nash equilibrium. If all three players announce 1, then no player can deviate and increase her payoff; thus (1, 1, 1) is a Nash equilibrium. Now consider an action profile in which not all three integers are the same; denote the highest by k∗ . • Suppose only one player names k∗ ; denote the other integers named by k 1 and k 2 , with k 1 ≥ k 2 . The average of the three integers is 13 (k∗ + k 1 + k 2 ), so that 23 of the average is 29 (k∗ + k 1 + k 2 ). If k 1 ≥ 29 (k∗ + k 1 + k 2 ) then k∗ is further from 23 of the average than is k 1 , and hence does not win. If k 1 < 2 ∗ 2 ∗ ∗ 9 (k + k 1 + k 2 ) then the difference between k and 3 of the average is k − 2 ∗ 7 ∗ 2 2 2 9 (k + k 1 + k 2 ) = 9 k − 9 k 1 − 9 k 2 , while the difference between k 1 and 3 2 ∗ 2 ∗ 7 2 of the average is 9 (k + k 1 + k 2 ) − k 1 = 9 k − 9 k 1 + 9 k 2 . The difference between the former and the latter is 59 k∗ + 59 k 1 − 49 k 2 > 0, so k 1 is closer to 23 of the average than is k∗ . Hence the player who names k∗ does not win, and is better off naming k 2 , in which case she obtains a share of the prize. Thus no such action profile is a Nash equilibrium. • Suppose two players name k∗ , and the third player names k < k∗ . The average of the three integers is then 13 (2k∗ + k), so that 23 of the average is 8 Chapter 2. Nash Equilibrium 4 ∗ 9k + 29 k. We have 49 k∗ + 29 k < 12 (k∗ + k) (since 49 < 12 and 29 < 12 ), so that the player who names k is the sole winner. Thus either of the other players can switch to naming k and obtain a share of the prize rather obtaining nothing. Thus no such action profile is a Nash equilibrium. We conclude that there is only one Nash equilibrium of this game, in which all three players announce the number 1. (This game is studied experimentally by Nagel (1995).) 32.2 Voter participation a. For k = m = 1 the game is shown in Figure 8.1. It is the same, except for the names of the actions, as the Prisoner’s Dilemma. A supporter abstain vote B supporter abstain vote 1, 1 0, 2 − c 2 − c, 0 1 − c, 1 − c Figure 8.1 The game of voter participation in Exercise 32.2. b. For k = m, denote the number of citizens voting for A by n A and the number voting for B by n B . The cases in which n A ≤ n B are symmetric with those in which n A ≥ n B ; I restrict attention to the latter. n A = n B = k (all citizens vote): A citizen who switches from voting to abstaining causes the candidate she supports to lose rather than tie, reducing her payoff from 1 − c to 0. Since c < 1, this situation is a Nash equilibrium. n A = n B < k (not all citizens vote; the candidates tie): A citizen who switches from abstaining to voting causes the candidate she supports to win rather than tie, increasing her payoff from 1 to 2 − c. Thus this situation is not a Nash equilibrium. n A = n B + 1 or n B = n A + 1 (a candidate wins by one vote): A supporter of the losing candidate who switches from abstaining to voting causes the candidate she supports to tie rather than lose, increasing her payoff from 0 to 1 − c. Thus this situation is not a Nash equilibrium. n A ≥ n B + 2 or n B ≥ n A + 2 (a candidate wins by two or more votes): A supporter of the winning candidate who switches from voting to abstaining does not affect the outcome, so such a situation is not a Nash equilibrium. We conclude that the game has a unique Nash equilibrium, in which all citizens vote. Chapter 2. Nash Equilibrium 9 c. If k < m then a similar logic shows that there is no Nash equilibrium. n A = n B ≤ k: A supporter of B who switches from abstaining to voting causes B to win rather than tie, increasing her payoff from 1 to 2 − c. Thus this situation is not a Nash equilibrium. n A = n B + 1 or n B = n A + 1: A supporter of the losing candidate who switches from abstaining to voting causes the candidates to tie, increasing her payoff from 0 to 1 − c. Thus this situation is not a Nash equilibrium. n A ≥ n B + 2 or n B ≥ n A + 2: A supporter of the winning candidate who switches from voting to abstaining does not affect the outcome, so such a situation is not a Nash equilibrium. 32.3 Choosing a route A strategic game that models this situation is: Players The four people. Actions The set of actions of each person is {X, Y} (the route via X and the route via Y). Preferences Each player’s payoff is the negative of her travel time. In every Nash equilibrium, two people take each route. (In any other case, a person taking the more popular route is better off switching to the other route.) For any such action profile, each person’s travel time is either 29.9 or 30 minutes (depending on the route they take). If a person taking the route via X switches to the route via Y her travel time becomes 12 + 21.8 = 33.8 minutes; if a person taking the route via Y switches to the route via X her travel time becomes 22 + 12 = 34 minutes. For any other allocation of people to routes, at least one person can decrease her travel time by switching routes. Thus the set of Nash equilibria is the set of action profiles in which two people take the route via X and two people take the route via Y. Now consider the situation after the road from X to Y is built. There is no equilibrium in which the new road is not used, by the following argument. Because the only equilibrium before the new road is built has two people taking each route, the only possibility for an equilibrium in which no one uses the new road is for two people to take the route A–X–B and two to take A–Y–B, resulting in a total travel time for each person of either 29.9 or 30 minutes. However, if a person taking A– X–B switches to the new road at X and then takes Y–B her total travel time becomes 9 + 7 + 12 = 28 minutes. I claim that in any Nash equilibrium, one person takes A–X–B, two people take A–X–Y–B, and one person takes A–Y–B. For this assignment, each person’s travel time is 32 minutes. No person can change her route and decrease her travel time, by the following argument. 10 Chapter 2. Nash Equilibrium • If the person taking A–X–B switches to A–X–Y–B, her travel time increases to 12 + 9 + 15 = 36 minutes; if she switches to A–Y–B her travel time increases to 21 + 15 = 36 minutes. • If one of the people taking A–X–Y–B switches to A–X–B, her travel time increases to 12 + 20.9 = 32.9 minutes; if she switches to A–Y–B her travel time increases to 21 + 12 = 33 minutes. • If the person taking A–Y–B switches to A–X–B, her travel time increases to 15 + 20.9 = 35.9 minutes; if she switches to A–X–Y–B, her travel time increases to 15 + 9 + 12 = 36 minutes. For every other allocation of people to routes at least one person can switch routes and reduce her travel time. For example, if one person takes A–X–B, one person takes A–X–Y–B, and two people take A–Y–B, then the travel time of those taking A–Y–B is 21 + 12 = 33 minutes; if one of them switches to A–X–B then her travel time falls to 12 + 20.9 = 32.9 minutes. Or if one person takes A–Y–B, one person takes A–X–Y–B, and two people take A–X–B, then the travel time of those taking A–X–B is 12 + 20.9 = 32.9 minutes; if one of them switches to A–X–Y–B then her travel time falls to 12 + 8 + 12 = 32 minutes. Thus in the equilibrium with the new road every person’s travel time increases, from either 29.9 or 30 minutes to 32 minutes. 35.1 Finding Nash equilibria using best response functions a. The Prisoner’s Dilemma and BoS are shown in Figure 10.1; Matching Pennies and the two-player Stag Hunt are shown in Figure 10.2. Quiet 2 ,2 3∗ , 0 Quiet Fink Fink 0 , 3∗ 1∗ , 1∗ Bach 2∗ , 1∗ 0 ,0 Bach Stravinsky Prisoner’s Dilemma Stravinsky 0 ,0 1∗ , 2∗ BoS Figure 10.1 The best response functions in the Prisoner’s Dilemma (left) and in BoS (right). Head Tail Head 1∗ , −1 −1 , 1∗ Tail −1 , 1 ∗ 1∗ , −1 Matching Pennies Stag Hare Stag 2∗ , 2∗ 1 ,0 Hare 0 ,1 1∗ , 1∗ Stag Hunt Figure 10.2 The best response functions in Matching Pennies (left) and the Stag Hunt (right). b. The best response functions are indicated in Figure 11.1. The Nash equilibria are (T, C), (M, L), and (B, R). Chapter 2. Nash Equilibrium 11 T M B L 2 ,2 3∗ , 1∗ 1 , 0∗ C 1∗ , 3∗ 0 ,0 0 , 0∗ R 0∗ , 1 0∗ , 0 0∗ , 0∗ Figure 11.1 The game in Exercise 35.1. 36.1 Constructing best response functions The analogue of Figure 36.2 is given in Figure 11.2. R A2 C L T M A1 B Figure 11.2 The players’ best response functions for the game in Exercise 36.1b. Player 1’s best responses are indicated by circles, and player 2’s by dots. The action pairs for which there is both a circle and a dot are the Nash equilibria. 36.2 Dividing money For each amount named by one of the players, the other player’s best responses are given in the following table. Other player’s action 0 1 2 3 4 5 6 7 8 9 10 Sets of best responses {10} {9, 10} {8, 9, 10} {7, 8, 9, 10} {6, 7, 8, 9, 10} {5, 6, 7, 8, 9, 10} {5, 6} {6} {7} {8} {9} 12 Chapter 2. Nash Equilibrium The best response functions are illustrated in Figure 12.1 (circles for player 1, dots for player 2). From this figure we see that the game has four Nash equilibria: (5, 5), (5, 6), (6, 5), and (6, 6). 10 9 8 7 6 A2 5 4 3 2 1 0 5 6 7 8 9 10 0 1 2 3 4 A1 Figure 12.1 The players’ best response functions for the game in Exercise 36.2. 39.1 Strict and nonstrict Nash equilibria Only the Nash equilibrium (a∗1 , a∗2 ) is strict. For each of the other equilibria, player 2’s ≤ a2 ≤ a∗∗ action a2 satisfies a∗∗∗ 2 2 , and for each such action player 1 has multiple best responses, so that her payoff is the same for a range of actions, only one of which is such that (a1 , a2 ) is a Nash equilibrium. 40.1 Finding Nash equilibria using best response functions First find the best response function of player 1. For any fixed value of a2 , player 1’s payoff function a1 (a2 − a1 ) is a quadratic in a1 . The coefficient of a21 is negative and the function is zero at a1 = 0 and at a1 = a2 . Thus, using the symmetry of quadratic functions, b1 (a2 ) = 12 a2 . Now find the best response function of player 2. For any fixed value of a1 , player 2’s payoff function a2 (1 − a1 − a2 ) is a quadratic in a2 . The coefficient on a22 is negative and the function is zero at a2 = 0 and at a2 = 1 − a1 . Thus if a1 ≤ 1 we have b2 (a1 ) = 12 (1 − a1 ) and if a1 > 1 we have b2 (a1 ) = 0. The best response functions are shown in Figure 13.1. A Nash equilibrium is a pair (a∗1 , a∗2 ) such that a∗1 = b1 (a∗2 ) and a∗2 = b2 (a∗1 ). From the figure we see that there is a unique Nash equilibrium, with a∗1 < 1. Thus Chapter 2. Nash Equilibrium 13 ↑ a2 b1 (a2 ) 1 2∗ a2 0 b2 (a1 ) 1 a1 → a∗1 Figure 13.1 The best response functions for the game in Exercise 40.1. in this equilibrium a∗1 = 12 a∗2 and a∗2 = 12 (1 − a∗1 ). Hence a∗1 = 14 (1 − a∗1 ), or 5a∗1 = 1, or a∗1 = 15 . Hence a∗2 = 25 . Thus the game has a unique Nash equilibrium, ( 15 , 25 ). 40.2 A joint project A strategic game that models this situation is: Players The two people. Actions The set of actions of each person i is the set of effort levels (the set of numbers xi with 0 ≤ xi ≤ 1). Preferences Person i’s payoff to the action pair (x1 , x2 ) is 1 2 f (x1 , x2 ) − c(xi ). a. Assume that f (x1 , x2 ) = 3x 1 x2 and c(xi ) = xi2 . To find the Nash equilibria of the game, first find the players’ best response functions. Player 1’s best response to x2 is the action x1 that maximizes 32 x1 x2 − x 12 , or x1 ( 32 x2 − x1 ). This function is a quadratic that is zero when x1 = 0 and when x 1 = 32 x2 . The coefficient of x12 is negative, so the maximum of the function occurs at x1 = 34 x2 . Thus player 1’s best response function is b1 (x2 ) = 34 x2 . Similarly, player 2’s best response function is b2 (x1 ) = 34 x1 . The best response functions are shown in Figure 14.1. In a Nash equilibrium (x1∗ , x2∗ ) we have x1∗ = b1 (x2∗ ) and x 2∗ = b2 (x1∗ ), or x1∗ = 3 ∗ 3 ∗ 9 ∗ ∗ ∗ ∗ 4 x 2 and x 2 = 4 x 1 . Substituting x 2 in the first equation we obtain x 1 = 16 x 1 , so that x1∗ = 0. Thus x2∗ = 0. We conclude that the game has a unique Nash equilibrium, (x1∗ , x 2∗ ) = (0, 0). In this equilibrium, both players’ payoffs are zero. If each player i chooses xi = 1 then the total output is 3, and each player’s payoff is 32 − 1 = 12 , rather than 0 as in the Nash equilibrium. 14 Chapter 2. Nash Equilibrium 1 ↑ x2 b1 (x2 ) b2 (x1 ) x1 → 1 0 Figure 14.1 The best response functions for the game in Exercise 40.2a. b. When f (x1 , x2 ) = 4x 1 x2 and c(xi ) = xi , player 1’s payoff function is 2x1 x2 − x1 = x1 (2x 2 − 1). Thus if x2 < 12 her best response is x1 = 0, if x2 = 12 then all values of x1 are best responses, and if x2 > 12 her best response is x1 = 1. That is, player 1’s best response function is if x 2 < 12 0 b1 (x2 ) = {x 1 : 0 ≤ x1 ≤ 1} if x2 = 12 1 if x2 > 12 . Player 2’s best response function is the same. (That is, b2 (x) = b1 (x) for all x.) The best response functions are shown in Figure 14.2. 1 ↑ x2 b2 (x1 ) b1 (x2 ) 0 x1 → 1 Figure 14.2 The best response functions for the game in Exercise 40.2b. We see that the game has three Nash equilibria, (0, 0), ( 12 , 12 ), and (1, 1). The players’ payoffs at these equilibria are (0, 0), (0, 0), and (1, 1). There is no pair of effort levels that yields both players payoffs higher than 1, but there are pairs of effort levels that yield both players payoffs higher than 0, for example (1, 1), which yields the payoffs (1, 1). Chapter 2. Nash Equilibrium 15 42.1 Contributing to a public good The best response of player 1 to the contribution c2 of player 2 is the value of c1 that maximizes player 1’s payoff w + c2 + (w − c1 )(c1 + c2 ). This function is a quadratic in c1 (remember that w + c2 is a constant). The coefficient of c21 is negative, and the value of the function is equal to w + c2 when c1 = w and when c1 = −c2 . Thus the function attains a maximum at c1 = 12 (w − c2 ). We conclude that player 1’s best response function is b1 (c2 ) = 12 (w − c2 ). Player 2’s best response function is similarly b2 (c1 ) = 12 (w − c1 ). A Nash equilibrium is a pair (c∗1 , c∗2 ) such that c∗1 = b1 (c∗2 ) and c∗2 = b2 (c∗1 ), so that c∗1 = 12 (w − c∗2 ) = 12 (w − 12 (w − c∗1 )) = 14 w + 14 c∗1 and hence c∗1 = 13 w. Substituting this value into player 2’s best response function we get c∗2 = 13 w. We conclude that the game has a unique Nash equilibrium (c∗1 , c∗2 ) = ( 13 w, 13 w), in which each person contributes one third of her wealth to the public good. In this equilibrium each player’s payoff is 43 w + 49 w2 . If each player contributes 1 3 1 2 4 4 2 2 w to the public good then her payoff is 2 w + 2 w , which exceeds 3 w + 9 w for all w (since 32 > 43 and 12 > 49 ). When there are n players the payoff function of player 1 is w − c1 + c1 + c2 + · · · + cn + (w − c1 )(c1 + c2 + · · · + cn ) = w + c2 + · · · + cn + (w − c1 )(c1 + c2 + · · · + cn ). This function is a quadratic in c1 . The coefficient of c21 is negative, and the value of the function is equal to w + c2 + · · · + cn when c1 = w and when c1 = −c2 − c3 − · · · − cn . Thus the function attains a maximum at c1 = 12 (w − c2 − c3 − · · · − cn ). We conclude that player 1’s best response function is b1 (c−1 ) = 12 (w − c2 − c3 − · · · − cn ) where c−1 is the list of the contributions of the players other than 1. Similarly, any player i’s best response function is bi (c−i ) = 12 (w − (c1 + c2 + · · · + cn ) + ci ). A Nash equilibrium is an action profile (c∗1 , . . . , c∗n ) such that c∗i = bi (c∗−i ) for all i. We can write the condition c∗1 = b1 (c∗−1 ) as 2c∗1 = w − c∗2 − c∗3 − · · · − c∗n , or w = 2c∗1 + c∗2 + c∗3 + · · · + c∗n . 16 Chapter 2. Nash Equilibrium Writing the other conditions c∗i = bi (c∗−i ) similarly, we obtain the system of equations w = 2c∗1 + c∗2 + c∗3 + · · · + c∗n w = .. . c∗1 + 2c∗2 + c∗3 + · · · + c∗n w = c∗1 + c∗2 + c∗3 + · · · + 2c∗n Subtracting the second equation from the first we conclude that c∗1 = c∗2 . Similarly subtracting each equation from the next we deduce that c∗i is the same for all i. Denote the common value by c∗ . From any of the equations we deduce that w = (n + 1)c∗ . Hence c∗ = w/(n + 1). In conclusion, when there are n players the game has a unique Nash equilibrium (c∗1 , . . . , c∗n ) = (w/(n + 1), . . . , w/(n + 1)). The total amount contributed in this equilibrium is nw/(n + 1), which increases as n increases, approaching w as n increases without bound. Player 1’s payoff in the equilibrium is w + (n − 1)w/(n + 1)+(nw/(n + 1))2 . As n increases without bound, this payoff increases, approaching 2w + w2 . If each player contributes 12 w to the public good, each player’s payoff is w + 12 (n − 1)w + n(w/2)2 , which increases without bound as n increases without bound. 45.2 Strict equilibria and dominated actions For player 1, T is weakly dominated by M, and strictly dominated by B. For player 2, no action is weakly or strictly dominated. The game has a unique Nash equilibrium, (M, L). This equilibrium is not strict. (When player 2 choose L, B yields player 1 the same payoff as does M.) 46.1 Nash equilibrium and weakly dominated actions The only Nash equilibrium of the game in Figure 16.1 is (T, L). The action T is weakly dominated by M and the action L is weakly dominated by C. (There are of course many other games that satisfy the conditions.) T M B L 1, 1 1, 0 0, 0 C 0, 1 2, 1 1, 1 R 0, 0 1, 2 2, 0 Figure 16.1 A game with a unique Nash equilibrium, in which both players’ equilibrium actions are weakly dominated. (The unique Nash equilibrium is (T, L).) Chapter 2. Nash Equilibrium 17 47.1 Voting First consider an action profile in which the winner receives one more vote than the loser and at least one citizen who votes for the winner prefers the loser to the winner. Any citizen who votes for the winner and prefers the loser to the winner can, by switching her vote, cause her favorite candidate to win rather than lose. Thus no such action profile is a Nash equilibrium. Next consider an action profile in which the winner receives one more vote than the loser and all citizens who vote for the winner prefer the winner to the loser. Because a majority of citizens prefer A to B, the winner in any such case must be A. No citizen who prefers A to B can induce a better outcome by changing her vote, since her favorite candidate wins. Now consider a citizen who prefers B to A. By assumption, every such citizen votes for B; a change in her vote has no effect on the outcome (A still wins). Thus every such action profile is a Nash equilibrium. Finally consider an action profile in which the winner receives at least three more votes than the loser. In this case no change in any citizen’s vote has any effect on the outcome. Thus every such profile is a Nash equilibrium. In summary, the Nash equilibria are: any action profile in which A receives one more vote than B and all the citizens who vote for A prefer A to B, and any action profile in which the winner receives at least three more votes than the loser. The only equilibrium in which no citizen uses a weakly dominated action is that in which every citizen votes for her favorite candidate. 47.2 Voting between three candidates Fix some citizen, say i; suppose she prefers A to B to C. By the argument in the text, citizen i’s voting for C is weakly dominated by her voting for A (and by her voting for B). Her voting for B is clearly not weakly dominated by her voting for C. I now argue that her voting for B is not weakly dominated by her voting for A. Suppose that the other citizens’ votes are equally divided between B and C; no one votes for A. Then if citizen i votes for A the outcome is a tie between B and C, while if she votes for B the outcome is that B wins. Thus for this configuration of the other citizens’ votes, citizen i is better off voting for B than she is voting for A. Thus her voting for B is not weakly dominated by her voting for A. Now fix some citizen, say i, and consider the candidate she ranks in the middle, say candidate B. The action profile in which all citizens vote for B is a Nash equilibrium. (No citizen’s changing her vote affects the outcome.) In this equilibrium, citizen i does not vote for her favorite candidate, but the action she takes is not weakly dominated. (Other Nash equilibria also satisfy the conditions in the exercise.) 18 Chapter 2. Nash Equilibrium 47.3 Approval voting First I argue that any action ai of player i that includes a vote for i’s least preferred candidate, say candidate k, is weakly dominated by the action ai that differs from ai only in that candidate k does not receive a vote in ai . For any list a−i of the other players’ actions, the outcome of (ai , a−i ) differs from that of (ai , a−i ) only in that the total number of votes received by candidate k is one less in (ai , a−i ) than it is in (ai , a−i ). There are two possible implications for the winners of the election, depending on a−i : either the set of winners is the same in (ai , a−i ) as it is in (ai , a−i ), or candidate k is a winner in (ai , a−i ) but not in (ai , a−i ). Because candidate k is player i’s least preferred candidate, ai thus weakly dominates ai . I now argue that any action ai of player i that excludes a vote for i’s most preferred candidate, say candidate 1, is weakly dominated by the action ai that differs from ai only in that candidate 1 receives a vote in ai . The argument is symmetric with the one in the previous paragraph. For any list a−i of the other players’ actions, the outcome of (ai , a−i ) differs from that of (ai , a−i ) only in that the total number of votes received by candidate 1 is one more in (ai , a−i ) than it is in (ai , a−i ). There are two possible implications for the winners of the election, depending on a−i : either the set of winners is the same in (ai , a−i ) as it is in (ai , a−i ), or candidate 1 is a winner in (ai , a−i ) but not in (ai , a−i ). Because candidate 1 is player i’s most preferred candidate, ai thus weakly dominates ai . Finally I argue that if citizen i prefers candidate 1 to candidate 2 to . . . to candidate k then the action ai that consists of votes for candidates 1 and k − 1 is not weakly dominated. • The action ai is not weakly dominated by any action that excludes votes for either candidate 1 or candidate k − 1 (or both). Suppose ai excludes a vote for candidate 1. Then if the total votes by the other citizens for candidates 1 and 2 are equal, and the total votes for all other candidates are less, then citizen i’s taking the action ai leads candidate 1 to win, while the action ai leads to at best (from the point of view of citizen i) a tie between candidates 1 and 2. Thus ai does not weakly dominate ai . Similarly, suppose that ai excludes a vote for candidate k − 1. Then if the total votes by the other citizens for candidates k − 1 and k are equal, while the total votes for all other candidates are less, then citizen i’s taking the action ai leads candidate k − 1 to win, while the action ai leads to at best (from the point of view of citizen i) a tie between candidates k − 1 and k. • Now let ai be an action that includes votes for both candidate 1 and candidate k − 1, and also for at least one other candidate, say candidate j. Suppose that the total votes by the other citizens for candidates 1 and j are equal, and the total votes for all other candidates are less. Then citizen i’s taking the action ai leads candidate 1 to win, while the action ai leads to at best (from the point of view of citizen i) a tie between candidates 1 and j. Thus ai does not weakly dominate ai . Chapter 2. Nash Equilibrium 19 49.1 Other Nash equilibria of the game modeling collective decision-making Denote by i the player whose favorite policy is the median favorite policy. The set of Nash equilibria includes every action profile in which (i) i’s action is her favorite policy x i∗ , (ii) every player whose favorite policy is less than xi∗ names a policy equal to at most xi∗ , and (iii) every player whose favorite policy is greater than xi∗ names a policy equal to at least xi∗ . To show this, first note that the outcome is xi∗ , so player i cannot induce a better outcome for herself by changing her action. Now, if a player whose favorite position is less than xi∗ changes her action to some x < xi∗ , the outcome does not change; if such a player changes her action to some x > xi∗ then the outcome either remains the same (if some player whose favorite position exceeds xi∗ names x i∗ ) or increases, so that the player is not better off. A similar argument applies to a player whose favorite position is greater than xi∗ . The set of Nash equilibria also includes, for any positive integer k ≤ n, every action profile in which k players name the median favorite policy xi∗ , at most 12 (n − 3) players name policies less than xi∗ , and at most 12 (n − 3) players name policies greater than xi∗ . (In these equilibria, the favorite policy of a player who names a policy less than x i∗ may be greater than xi∗ , and vice versa. The conditions on the numbers of players who name policies less than xi∗ and greater than xi∗ ensure that no such player can, by naming instead her favorite policy, move the median policy closer to her favorite policy.) Any action profile in which all players name the same, arbitrary, policy is also a Nash equilibrium; the outcome is the common policy named. More generally, any profile in which at least three players name the same, arbitrary, policy x, at most (n − 3)/2 players name a policy less than x, and at most (n − 3)/2 players name a policy greater than x is a Nash equilibrium. (In both cases, no change in any player’s action has any effect on the outcome.) 49.2 Another mechanism for collective decision-making When the policy chosen is the mean of the announced policies, player i’s announcing her favorite policy does not weakly dominate all her other actions. For example, if there are three players, the favorite policy of player 1 is 0.3, and the other players both announce the policy 0, then player 1 should announce the policy 0.9, which leads to the policy 0.3 (= (0 + 0 + 0.9)/3) being chosen, rather than 0.3, which leads to the policy 0.1. 50.1 Symmetric strategic game The games in Exercise 29.1, Example 37.1, and Figure 46.1 (both games) are symmetric. The game in Exercise 40.1 is not symmetric. The game in Section 2.8.4 is symmetric if and only if u1 = u2 . 20 Chapter 2. Nash Equilibrium 51.1 Equilibrium for pairwise interactions in a single population The Nash equilibria are (A, A), (A, C), and (C, A). Only the equilibrium (A, A) is relevant if the game is played between the members of a single population—this equilibrium is the only symmetric equilibrium. Draft of solutions to exercises in chapter of An introduction to game theory by Martin J. Osborne

[email protected]; www.chass.utoronto.ca/~osborne/index.html Version: 00/11/6. c 1995–2000 by Martin J. Osborne. All rights reserved. No part of this book may be reCopyright produced in any form by any electronic or mechanical means (including photocopying, recording, or information storage and retrieval) without permission in writing from Martin J. Osborne. On request, permission to make one copy for each student will be granted to instructors who wish to use the book in a course, on condition that copies be sold at a price not more than the cost of duplication. 3 Nash Equilibrium: Illustrations 57.1 Cournot’s duopoly game with linear inverse demand and diﬀerent unit costs Following the analysis in the text, the best response function of firm 1 is 1 b1 (q2 ) = 2 (α − c1 − q2 ) if q2 ≤ α − c1 0 otherwise while that of firm 2 is b2 (q1 ) = 1 2 (α − c 2 − q1 ) 0 if q1 ≤ α − c2 otherwise. To find the Nash equilibrium, first plot these two functions. Each function has the same general form as the best response function of either firm in the case studied in the text. However, the fact that c1 = c2 leads to two qualitatively different cases when we combine the two functions to find a Nash equilibrium. If c1 and c2 do not differ very much then the functions in the analogue of Figure 56.2 intersect at a pair of outputs that are both positive. If c1 and c2 differ a lot, however, the functions intersect at a pair of outputs in which q1 = 0. Precisely, if c1 ≤ 12 (α + c2 ) then the downward-sloping parts of the best response functions intersect (as in Figure 56.2), and the game has a unique Nash equilibrium, given by the solution of the two equations q1 = q2 = This solution is (q∗1 , q∗2 ) = 1 2 (α 1 2 (α 1 3 (α − 2c 1 − c 1 − q2 ) − c2 − q1 ). + c2 ), 13 (α − 2c2 + c1 ) . If c1 > 12 (α + c2 ) then the downward-sloping part of firm 1’s best response function lies below the downward-sloping part of firm 2’s best response function (as in Figure 22.1), and the game has a unique Nash equilibrium, (q∗1 , q∗2 ) = (0, 12 (α − c2 )). In summary, the game always has a unique Nash equilibrium, defined as follows: 1 (α − 2c1 + c2 ), 1 (α − 2c2 + c1 ) if c1 ≤ 12 (α + c2 ) 3 3 0, 1 (α − c2 ) if c1 > 12 (α + c2 ). 2 21 22 Chapter 3. Nash Equilibrium: Illustrations ↑ q2 α−c 2 2 (q∗1 , q∗2 ) α − c1 b2 (q1 ) b1 (q2 ) 0 α − c2 α−c 1 2 q1 → Figure 22.1 The best response functions in Cournot’s duopoly game under the assumptions of Exercise 57.1 when α − c1 < 12 (α − c2 ). The unique Nash equilibrium in this case is (q ∗1 , q ∗2 ) = (0, 12 (α − c2 )). The output of firm 2 exceeds that of firm 1 in every equilibrium. If c2 decreases then firm 2’s output increases and firm 1’s output either falls, if c1 ≤ 12 (α + c2 ), or remains equal to 0, if c1 > 12 (α + c2 ). The total output increases and the price falls. 57.2 Cournot’s duopoly game with linear inverse demand and a quadratic cost function Firm 1’s profit is π1 (q1 , q2 ) = or q1 (α − q1 − q2 ) − q21 −q21 π1 (q1 , q2 ) = q1 (α − 2q1 − q2 ) −q21 if q1 + q2 ≤ α if q1 + q2 > α if q1 + q2 ≤ α if q1 + q2 > α. When it is positive, this function is a quadratic in q1 that is zero at q1 = 0 and at q1 = (α − q2 )/2. Thus firm 1’s best response function is b1 (q2 ) = 1 4 (α 0 − q2 ) if q2 ≤ α if q2 > α. Since the firms’ cost functions are the same, firm 2’s best response function is the same as firm 1’s: b2 (q) = b1 (q) for all q. The firms’ best response functions are shown in Figure 23.1. Solving the two equations q∗1 = b1 (q∗2 ) and q∗2 = b2 (q∗1 ) we find that there is a unique Nash equilibrium, in which the output of firm i (i = 1, 2) is q∗i = 15 α. Chapter 3. Nash Equilibrium: Illustrations 23 ↑ q2 α b1 (q2 ) 1 4α 0 (q∗1 , q∗2 ) b2 (q1 ) α q1 → 1 4α Figure 23.1 The best response functions in Cournot’s duopoly game with linear inverse demand and a quadratic cost function, as in Exercise 57.2. The unique Nash equilibrium is (q ∗1 , q ∗2 ) = ( 51 α, 15 α). 57.3 Cournot’s duopoly game with linear inverse demand and a ﬁxed cost Firm i’s payoff function is 0 qi (P(q1 + q2 ) − c) − f if qi = 0 if qi > 0. As before firm 1’s best response to q2 is (α − c − q2 )/2 if firm 1’s profit is nonnegative for this output; otherwise its best response is the output of zero. Firm 1’s profit when it produces (α − c − q2 )/2 and firm 2 produces q2 is α − c − q2 2 α−c− which is nonnegative if α − c − q2 − q2 2 −f = α − c − q2 2 2 − f, α − c − q2 2 > f, 2 or if q2 ≤ α − c − 2 f . Let q = α − c − 2 f . Then firm 1’s best response function is 1 if q2 < q 2 (α − c − q2 ) b1 (q2 ) = {0, 12 (α − c − q2 )} if q2 = q 0 if q2 > q. (If q2 = q then firm 1’s profit is zero whether it produces the output 12 (α − c − q2 ) or the output 0; both outputs are optimal.) Thus firm 1’s best response function has a “jump”: for outputs of firm 2 slightly less than q firm 1 wants to produce a positive output (and earn a small profit), while for outputs of firm 2 slightly greater than q it wants to produce an output of zero. 24 Chapter 3. Nash Equilibrium: Illustrations Firm 2’s cost function is the same as firm 1’s, so its best response function is the same. Because of the jumps in the best response functions, there are four qualitatively different cases, depending on the value of f . If f is small enough that q > 12 (α − c) (or, equivalently, f < (α − c)2 /16) then the best response functions take the form given in Figure 24.1. In this case the existence of the fixed cost has no impact on the equilibrium, which remains (q∗1 , q∗2 ) = ( 13 (α − c), 13 (α − c)). ↑ q2 α−c q α−c 2 α−c 3 0 b1 (q2 ) (q∗1 , q∗2 ) b2 (q1 ) α−c 3 α−c 2 q α−c q1 → Figure 24.1 The best response functions in Cournot’s duopoly game when the inverse demand function is P(Q) = α − Q (where this is positive) and the cost function of each firm is f + cq, with f < (α − c) 2 /16. The unique Nash equilibrium is (q ∗1 , q ∗2 ) = ( 31 (α − c), 13 (α − c)) (as in the case in which f = 0). As f increases, the point at which the best response functions jump moves closer to the origin. Eventually q enters the range from 13 (α − c) to 12 (α − c) (which implies that (α − c)2 /16 < f < (α − c)2 /9), in which case the best response functions take the forms shown in the left panel of Figure 25.1. In this case there are three Nash equilibria: (0, 12 (α − c)), ((α − c)/3, (α − c)/3), and ( 12 (α − c), 0). As f increases further, there is a point at which q becomes less than 13 (α − c) but is still positive (implying that (α − c)2 /9 < f < (α − c)2 /4), so that the best response functions take the forms shown in the right panel of Figure 25.1. In this case there are two Nash equilibria: (0, 12 (α − c)) and ( 12 (α − c), 0). Finally, if f is extremely large then a firm does not want to produce any output even if the other firm produces no output. This occurs when f > (α − c)2 /4; the unique Nash equilibrium in this case is (0, 0). Chapter 3. Nash Equilibrium: Illustrations ↑ q2 ↑ q2 α−c 2 q α−c 3 (0, 12 (α − c)) b1 (q2 ) α−c 2 ( 13 (α − c), 13 (α − c)) b2 (q1 ) 0 25 α−c 3 ( 12 (α − c), 0) q1 → q α−c 2 (0, 12 (α − c)) b1 (q2 ) α−c 3 b2 (q1 ) q 0 q α−c 3 ( 12 (α − c), 0) α−c q1 → 2 Figure 25.1 The best response functions in Cournot’s duopoly game when the inverse demand function is P(Q) = α − Q (where this is positive) and the cost function of each firm is f + cq, with (α − c) 2 /16 < f < (α − c) 2 /9 (left panel) and f > (α − c) 2 /9 (right panel). In the first case the game has three Nash equilibria: (0, 12 (α − c)), ( 13 (α − c), 13 (α − c)), and ( 12 (α − c), 0). In the second case it has two Nash equilibria: (0, 12 (α − c)) and ( 21 (α − c), 0). 58.2 Nash equilibrium of Cournot’s duopoly game and the collusive outcome The firms’ total profit is (q1 + q2 )(α − c − q1 − q2 ), or Q(α − c − Q), where Q denotes total output. This function is a quadratic in Q that is zero when Q = 0 and when Q = α − c, so that its maximizer is Q∗ = 12 (α − c). If each firm produces 14 (α − c) then its profit is 18 (α − c)2 . This profit exceeds its Nash equilibrium profit of 19 (α − c)2 . If one firm produces Q∗ /2, the other firm’s best response is bi (Q∗ /2) = 12 (α − c − 14 (α − c)) = 38 (α − c). That is, if one firm produces Q∗ /2, the other firm wants to produce more than Q∗ /2. 58.1 Variant of Cournot’s game, with market-share maximizing ﬁrms Let firm 1 be the market-share maximizing firm. If q2 > α − c, there is no output of firm 1 for which its profit is nonnegative. Thus its best response to such an output of firm 2 is q1 = 0. If q2 ≤ α − c then firm 1 wants to choose its output q1 large enough that the price is c (and hence its profit is zero). Thus firm 1’s best response to such a value of q2 is q1 = α − c − q2 . We conclude that firm 1’s best response function is α − c − q2 if q2 ≤ α − c b1 (q2 ) = 0 if q2 > α − c. Firm 2’s best response function is the same as in Section 3.1.3, namely (α − c − q2 )/2 if q2 ≤ α − c b2 (q1 ) = 0 if q2 > α − c. These best response functions are shown in Figure 26.1. The game has a unique 26 Chapter 3. Nash Equilibrium: Illustrations Nash equilibrium, (q∗1 , q∗2 ) = (α − c, 0), in which firm 2 does not operate. (The price is c, and firm 1’s profit is zero.) ↑ q2 α−c b1 (q2 ) α−c 2 b2 (q1 ) 0 α−c q1 → Figure 26.1 The best response functions in a variant of Cournot’s duopoly game in which in which the inverse demand function is P(Q) = α − Q (where this is positive) and the cost function of each firm is cq, and firm 1 maximizes its market share, rather than its profit. The unique Nash equilibrium is (q ∗1 , q ∗2 ) = (α − c, 0). If both firms maximize their market shares, then the downward-sloping parts of their best response functions coincide in the analogue of Figure 26.1. Thus every pair (q1 , q2 ) with q1 + q2 = α − c is a Nash equilibrium. 59.1 Cournot’s game with many ﬁrms Firm 1’s payoff function is q1 (α − c − q1 − q2 − · · · − qn ) −cq1 if q1 + q2 + · · · + qn ≤ α if q1 + q2 + · · · + qn > α. As in the case of two firms, this function is a quadratic in q1 where it is positive, and is zero when q1 = 0 and when q1 = α − c − q2 − · · · − qn . Thus firm 1’s best response function is (α − c − q2 − · · · − qn ) /2 if q2 + · · · + qn ≤ α − c b1 (q−1 ) = 0 if q2 + · · · + qn > α − c. (Recall that q−1 stands for the list of the outputs of all the firms except firm 1.) The best response functions of every other firm is the same. Chapter 3. Nash Equilibrium: Illustrations 27 The conditions for (q∗1 , . . . , q∗n ) to be a Nash equilibrium are q∗1 = b1 (q∗−1 ) q∗2 = b2 (q∗−2 ) .. . q∗n = b2 (q∗−n ) or, in an equilibrium in which all the firms’ outputs are positive, q∗1 = 1 2 (α 1 2 (α − c − q∗2 − q∗3 − · · · − q∗n ) q∗n = 1 2 (α − c − q∗1 − q∗2 − · · · − q∗n−1 ). q∗2 = .. . − c − q∗1 − q∗3 − · · · − q∗n ) We can write these equations as 0 = α − c − 2q∗1 − q∗2 − · · · − q∗n−1 − q∗n 0 = α − c − q∗1 − 2q∗2 − · · · − q∗n−1 − q∗n .. . 0 = α − c − q∗1 − q∗2 − · · · − q∗n−1 − 2q∗n . If we subtract the second equation from the first we obtain 0 = −q∗1 + q∗2 , or q∗1 = q∗2 . Similarly subtracting the third equation from the second we conclude that q∗2 = q∗3 , and continuing with all pairs of equations we deduce that q∗1 = q∗2 = · · · = q∗n . Let the common value of the firms’ outputs be q∗ . Then each equation is 0 = α − c − (n + 1)q∗ , so that q∗ = (α − c)/(n + 1). In summary, the game has a unique Nash equilibrium, in which the output of every firm i is (α − c)/(n + 1). The price at this equilibrium is α − n(α − c)/(n + 1), or (α + nc)/(n + 1). As n increases this price decreases, approaching c as n increases without bound: α/(n + 1) decreases to 0 and nc/(n + 1) decreases to c. 60.1 Nash equilibrium of Cournot’s game with small ﬁrms • If P(Q∗ ) < p then every firm producing a positive output makes a negative profit, and can increase its profit (to 0) by deviating and producing zero. • If P(Q∗ + q) > p, take a firm that is either producing no output, or an arbitrarily small output. (Such a firm exists, since demand is finite.) Such a firm earns a profit of either zero or arbitrarily close to zero. If it deviates and chooses the output q then total output changes to at most Q∗ + q, so that the price still exceeds p (since P(Q∗ + q) > p). Hence the deviant makes a positive profit. 28 Chapter 3. Nash Equilibrium: Illustrations 61.1 Interaction among resource-users The game is given as follows. Players The firms. Actions Each firm’s set of actions is the set of all nonnegative numbers (representing the amount of input it uses). Preferences The payoff of each firm i is xi (1 − (x 1 + · · · + xn )) 0 if x 1 + · · · + xn ≤ 1 if x1 + · · · + xn > 1. This game is the same as that in Exercise 59.1 for c = 0 and α = 1. Thus it has a unique Nash equilibrium, (x1 , . . . , x n ) = (1/(n + 1), . . . , 1/(n + 1)). In this Nash equilibrium, each firm’s output is (1/(n + 1))(1 − n/(n + 1)) = 1/(n + 1)2 . If x i = 1/(2n) for i = 1, . . . , n then each firm’s output is 1/(4n), which exceeds 1/(n + 1)2 for n ≥ 2. (We have 1/(4n) − 1/(n + 1)2 = (n − 1)2 /(4n(n + 1)2 ) > 0 for n ≥ 2.) 65.1 Bertrand’s duopoly game with constant unit cost The pair (c, c) of prices remains a Nash equilibrium; the argument is the same as before. Further, as before, there is no other Nash equilibrium. The argument needs only very minor modification. For an arbitrary function D there may exist no monopoly price p m ; in this case, if pi > c, p j > c, pi ≥ p j , and D(p j ) = 0 then firm i can increase its profit by reducing its price slightly below p (for example). 65.2 Bertrand’s duopoly game with discrete prices Yes, (c, c) is still a Nash equilibrium, by the same argument as before. In addition, (c + 1, c + 1) is a Nash equilibrium (where c is given in cents). In this equilibrium both firms’ profits are positive. If either firm raises its price or lowers it to c, its profit becomes zero. If either firm lowers its price below c, its profit becomes negative. No other pair of prices is a Nash equilibrium, by the following argument, similar to the argument in the text for the case in which a price can be any nonnegative number. • If p i < c then the firm whose price is lowest (or either firm, if the prices are the same) can increase its profit (to zero) by raising its price to c. • If pi = c and p j ≥ c + 1 then firm i can increase its profit from zero to a positive amount by increasing its price to c + 1. Chapter 3. Nash Equilibrium: Illustrations 29 • If p i > p j ≥ c + 1 then firm i can increase its profit (from zero) by lowering its price to c + 1. • If p i = p j ≥ c + 2 and p j < α then either firm can increase its profit by lowering its price by one cent. (If firm i does so, its profit changes from 1 2 (p i − c)(α − p i ) to (p i − 1 − c)(α − p i + 1) = (p i − 1 − c)(α − p i ) + p i − 1 − c. We have pi − 1 − c ≥ 12 (pi − c) and p i − 1 − c > 0, since p i ≥ c + 2.) • If pi = p j ≥ c + 2 and p j ≥ α then either firm can increase its profit by lowering its price to pm . 66.1 Bertrand’s oligopoly game Consider a profile (p1 , . . . , pn ) of prices in which pi ≥ c for all i and at least two prices are equal to c. Every firm’s profit is zero. If any firm raises its price its profit remains zero. If a firm charging more than c lowers its price, but not below c, its profit also remains zero. If a firm lowers its price below c then its profit is negative. Thus any such profile is a Nash equilibrium. To show that no other profile is a Nash equilibrium, we can argue as follows. • If some price is less than c then the firm charging the lowest price can increase its profit (to zero) by increasing its price to c. • If exactly one firm’s price is equal to c then that firm can increase its profit by raising its price a little (keeping it less than the next highest price). • If all firms’ prices exceed c then the firm charging the highest price can increase its profit by lowering its price to some price between c and the lowest price being charged. 66.2 Bertrand’s duopoly game with diﬀerent unit costs a. If all consumers buy from firm 1 when both firms charge the price c2 , then (p1 , p2 ) = (c2 , c2 ) is a Nash equilibrium by the following argument. Firm 1’s profit is positive, while firm 2’s profit is zero (since it serves no customers). • If firm 1 increases its price, its profit falls to zero. • If firm 1 reduces its price, say to p, then its profit changes from (c2 − c1 )(α − c2 ) to (p − c1 )(α − p). Since c2 is less than the maximizer of (p − c1 )(α − p), firm 1’s profit falls. • If firm 2 increases its price, its profit remains zero. • If firm 2 decreases its price, its profit becomes negative (since its price is less than its unit cost). 30 Chapter 3. Nash Equilibrium: Illustrations Under this rule no other pair of prices is a Nash equilibrium, by the following argument. • If p i < c1 for i = 1, 2 then the firm with the lower price (or either firm, if the prices are the same) can increase its profit (to zero) by raising its price above that of the other firm. • If p1 > p2 ≥ c2 then firm 2 can increase its profit by raising its price a little. • If p2 > p1 ≥ c1 then firm 1 can increase its profit by raising its price a little. • If p2 ≤ p1 and p2 < c2 then firm 2’s profit is negative, so that it can increase its profit by raising its price. • If p1 = p2 > c2 then at least one of the firms is not receiving all of the demand, and that firm can increase its profit by lowering its price a little. b. Now suppose that the rule for splitting up the customers when the prices are equal specifies that firm 2 receives some customers when both prices are c2 . By the argument for part a, the only possible Nash equilibrium is (p1 , p2 ) = (c2 , c2 ). (The argument in part a that every other pair of prices is not a Nash equilibrium does not use the fact that customers are split equally when (p1 , p2 ) = (c2 , c2 ).) But if (p1 , p2 ) = (c2 , c2 ) and firm 2 receives some customers, firm 1 can increase its profit by reducing its price a little and capturing the entire market. 67.1 Bertrand’s duopoly game with ﬁxed costs At the pair of prices (p, p), both firms’ profits are zero. (Firm 1 receives all the demand and obtains the profit (p − c)(α − p) − f = 0, and firm 2 receives no demand.) This pair of prices is a Nash equilibrium by the following argument. • If either firm raises its price its profit remains zero (it receives no customers). • If either firm lowers its price then it receives all the demand and earns a negative profit (since f is less than the maximum of (p − c)(α − p)). No other pair of prices (p1 , p2 ) is a Nash equilibrium, by the following argument. • If p1 = p2 < p then firm 1’s profit is negative; firm 1 can increase its profit by raising its price. • If p1 = p2 > p then firm 2’s profit is zero; firm 2 can obtain a positive profit by lowering its price a little. • If pi < p j and firm i’s profit is positive then firm j can increase its profit from zero to almost the current level of i’s profit by changing its price to be slightly less than pi . Chapter 3. Nash Equilibrium: Illustrations 31 • If pi < p j and firm i’s profit is zero then firm i can earn a positive profit by raising its price a little. • If pi < p j and firm i’s profit is negative then firm i can increase its profit to zero by raising its price above p j . 72.1 Electoral competition with asymmetric voters’ preferences The unique Nash equilibrium remains (m, m); the direct argument is exactly the same as before. (The dividing line between the supporters of two candidates with different positions changes. If xi < x j , for example, the dividing line is 13 xi + 23 x j rather than 12 (xi + x j ). The resulting change in the best response functions does not affect the Nash equilibrium.) 72.2 Electoral competition with three candidates If a single candidate enters, then either of the remaining candidates can enter and either win outright or tie for first place. Thus there is no Nash equilibrium in which a single candidate enters. In any Nash equilibrium in which more than one candidate enters, all the candidates that enter tie for first place, since if they do not then some candidate loses, and hence can do better by staying out of the race. If two candidates enter, then by the argument in the text for the case in which there are two candidates, each takes the position m. But then the third candidate can enter and win outright. Thus there is no Nash equilibrium in which two candidates enter. If all three candidates enter and choose the same position, each candidate receives one third of the votes. If the common position is equal to m then any candidate can win outright (obtaining close to one-half of the votes) by moving slightly to one side of m. If the common position is different from m then any candidate can win outright (obtaining more than one-half of the votes) by moving to m. Thus there is no Nash equilibrium in which all three candidates enter and choose the same position. If all three candidates enter and do not all choose the same position then they all tie for first place, by the second argument. At least one candidate (i) does not share her position with any other candidate and (ii) is an extremist (her position is not between the positions of the other candidates). This candidate can move slightly closer to the other candidates and win outright. Thus there is no Nash equilibrium in which all three candidates enter and not all of them choose the same position. We conclude that the game has no Nash equilibrium. 32 Chapter 3. Nash Equilibrium: Illustrations 72.3 Electoral competition in two districts The game has a unique equilibrium, in which the both candidates choose the position m1 (the median favorite position in the district with the most electoral college votes). The outcome is a tie. The following argument shows that this pair of positions is a Nash equilibrium. If a candidate deviates to a position less than m1 , she loses in district 1 and wins in district 2, and thus loses overall. If a candidate deviates to a position greater than m1 , she loses in both districts. To see that there is no other Nash equilibrium, first consider a pair of positions for which candidate 1 loses in district 1, and hence loses overall. By deviating to m1 , she either wins in district 1, and hence wins overall, or, if candidate 2’s position is m1 , ties in district 1, and ties overall. Thus her deviation induces an outcome she prefers. The same argument applies to candidate 2, so that in any equilibrium the candidates tie in district 1. Now, if the candidates’ positions are either different, or the same and different from m1 , either candidate can win outright rather than tying for first place by moving to m1 . Thus there is a single equilibrium, in which both candidates’ positions are m1 . 73.1 Electoral competition between candidates who care only about the winning position First consider a pair (x1 , x2 ) of positions for which either x1 < m and x2 < m, or x1 > m and x2 > m. • If x1 = x2 and the winner’s position is different from her favorite position then the winner can move slightly closer to her favorite position and still win. • If x1 = x2 and the winner’s position is equal to her favorite position then the other candidate can move to m, which is closer to her favorite position than the winner’s position, and win. • If x1 = x 2 < m then the candidate whose favorite position exceeds m can move to m and cause the winning position to be m rather than x1 = x2 . • If x1 = x2 > m then the candidate whose favorite position is less than m can move to m and cause the winning position to be m rather than x1 = x2 . Now suppose the candidates’ positions are on opposite sides of m: either x1 < m < x2 , or x2 < m < x1 . • If each candidate’s position is on the same side of m as her favorite position and one candidate wins outright, then the loser can win outright by moving to m, which she prefers to the position of the other candidate. Chapter 3. Nash Equilibrium: Illustrations 33 • If each candidate’s position is on the same side of m as her favorite position and the candidates tie for first place, then by moving slightly closer to m either candidate can win. If her movement is small enough she prefers her new position to the previous compromise 12 (x1 + x2 ) (= m). • If each candidate’s position is on the opposite side of m to her favorite position then the winner, or either player in the case of a tie, can move to her favorite position and either win outright or cause the winning position to be the other candidate’s position, in both cases improving the outcome from her point of view. Now suppose that x1 = m and x2 < m. If x1∗ < m then candidate 1 is better off choosing a slightly smaller value of x1 (in which case she still wins). If x1∗ > m then candidate 1 is better off choosing a slightly larger value of x1 (in which case she still wins). Thus (x1 , x2 ) is not a Nash equilibrium. A similar argument applies to pairs (x1 , x2 ) for which x1 = m and x2 > m, and for which x1 = m and x2 = m. Finally, if (x1 , x2 ) = (m, m), then the candidates tie. If either candidate changes her position then she loses, and the winning position does not change. Thus this pair of positions is a Nash equilibrium. 73.2 Citizen-candidates If b ≤ 2c then the game has a Nash equilibrium in which a single citizen, with favorite position m, stands as a candidate. Another citizen with the same favorite position who stands obtains the payoff 12 b − c, as opposed to the payoff of 0 if she does not stand. Given b ≤ 2c, it is optimal for any such citizen not to stand. A citizen with any other favorite position who stands loses, and hence is worse off than if she does not stand. If two citizens with favorite position m become candidates, each candidate’s payoff is 12 b − c; if one withdraws then she obtains the payoff of 0, so for equilibrium we require b ≥ 2c. Now consider a citizen whose favorite position is close to m. If she enters she wins outright, obtaining the payoff b − c. Since b ≥ 2c, this payoff is positive, and hence exceeds her payoff if she does not stand (which is negative, since the winner’s position is then different from her favorite position). Thus there is no equilibrium in which two citizens with favorite position m stand as candidates. Now consider the possibility of an equilibrium in which two citizens with favorite positions different from m stand as candidates. For an equilibrium the candidates must tie, otherwise one loses, and can do better by withdrawing. Thus the positions, say x 1 and x2 , must satisfy 12 (x1 + x2 ) = m. If x1 and x2 are close enough to m then any other citizen loses if she becomes a candidate. Thus there are equilibria in which two citizens with positions symmetric about m, and sufficiently close to m, become candidates. 34 Chapter 3. Nash Equilibrium: Illustrations 74.1 Electoral competition for more general preferences a. If x ∗ is a Condorcet winner then for any y = x ∗ a majority of voters prefer x ∗ to y, so y is not a Condorcet winner. Thus there is no more than one Condorcet winner. b. Suppose that one of the remaining voters prefers y to z to x, and the other prefers z to x to y. For each position there is another position preferred by a majority of voters, so no position is a Condorcet winner. c. Now suppose that x ∗ is a Condorcet winner. Then the strategic game described the exercise has a unique Nash equilibrium in which both candidates choose x ∗ . This pair of actions is a Nash equilibrium because if either candidate chooses a different position she loses. For any other pair of actions either one candidate loses, in which case that candidate can deviate to the position x ∗ and at least tie, or the candidates tie at a position different from x ∗ , in which case either of them can deviate to x ∗ and win. If there is no Condorcet winner then for every position there is another position preferred by a majority of voters. Thus for every pair of distinct positions the loser can deviate and win, and for every pair of identical positions either candidate can deviate and win. Thus there is no Nash equilibrium. 75.1 Competition in product characteristics Suppose there are two firms. If the products are different, then either firm increases its market share by making its product more similar to that of its rival. Thus in every possible equilibrium the products are the same. But if x1 = x2 = m then each firm’s market share is 50%, while if it changes its product to be closer to m then its market share rises above 50%. Thus the only possible equilibrium is (x1 , x2 ) = (m, m). This pair of positions is an equilibrium, since each firm’s market share is 50%, and if either firm changes its product its market share falls below 50%. Now suppose there are three firms. If all firms’ products are the same, each obtains one-third of the market. If x1 = x 2 = x 3 = m then any firm, by changing its product a little, can obtain close to one-half of the market. If x1 = x2 = x3 = m then any firm, by changing its product a little, can obtain more than one-half of the market. If the firms’ products are not all the same, then at least one of the extreme products is different from the other two products, and the firm that produces it can increase its market share by making it more similar to the other products. Thus when there are three firms there is no Nash equilibrium. 76.1 Direct argument for Nash equilibria of War of Attrition • If t1 = t2 then either player can increase her payoff by conceding slightly later (in which case she obtains the object for sure, rather than getting it with Chapter 3. Nash Equilibrium: Illustrations 35 probability 12 ). • If 0 < ti < t j then player i can increase her payoff by conceding at 0. • If 0 = ti < t j < vi then player i can increase her payoff (from 0 to almost vi − t j > 0) by conceding slightly after t j . Thus there is no Nash equilibrium in which t1 = t2 , 0 < ti < t j , or 0 = ti < t j < v i (for i = 1 and j = 2, or i = 2 and j = 1). The remaining possibility is that 0 = ti < t j and t j ≥ vi for i = 1 and j = 2, or i = 2 and j = 1. In this case player i’s payoff is 0, while if she concedes later her payoff is negative; player j’s payoff is v j , her highest possible payoff in the game. 77.1 Variant of War of Attrition The game is Players The two parties to the dispute. Actions Each player’s set of actions is the set of possible concession times (nonnegative numbers). Preferences Player i’s preferences are represented by the payoff function if ti < t j 0 1 (v − ti ) if ti = t j ui (t1 , t2 ) = 2 i vi − t j if ti > t j . where j is the other player. Three representative cross-sections of player i’s payoff function are shown in Figure 35.1. ↑ ui 0 ↑ ui tj vi 0 ↑ ui t j = vi ti → t j < vi 0 vi ti → t j = vi tj ti → t j > vi Figure 35.1 Three cross-sections of player i’s payoff function in the variant of the War of Attrition in Exercise 77.1. From this figure we deduce that the best response function of player i is if t j < vi {ti : ti > t j } if t j = vi Bi (t j ) = {ti : ti ≥ 0} {ti : 0 ≤ ti < t j } if t j > vi . 36 Chapter 3. Nash Equilibrium: Illustrations ↑ t2 ↑ t2 B1 (t2 ) B2 (t1 ) v1 v2 v1 0 t1 → v2 0 t1 → Figure 36.1 The players’ best response functions in the variant of the War of Attrition in Exercise 77.1 for v1 > v2 . Player 1’s best response function is in the left panel; player 2’s is in the right panel. (The sloping edges are excluded.) The best response functions are shown in Figure 36.1 for a case in which v1 > v2 . Superimposing the two best response functions, we see that if v1 > v2 then the set of Nash equilibrium action pairs is the union of the shaded regions in Figure 36.2, namely the set of all pairs (t1 , t2 ) such that either t1 ≤ v2 and t2 ≥ v1 , or t1 ≥ v2 , t1 > t2 , and t2 ≤ v1 . ↑ t2 v1 v2 0 v2 v1 t1 → Figure 36.2 The set of Nash equilibria of the variant of the War of Attrition in Exercise 77.1 when v1 > v2 . 78.1 Timing product release A strategic game that models this situation is: Chapter 3. Nash Equilibrium: Illustrations 37 Players The two firms Actions The set of actions of each player is the set of possible release times, which we can take to be the set of numbers t for which 0 ≤ t ≤ T. Preferences Each firm’s preferences are represented by its market share; the market share of firm i when it releases its product at time ti and its rival releases its product at time t j is h(ti ) if ti < t j if ti = t j if ti > t j . 1 2 1 − h(t j ) Three representative cross-sections of firm i’s payoff function are shown in Figure 37.1. ↑ ui ↑ ui ↑ ui 1 2 1 2 1 2 0 tj h(t j ) < ti → 0 1 2 tj h(t j ) = ti → 1 2 t j ti → 0 h(t j ) > 1 2 Figure 37.1 Three cross-sections of firm i’s payoff function in the game in Exercise 78.1. From the payoff function we see that if t j is such that h(t j ) < 12 then the set of firm i’s best responses is the set of release times after t j . If t j is such that h(t j ) = 12 then the set of firm i’s best responses is the set of release times greater than or equal to t j . If t j is such that h(t j ) > 12 then firm i wants to release its product just before t j . Since there is no latest time before t j , firm i has no best response in this case. (It has good responses, but none is optimal.) Denoting the time t for which h(t) = 12 by t∗ , the firms’ best response functions are shown in Figure 38.1. Combining the best response functions we see that the game has a unique Nash equilibrium, in which both firms release their products at the time t∗ (where h(t∗ ) = 12 ). 78.2 A ﬁght The game is defined as follows. Players The two people. Actions The set of actions of each player i is the set of amounts of the resource that player i can devote to fighting (the set of numbers yi with 0 ≤ yi ≤ 1). 38 Chapter 3. Nash Equilibrium: Illustrations ↑ t2 ↑ t2 t∗ t∗ B2 (t1 ) B1 (t2 ) 0 t∗ t1 → 0 t∗ t1 → Figure 38.1 The firms’ best response functions in the game in Exercise 78.1. Firm 1’s best response function is in the left panel; firm 2’s is in the right panel. Preferences The preferences of player i are represented by the payoff function if yi > y j f (y 1 , y2 ) 1 f (y 1 , y2 ) if y1 = y2 ui (y 1 , y2 ) = 2 0 if yi < y j . If yi < y j then player j can increase her payoff by reducing y j a little, keeping it greater than yi (output increases, and she still wins). So no action profile in which y1 = y2 is a Nash equilibrium. If y1 = y2 < 1 then either player i can increase her payoff by increasing yi to slightly above y j (output falls a little, but i’s share of it increases from 12 to 1). So no action profile in which y1 = y2 < 1 is a Nash equilibrium. The only action profile that remains is (y1 , y2 ) = (1, 1). This profile is a Nash equilibrium: each player’s payoff is 0, and remains 0 if she reduces the amount of the resource she devotes to fighting (given the other player’s action). 82.1 Nash equilibrium of second-price sealed-bid auction The action profile (v n , 0, . . . , 0, v1 ) is a Nash equilibrium of a second-price sealedbid auction, by the following argument. • If player 1 increases her bid she wins and obtains the payoff 0, equal to her current payoff. If she reduces her bid her payoff also remains 0. • If player n increases her bid or reduces it to a level greater than vn then the outcome does not change. If she reduces her bid to vn or less then she loses, and her payoff remains 0. • If any other player increases her bid, either the outcome remains the same or the player wins and pays the price v1 , thus obtaining a negative payoff. 83.1 Second-price sealed-bid auction with two bidders If player 2’s bid b2 is less than v1 then any bid of b2 or more is a best response of player 1 (she wins and pays the price b2 ). If player 2’s bid is equal to v1 then every Chapter 3. Nash Equilibrium: Illustrations 39 bid of player 1 yields her the payoff zero (either she wins and pays v1 , or she loses), so every bid is a best response. If player 2’s bid b2 exceeds v 1 then any bid of less than b2 is a best response of player 1. (If she bids b2 or more she wins, but pays the price b2 > v1 , and hence obtains a negative payoff.) In summary, player 1’s best response function is if b2 < v1 {b1 : b1 ≥ b2 } if b2 = v1 B1 (b2 ) = {b1 : b1 ≥ 0} {b1 : 0 ≤ b1 < b2 } if b2 > v1 . By similar arguments, player 2’s best response function is if b1 < v2 {b2 : b2 > b1 } B2 (b1 ) = {b2 : b2 ≥ 0} if b1 = v2 . {b2 : 0 ≤ b2 ≤ b1 } if b1 > v2 . These best response functions are shown in Figure 39.1. ↑ b2 ↑ b2 B1 (b2 ) v1 v1 v2 v2 0 v2 v1 B2 (b1 ) b1 → v2 v1 b1 → Figure 39.1 The players’ best response functions in a two-player second-price sealed-bid auction (Exercise 83.1). Player 1’s best response function is in the left panel; player 2’s is in the right panel. (Only the edges marked by a black line are included.) Superimposing the best response functions, we see that the set of Nash equilibria is the shaded set in Figure 40.1, namely the set of pairs (b1 , b2 ) such that either b1 ≤ v2 and b2 ≥ v1 or b1 ≥ v2 , b1 ≥ b2 , and b2 ≤ v1 . 84.1 Nash equilibrium of ﬁrst-price sealed-bid auction The profile (b1 , . . . , bn ) = (v2 , v2 , v3 , . . . , vn ) is a Nash equilibrium by the following argument. 40 Chapter 3. Nash Equilibrium: Illustrations ↑ b2 v1 v2 0 v2 v1 b1 → Figure 40.1 The set of Nash equilibria of a two-player second-price sealed-bid auction (Exercise 83.1). • If player 1 raises her bid she still wins, but pays a higher price and hence obtains a lower payoff. If player 1 lowers her bid then she loses, and obtains the payoff of 0. • If any other player changes her bid to any price at most equal to v2 the outcome does not change. If she raises her bid above v2 she wins, but obtains a negative payoff. 85.1 First-price sealed-bid auction A profile of bids in which the two highest bids are not the same is not a Nash equilibrium because the player naming the highest bid can reduce her bid slightly, continue to win, and pay a lower price. By the argument in the text, in any equilibrium player 1 wins the object. Thus she submits one of the highest bids. If the highest bid is less than v2 , then player 2 can increase her bid to a value between the highest bid and v2 , win, and obtain a positive payoff. Thus in an equilibrium the highest bid is at least v2 . If the highest bid exceeds v1 , player 1’s payoff is negative, and she can increase this payoff by reducing her bid. Thus in an equilibrium the highest bid is at most v1 . Finally, any profile (b1 , . . . , bn ) of bids that satisfies the conditions in the exercise is a Nash equilibrium by the following argument. • If player 1 increases her bid she continues to win, and reduces her payoff. If player 1 decreases her bid she loses and obtains the payoff 0, which is at most her payoff at (b1 , . . . , bn ). • If any other player increases her bid she either does not affect the outcome, or wins and obtains a negative payoff. If any other player decreases her bid she does not affect the outcome. Chapter 3. Nash Equilibrium: Illustrations 41 86.1 Third-price auction a. The argument that a bid of vi weakly dominates any lower bid is the same as for a second-price auction. Now compare bids of bi > vi and v i . Suppose that one of the other players’ bids is between vi and bi and all the remaining bids are less than vi . If player i bids vi she loses, and obtains the payoff of 0. If she bids bi she wins, and pays the third highest bid, which is less than vi . Thus she is better off bidding bi than she is bidding vi . b. Each player’s bidding her valuation is not a Nash equilibrium because player 2 can deviate and bid more than v1 and obtain the object at the price v3 instead of not obtaining the object. c. Any action profile in which every player bids b, where v2 ≤ b ≤ v1 is a Nash equilibrium. (Player 1’s changing her bid has no effect on her payoff. If any other player raises her bid then she wins and pays b, obtaining a nonpositive payoff; if any other player lowers her bid the outcome does not change.) Any action profile in which player 1’s bid b1 satisfies v2 ≤ b1 ≤ v1 , every other player’s bid is at most b1 , and at least two other players’ bids are at least v2 is also a Nash equilibrium. 88.3 Lobbying as an auction First-price auction In the action pair, each interest group’s payoff is −100. Consider group A. If it raises the price it will pay for y, then the government still chooses y, and A is worse off. If it lowers the price it will pay for y, then the government chooses z and A’s payoff remains −100. Now suppose it changes its bid from y to x and bids p. If p < 103, then the government chooses z and A’s payoff remains −100. If p ≥ 103, then the government chooses x and A’s payoff is at most −103. Group A cannot increase its payoff by changing its bid from y to z, for similar reasons. A similar argument applies to group B’s bid. Menu auction In the action pair, each group’s payoff is −3. Consider group A. If it changes its bids then either the outcome remains x and it pays at least 3, so that its payoff is at most −3, or the outcome becomes y and it pays at least 6, in which case its payoff is at most −3, or the outcome becomes z and it pays at least 0, in which case its payoff is at most −100. (Note that if it reduces its bids for both x and y then z is chosen.) Thus no change in its bids increases its payoff. Similar considerations apply to group B’s bid. 42 Chapter 3. Nash Equilibrium: Illustrations 87.1 Multi-unit auctions Discriminatory auction To show that the action of bidding vi and wi is not dominant for player i, we need only find actions for the other players and alternative bids for player i such that player i’s payoff is higher under the alternative bids than it is under the vi and wi , given the other players’ actions. Suppose that each of the other players submits two bids of 0. Then if player i submits one bid between 0 and vi and one bid between 0 and wi she still wins two units, and pays less than when she bids vi and wi . Uniform-price auction Suppose that some bidder other than i submits one bid between wi and vi and one bid of 0, and all the remaining bidders submit two bids of 0. Then bidder i wins one unit, and pays the price wi . If she replaces her bid of wi with a bid between 0 and wi then she pays a lower price, and hence is better off. Vickrey auction Suppose that player i bids vi and wi . Consider separately the cases in which the bids of the players other than i are such that player i wins 0, 1, and 2 units. Player i wins 0 units: In this case the second highest of the other players’ bids is at least vi , so that if player i changes her bids so that she wins one or more units, for any unit she wins she pays at least vi . Thus no change in her bids increases her payoff from its current value of 0 (and some changes lower her payoff). Player i wins 1 unit: If player i raises her bid of vi then she still wins one unit and the price remains the same. If she lowers this bid then either she still wins and pays the same price, or she does not win any units. If she raises her bid of wi then either the outcome does not change, or she wins a second unit. In the latter case the price she pays is the previously-winning bid she beat, which is at least wi , so that her payoff either remains zero or becomes negative. Player i wins 2 units: Player i’s raising either of her bids has no effect on the outcome; her lowering a bid either has no effect on the outcome or leads her to lose rather than to win, leading her to obtain the payoff of zero. 88.1 Waiting in line The situation is modeled by a variant of a discriminatory multi-unit auction in which 100 units are available, and each person attaches a positive value only to one unit and submits a bid for only one unit. We can argue along the lines of Exercise 85.1. • The first 100 people to arrive must do so at the same time. If not, at least one of them could arrive a little later and still be in the first 100. Chapter 3. Nash Equilibrium: Illustrations 43 • The first 100 people to arrive must be persons 1 through 100. Suppose, to the contrary, that one of these people is person i with i ≥ 101, and person j with j ≤ 100 is not in the group that arrives first. Then the common waiting time of the first 100 must be at most v101 , otherwise person i obtains a negative payoff. But then person j can deviate and arrive slightly earlier than the group of 100, and obtain a positive payoff. • The common waiting time of the first 100 people must be at least v101 . If not, then person 101 could arrive slightly before the first 100 and obtain a positive payoff. • The common waiting time of the first 100 people must be at most v100. If not, then person 100 obtains a negative payoff, while by arriving later her payoff is zero. • At least one person i with i ≥ 101 arrives at the same time as the first 100 people. If not, then any person i with i ≤ 100 can arrive slightly later and still be one of the first 100 to arrive. This argument shows that in a Nash equilibrium persons 1 through 100 choose the same waiting time t∗ with v101 ≤ t∗ ≤ v100 , all the remaining people choose waiting times of at most t∗ , and at least one of the remaining people chooses a waiting time equal to t∗ . Any such action profile is a Nash equilibrium: any person i with i ≤ 100 obtains a smaller payoff if she arrives earlier and a payoff of zero if she arrives later. Any person i with i ≥ 101 obtains a negative payoff if she arrives before the first 100 people and a payoff of zero if she arrives at or after the first 100 people. Thus the set of Nash equilibria is the set of action profiles (t1 , . . . , t200 ) in which t1 = · · · = t100, this common waiting time, say t∗ , satisfies v101 ≤ t∗ ≤ v100, ti ≥ t∗ for all i ≥ 101, and t j = t∗ for some j ≥ 101. When goods are rationed by line-ups in the world, people in general do not all arrive at the same time. The feature missing from the model that seems to explain the dispersion in arrival times is uncertainty on the part of each player about the other players’ valuations. 88.2 Internet pricing The situation may be modeled as a multi-unit auction in which k units are available, and each player attaches a positive value to only one unit and submits a bid for only one unit. The k highest bids win, and each winner pays the (k + 1)st highest bid. By a variant of the argument for a second-price auction, in which “highest of the other players’ bids” is replaced by “highest rejected bid”, each player’s action of bidding her value is weakly dominates all her other actions. 44 Chapter 3. Nash Equilibrium: Illustrations 94.3 Alternative standards of care under negligence with contributory negligence First consider the case in which X1 = aˆ 1 and X2 ≤ aˆ 2 . The pair ( aˆ 1 , aˆ 2 ) is a Nash equilibrium by the following argument. If a2 = aˆ 2 then the victim’s level of care is sufficient (at least X2 ), so that the injurer’s payoff is given by (91.1) in the text. Thus the argument that the injurer’s action aˆ 1 is a best response to aˆ 2 is exactly the same as the argument for the case X2 = aˆ 2 in the text. Since X1 is the same as before, the victim’s payoff is the same also, so that by the argument in the text the victim’s best response to aˆ 1 is aˆ 2 . Thus ( aˆ 1 , aˆ 2 ) is a Nash equilibrium. To show that ( aˆ 1 , aˆ 2 ) is the only Nash equilibrium of the game, we study the players’ best response functions. First consider the injurer’s best response function. As in the text, we split the analysis into three cases. a2 < X2 : In this case the injurer does not have to pay any compensation, regardless of her level of care; her payoff is −a1 , so that her best response is a1 = 0. a2 = X2 : In this case the injurer’s best response is aˆ 1 , as argued when showing that ( aˆ 1 , aˆ 2 ) is a Nash equilibrium. a2 > X2 : In this case the injurer’s best response is at most aˆ 1 , since her payoff is equal to −a1 for larger values of a1 . Thus the injurer’s best response takes a form like that shown in the left panel of Figure 44.1. (In fact, b1 (a2 ) = aˆ 1 for X2 ≤ a2 ≤ aˆ 2 , but the analysis depends only on the fact that b1 (a2 ) ≤ aˆ 1 for a2 > X2 .) ↑ a2 b1 (a2 ) aˆ 2 X2 0 ↑ a2 X2 aˆ 1 a1 → 0 b2 (a1 ) ? aˆ 1 a1 → Figure 44.1 The players’ best response functions under the rule of negligence with contributory negligence when X1 = aˆ 1 and X2 = aˆ 2 . Left panel: the injurer’s best response function b1 . Right panel: the victim’s best response function b2 . (The position of the victim’s best response function for a1 > aˆ 1 is not significant, and is not determined in the solution.) is Now consider the victim’s best response function. The victim’s payoff function −a2 if a1 < aˆ 1 and a2 ≥ X2 u2 (a1 , a2 ) = −a2 − L(a1 , a2 ) if a1 ≥ aˆ 1 or a2 < X2 . Chapter 3. Nash Equilibrium: Illustrations 45 As before, for a1 < aˆ 1 we have −a2 − L(a1 , a2 ) < − aˆ 2 for all a2 , so that the victim’s best response is X2 . As in the text, the nature of the victim’s best responses to levels of care a1 for which a1 > aˆ 1 are not significant. Combining the two best response functions we see that ( aˆ 1 , aˆ 2 ) is the unique Nash equilibrium of the game. Now consider the case in which X1 = M and a2 = aˆ 2 , where M ≥ aˆ 1 . The injurer’s payoff is −a1 − L(a1 , a2 ) if a1 < M and a2 ≥ aˆ 2 u1 (a1 , a2 ) = −a1 if a1 ≥ M or a2 < aˆ 2 . Now, the maximizer of −a1 − L(a1 , aˆ 2 ) is aˆ 1 (see the argument following (91.1) in the text), so that if M is large enough then the injurer’s best response to aˆ 2 is aˆ 1 . As before, if a2 < aˆ 2 then the injurer’s best response is 0, and if a2 > aˆ 2 then the injurer’s payoff decreases for a1 > M, so that her best response is less than M. The injurer’s best response function is shown in the left panel of Figure 45.1. ↑ a2 ↑ a2 b1 (a2 ) aˆ 2 0 aˆ 2 aˆ 1 M a1 → 0 b2 (a1 ) ? aˆ 1 M a1 → Figure 45.1 The players’ best response functions under the rule of negligence with contributory negligence when (X1 , X2 ) = (M, aˆ 2 ), with M ≥ aˆ 1 . Left panel: the injurer’s best response function b1 . Right panel: the victim’s best response function b2 . (The position of the victim’s best response function for a1 > M is not significant, and is not determined in the text.) The victim’s payoff is u2 (a1 , a2 ) = −a2 −a2 − L(a1 , a2 ) if a1 < M and a2 ≥ aˆ 2 if a1 ≥ M or a2 < aˆ 2 . If a1 ≤ aˆ 1 then the victim’s best response is aˆ 2 by the same argument as the one in the text. If a1 is such that aˆ 1 < a1 < M then the victim’s best response is at most aˆ 2 (since her payoff is decreasing for larger values of a2 ). This information about the victim’s best response function is recorded in the right panel of Figure 45.1; it is sufficient to deduce that ( aˆ 1 , aˆ 2 ) is the unique Nash equilibrium of the game. 94.4 Equilibrium under strict liability In this case the injurer’s payoff is −a1 − L(a1 , a2 ) and the victim’s is −a2 for all (a1 , a2 ). Thus the victim’s optimal action is 0, regardless of the injurer’s action. 46 Chapter 3. Nash Equilibrium: Illustrations (The victim takes no care, given that, regardless of her level of care, the injurer is obliged to compensate her for any loss.) Thus in a Nash equilibrium the injurer chooses the level of care that maximizes −a1 − L(a1 , 0) and the victim chooses a2 = 0. If the function −a1 − L(a1 , 0) has a unique maximizer then the game has a unique Nash equilibrium; if there are multiple maximizers then the game has many Nash equilibria, though the players’ payoffs are the same in all the equilibria. The relation between aˆ 1 and the equilibrium value of a1 depends on the character of L(a1 , a2 ). If, for example, L decreases more sharply as a1 increases when a2 = 0 than when a2 is positive, the equilibrium value of a1 exceeds aˆ 1 . Draft of solutions to exercises in chapter of An introduction to game theory by Martin J. Osborne

[email protected]; www.chass.utoronto.ca/~osborne/index.html Version: 00/11/6. c 1995–2000 by Martin J. Osborne. All rights reserved. No part of this book may be reCopyright produced in any form by any electronic or mechanical means (including photocopying, recording, or information storage and retrieval) without permission in writing from Martin J. Osborne. On request, permission to make one copy for each student will be granted to instructors who wish to use the book in a course, on condition that copies be sold at a price not more than the cost of duplication. 4 Mixed strategy equilibrium 99.1 Variant of Matching Pennies The analysis is the same as for Matching Pennies. There is a unique steady state, in which each player chooses each action with probability 12 . 104.1 Extensions of BoS with vNM preferences In the first case, when player 1 is indifferent between going to her less preferred concert in the company of player 2 and the lottery in which with probability 12 she and player 2 go to different concerts and with probability 12 they both go to her more preferred concert, the Bernoulli payoffs that represent her preferences satisfy the condition u1 (S, S) = 12 u1 (S, B) + 12 u1 (B, B). If we choose u1 (S, B) = 0 and u1 (B, B) = 2, then u1 (S, S) = 1. Similarly, for player 2 we can set u2 (B, S) = 0, u2 (S, S) = 2, and u2 (B, B) = 1. Thus the Bernoulli payoffs in the left panel of Figure 47.1 are consistent with the players’ preferences. In the second case, when player 1 is indifferent between going to her less preferred concert in the company of player 2 and the lottery in which with probability 34 she and player 2 go to different concerts and with probability 14 they both go to her more preferred concert, the Bernoulli payoffs that represent her preferences satisfy the condition u1 (S, S) = 34 u1 (S, B) + 14 u1 (B, B). If we choose u1 (S, B) = 0 and u1 (B, B) = 2 (as before), then u1 (S, S) = 12 . Similarly, for player 2 we can set u2 (B, S) = 0, u2 (S, S) = 2, and u2 (B, B) = 12 . Thus the Bernoulli payoffs in the right panel of Figure 47.1 are consistent with the players’ preferences. Bach Stravinsky Bach 2, 1 0, 0 Stravinsky 0, 0 1, 2 Bach Stravinsky Bach 2, 12 0, 0 Figure 47.1 The Bernoulli payoffs for two extensions of BoS. 47 Stravinsky 0, 0 1 2, 2 48 Chapter 4. Mixed strategy equilibrium 107.1 Expected payoﬀs For BoS, player 1’s expected payoff is shown in Figure 48.1. ↑ Player 1’s expected payoff 2 q=1 1 q= 1 2 1 2 1 q=0 p→ 1 0 Figure 48.1 Player 1’s expected payoff as a function of the probability p that she assigns to B in BoS, when the probability q that player 2 assigns to B is 0, 12 , and 1. For the game in Figure 19.1 in the book, player 1’s expected payoff is shown in Figure 48.2. ↑ Player 1’s expected payoff 2 3 q=1 3 2 q= 1 1 2 q=0 0 p→ 1 Figure 48.2 Player 1’s expected payoff as a function of the probability p that she assigns to Refrain in the game in Figure 19.1 in the book, when the probability q that player 2 assigns to Refrain is 0, 12 , and 1. 108.1 Examples of best responses For BoS: for q = 0 player 1’s unique best response is p = 0 and for q = 12 and q = 1 her unique best response is p = 1. For the game in Figure 19.1: for q = 0 player 1’s unique best response is p = 0, for q = 12 her set of best responses is the set of all her mixed strategies (all values of p), and for q = 1 her unique best response is p = 1. Chapter 4. Mixed strategy equilibrium 49 111.1 Mixed strategy equilibrium of Hawk–Dove Denote by ui a payoff function whose expected value represents player i’s preferences. The conditions in the problem imply that for player 1 we have u1 (Passive, Passive) = 12 u1 (Aggressive, Aggressive) + 12 u1 (Aggressive, Passive) and u1 (Passive, Aggressive) = 23 u1 (Aggressive, Aggressive) + 13 u1 (Passive, Passive). Given u1 (Aggressive, Aggressive) = 0 and u1 (Passive, Aggressive = 1, we have u1 (Passive, Passive) = 12 u1 (Aggressive, Passive) and 1 = 13 u1 (Passive, Passive), so that u1 (Passive, Passive) = 3 and u1 (Aggressive, Passive) = 6. Similarly, u2 (Passive, Passive) = 3 and u2 (Passive, Aggressive) = 6. Thus the game is given in the left panel of Figure 49.1. The players’ best response functions are shown in the right panel. The game has three mixed strategy Nash equilibria: ((0, 1), (1, 0)), (( 34 , 14 ), ( 34 , 14 )), and ((1, 0), (0, 1)). ↑ 1 q 3 4 Aggressive Passive Aggressive 0, 0 1, 6 B1 Passive 6, 1 3, 3 B2 0 3 4 1 p→ Figure 49.1 An extension of Hawk–Dove (left panel) and the players’ best response functions when randomization is allowed in this game (right panel). The probability that player 1 assigns to Aggressive is p and the probability that player 2 assigns to Aggressive is q. The disks indicate the Nash equilibria (two pure, one mixed). 50 Chapter 4. Mixed strategy equilibrium 111.2 Games with mixed strategy equilibria The best response functions for the left game are shown in the left panel of Figure 50.1. We see that the game has a unique mixed strategy Nash equilibrium (( 14 , 34 ), ( 23 , 13 )). The best response functions for the right game are shown in the right panel of Figure 50.1. We see that the mixed strategy Nash equilibria are ((0, 1), (1, 0)) and any ((p, 1 − p), (0, 1)) with 12 ≤ p ≤ 1. ↑ 1 q ↑ 1 q 2 3 B1 B1 B2 B2 0 1 4 1 p→ 0 1 2 1 p→ Figure 50.1 The players’ best response functions in the left game (left panel) and right game (right panel) in Exercise 111.2. The probability that player 1 assigns to T is p and the probability that player 2 assigns to L is q. The disks and the heavy line indicate Nash equilibria. 112.1 A coordination game The best response functions are shown in Figure 51.1. From the figure we see that the game has three mixed strategy Nash equilibria, ((1, 0), (1, 0)) (the pure strategy equilibrium (No effort, No effort)), ((0, 1), (0, 1)) (the pure strategy equilibrium (Effort, Effort)), and ((1 − c, c), (1 − c, c)). An increase in c has no effect on the pure strategy equilibria, and increases the probability that each player chooses to exert effort in the mixed strategy equilibrium (because this probability is precisely c). The pure Nash equilibria are not affected by the cost of effort because a change in c has no effect on the players’ rankings of the four outcomes. An increase in c reduces a player’s payoff to the action Effort, given the other player’s mixed strategy; the probability the other player assigns to Effort must increase in order to keep the player indifferent between No effort and Effort, as required in an equilibrium. 112.2 Swimming with sharks As argued in the question, if you swim today, your expected payoff is −πc + 2(1 − π), regardless of your friend’s action. If you do not swim today and your friend Chapter 4. Mixed strategy equilibrium ↑ 1 q 1−c 51 B1 B2 0 1−c 1 p→ Figure 51.1 The players’ best response functions in the coordination game in Exercise 112.1. The probability that player 1 assigns to No effort is p and the probability that player 2 assigns to No effort is q. The disks indicate the Nash equilibria (two pure, one mixed). does, then with probability π your friend is attacked and you do not swim tomorrow, and with probability 1 − π your friend is not attacked and you do swim tomorrow. Thus your expected payoff in this case is π · 0 + (1 − π) · 1 = 1 − π. If neither of you swims today then your expected payoff if you swim tomorrow is π(−c) + (1 − π) · 1 = −πc + 1 − π; if this is negative you prefer to stay on the beach tomorrow, getting a payoff of 0, and if it is positive you prefer to swim tomorrow, getting a payoff of −πc + 1 − π. The game is given in Figure 51.2. Swim today Wait Swim today −πc + 2(1 − π), −πc + 2(1 − π) 1 − π, −πc + 2(1 − π) Wait −πc + 2(1 − π), 1 − π max{0, −πc + 1 − π}, max{0, −πc + 1 − π} Figure 51.2 Swimming with sharks. To find the mixed strategy Nash equilibria, first note that if −πc + 1 − π > 0, or c < (1 − π)/π, then Swim today is the best response to both Swim today and Wait. Thus in this case there is a unique mixed strategy Nash equilibrium, in which both players choose Swim today. At the other extreme, if −πc + 2(1 − π) < 0, or c > 2(1 − π)/π, then Wait is the best response to both Swim today and Wait. Thus in this case there is a unique mixed strategy Nash equilibrium, in which neither of you swims today, and consequently neither of you swims tomorrow. In the intermediate case in which 0 < −πc + 2(1 − π) < 1 − π, or (1 − π)/π < c < 2(1 − π)/π, the best response to Swim today is Wait and the best response to Wait is Swim today. Denoting by q the probability that player 2 chooses Swim today, player 1’s expected payoff to Swim today is −πc + 2(1 − π) and her expected payoff to Wait is q(1 − π). (Because −πc + 2(1 − π) < 1 − π, we have −πc + 1 − π < 0, so that each player’s payoff if both players Wait is 0.) Thus player 1’s expected 52 Chapter 4. Mixed strategy equilibrium payoffs to her two actions are equal if and only if −πc + 2(1 − π) = q(1 − π), or q = [−πc + 2(1 − π)]/(1 − π). The same calculation implies that player 2’s expected payoffs to her two actions are equal if and only if the probability that player 1 assigns to Swim today is [−πc + 2(1 − π)]/(1 − π) = 2 − πc/(1 − π). We conclude that if (1 − π)/π < c < 2(1 − π)/π then the game has a unique mixed strategy Nash equilibrium, in which each person swims today with probability 2 − πc/(1 − π). If c = (1 − π)/π the payoffs simplify to those given in the left panel of Figure 52.1. The set of mixed strategy Nash equilibria in this case is the set of all mixed strategy pairs ((p, 1 − p), (q, 1 − q)) for which either p = 1 or q = 1. If c = 2(1 − π)/π the payoffs simplify to those given in the right panel of Figure 52.1. The set of mixed strategy Nash equilibria in this case is the set of all mixed strategy pairs ((p, 1 − p), (q, 1 − q)) for which either p = 0 or q = 0. Swim Wait Swim 1 − π, 1 − π 1 − π, 1 − π Wait 1 − π, 1 − π 0, 0 Swim Wait Swim 0, 0 1 − π, 0 Wait 0, 1 − π 0, 0 Figure 52.1 The game if Figure 51.2 for c = (1 − π)/π (left panel) and c = 2(1 − π)/π (right panel). If you were alone your expected payoff to swimming on the first day would be −πc + 2(1 − π); your expected payoff to staying out of the water on the first day and acting optimally on the second day would be max{0, −πc + 1 − π}. Thus if −πc + 2(1 − π) > 0, or c < 2(1 − π)/π, you swim on the first day (and stay out of the water on the second day if you get attacked on the first day), and if c > 2(1 − π)/π you stay out of the water on both days. In the presence of your friend, you also swim on the first day only if c < (1 − π)/π. If (1 − π)/π < c < 2(1 − π)/π you do not swim for sure on the first day as you would if you were alone, but rather swim with probability less than one. That is, the presence of your friend decreases the probability of your swimming on the first day when c lies in this range. (For other values of c your decision is the same whether or not you are alone.) 115.1 Choosing numbers a. To show that the pair of mixed strategies in the question is a mixed strategy equilibrium, it suffices to verify the conditions in Proposition 113.2. Thus, given that each player’s strategy specifies a positive probability for every action, it suffices to show that each action of each player yields the same expected payoff. Player 1’s expected payoff to each pure strategy is 1/K, because with probability 1/K player 2 chooses the same number, and with probability 1 − 1/K player 2 chooses a different number. Similarly, player 2’s Chapter 4. Mixed strategy equilibrium 53 expected payoff to each pure strategy is −1/K, because with probability 1/K player 1 chooses the same number, and with probability 1 − 1/K player 2 chooses a different number. Thus the pair of strategies is a mixed strategy Nash equilibrium. b. Let (p∗ , q∗ ) be a mixed strategy equilibrium, where p∗ and q∗ are vectors, the jth components of which are the probabilities assigned to the integer j by each player. Given that player 2 uses the mixed strategy q∗ , player 1’s expected payoff if she chooses the number k is q∗k . Hence if p∗k > 0 then (by the first condition in Proposition 113.2) we need q∗k ≥ q∗j for all j, so that, in particular, q∗k > 0 (q∗j cannot be zero for all j!). But player 2’s expected payoff if she chooses the number k is −pk , so given q∗k > 0 we need p∗k ≤ p∗j for all j (again by the first condition in Proposition 113.2), and, in particular, p∗k ≤ 1/K (p∗j cannot exceed 1/K for all j!). We conclude that any probability p∗k that is positive must be at most 1/K. The only possibility is that p∗k = 1/K for all k. A similar argument implies that q∗k = 1/K for all k. 115.2 Silverman’s game The game has no pure strategy Nash equilibrium in which the players’ integers are the same because either player can increase her payoff from 0 to 1 by naming the next higher integer. It has no Nash equilibrium in which the players’ integers are different because the losing player (the player whose payoff is −1) can increase her payoff to 1 by changing her integer to be one more than the other player’s integer. Thus the game has no pure strategy Nash equilibrium. To show that the pair of mixed strategies in the question is a mixed strategy equilibrium, it suffices to verify the conditions in Proposition 113.2. That is, it suffices to show that for each player, each action to which the player’s mixed strategy assigns positive probability yields the player the same expected payoff, and every other action yields her a payoff at most as large. The game is symmetric and the players’ strategies are the same, so we need to make an argument only for one player. Suppose player 2 uses the mixed strategy in the question. Player 1’s expected payoffs to her actions are as follows: 1: 1 3 ·0+ 1 3 · (−1) + 2: 1 3 ·1+ 1 3 ·0+ 1 3 3 or 4: 5: 1 3 · (−1) + · (−1) + 6–14: 1 3 1 3 1 3 15 or more: 1 3 1 3 · 1 = 0. · (−1) = 0. 1 3 ·1+ ·1+ · (−1) + 1 3 1 3 1 3 · (−1) = − 13 . · 0 = 0. · (−1) + · (−1) + 1 3 1 3 · 1 = − 13 . · (−1) + 1 3 · (−1) = −1. Thus the pair of strategies is a mixed strategy equilibrium. 54 Chapter 4. Mixed strategy equilibrium 115.3 Voter participation I verify that the conditions in Proposition 113.2 are satisfied. First consider a supporter of candidate A. If she votes then candidate A ties if all k − 1 of her comrades vote, an event with probability pk−1 , and otherwise candidate A loses. Thus her expected payoff is pk−1 − c. If she abstains, then candidate A surely loses, so her payoff is 0. Thus in an equilibrium in which 0 < p < 1 the first condition in Proposition 113.2 implies that p k−1 = c, or p = c1/(k−1). Now consider a supporter of candidate B who votes. With probability pk all of the supporters of candidate A vote, in which case the election is a tie; with probability 1 − pk at least one of the supporters of candidate A does not vote, in which case candidate B wins. Thus the expected payoff of a supporter of candidate B who votes is p k + 2(1 − pk ) − c. If the supporter of candidate B switches to abstaining, then • candidate B loses if all supporters of candidate A vote, an event with probability p k • candidate B ties if exactly k − 1 supporters of candidate A vote, an event with probability kp k−1 (1 − p) • candidate B wins if fewer than k − 1 supporters of candidate A vote, an event with probability 1 − p k − kp k−1 (1 − p). Thus a supporter of candidate B who switches from voting to abstaining obtains an expected payoff of kpk−1 (1 − p) + 2(1 − pk − kp k−1 (1 − p)) = 2 − (2 − k)pk − kp k−1 . Hence in order for it to be optimal for such a citizen to vote (i.e. in order for the second condition in Proposition 113.2 to be satisfied), we need pk + 2(1 − pk ) − c ≥ 2 − (2 − k)pk − kp k−1 , or kpk−1 (1 − p) + pk ≥ c. Finally, consider a supporter of candidate B who abstains. With probability pk all the supporters of candidate A vote, in which case the candidates tie; with probability 1 − pk at least one of the supporters of candidate A does not vote, in which Chapter 4. Mixed strategy equilibrium 55 case candidate B wins. Thus the expected payoff of a supporter of candidate B who abstains is p k + 2(1 − pk ). If this citizen instead votes, candidate B surely wins (she gets k + 1 votes, while candidate A gets at most k). Thus the citizen’s expected payoff is 2 − c. Hence in order for the citizen to wish to abstain, we need p k + 2(1 − pk ) ≥ 2 − c or c ≥ pk . In summary, for equilibrium we need p = c1/(k−1) and pk ≤ c ≤ kpk−1 (1 − p) + pk . Given p = c1/(k−1), c = p k−1 , so that the two inequalities are satisfied. Thus p = c1/(k−1) defines an equilibrium. As c increases, the probability p, and hence the expected number of voters, increases. 115.4 Defending territory (The solution to this problem, which corrects an error in Shubik (1982, 226), is due to Nick Vriend.) The game is shown in Figure 55.1, where each action (x, y) gives the number x of divisions allocated to the first pass and the number y allocated to the second pass. (3, 0) (2, 1) General A (1, 2) (0, 3) (2, 0) 1, −1 1, −1 −1, 1 −1, 1 General B (1, 1) (0, 2) −1, 1 −1, 1 1, −1 −1, 1 1, −1 1, −1 −1, 1 1, −1 Figure 55.1 The game in Exercise 115.4. Denote a mixed strategy of A by (p1 , p2 , p3 , p4 ) and a mixed strategy of B by (q1 , q2 , q3 ). First I argue that in every equilibrium q2 = 0. If q2 > 0 then A’s expected payoff to (3, 0) is less than her expected payoff to (2, 1), and her expected payoff to (0, 3) is less than her expected payoff to (1, 2), so that p1 = p4 = 0. But then B’s 56 Chapter 4. Mixed strategy equilibrium expected payoff to at least one of her actions (2, 0) and (0, 2) exceeds her expected payoff to (1, 1), contradicting q2 > 0. Now I argue that in every equilibrium q1 = q3 = 0. Given q2 = 0 we have q3 = 1 − q1 , and A’s payoffs are 2q1 − 1 to (3, 0) and to (2, 1), and 1 − 2q1 to (1, 2) and (0, 3). Thus if q1 < 12 then in any equilibrium we have p1 = p2 = 0. Then B’s action (2, 0) yields her a higher payoff than does (0, 2), so that in any equilibrium q1 = 1. But then A’s actions (3, 0) and (2, 1) both yield higher payoffs than do (1, 2) and (0, 3), contradicting p 1 = p2 = 0. Similarly, q1 > 12 is inconsistent with equilibrium. Hence in any equilibrium q1 = q3 = 12 . Now, given q1 = q3 = 12 , A’s payoffs to her four actions are all equal. Thus ((p1 , p2 , p3 , p4 ), (q1 , q2 , q3 )) is a Nash equilibrium if and only if B’s payoff to (2, 0) is the same as her payoff to (0, 2), and this payoff is at least her payoff to (1, 1). The first condition is −p1 − p2 + p3 + p4 = p1 + p2 − p3 − p4 , or p1 + p2 = p3 + p4 = 12 . Thus B’s payoff to (2, 0) and to (0, 2) is zero, and the second condition is p 1 − p 2 − p3 + p4 ≤ 0, or p1 + p 4 ≤ 12 (using p1 + p2 + p3 + p 4 = 1). We conclude that the set of mixed strategy Nash equilibria of the game is the set of strategy pairs ((p1 , 12 − p1 , 12 − p4 , p4 ), ( 12 , 0, 12 )) with p1 + p4 ≤ 12 . In this equilibrium general A splits her resources between the two passes with probability at least 12 (p 2 + p3 = 12 − p 1 + 12 − p4 = 1 − (p1 + p4 ) ≥ 12 ) while general B concentrates all of her resources in one or other of the passes (with equal probability). 118.1 Strictly dominated actions Denote the probability that player 1 assigns to T by p and the probability she assigns to M by r (so that the probability she assigns to B is 1 − p − r). A mixed strategy of player 1 strictly dominates T if and only if p + 4r > 1 and p + 3(1 − p − r) > 1, or if and only if 1 − 4r < p < 1 − 32 r. For example, the mixed strategies ( 14 , 14 , 12 ) and (0, 14 , 34 ) both strictly dominate T. 119.1 Eliminating dominated actions when ﬁnding equilibria Player 2’s action L is strictly dominated by the mixed strategy that assigns probability 14 to M and probability 34 to R (for example), so that we can ignore the action L. The players’ best response functions in the reduced game in which player 2’s actions are M and R are shown in Figure 57.1. We see that the game has a single mixed strategy Nash equilibrium, namely (( 23 , 13 ), (0, 12 , 12 )). 124.1 Equilibrium in the expert diagnosis game When E = rE + (1 − r)I the consumer is indifferent between her two actions when p = 0, so that her best response function has a vertical segment at p = 0. Chapter 4. Mixed strategy equilibrium 57 ↑ 1 q B2 1 2 B1 0 2 3 1 p→ Figure 57.1 The players’ best response functions in the game in Figure 119.1 after player 2’s action L has been eliminated. The probability assigned by player 1 to T is p and the probability assigned by player 2 to M is q. The best response function of player 1 is black and that of player 2 is gray. The disk indicates the unique Nash equilibrium. Referring to Figure 123.1 in the text, we see that the set of mixed strategy Nash equilibria correspond to p = 0 and π/π ≤ q ≤ 1. 125.1 Incompetent experts The payoffs are given in Figure 57.2. (The actions are the same as those in the game in which every expert is fully competent.) H D A π, −rE − (1 − r)[sI + (1 − s)E] rπ + (1 − r)[sπ + (1 − s)π], −E R (1 − r)sπ, −rE − (1 − r)[sI + (1 − s)I ] 0, −rE − (1 − r)I Figure 57.2 A game between a consumer with a problem and a not-fully-competent expert. Following the method in the text for the case s = 1, we find that in the case E > rE + (1 − r)I there is a unique mixed strategy equilibrium, in which the probability the expert’s strategy assigns to H is p∗ = E − [rE + (1 − r)I ] (1 − r)s(E − I ) and the probability the consumer’s strategy assigns to A is q∗ = π . π We see that q∗ is independent of s. That is, the degree of competence has no effect on consumer behavior: consumers do not become more, or less, wary. The fraction of experts who are honest is a decreasing function of s, so that greater incompetence (smaller s) leads to a higher fraction of honest experts: incompetence 58 Chapter 4. Mixed strategy equilibrium breeds honesty! The intuition is that when experts become less competent, the potential gain from ignoring their advice increases (since I < E), so that they need to be more honest to attract business. 125.2 Choosing a seller The game is given in Figure 58.1. Buyer 1 Seller 1 Seller 2 Buyer 2 Seller 1 Seller 2 1 1 1 − p1 , 1 − p2 2 (1 − p 1 ), 2 (1 − p 1 ) 1 1 − p2 , 1 − p1 (1 − p2 ), 12 (1 − p2 ) 2 Figure 58.1 The game in Exercise 125.2. The character of its equilibria depend on the value of (p1 , p2 ). If p1 = p2 = 1 every pair ((π1 , 1 − π1 ), ((π2 , 1 − π2 )) is a mixed strategy equilibrium (where πi is the probability of buyer i’s choosing seller 1) is a equilibrium. Now suppose that at least one price is less than 1. • If 12 (1 − p2 ) > 1 − p1 (i.e. p2 < 2p 1 − 1), each buyer’s action of approaching seller 2 strictly dominates her action of approaching seller 1. Thus the game has a unique mixed strategy equilibrium, in which both buyers use a pure strategy: each approaches seller 2. • If 12 (1 − p2 ) = 1 − p 1 (i.e. p 2 = 2p1 − 1), every mixed strategy is a best response of a buyer to the other buyer’s approaching seller 2, and the pure strategy of approaching seller 2 is the unique best response to the other buyer’s using any other strategy. Thus ((π1 , 1 − π1 ), ((π2 , 1 − π2 )) is a mixed strategy equilibrium if and only if either π1 = 0 or π2 = 0. • If 12 (1 − p 1 ) > 1 − p2 (i.e. p 2 > 12 (1 + p 1 )), each buyer’s action of approaching seller 1 strictly dominates her action of approaching seller 2. Thus the game has a unique mixed strategy equilibrium, in which both buyers use a pure strategy: each approaches seller 1. • If 12 (1 − p 1 ) = 1 − p 2 (i.e. p2 = 12 (1 + p 1 )), every mixed strategy is a best response of a buyer to the other buyer’s strategy of approaching seller 1, and the pure strategy of approaching seller 1 is the unique best response to any other strategy of the other buyer. Thus ((π1 , 1 − π1 ), ((π2 , 1 − π2 )) is a mixed strategy equilibrium if and only if either π1 = 1 or π2 = 1. • For the case 2p1 − 1 < p2 < 12 (1 + p 1 ), a buyer’s expected payoff to choosing each seller is the same when 1 2 (1 − p1 )π + (1 − p1 )(1 − π) = (1 − p2 )π + 12 (1 − p 2 )(1 − π), Chapter 4. Mixed strategy equilibrium 59 where π is the probability that the other buyer chooses seller 1, or when π= 1 − 2p 1 + p 2 . 2 − p1 − p2 The players’ best response functions are shown in Figure 59.1. We see that the game has three mixed strategy equilibria: two pure equilibria in which the buyers approach different sellers, and one mixed strategy equilibrium in which each buyer approaches seller 1 with probability (1 − 2p1 + p2 )/(2 − p1 − p2 ). ↑1 π2 Buyer 1 1−2p 1 +p 2 2−p 1 −p 2 Buyer 2 0 1−2p 1 +p 2 2−p 1 −p 2 1 π1 → Figure 59.1 The players’ best response functions in the game in Exercise 125.2. The probability with which buyer i approaches seller 1 is πi . The three main cases are illustrated in Figure 60.1. If the prices are relatively close, there are two pure strategy equilibria, in which the buyers choose different sellers, and a symmetric mixed strategy equilibrium in which both buyers approach seller 1 with the same probability. If seller 2’s price is high relative to seller 1’s, there is a unique equilibrium, in which both buyers approach seller 1. If seller 1’s price is high relative to seller 2’s, there is a unique equilibrium, in which both buyers approach seller 2. 127.2 Approaching cars The game has three Nash equilibria: (Stop, Continue), (Continue, Stop), and a mixed strategy equilibrium in which each player chooses Stop with probability 1− . 2− Only the mixed strategy equilibrium is symmetric; the expected payoff of each player in this equilibrium is 2(1 − )/(2 − ). 60 Chapter 4. Mixed strategy equilibrium 1 ↑ p2 Pure equilibrium: both buyers approach 1) seller 1 +p 1 (1 2p Two pure equilibria (buyers approach different sellers) and one symmetric mixed equilibrium 1 −1 2 Pure equilibrium: both buyers approach seller 2 p1 → 0 1 Figure 60.1 Equilibria of the game in Exercise 125.2 as a function of the sellers’ prices. The modified game also has a unique symmetric equilibrium. In this equilibrium each player chooses Stop with probability 1−+δ 2− if δ ≤ 1 and chooses Stop with probability 1 if δ ≥ 1. The expected payoff of each player in this equilibrium is (2(1 − ) + δ)/(2 − ) if δ ≤ 1 and 1 if δ ≥ 1, both of which are larger than her payoff in the original game (given δ > 0). After reeducation, each driver’s payoffs to stopping stay the same, while those to continuing fall. Thus if the behavioral norm (the probability of stopping) were to remain the same, every driver would find it beneficial to stop. Equilibrium is restored only if enough drivers switch to Stop, raising everyone’s expected payoff. (Each player’s expected payoff in a mixed strategy Nash equilibrium is her expected payoff to choosing Stop, which is p + (1 − )(1 − p), where p is the probability of a player’s choosing Stop.) 128.1 Bargaining The game is given in Figure 61.1. By inspection it has a single symmetric pure strategy Nash equilibrium, (10, 10). Now consider situations in which the common mixed strategy assigns positive probability to two actions. Suppose that player 2 assigns positive probability only Chapter 4. Mixed strategy equilibrium 0 2 4 6 8 10 0 5, 5 6, 4 7, 3 8, 2 9, 1 10, 0 2 4, 6 5, 5 6, 4 7, 3 8, 2 0, 0 61 4 3, 7 4, 6 5, 5 6, 4 0, 0 0, 0 6 2, 8 3, 7 4, 6 0, 0 0, 0 0, 0 8 1, 9 2, 8 0, 0 0, 0 0, 0 0, 0 10 0, 10 0, 0 0, 0 0, 0 0, 0 0, 0 Figure 61.1 A bargaining game. to 0 and 2. Then player 1’s payoff to her action 4 exceeds her payoff to either 0 or 2. Thus there is no symmetric equilibrium in which the actions assigned positive probability are 0 and 2. By a similar argument we can rule out equilibria in which the actions assigned positive probability are any pair except 2 and 8, or 4 and 6. If the actions to which player 2 assigns positive probability are 2 and 8 then player 1’s expected payoffs to 2 and 8 are the same if the probability player 2 assigns to 2 is 25 (and the probability she assigns to 8 is 35 ). Given these probabilities, player 1’s expected payoff to her actions 2 and 8 is 16 5 , and her expected payoff to . Thus the pair of mixed strategies in which every every other action is less than 16 5 player assigns probability 25 to 2 and 35 to 8 is a symmetric mixed strategy Nash equilibrium. Similarly, the game has a symmetric mixed strategy equilibrium (α∗ , α∗ ) in which α∗ assigns probability 45 to the demand of 4 and probability 15 to the demand of 6. In summary, the game has three symmetric mixed strategy Nash equilibria in which each player’s strategy assigns positive probability to at most two actions: one in which probability 1 is assigned to 10, one in which probability 25 is assigned to 2 and probability 35 is assigned to 8, and one in which probability 45 is assigned to 4 and probability 15 is assigned to 6. 130.1 Contributing to a public good In a mixed strategy equilibrium each player obtains the same expected payoff whether or not she contributes. A player’s contribution makes a difference to the outcome only if exactly k − 1 of the other players contribute. Thus the difference between the expected benefit of contributing and that of not contributing is vQn−1,k−1(p) − c, which must be 0 in a mixed strategy equilibrium. For v = 1, n = 4, k = 2, and c = 38 this equilibrium condition is Q3,1 (p) = 38 . 62 Chapter 4. Mixed strategy equilibrium Now, Q3,1 (p) = 3p(1 − p)2 , so an equilibrium value of p satisfies 3p(1 − p)2 = 38 , or p3 − 2p 2 + p − 1 8 = 0, or (p − 12 )(p2 − 32 p + 14 ) = 0. Thus p = 12 or p = 34 − 12 54 ≈ 0.19. (The other root of the quadratic is greater than one, and thus not meaningful as a solution of the problem.) We conclude that the game has two symmetric mixed strategy Nash equilibria: one in which the common probability is 12 and one in which this probability is 3 1 5 − 4 2 4. 133.1 Best response dynamics in Cournot’s duopoly game The best response functions of both firms are the same, so if the firms’ outputs are initially the same, they are the same in every period: qt1 = qt2 for every t. For each period t, we thus have qti = 12 (α − c − qti ). Given that q1i = 0 for i = 1, 2, solving this first-order difference equation we have qti = 13 (α − c)[1 − (− 12 )t−1 ] for each period t. When t is large, qti is close to 13 (α − c), a firm’s equilibrium output. 5 (α − In the first few periods, these outputs are 0, 12 (α − c), 14 (α − c), 38 (α − c), 16 c). 133.2 Best response dynamics in Bertrand’s duopoly game If pi > c + 1 then firm j has a unique best response, equal to the lesser of pi − 1 and the monopoly price. Thus if both prices initially exceed c + 1 then for every period t in which at least one price exceeds c + 1 the maximal price in period t + 1 is (i) less than the maximal price in period t and (ii) at least c + 1. Thus the process converges to the Nash equilibrium (c + 1, c + 1). If pi = c then all prices p j ≥ c are best responses. Thus if the pair of prices is initially (c, c), many subsequent sequences of prices are consistent with best response dynamics. We can divide the sequences into three cases. • Both prices are equal to c in every subsequent period. • In some period both prices are at least c + 1, in which case eventually the Nash equilibrium (c + 1, c + 1) is reached (by the analysis for the first part of the exercise). Chapter 4. Mixed strategy equilibrium 63 • In every period one of the prices is equal to c, while the other price is greater than c; the identity of the firm charging c changes from period to period. The pairs of prices eventually alternate between (c, c + 1) and (c + 1, c) (neither of which are Nash equilibria). 136.1 Finding all mixed strategy equilibria of two-player games Left game: • There is no equilibrium in which each player’s mixed strategy assigns positive probability to a single action (i.e. there is no pure equilibrium). • Consider the possibility of an equilibrium in which one player assigns probability 1 to a single action while the other player assigns positive probability to both her actions. For neither action of player 1 is player 2’s payoff the same for both her actions, and for neither action of player 2 is player 1’s payoff the same for both her actions, so there is no mixed strategy equilibrium of this type. • Consider the possibility of a mixed strategy equilibrium in which each player assigns positive probability to both her actions. Denote by p the probability player 1 assigns to T and by q the probability player 2 assigns to L. For player 1’s expected payoff to her two actions to be the same we need 6q = 3q + 6(1 − q), or q = 23 . For player 2’s expected payoff to her two actions to be the same we need 2(1 − p) = 6p, or p = 14 . We conclude that the game has a unique mixed strategy equilibrium, (( 14 , 34 ), ( 23 , 13 )). Right game: • By inspection, (T, R) and (B, L) are the pure strategy equilibria. • Consider the possibility of a mixed strategy equilibrium in which one player assigns probability 1 to a single action while the other player assigns positive probability to both her actions. – {T} for player 1, {L, R} for player 2: no equilibrium, because player 2’s payoffs to (T, L) and (T, R) are not the same. – {B} for player 1, {L, R} for player 2: no equilibrium, because player 2’s payoffs to (B, L) and (B, R) are not the same. – {T, B} for player 1, {L} for player 2: no equilibrium, because player 1’s payoffs to (T, L) and (B, L) are not the same. 64 Chapter 4. Mixed strategy equilibrium – {T, B} for player 1, {R} for player 2: player 1’s payoffs to (T, R) and (B, R) are the same, so there is an equilibrium in which player 1 uses T with probability p if player 2’s expected payoff to R, which is 2p + 1 − p, is at least her expected payoff to L, which is p + 2(1 − p). That is, the game has equilibria in which player 1’s mixed strategy is (p, 1 − p), with p ≥ 12 , and player 2 uses R with probability 1. • Consider the possibility of an equilibrium in which both players assign positive probability to both their actions. Denote by q the probability that player 2 assigns to L. For player 1’s expected payoffs to T and B to be the same we need 0 = 2q, or q = 0, so there is no equilibrium in which both players assign positive probability to both their actions. In summary, the mixed strategy equilibria of the game are ((0, 1), (1, 0)) (i.e. the pure equilibrium (B, L)) and ((p, 1 − p), (0, 1)) for 12 ≤ p ≤ 1 (of which one equilibrium is the pure equilibrium (T, R)). 138.1 Finding all mixed strategy equilibria of a two-player game By inspection, (T, R) and (B, L) are pure strategy equilibria. Now consider the possibility of an equilibrium in which player 1’s strategy is pure while player 2’s strategy assigns positive probability to two or more actions. • If player 1’s strategy is T then player 2’s payoffs to M and R are the same, and her payoff to L is less, so an equilibrium in which player 2 randomizes between M and R is possible. In order that T be optimal we need 1 − q ≥ q, or q ≤ 12 , where q is the probability player 2’s strategy assigns to M. Thus every mixed strategy pair ((1, 0), (0, q, 1 − q)) in which q ≤ 12 is a mixed strategy equilibrium. • If player 1’s strategy is B then player 2’s payoffs to L and R are the same, and her payoff to M is less, so an equilibrium in which player 2 randomizes between L and R is possible. In order that B be optimal we need 2q + 1 − q ≤ 3q, or q ≥ 12 , where q is the probability player 2’s strategy assigns to L. Thus every mixed strategy pair ((0, 1), (q, 0, 1 − q)) in which q ≥ 12 is a mixed strategy equilibrium. Now consider the possibility of an equilibrium in which player 2’s strategy is pure while player 1’s strategy assigns positive probability to both her actions. For each action of player 2, player 1’s two actions yield her different payoffs, so there is no equilibrium of this sort. Next consider the possibility of an equilibrium in which both player 1’s and player 2’s strategies assign positive probability to two actions. Denote by p the probability player 1’s strategy assigns to T. There are three possibilities for the pair of player 2’s actions that have positive probability. Chapter 4. Mixed strategy equilibrium 65 L and M: For an equilibrium we need player 2’s expected payoff to L to be equal to her expected payoff to M and at least her expected payoff to R. That is, we need 2 = 3p + 1 − p ≥ 3p + 2(1 − p). The inequality implies that p = 1, so that player 1’s strategy assigns probability zero to B. Thus there is no equilibrium of this type. L and R: For an equilibrium we need player 2’s expected payoff to L to be equal to her expected payoff to R and at least her expected payoff to M. That is, we need 2 = 3p + 2(1 − p) ≥ 3p + 1 − p. The equation implies that p = 0, so there is no equilibrium of this type. M and R: For an equilibrium we need player 2’s expected payoff to M to be equal to her expected payoff to R and at least her expected payoff to L. That is, we need 3p + 1 − p = 3p + 2(1 − p) ≥ 2. The equation implies that p = 1, so there is no equilibrium of this type. The final possibility is that there is an equilibrium in which player 1’s strategy assigns positive probability to both her actions and player 2’s strategy assigns positive probability to all three of her actions. Let p be the probability player 1’s strategy assigns to T. Then for player 2’s expected payoffs to her three actions to be equal we need 2 = 3p + 1 − p = 3p + 2(1 − p). For the first equality we need p = 12 , violating the second equality. That is, there is no value of p for which player 2’s expected payoffs to her three actions are equal, and thus no equilibrium in which she chooses each action with positive probability. We conclude that the mixed strategy equilibria of the game are the strategy pairs of the forms ((1, 0), (0, q, 1 − q)) for 0 ≤ q ≤ 12 (q = 0 is the pure equilibrium (T, R)) and ((0, 1), (q, 0, 1 − q)) for 12 ≤ q ≤ 1 (q = 1 is the pure equilibrium (B, L)). 138.2 Rock, paper, scissors The game is shown in Figure 65.1. Rock Paper Scissors Rock 0, 0 1, −1 −1, 1 Paper −1, 1 0, 0 1, −1 Scissors 1, −1 −1, 1 0, 0 Figure 65.1 Rock, paper, scissors 66 Chapter 4. Mixed strategy equilibrium By inspection the game has no pure strategy equilibrium, and no mixed strategy equilibrium in which one player’s strategy is pure and the other’s is strictly mixed. In the remaining possibilities both players use at least two actions with positive probability. Suppose that player 1’s mixed strategy assigns positive probability to Rock and to Paper. Then player 2’s expected payoff to Paper exceeds her expected payoff to Rock, so in any such equilibrium player 2 must assign positive probability only to Paper and Scissors. Player 1’s expected payoffs to Rock and Paper are equal only if player 2 assigns probability 23 to Paper and probability 13 to Scissors. But then player 1’s expected payoff to Scissors exceeds her expected payoffs to Rock and Paper. So there is no mixed strategy equilibrium in which player 1 assigns positive probability only to Rock and to Paper. Given the symmetry of the game, the same argument implies that there is no equilibrium in which player 1 assigns positive probability to only two actions, nor any equilibrium in which player 2 assigns positive probability to only two actions. The remaining possibility is that each player assigns positive probability to all three of her actions. Denote the probabilities player 1 assigns to her three actions by (p 1 , p2 , p3 ) and the probabilities player 2 assigns to her three actions by (q1 , q2 , q3 ). Player 1’s actions all yield her the same expected payoff if and only if there is a value of c for which −q2 + q3 = c q1 −q1 + q2 − q3 = c = c. Adding the three equations we deduce c = 0, and hence q1 = q2 = q3 = 13 . A similar calculation for player 2 yields p1 = p2 = p3 = 13 . In conclusion, the game has a unique mixed strategy equilibrium, in which each player uses the strategy ( 13 , 13 , 13 ). Each player’s equilibrium payoff is 0. In the modified game in which player 1 is prohibited from using the action Scissors, player 2’s action Rock is strictly dominated. The remaining game has a unique mixed strategy equilibrium, in which player 1 chooses Rock with probability 13 and Paper with probability 23 , and player 2 chooses Paper with probability 23 and Scissors with probability 13 . The equilibrium payoff of player 1 is − 13 and that of player 2 is 1 3. 139.1 Election campaigns A strategic game that models the situation is shown in Figure 67.1, where action k means devote resources to locality k. By inspection the game has no pure strategy equilibrium and no equilibrium in which one player’s strategy is pure and the other is strictly mixed. (For each action of each player, the other player has a single best action.) Chapter 4. Mixed strategy equilibrium 1 Party A 2 3 67 1 0, 0 a2 , −a2 a3 , −a3 Party B 2 a1 , −a1 0, 0 a3 , −a3 3 a1 , −a1 a2 , −a2 0, 0 Figure 67.1 The game in Exercise 139.1. Now consider the possibility of an equilibrium in which party A assigns positive probability to exactly two actions. There are three possible pairs of actions. Throughout the argument I denote the probability party A’s strategy assigns to her action i by pi , and the probability party B’s strategy assigns to her action i by qi . 1 and 2: Party B’s action 3 is strictly dominated by her mixed strategy that assigns probability 12 to each of her actions 1 and 2, so that we can eliminate it from consideration. For party A’s actions 1 and 2 to yield the same expected payoff we need q2 a1 = q1 a2 , or, given q2 = 1 − q1 , q1 = a1 /(a1 + a2 ). For party B’s actions 1 and 2 to yield the same expected payoff we similarly need p1 = a2 /(a1 + a2 ). Finally, for party A’s expected payoff to her action 3 to be no more than her expected payoff to her other two actions, we need a3 ≤ a1 a2 . a1 + a2 We conclude that if a3 ≤ a1 a2 /(a1 + a2 ) (or equivalently a1 a3 + a2 a3 ≤ a1 a2 ) then the game has a mixed strategy equilibrium a1 a2 a1 a2 , ,0 , , ,0 . (67.1) a1 + a2 a1 + a2 a1 + a2 a1 + a2 1 and 3: Party B’s action 2 is strictly dominated her mixed strategy that assigns probability 12 to each of her actions 1 and 3, so that we can eliminate it from consideration. But then party A’s action 2 strictly dominates her action 3, so there is no equilibrium in which she assigns positive probability to action 3. Thus there is no equilibrium of this type. 2 and 3: For similar reasons, there is no equilibrium of this type. The remaining possibility is that there is an equilibrium in which each player assigns positive probability to all three of her actions. In order that party A’s actions yield the same expected payoff we need a1 (q2 + q3 ) = a2 (q1 + q3 ) = a3 (q1 + q2 ), or, using q1 + q2 + q3 = 1, q1 = a1 a2 + a1 a3 − a2 a3 , a1 a2 + a1 a3 + a2 a3 q2 = a1 a2 − a1 a3 + a2 a3 , a1 a2 + a1 a3 + a2 a3 q3 = −a1 a2 + a1 a3 + a2 a3 . a1 a2 + a1 a3 + a2 a3 (67.2) 68 Chapter 4. Mixed strategy equilibrium For these three numbers to be positive we need a1 a2 + a1 a3 − a2 a3 > 0, a1 a2 − a1 a3 + a2 a3 > 0, −a1 a2 + a1 a3 + a2 a3 > 0. Since a1 > a2 > a3 , these inequalities are satisfied if and only if a1 a3 + a2 a3 > a1 a2 . Similarly, in order that party B’s actions yield the same expected payoff we need p1 = a2 a3 , a1 a2 + a1 a3 + a2 a3 p2 = a1 a3 , a1 a2 + a1 a3 + a2 a3 p3 = a1 a2 . a1 a2 + a1 a3 + a2 a3 (68.1) These three numbers are positive, given ai > 0 for all i. Thus if a1 a3 + a2 a3 > a1 a2 there is an equilibrium in which player 1’s mixed strategy is (p1 , p2 , p3 ) and player 2’s mixed strategy is (q1 , q2 , q3 ). In summary, • if (a1 + a2 )a3 ≤ a1 a2 then the game has a unique mixed strategy equilibrium given by (67.1) • if (a1 + a2 )a3 > a1 a2 then the game has a unique mixed strategy equilibrium given by (67.2) and (68.1). That is, if the first two localities are sufficiently more valuable than the third then both parties concentrate all their efforts on these two localities, while otherwise they both randomize between all three localities. 139.2 A three-player game By inspection the game has two pure strategy equilibria, namely (A, A, A) and (B, B, B). Now consider the possibility of an equilibrium in which one or more of the players’ strategies is pure, and at least one is strictly mixed. If player 1 uses the action A and player 2 uses a strictly mixed strategy then player 3’s uniquely best action is A, in which case player 2’s uniquely best action is A. Thus there is no equilibrium in which player 1 uses the action A and at least one of the other players randomizes. By similar arguments, there is no equilibrium in which player 1 uses the action B and at least one of the other players randomizes, or indeed any equilibrium in which some player’s strategy is pure while some other player’s strategy is mixed. The remaining possibility is that there is an equilibrium in which each player’s strategy assigns positive probability to each of her actions. Denote the probabilities that players 1, 2, and 3 assign to A by p, q, and r respectively. In order that player 1’s expected payoffs to her two actions be the same we need qr = 4(1 − q)(1 − r). Chapter 4. Mixed strategy equilibrium 69 Similarly, for player 2’s and player 3’s expected payoffs to their two actions to be the same we need pr = 4(1 − p)(1 − r) and pq = 4(1 − p)(1 − q). The unique solution of these three equations is p = q = r = 23 (isolate r in the second equation and q in the third equation, and substitute into the first equation). We conclude that the game has three mixed strategy equilibria: ((1, 0), (1, 0), (1, 0)) (i.e. the pure strategy equilibrium (A, A, A)), ((0, 1), (0, 1), (0, 1)) (i.e. the pure strategy equilibrium (B, B, B)), and (( 23 , 13 ), ( 23 , 13 ), ( 23 , 13 )). 143.1 All-pay auction with many bidders Denote the common mixed strategy by F. Look for an equilibrium in which the largest value of z for which F(z) = 0 is 0 and the smallest value of z for which F(z) = 1 is z = K. A player who bids ai wins if and only if the other n − 1 players all bid less than she does, an event with probability (F(ai ))n−1 . Thus, given that the probability that she ties for the highest bid is zero, her expected payoff is (K − ai )(F(ai ))n−1 + (−ai )(1 − (F(ai ))n−1 ). Given the form of F, for an equilibrium this expected payoff must be constant for all values of ai with 0 ≤ ai ≤ K. That is, for some value of c we have K(F(ai ))n−1 − ai = c for all 0 ≤ ai ≤ K. For F(0) = 0 we need c = 0, so that F(ai ) = (ai /K)1/(n−1) is the only candidate for an equilibrium strategy. The function F is a cumulative probability distribution on the interval from 0 to K because F(0) = 0, F(K) = 1, and F is increasing. Thus F is indeed an equilibrium strategy. We conclude that the game has a mixed strategy Nash equilibrium in which each player randomizes over all her actions according to the probability distribution F(ai ) = (ai /K)1/(n−1); each player’s equilibrium expected payoff is 0. Each player’s mean bid is K/n. 143.2 Bertrand’s duopoly game Denote the common mixed strategy by F. If firm 1 charges p it earns a profit only if the price charged by firm 2 exceeds p, an event with probability 1 − F(p). Thus firm 1’s expected profit is (1 − F(p))(p − c)D(p). This profit is constant, equal to B, over some range of prices, if F(p) = 1 − B/((p − c)D(p)) over this range of prices. Because (p − c)D(p) increases without bound as 70 Chapter 4. Mixed strategy equilibrium p increases without bound, for any value of B the number F(p) approaches 1 as p increases without bound. Further, for any B > 0, there exists some p > c such that (p − c)D(p) = B, so that F(p) = 0. Finally, because (p − c)D(p) is an increasing function, so is F. Thus F is a cumulative probability distribution function. We conclude that for any p > c, the game has a mixed strategy equilibrium in which each firm’s mixed strategy is given by if p < p 0 (p − c)D(p) F(p) = 1− if p ≥ p. (p − c)D(p) 144.2 Preferences over lotteries The first piece of information about the decision-maker’s preferences among lotteries is consistent with her preferences being represented by the expected value of a payoff function. For example, set u(a1 ) = 0, u(a2 ) = 1, and u(a3 ) = 13 (or any number between 12 and 14 ). The second piece of information about the decision-maker’s preferences is not consistent with these preferences being represented by the expected value of a payoff function, by the following argument. For consistency with the information about the decision-maker’s preferences among the four lotteries, we need 0.4u(a1 ) + 0.6u(a3 ) > 0.5u(a2 ) + 0.5u(a3 ) > 0.3u(a1 ) + 0.2u(a2 ) + 0.5u(a3 ) > 0.45u(a1 ) + 0.55u(a3 ). The first inequality implies u(a2 ) < 0.8u(a1 ) + 0.2u(a3 ) and the last inequality implies u(a2 ) > 0.75u(a1 ) + 0.25u(a3 ). Because u(a1 ) < u(a3 ), we have 0.75u(a1 ) + 0.25u(a3 ) > 0.8u(a1 ) + 0.2u(a3 ), so that the two inequalities are incompatible. 146.2 Normalized vNM payoﬀ functions Let a be the best outcome according to her preferences and let a be the worse outcome. Let η = −u(a)/(u(a) − u(a)) and θ = 1/(u(a) − u(a)) > 0. Lemma 145.1 implies that the function v defined by v(x) = η + θu(x) represents the same preferences as does u; we have v(a) = 0 and v(a) = 1. 147.1 Games equivalent to the Prisoner’s Dilemma The left-hand game is not equivalent, by the following argument. Using either player’s payoffs, for equivalence we need η and θ > 0 such that 0 = η + θ · 0, 2 = η + θ · 1, 3 = η + θ · 2, and 4 = η + θ · 3. From the first equation we have η = 0 and hence from the second we have θ = 2. But these values do not satisfy the last two equations. (Alternatively, note that in Chapter 4. Mixed strategy equilibrium 71 the game in the left panel of Figure 104.1, player 1 is indifferent between (D, D) and the lottery in which (C, D) occurs with probability 12 and (D, C) occurs with probability 12 , while in the left-hand game in Figure 148.1 she is not.) The right-hand game is equivalent, by the following argument. For the equivalence of player 1’s payoffs, we need η and θ > 0 such that 0 = η + θ · 0, 3 = η + θ · 1, 6 = η + θ · 2, and 9 = η + θ · 3. The first two equations yield η = 0 and θ = 3; these values satisfy the second two equations. A similar argument for player 2’s payoffs yields η = −4 and θ = 2. Draft of solutions to exercises in chapter of An introduction to game theory by Martin J. Osborne

[email protected]; www.chass.utoronto.ca/~osborne/index.html Version: 00/11/6. c 1995–2000 by Martin J. Osborne. All rights reserved. No part of this book may be reCopyright produced in any form by any electronic or mechanical means (including photocopying, recording, or information storage and retrieval) without permission in writing from Martin J. Osborne. On request, permission to make one copy for each student will be granted to instructors who wish to use the book in a course, on condition that copies be sold at a price not more than the cost of duplication. 5 Extensive games with perfect information: Theory 154.2 Examples of extensive games with perfect information a. The game is given in Figure 73.1. 1 C D 2 2 E G F 1, 0 3, 2 H 2, 3 0, 1 Figure 73.1 The game in Exercise 154.2a. b. The game is specified as follows. Players 1 and 2. Terminal histories (C, E, G), (C, E, H), (C, F), D. Player function P(∅) = 1, P(C) = 2, P(C, E) = 1. Preferences Player 1 prefers (C, F) to D to (C, E, G) to (C, E, H); player 2 prefers (C, E, G) to (C, F) to (C, E, H), and is indifferent between this outcome and D. c. The game in shown in Figure 73.2, where the order of the payoffs is Karl, Rosa, Ernesto. K R E R B E H E B 1, 2, 1 E H 0, 0, 0 B 0, 0, 0 B H H B R H B 2, 1, 2 1, 2, 1 R 0, 0, 0 Figure 73.2 The game in Exercise 154.2c. 73 0, 0, 0 H 2, 1, 2 74 Chapter 5. Extensive games with perfect information: Theory 159.1 Strategies in extensive games In the entry game, the challenger moves only at the start of the game, where it has two actions, In and Out. Thus it has two strategies, In and Out. The incumbent moves only after the history In, when it has two actions, Acquiesce and Fight. Thus it also has two strategies, Acquiesce and Fight. In the game in Exercise 154.2c, Rosa moves after the histories R (Karl chooses her to move first), (E, B) (Karl chooses Ernesto to move first, and Ernesto chooses B), and (E, H) (Karl chooses Ernesto to move first, and Ernesto chooses H). In each case Rosa has two actions, B and H. Thus she has eight strategies. Each strategy takes the form (x, y, z), where each of x, y, and z are either B or H; the strategy (x, y, z) means that she chooses x after the history R, y after the history (E, B), and z after the history (E, H). 161.1 Nash equilibria of extensive games The strategic form of the game in Exercise 154.2a is given in Figure 74.1. C D EG 1, 0 2, 3 EH 1, 0 0, 1 FG 3, 2 2, 3 FH 3, 2 0, 1 Figure 74.1 The strategic form of the game in Exercise 154.2a. The Nash equilibria of the game are (C, FG), (C, FH), and (D, EG). The strategic form of the game in Figure 158.1 is given in Figure 74.2. CG CH DG DH E 1, 2 0, 0 2, 0 2, 0 F 3, 1 3, 1 2, 0 2, 0 Figure 74.2 The strategic form of the game in Figure 158.1. The Nash equilibria of the game are (CH, F), (DG, E), and (DH, E). 161.2 Voting by alternating veto The following extensive game models the situation. Players The two people. / ,Y / ), (X / ,Z / ), (Y / ,X / ), (Y / ,Z / ), (Z / ,X / ), and (Z / ,Y / ) (where A / means Terminal histories (X veto A). Chapter 5. Extensive games with perfect information: Theory 75 / ) = P(Y / ) = P(Z / ) = 2. P(∅) = 1 and P(X Player function Preferences Person 1’s preferences are represented by the payoff function u1 / ,Z / ) = u1 (Z / ,Y / ) = 2 (both of these terminal histories result in for which u1 (Y / ,Z / ) = u1 (Z / ,X / ) = 1, and u1 (X / ,Y / ) = u1 (Y / ,X / ) = 0. X’s being chosen), u1 (X Person 2’s preferences are represented by the payoff function u2 for which / ,Y / ) = u2 (Y / ,X / ) = 2, u2 (X / ,Z / ) = u2 (Z / ,X / ) = 1, and u2 (Y / ,Z / ) = u2 (Z / ,Y /)= u2 (X 0. This game is shown in Figure 75.1. 1 / X / Y 2 / Y / Z 0, 2 / X 1, 1 0, 2 / Z 2 2 / Z / X 2, 0 / Y 1, 1 2, 0 Figure 75.1 An extensive game that models the alternate strikeoff method of selecting an arbitrator, as specified in Exercise 161.2. /B /C / is person 2’s The strategic form of the game is given in Figure 75.2 (where A strategy in which it vetoes A if person 1 vetoes X, B if person 1 vetoes Y, and C if / ,Y /X /X / ) and (Z / ,Z /X /X / ). person 1 vetoes Z). Its Nash equilibria are (Z / X / Y / Z /X /X / Y 0, 2 0, 2 1, 1 /X /Y / Y 0, 2 0, 2 2, 0 /Z /X / Y 0, 2 2, 0 1, 1 /Z /Y / Y 0, 2 2, 0 2, 0 /X /X / Z 1, 1 0, 2 1, 1 /X /Y / Z 1, 1 0, 2 2, 0 /Z /X / Z 1, 1 2, 0 1, 1 /Z /Y / Z 1, 1 2, 0 2, 0 Figure 75.2 The strategic form of the game in Figure 75.1. 163.1 Subgames The subgames of the game in Exercise 154.2c are the whole game and the six games in Figure 76.1. 166.2 Checking for subgame perfect equilibria The Nash equilibria (CH, F) and (DH, E) are not subgame perfect equilibria: in the subgame following the history (C, E), player 1’s strategies CH and DH induce the strategy H, which is not optimal. The Nash equilibrium (DG, E) is a subgame perfect equilibrium: (a) it is a Nash equilibrium, so player 1’s strategy is optimal at the start of the game, given 76 Chapter 5. Extensive games with perfect information: Theory R B E H E E B H 1, 2, 1 0, 0, 0 B E B 1, 2, 1 2, 1, 2 B 1, 2, 1 E H 0, 0, 0 B 0, 0, 0 H H B R H 0, 0, 0 B R 0, 0, 0 0, 0, 0 R H 2, 1, 2 B 1, 2, 1 H 2, 1, 2 R H 0, 0, 0 B 0, 0, 0 H 2, 1, 2 Figure 76.1 The proper subgames of the game in Exercise 154.2c. player 2’s strategy, (b) in the subgame following the history C, player 2’s strategy E induces the strategy E, which is optimal given player 1’s strategy, and (c) in the subgame following the history (C, E), player 1’s strategy DG induces the strategy G, which is optimal. 171.2 Finding subgame perfect equilibria The game in Exercise 154.2a has a unique subgame perfect equilibrium, (C, FG). The game in Exercise 154.2c has a unique subgame perfect equilibrium in which Karl’s strategy is E, Rosa’s strategy is to choose B after the history R, B after the history (E, B), and H after the history (E, H), and Ernesto’s strategy is to chooses B after the history (R, B), H after the history (R, H), and H after the history E. (The outcome is that Karl chooses Ernesto to move first, he chooses H, and then Rosa chooses H.) The game in Figure 171.1 has six subgame perfect equilibria: (C, EG), (D, EG), (C, EH), (D, FG), (C, FH), (D, FH). 171.3 Voting by alternating veto / ,Y /X /X / ). The outcome is that The game has a unique subgame perfect equilibrium (Z action Y is taken. / ,Z /X /X / ) (see Exercise 161.2) is not a subgame perThus the Nash equilibrium (Z fect equilibrium. However, this equilibrium generates the same outcome as the unique subgame perfect equilibrium. If player 2 prefers Y to X to Z then in the unique subgame perfect equilibrium of the game in which player 1 moves first the outcome is that X is chosen, while in the unique subgame perfect equilibrium of the game in which player 2 moves first the outcome is that Y is chosen. (For all other strict preferences of player 2 (i.e. Chapter 5. Extensive games with perfect information: Theory 77 preferences in which player 2 is not indifferent between any pair of policies) the outcome of the subgame perfect equilibria of the two games are the same.) 171.4 Burning a bridge An extensive game that models the situation has the same structure as the entry game in Figure 154.1 in the book. The challenger is army 1, the incumbent army 2. The action In corresponds to attacking; Acquiesce corresponds to retreating. The game has a single subgame perfect equilibrium, in which army 1 attacks, and army 2 retreats. If army 2 burns the bridge, the game has a single subgame perfect equilibrium in which army 1 does not attack. 172.1 Sharing heterogeneous objects Let n = 2 and k = 3, and call the objects a, b, and c. Suppose that the values person 1 attaches to the objects are 3, 2, and 1 respectively, while the values player 2 attaches are 1, 3, 2. If player 1 chooses a on the first round, then in any subgame perfect equilibrium player 2 chooses b, leaving player 1 with c on the second round. If instead player 1 chooses b on the first round, in any subgame perfect equilibrium player 2 chooses c, leaving player 1 with a on the second round. Thus in every subgame perfect equilibrium player 1 chooses b on the first round (though she values a more highly.) Now I argue that for any preferences of the players, G(2, 3) has a subgame perfect equilibrium of the type described in the exercise. For any object chosen by player 1 in round 1, in any subgame perfect equilibrium player 2 chooses her favorite among the two objects remaining in round 2. Thus player 2 never obtains the object she least prefers; in any subgame perfect equilibrium, player 1 obtains that object. Player 1 can ensure she obtains her more preferred object of the two remaining by choosing that object on the first round. That is, there is a subgame perfect equilibrium in which on the first round player 1 chooses her more preferred object out of the set of objects excluding the object player 2 least prefers, and on the last round she obtains x3 . In this equilibrium, player 2 obtains the object less preferred by player 1 out of the set of objects excluding the object player 2 least prefers. That is, player 2 obtains x2 . (Depending on the players’ preferences, the game also may have a subgame perfect equilibrium in which player 1 chooses x3 on the first round.) 172.2 An entry game with a financially-constrained firm a. Consider the last period, after any history. If the incumbent chooses to fight, the challenger’s best action is to exit, in which case both firms obtain the 78 Chapter 5. Extensive games with perfect information: Theory profit zero. If the incumbent chooses to cooperate, the challenger’s best action is to stay in, in which case both firms obtain the profit C > 0. Thus the incumbent’s best action at the start of the period is to cooperate. Now consider period T − 1. Regardless of the outcome in this period, the incumbent will cooperate in the last period, and the challenger will stay in (as we have just argued). Thus each player’s action in the period affects its payoff only because it affects its profit in the period. Thus by the same argument as for the last period, in period T − 1 the incumbent optimally cooperates, and the challenger optimally stays in if the incumbent cooperates. If, in period T − 1, the incumbent fights, then the challenger also optimally stays in, because in the last period it obtains C > F. Working back to the start of the game, using the same argument in each period, we conclude that in every period before the last the incumbent cooperates and the challenger stays in regardless of the incumbent’s action. Given C > f , the challenger optimally enters at the start of the game. That is, the game has a unique subgame perfect equilibrium, in which • the challenger enters at the start of the game, exits in the last period if the challenger fights in that period, and stays in after every other history after which it moves • the incumbent cooperates after every history after which it moves. The incumbent’s payoff in this equilibrium is TC and the challenger’s payoff is TC − f . b. First consider the incumbent’s action after the history in which the challenger enters, the incumbent fights in the first T − 2 periods, and in each of these periods the challenger stays in. Denote this history h T−2 . If the incumbent fights after h T−2 , the challenger exits (it has no alternative), and the incumbent’s profit in the last period is M. If the incumbent cooperates after h T−2 then by the argument for the game in part a, the challenger stays in, and in the last period the incumbent also cooperates and the challenger stays in. Thus the incumbent’s payoff if it cooperates after the history h T−2 is 2C. Because M > 2C, we conclude that the incumbent fights after the history h T−2 . Now consider the incumbent’s action after the history in which the challenger enters, the incumbent fights in the first T − 3 periods, and in each period the challenger stays in. Denote this history h T−3 . If the incumbent fights after h T−3 , we know, by the previous paragraph, that if the challenger stays in then the incumbent will fight in the next period, driving the challenger out. Thus the challenger will obtain an additional profit of −F if it stays in and 0 if it exits. Consequently the challenger exits if the incumbent fights after h T−3 , making a fight by the incumbent optimal (it yields the incumbent the additional profit 2M). Chapter 5. Extensive games with perfect information: Theory 79 Working back to the first period we conclude that the incumbent fights and the challenger exits. Thus the challenger’s optimal action at the start of the game is to stay out. In summary, the game has a unique subgame perfect equilibrium, in which • the challenger stays out at the start of the game, exits after any history in which the incumbent fought in every period, exits in the last period if the incumbent fights in that period, and stays in after every other history. • the incumbent fights after the challenger enters and after any history in which it has fought in every period, and cooperates after every other history. The incumbent’s payoff in this equilibrium is TM and the challenger’s payoff is 0. 173.2 Dollar auction The game is shown in Figure 80.1. It has four subgame perfect equilibria. In all the equilibria player 2 passes after player 1 bids $2. After other histories the actions in the equilibria are as follows. • Player 1 bids $3 after the history ($1, $2), player 2 passes after the history $1, and player 1 bids $1 at the start of the game. • Player 1 passes after the history ($1, $2), player 2 passes after the history $1, and player 1 bids $1 at the start of the game. • Player 1 passes after the history ($1, $2), player 2 bids $2 after the history $1, and player 1 passes at the start of the game. • Player 1 passes after the history ($1, $2), player 2 bids $2 after the history $1, and player 1 bids $2 at the start of the game. There are three subgame perfect equilibrium outcomes: player 1 passes at the start of the game (player 2 gets the object without making any payment), player 1 bids $1 and then player 2 passes (player 1 gets the object for $1), and player 1 bids $2 and then player 2 passes (player 1 gets the object for $2). 174.2 Firm–union bargaining a. The following extensive game models the situation. Players The firm and the union. Terminal histories All sequences of the form (w, Y, L) and (w, N) for nonnegative numbers w and L (where w is a wage, Y means accept, N means reject, and L is the number of workers hired). 80 Chapter 5. Extensive games with perfect information: Theory p 1 $3 0, 2 $1 1, 0 −1, 0 p2 $2 $2 2 p $3 p1 $3 −1, 0 −1, −1 0, 0 $3 −2, −1 −1, −2 Figure 80.1 The extensive form of the dollar auction for w = 3 and v = 2. A pass is denoted p. Player function P(∅) is the union, and, for any nonnegative number w, P(w) and P(w, Y) are the firm. Preferences The firm’s preferences are represented by its profit, and the union’s preferences are represented by the value of wL (which is zero after any history (w, N)). b. First consider the subgame following a history (w, Y), in which the firm accepts the wage demand w. In a subgame perfect equilibrium, the firm chooses L to maximize its profit, given w. For L ≤ 50 this profit is L(100 − L) − wL, or L(100 − w − L). This function is a quadratic in L that is zero when L = 0 and when L = 100 − w and reaches a maximum in between. Thus the value of L that maximizes the firm’s profit is 12 (100 − w) if w ≤ 100, and 0 if w > 100. Given the firm’s optimal action in such a subgame, consider the subgame following a history w, in which the firm has to decide whether to accept or reject w. For any w the firm’s profit, given its subsequent optimal choice of L, is nonnegative; if w < 100 this profit is positive, while if w ≥ 100 it is 0. Thus in a subgame perfect equilibrium, the firm accepts any demand w < 100 and either accepts or rejects any demand w ≥ 100. Finally consider the union’s choice at the beginning of the game. If it chooses w < 100 then the firm accepts and chooses L = (100 − w)/2, yielding the union a payoff of w(100 − w)/2. If it chooses w > 100 then the firm either accepts and chooses L = 0 or rejects; in both cases the union’s payoff is 0. Thus the best value of w for the union is the number that maximizes w(100 − w)/2. This function is a quadratic that is zero when w = 0 and when w = 100 and reaches a maximum in between; thus its maximizer is w = 50. In summary, in a subgame perfect equilibrium the union’s strategy is w = 50, and the firm’s strategy accepts any demand w < 100 and chooses L = (100 − w)/2, and either rejects a demand w ≥ 100 or accepts such a demand and chooses L = 0. The outcome of any equilibrium is that the union demands Chapter 5. Extensive games with perfect information: Theory 81 w = 50 and the firm chooses L = 25. c. Yes. In any subgame perfect equilibrium the union’s payoff is (50)(25) = 1250 and the firm’s payoff is (25)(75) − (50)(25) = 625. Thus both parties are better off at the outcome (w, L) than they are in the unique subgame perfect equilibrium if and only if L ≤ 50 and wL > 1250 L(100 − L) − wL > 625 or L ≥ 50 and wL > 1250 2500 − wL > 625. These conditions are satisfied for a nonempty set of pairs (w, L). For example, if L = 50 the conditions are satisfied by 25 < w < 37.5; if L = 100 they are satisfied by 12.5 < w < 18.75. d. There are many Nash equilibria in which the firm “threatens” to reject high wage demands. In one such Nash equilibrium the firm threatens to reject any positive wage demand. In this equilibrium the union’s strategy is w = 0, and the firm’s strategy rejects any demand w > 0, and accepts the demand w = 0 and chooses L = 50. (The union’s payoff is 0 no matter what demand it makes; given w = 0, the firm’s optimal action is L = 50.) 175.1 The “rotten kid theorem” The situation is modeled by the following extensive game. Players The parent and the child. Terminal histories The set of sequences (a, t), where a (an action of the child) and t (a transfer from the parent to the child) are numbers. Player function P(∅) is the child, P(a) is the parent for every value of a. Preferences The child’s preferences are represented by the payoff function c(a) + t and the parent’s preferences are represented by the payoff function min{p(a) − t, c(a) + t}. To find the subgame perfect equilibria of this game, first consider the parent’s optimal actions in the subgames of length 1. Consider the subgame following the choice of a by the child. We have p(a) > c(a) (by assumption), so if the parent makes no transfer her payoff is c(a). If she transfers $1 to the child then her payoff increases to c(a) + 1. As she increases the transfer her payoff increases until p(a) − t = c(a) + t; that is, until t = 12 (p(a) − c(a)). (If she increases the transfer any 82 Chapter 5. Extensive games with perfect information: Theory more, she has less money than her child.) Thus the parent’s optimal action in the subgame following the choice of a by the child is t = 12 (p(a) − c(a)). Now consider the whole game. Given the parent’s optimal action in each subgame, a child who chooses a receives the payoff c(a) + 12 (p(a) − c(a)) = 12 (p(a) + c(a)). Thus in a subgame perfect equilibrium the child chooses the action that maximizes p(a) + c(a), the sum of her own private income and her parent’s income. 175.2 Comparing simultaneous and sequential games a. Denote by (a∗1 , a∗2 ) a Nash equilibrium of the strategic game in which player 1’s payoff is maximal in the set of Nash equilibria. Because (a∗1 , a∗2 ) is a Nash equilibrium, a∗2 is a best response to a∗1 . By assumption, it is the only best response to a∗1 . Thus if player 1 chooses a∗1 in the extensive game, player 2 must choose a∗2 in any subgame perfect equilibrium of the extensive game. That is, by choosing a∗1 , player 1 is assured of a payoff of at least u1 (a∗1 , a∗2 ). Thus in any subgame perfect equilibrium player 1’s payoff must be at least u1 (a∗1 , a∗2 ). b. Suppose that A1 = {T, B}, A2 = {L, R}, and the payoffs are those given in Figure 82.1. The strategic game has a unique Nash equilibrium, (T, L), in which player 2’s payoff is 1. The extensive game has a unique subgame perfect equilibrium, (B, LR) (where the first component of player 2’s strategy is her action after the history T and the second component is her action after the history B). In this subgame perfect equilibrium player 2’s payoff is 2. T B L 1, 1 0, 0 R 3, 0 2, 2 Figure 82.1 The payoffs for the example in Exercise 175.2a. c. Suppose that A1 = {T, B}, A2 = {L, R}, and the payoffs are those given in Figure 83.1. The strategic game has a unique Nash equilibrium, (T, L), in which player 2’s payoff is 2. A subgame perfect equilibrium of the extensive game is (B, RL) (where the first component of player 2’s strategy is her action after the history T and the second component is her action after the history B). In this subgame perfect equilibrium player 1’s payoff is 1. (If you read Chapter 4, you can find the mixed strategy Nash equilibria of the strategic game; in all these equilibria, as in the pure strategy Nash equilibrium, player 1’s expected payoff exceeds 1.) Chapter 5. Extensive games with perfect information: Theory T B L 2, 2 1, 1 83 R 0, 2 3, 0 Figure 83.1 The payoffs for the example in Exercise 175.2b. 176.1 Subgame perfect equilibria of ticktacktoe Player 2 puts her O in the center. If she does so, each player has a strategy that guarantees at least a draw in the subgame. Player 1 guarantees at least a draw by next marking one of the two squares adjacent to her first X and then subsequently completing a line of X’s, if possible, or, if not possible, blocking a line of O’s, if necessary, or, if not necessary, moving arbitrarily. Player 2 guarantees at least a draw as follows. • If player 1’s second X is adjacent to her first X or is in a corner not diagonally opposite player 1’s first X, player 2 should, on each move, either complete a line of O’s, if possible, or, if not possible, block a line of X’s, if necessary, or, if not necessary, move arbitrarily. • If player 1’s second X is in some other square then player 2 should, on her second move, mark one of the corners not diagonally opposite player 1’s first X, and then, on each move, either complete a line of O’s, if possible, or, if not possible, block a line of X’s, if necessary, or, if not necessary, move arbitrarily. For each of player 2’s other opening moves, player 1 has a strategy in the subgame that wins, as follows. • Suppose player 2 marks the corner diagonally opposite player 1’s first X. If player 1 next marks another corner, player 2 must next mark the square between player 1’s two X’s; by marking the remaining corner, player 1 wins on her next move. • Suppose player 2 marks one of the other corners. If player 1 next marks the corner diagonally opposite her first X, player 2 must mark the center, then player 1 must mark the remaining corner, leading her to win on her next move. • Suppose player 2 marks one of the two squares adjacent to player 1’s X. If player 1 next marks the center, player 2 must mark the corner opposite player 1’s first X, in which case player 1 can mark the other square adjacent to her first X, leading her to win on her next move. • Suppose player 2 marks one of the other squares, other than the center. If player 1 next marks the center, player 2 must mark the corner opposite player 1’s first X, in which case player 1 can mark the corner that blocks a row of O’s, leading her to win on her next move. 84 Chapter 5. Extensive games with perfect information: Theory 176.2 Toetacktick The following strategy leads to either a draw or a win for player 1: mark the central square initially, and on each subsequent move mark the square symmetrically opposite the one just marked by the second player. 177.1 Three Men’s Morris, or Mill Number the squares 1 through 9, starting at the top left, working across each row. The following strategy of player 1 guarantees she wins, so that the subgame perfect equilibrium outcome is that she wins. First player 1 chooses the central square (5). • Suppose player 2 then chooses a corner; take it to be square 1. Then player 1 chooses square 6. Now player 2 must choose square 4 to avoid defeat; player 1 must choose square 7 to avoid defeat; and then player 2 must choose square 3 to avoid defeat (otherwise player 1 can move from square 6 to square 3 on her next turn). If player 1 now moves from square 6 to square 9, then whatever player 2 does she can subsequently move her counter from square 5 to square 8 and win. • Suppose player 2 then chooses a noncorner; take it to be square 2. Then player 1 chooses square 7. Now player 2 must choose square 3 to avoid defeat; player 1 must choose square 1 to avoid defeat; and then player 2 must choose square 4 to avoid defeat (otherwise player 1 can move from square 5 to square 4 on her next turn). If player 1 now moves from square 7 to square 8, then whatever player 2 does she can subsequently move from square 8 to square 9 and win. Draft of solutions to exercises in chapter of An introduction to game theory by Martin J. Osborne

[email protected]; www.chass.utoronto.ca/~osborne/index.html Version: 00/11/6. c 1995–2000 by Martin J. Osborne. All rights reserved. No part of this book may be reCopyright produced in any form by any electronic or mechanical means (including photocopying, recording, or information storage and retrieval) without permission in writing from Martin J. Osborne. On request, permission to make one copy for each student will be granted to instructors who wish to use the book in a course, on condition that copies be sold at a price not more than the cost of duplication. 6 Extensive Games with Perfect Information: Illustrations 180.1 Nash equilibria of the ultimatum game For every amount x there are Nash equilibria in which person 1 offers x. For example, for any value of x there is a Nash equilibrium in which person 1’s strategy is to offer x and person 2’s strategy is to accept x and any offer more favorable, and reject every other offer. (Given person 2’s strategy, person 1 can do no better than offer x. Given person 1’s strategy, person 2 should accept x; whether person 2 accepts or rejects any other offer makes no difference to her payoff, so that rejecting all less favorable offers is, in particular, optimal.) 180.2 Subgame perfect equilibria of the ultimatum game with indivisible units In this case each player has finitely many actions, and for both possible subgame perfect equilibrium strategies of player 2 there is an optimal strategy for player 1. If player 2 accepts all offers then player 1’s best strategy is to offer 0, as before. If player 2 accepts all offers except 0 then player 1’s best strategy is to offer one cent (which player 2 accepts). Thus the game has two subgame perfect equilibria: one in which player 1 offers 0 and player 2 accepts all offers, and one in which player 1 offers one cent and player 2 accepts all offers except 0. 180.3 Dictator game and impunity game Dictator game Person 2 has no choice; person 1 optimally chooses the offer 0. Impunity game The analysis of the subgames of length one is the same as it is in the ultimatum game. That is, in any subgame perfect equilibrium person 2 either accepts all offers, or accepts all positive offers and rejects 0. Now consider the whole game. Regardless of person 2’s behavior in the subgames, person 1’s best action is to offer 0. Thus the game has two subgame perfect equilibria. In both equilibria person 1 offers 0. In one equilibrium person 2 accepts all offers, and in the other equilibrium she accepts all positive offers and rejects 0. The outcome of the first equilibrium is that person 1 offers 0, which person 2 accepts; the outcome of the second equilibrium is that person 1 offers 0, which person 2 rejects. In both equilibria person 1’s payoff is c and person 2’s payoff is 0. 85 86 Chapter 6. Extensive Games with Perfect Information: Illustrations 181.1 Variant of ultimatum game and impunity game with equity-conscious players Ultimatum game First consider the optimal response of person 2 to each possible offer. If person 2 accepts an offer x her payoff is x − β 2 |(1 − x) − x|, while if she rejects an offer her payoff is 0. Thus she accepts an offer x if x − β 2 |(1 − x) − x| > 0, or x − β 2 |1 − 2x| > 0, (86.1) rejects an offer x if x − β 2 |1 − 2x| < 0, and is indifferent between accepting and rejecting if x − β 2 |1 − 2x| = 0. Which values of x satisfy (86.1)? Because of the absolute value in the expression, we can conveniently consider the cases x ≤ 12 and x > 12 separately. • For x ≤ 1 2 the condition is x − β 2 (1 − 2x) > 0, or x > β 2 /(1 + 2β 2 ). • For x ≥ 12 the condition is x + β 2 (1 − 2x) > 0, or x(1 − 2β 2 ) + β 2 > 0. The values of x that satisfy this inequality depend on whether β 2 is greater than or less than 12 . β 2 ≤ 12 : All values of x satisfy the inequality. β 2 > 12 : The inequality is x < β 2 /(2β 2 − 1) (the right-hand side of which is less than 1 only if β 2 > 1). In summary, person 2 accepts any offer x with β 2 /(1 + 2β 2 ) < x < β 2 /(2β 2 − 1), may accept or reject the offers β 2 /(1 + 2β 2 ) and β 2 /(2β 2 − 1), and rejects any offer x with x < β 2 /(1 + 2β 2 ) or x > β 2 /(2β 2 − 1). The shaded region of Figure 86.1 shows, for each value of β 2 , the set of offers that person 2 accepts. Note, in particular, that, for every value of β 2 , person 2 accepts the offer 12 . 1 ↑ x β 2 /(2β 2 − 1) Offers accepted by person 2 β 2 /(1 + 2β 2 ) 0 1 β2 → 2 Figure 86.1 The set of offers x that person 2 accepts for each value of β 2 ≤ 2 in the variant of the ultimatum game with equity-conscious players studied in Exercise 181.1. Chapter 6. Extensive Games with Perfect Information: Illustrations 87 Now consider person 1’s decision. Her payoff is 0 if her offer is rejected and 1 − x − β 1 |(1 − x) − x| = 1 − x − β 1 |1 − 2x| if it is accepted. We can conveniently separate the analysis into three cases. β 1 < 12 : Person 1’s payoff when her offer x is accepted is positive for 0 ≤ x < 1 and is decreasing in x. Thus person 1’s optimal offer is the smallest one that person 2 accepts. If person 2’s strategy rejects the offer β 2 /(1 + 2β 2 ), then as in the analysis of the original game when person 2’s strategy rejects 0, person 1 has no optimal response. Thus in any subgame perfect equilibrium person 2 accepts β 2 /(1 + 2β 2 ), and person 1 offers this amount. β 1 = 12 : Person 1’s payoff to an offer that is accepted is positive and constant from x = 0 to x = 12 , then decreasing. Thus if person 2 accepts the offer β 2 /(1 + 2β 2 ) then every offer x with β 2 /(1 + 2β 2 ) ≤ x ≤ 12 is optimal, while if person 2 rejects the offer β 2 /(1 + 2β 2 ) then every offer x with β 2 /(1 + 2β 2 ) < x ≤ 12 is optimal. β 1 > 12 : Person 1’s payoff to an offer that is accepted is increasing up to x = and then decreasing, and is positive at x = 12 , so that her optimal offer is (which person 2 accepts). 1 2 1 2 We conclude that the set of subgame perfect equilibria depends on the values of β 1 and β 2 , as follows. β 1 < 12 : the set of subgame perfect equilibria is the set of all strategy pairs for which • person 1 offers β 2 /(1 + 2β 2 ) • person 2 accepts all offers x with β 2 /(1 + 2β 2 ) ≤ x < β 2 /(2β 2 − 1), rejects all offers x with x < β 2 /(1 + 2β 2 ) or x > β 2 /(2β 2 − 1), and either accepts or rejects the offer β 2 /(2β 2 − 1). β 1 = 12 : the set of subgame perfect equilibria is the set of all strategy pairs for which • person 1’s offer x satisfies β 2 /(1 + 2β 2 ) ≤ x ≤ 1 2 • person 2 accepts all offers x with β 2 /(1 + 2β 2 ) < x < β 2 /(2β 2 − 1), rejects all offers x with x < β 2 /(1 + 2β 2 ) or x > β 2 /(2β 2 − 1), either accepts or rejects the offer β 2 /(2β 2 − 1), and either accepts or rejects the offer β 2 /(1 + 2β 2 ) unless person 1 makes this offer, in which case person 2 definitely accepts it. β 1 > 12 : the set of subgame perfect equilibria is the set of all strategy pairs for which • person 1 offers 1 2 88 Chapter 6. Extensive Games with Perfect Information: Illustrations • person 2 accepts all offers x with β 2 /(1 + 2β 2 ) < x < β 2 /(2β 2 − 1), rejects all offers x with x < β 2 /(1 + 2β 2 ) or x > β 2 /(2β 2 − 1), and either accepts or rejects the offer β 2 /(2β 2 − 1) and the offer β 2 /(1 + 2β 2 ). The subgame perfect equilibrium outcomes are: β 1 < 12 : person 1 offers β 2 /(1 + 2β 2 ), which person 2 accepts β 1 = 12 : person 1 makes an offer x that satisfies β 2 /(1 + 2β 2 ) ≤ x ≤ person 2 accepts this offer 1 2, and β 1 > 12 : person 1 offers 12 , which person 2 accepts. In particular, in all cases the offer made by person 1 in equilibrium is accepted by person 2. Impunity game First consider the optimal response of person 2 to each possible offer. If person 2 accepts an offer x her payoff is x − β 2 |(1 − x) − x|, while if she rejects an offer her payoff is −β 2 (1 − x). Thus she accepts an offer x if x − β 2 |(1 − x) − x| > −β 2 (1 − x), or x(1 − β 2 ) + β 2 (1 − |1 − 2x|) > 0, (88.1) rejects an offer x if x(1 − β 2 ) + β 2 (1 − |1 − 2x|) < 0, and is indifferent between accepting and rejecting if x(1 − β 2 ) + β 2 (1 − |1 − 2x|) = 0. As before, we can conveniently consider the cases x ≤ 12 and x > 12 separately. • For x ≤ 1 2 the condition is x(1 + β 2 ) > 0, or x > 0. • For x ≥ 12 the condition is x(1 − 3β 2 ) + 2β 2 > 0, which is satisfied by all values of x if β 2 ≤ 13 , and for all x with x < 2β 2 /(3β 2 − 1) if β 2 > 13 . In summary, person 2 accepts any offer x with 0 < x < 2β 2 /(3β 2 − 1), may accept or reject the offers 0 and 2β 2 /(3β 2 − 1), and rejects any offer x with x > 2β 2 /(3β 2 − 1). Now consider person 1. If she offers x, her payoff is 1 − x − β 1 |1 − 2x| if person 1 accepts x 1 − x − β 1 (1 − x) if person 1 rejects x. If β 1 < 12 then in both cases person 1’s payoff is decreasing in x; for x = 0 the payoffs are equal. Thus, given person 2’s optimal strategy, in any subgame perfect equilibrium person 1’s optimal offer is 0, which person 2 may accept or reject. If β 1 = 12 then person 1’s payoff when person 2 accepts x is constant from 0 to 1 2 , then decreases. Her payoff when person 2 rejects x is decreasing in x, and the two payoffs are equal when x = 0. Thus the optimal offers of person 1 are 0, which person 2 may accept or reject, and any x with 0 < x ≤ 12 , which person 2 accepts. If β 1 > 12 then person 1’s highest payoff is obtained when x = 12 , which person 2 accepts. Thus x = 12 is her optimal offer. Chapter 6. Extensive Games with Perfect Information: Illustrations 89 In summary, in all subgame perfect equilibria the strategy of person 2 accepts all offers x with 0 < x < 2β 2 /(3β 2 − 1), rejects all offers x with x > 2β 2 /(3β 2 − 1), and either accepts or rejects the offer 0 and the offer 2β 2 /(3β 2 − 1). Person 1’s offer depends on the value of β 1 and β 2 , as follows. β 1 < 12 : person 1 offers 0 β 1 = 12 : person 1’s offer x satisfies 0 ≤ x ≤ 1 2 β 1 > 12 : person 1 offers x = 12 . The subgame perfect equilibrium outcomes are: β 1 < 12 : person 1 offers 0, which person 2 may accept or reject β 1 = 12 : person 1 either offers 0, which person 2 either accepts or rejects, or makes an offer x that satisfies 0 < x ≤ 12 , which person 2 accepts β 1 > 12 : person 1 offers 12 , which person 2 accepts. In particular, if β 1 ≤ 12 there are equilibria in which person 1 offers 0, and person 2 rejects this offer. Comparison of subgame perfect equilibria of ultimatum and impunity games The equilibrium outcomes of the two games are the same unless 0 < β 1 ≤ 12 , or β 1 = 0 and β 2 > 0, in which case person 1’s offer in the ultimatum game is higher than her offer in the impunity game. 183.1 Bargaining over two indivisible objects An extensive game that models the situation is shown in Figure 89.1, where the action (x, 2 − x) of player 1 means that she keeps x objects and offers 2 − x objects to player 2. 1 (2, 0) 2 yes 2, 0 no 0, 0 (1, 1) yes 1, 1 (0, 2) 2 2 no 0, 0 yes 0, 2 no 0, 0 Figure 89.1 An extensive game that models the procedure described in Exercise 183.1 for allocating two identical indivisible objects between two people. Denote a strategy of player 2 by a triple abc, where a is the action (y or n, for yes or no) taken after the offer (2, 0), b is the action taken after the offer (1, 1), and c is the action taken after the offer (0, 2). The subgame perfect equilibria of the game are ((2, 0), yyy) (resulting in the division (2, 0)), and ((1, 1), nyy) (resulting in the division (1, 1)). 90 Chapter 6. Extensive Games with Perfect Information: Illustrations The strategic form of the game is given in Figure 90.1. Its Nash equilibria are ((2, 0), yyy), ((2, 0), yyn), ((2, 0), yny), ((2, 0), ynn), ((2, 0), nny), ((1, 1), nyy), ((1, 1), nyn), ((0, 2), nny), and ((2, 0), nnn). The first four equilibria result in the division (2, 0), the next two result in the division (1, 1), and the last two result in the divisions (0, 2) and (0, 0) respectively. (2, 0) (1, 1) (0, 2) yyy 2, 0 1, 1 0, 2 yyn 2, 0 1, 1 0, 0 yny 2, 0 0, 0 0, 2 ynn 2, 0 0, 0 0, 0 nyy 0, 0 1, 1 0, 2 nyn 0, 0 1, 1 0, 0 nny 0, 0 0, 0 0, 2 nnn 0, 0 0, 0 0, 0 Figure 90.1 The strategic form of the game in Figure 89.1 The outcomes (0, 2) and (0, 0) are generated by Nash equilibria but not by any subgame perfect equilibria. 183.2 Dividing a cake fairly a. If player 1 divides the cake unequally then player 2 chooses the larger piece. Thus in any subgame perfect equilibrium player 1 divides the cake into two pieces of equal size. b. In a subgame perfect equilibrium player 2 chooses P2 over P1 , so she likes P2 at least as much as P1 . To show that in fact she is indifferent between P1 and P2 , suppose to the contrary that she prefers P2 to P1 . I argue that in this case player 1 can slightly increase the size of P1 in such a way that player 2 still prefers the now-slightly-smaller P2 . Precisely, by the continuity of player 2’s preferences, there is a subset P of P2 , not equal to P2 , that player 2 prefers to its complement C \ P (the remainder of the cake). Thus if player 1 makes the division (C \ P, P), player 2 chooses P. The piece P1 is a subset of C \ P not equal to C \ P, so player 1 prefers C \ P to P1 . Thus player 1 is better off making the division (C \ P, P) than she is making the division (P1 , P2 ), contradicting the fact that (P1 , P2 ) is a subgame perfect equilibrium division. We conclude that in any subgame perfect equilibrium player 2 is indifferent between the two pieces into which player 1 divides the cake. I now argue that player 1 likes P1 as least as much as P2 . Suppose that, to the contrary, she prefers P2 to P1 . If she deviates and makes a division (P, C \ P) in which P is slightly bigger than P1 but still such that she prefers C \ P to P, then player 2, who is indifferent between P1 and P2 , chooses P, leaving C \ P for player 1, who prefers it to P and hence to P1 . Thus in any subgame perfect equilibrium player 1 likes P1 at least as much as P2 . To show that player 1 may strictly prefer P1 to P2 , consider a cake that is perfectly homogeneous except for the presence of a single cherry. Assume that player 2 values a piece of the cherry in exactly the same way that she Chapter 6. Extensive Games with Perfect Information: Illustrations 91 values a piece of the cake of the same size, while player 1 prefers a piece of the cherry to a piece of the cake of the same size. Then there is a subgame perfect equilibrium in which player 1 divides the cake equally, with one piece containing all of the cherry, and player 2 chooses the piece without the cherry. (In this equilibrium, as in all equilibria, player 2 is indifferent between the two pieces—but note that there is no subgame perfect equilibrium in which she chooses the piece with the cherry in it. A strategy pair in which she acts in this way is not an equilibrium, because player 1 can deviate and increase slightly the size of the cherryless piece of cake, inducing player 2 to choose that piece.) 183.3 Holdup game The game is defined as follows. Players Two people, person 1 and person 2. Terminal histories The set of all sequences (low, x, Z), where x is a number with 0 ≤ x ≤ c L (the amount of money that person 1 offers to person 2 when the pie is small), and (high, x, Z), where x is a number with 0 ≤ x ≤ c H (the amount of money that person 1 offers to person 2 when the pie is large) and Z is either Y (“yes, I accept”) or N (“no, I reject”). Player function 2 for all x. P(∅) = 2, P(low) = P(high) = 1, and P(low, x) = P(high, x) = Preferences Person 1’s preferences are represented by payoffs equal to the amounts of money she receives, equal to c L − x for any terminal history (low, x, Y) with 0 ≤ x ≤ c L , equal to c H − x for any terminal history (high, x, Y) with 0 ≤ x ≤ c H , and equal to 0 for any terminal history (low, x, N) with 0 ≤ x ≤ c L and for any terminal history (high, x, N) with 0 ≤ x ≤ c H . Person 2’s preferences are represented by payoffs equal to x − L for the terminal history (low, x, Y), x − H for the terminal history (high, x, Y), −L for the terminal history (low, x, N), and −H for the terminal history (high, x, N). 186.1 Stackelberg’s duopoly game with quadratic costs From Exercise 57.2, the best response function of firm 2 is the function b2 defined by 1 (α − q1 ) if q1 ≤ α b2 (q1 ) = 4 0 if q1 > α. Firm 1’s subgame perfect equilibrium strategy is the value of q1 that maximizes q1 (α − q1 − b2 (q1 )) − q21 , or q1 (α − q1 − 14 (α − q1 )) − q21 , or 14 q1 (3α − 7q1 ). 3 The maximizer is q1 = 14 α. 92 Chapter 6. Extensive Games with Perfect Information: Illustrations We conclude that the game has a unique subgame perfect equilibrium, in which 3 α and firm 2’s strategy is its best response function firm 1’s strategy is the output 14 b2 . The outcome of the subgame perfect equilibrium is that firm 1 produces q∗1 = 3 3 11 ∗ 14 α units of output and firm 2 produces q2 = b2 ( 14 α) = 56 α units. In a Nash equilibrium of Cournot’s (simultaneous-move) game each firm produces 15 α (see Exercise 57.2). Thus firm 1 produces more in the subgame perfect equilibrium of the sequential game than it does in the Nash equilibrium of Cournot’s game, and firm 2 produces less. 188.1 Stackelberg’s duopoly game with ﬁxed costs We have f < (α − c)2 /16 ( f = 4; (α − c)2 /16 = 9), so the best response function of firm 2 takes the form shown in Figure 24.1 (in the solution to Exercise 57.3). To determine the subgame perfect equilibrium we need to compare firm 1’s profit when it produces q = 8 units of output, so that firm 2 produces 0, with its profit when it produces the output that maximizes its profit on the positive part of firm 2’s best response function. If firm 1 produces 8 units of output and firm 2 produces 0, firm 1’s profit is 8(12 − 8) = 32. Firm 1’s best output on the positive part of firm 2’s best response function is 12 (α − c) = 6. If it produces this output then firm 2 produces 12 (α − c − q1 ) = 12 (12 − 6) = 3, and firm 1’s profit is 6(12 − 9) = 18. Thus firm 1’s profit is higher when it produces enough to induce firm 2 to produce zero. We conclude that the game has a unique subgame perfect equilibrium, in which firm 1’s strategy is to produce 8 units, and firm 2’s strategy is to produce 12 (α − c − q1 ) = 12 (12 − q1 ) units if firm 1 produces q1 < 8 and 0 if firm 1 produces q1 ≥ 8 units. 189.1 Sequential variant of Bertrand’s duopoly game a. Players The two firms. Terminal histories The set of all sequences (p1 , p2 ) of prices (where each pi is a nonnegative number). Player function P(∅) = 1 and P(p1 ) = 2 for all p1 . Preferences The payoff of each firm i to the terminal history (p1 , p2 ) is its profit if pi < p j (pi − c)D(pi ) 1 (p − c)D(p ) if pi = p j i 2 i 0 if p i > p j , where j is the other firm. b. A strategy of firm 1 is a price (e.g. the price c). A strategy of firm 2 is a function that associates a price with every price chosen by firm 1 (e.g. Chapter 6. Extensive Games with Perfect Information: Illustrations 93 s2 (p1 ) = p 1 − 1, the strategy in which firm 2 always charges 1 cent less than firm 1). c. First consider firm 2’s best responses to each price p1 chosen by firm 1. • If p1 < c, any price greater than p1 is a best response for firm 2. • If p1 = c, any price at least equal to c is a best response for firm 2. • If p1 = c + 1, firm 2’s unique best response is to set the same price. • If p 1 > c + 1, firm 2’s unique best response is to set the price min{pm , p1 − 1} (where pm is the monopoly price). Now consider the optimal action of firm 1. Given firm 2’s best responses, • if p1 < c, firm 1’s profit is positive • if p1 = c, firm 1’s profit is zero • if p1 = c + 1, firm 1’s profit is positive • if p 1 > c + 1, firm 1’s profit is zero. Thus the only price p1 for which there is a best response of firm 2 that leads to a positive profit for firm 1 is c + 1. We conclude that in every subgame perfect equilibrium firm 1’s strategy is p1 = c + 1, and firm 2’s strategy assigns to each price chosen by firm 1 one of its best responses, so that firm 2’s strategy takes the form k(p1 ) k s2 (p1 ) = c+1 min{pm , p1 − 1} if if if if p1 p1 p1 p1 < = = > c c c+1 c+1 where k(p1 ) > p1 for all p1 and k ≥ c. The outcome of every subgame perfect equilibrium is that both firms choose the price c + 1. 193.1 Three interest groups buying votes a. Consider the possibility of a subgame perfect equilibrium in which bill X passes. In any such equilibrium, groups Y and Z make no payments. But now given that Y makes no payments and that VX = VZ , group Z can match X’s payments to the two legislators to whom X’s payments are smallest, and gain the passage of bill Z. Thus there is no subgame perfect equilibrium in which bill X passes. Similarly there is no subgame perfect equilibrium in which bill Y passes. Thus in every subgame perfect equilibrium bill Z passes. 94 Chapter 6. Extensive Games with Perfect Information: Illustrations b. By making payments of more than 50 to each legislator, group X ensures that neither group Y nor group Z can profitably buy the passage of its favorite bill. (In any subgame perfect equilibrium, group X’s payments to each legislator are exactly 50.) Thus in every subgame perfect equilibrium the outcome is that bill X is passed. c. For any payments of group X that sum to at most 300, group Y can make payments that are (i) at least as high to at least two legislators and (ii) high enough that group Z cannot buy off more than one legislator. (Take the two legislators to whom group X pays the least. Let them be legislators 1 and 2, and denote group X’s payments x 1 and x2 ; suppose that x1 ≥ x2 . Group Y pays x1 + 1 to legislator 1 and 200 − x1 to legislator 2.) Thus in every subgame perfect equilibrium the outcome is that bill Y is passed. 193.2 Interest groups buying votes under supermajority rule a. However group X allocates payments summing to 700, group Y can buy off five legislators for at most 500. Thus in any subgame perfect equilibrium neither group makes any payment, and bill Y is passed. b. If group X pays each legislator 80 then group Y is indifferent between buying off five legislators, in which case bill Y is passed, and in making no payments, in which case bill X is passed. If group Y makes no payments then X is selected, and group X is better off than it is if it makes no payments. There is no subgame perfect equilibrium in which group Y buys off five legislators, because if it were to do so group X could pay each legislator slightly more than 80 to ensure the passage of bill X. Thus in every subgame perfect equilibrium group X pays each legislator 80, group Y makes no payments, and bill X is passed. c. If only a simple majority is required to pass a bill, in case a the outcome under majority rule is the same as it is when five votes are required. In case b, group X needs to pay each legislator 100 in order to prevent group Y from winning. If it does so, its total payments are less than VX , so doing so is optimal. Thus in this case the payment to each legislator is higher under majority rule. 193.3 Sequential positioning by two political candidates The following extensive game models the situation. Players The candidates. Terminal histories The set of all sequences (x1 , . . . , x n ), where xi is a position of candidate i (a number) for i = 1, . . . , n. Chapter 6. Extensive Games with Perfect Information: Illustrations 95 Player function P(∅) = 1, P(x1 ) = 2 for all x1 , P(x1 , x2 ) = 3 for all (x1 , x2 ), . . . , P(x1 , . . . , xn−1 ) = n for all (x1 , . . . , x n−1 ). Preferences Each candidate’s preferences are represented by a payoff function that assigns n to every terminal history in which she wins outright, k to every terminal history in which she ties for first place with n − k other candidates, for 1 ≤ k ≤ n − 1, and 0 to every terminal history in which she loses, where positions attract votes as in Hotelling’s model of electoral competition (Section 3.3). This game has a finite horizon, so we may use backward induction to find its subgame perfect equilibria. Suppose there are two candidates. First consider candidate 2’s best response to each strategy of candidate 1. Suppose candidate 1’s strategy is m. Then candidate 2 loses if she chooses any position different from m and ties with candidate 1 if she chooses m. Thus candidate 2’s best response to m is m. Now suppose candidate 1’s strategy is x1 = m. Then candidate 2 wins if she chooses any position between x1 and 2m − x1 ; thus every such position is a best response. Given candidate 2’s best responses, the best strategy for candidate 1 is m, leading to a tie. (Every other strategy of candidate 1 leads her to lose.) We conclude that in every subgame perfect equilibrium candidate 1’s strategy is m; candidate 2’s strategy chooses m after the history m and some position between x1 and 2m − x1 after any other history x1 . 193.4 Sequential positioning by three political candidates The following extensive game models the situation. Players The candidates. Terminal histories The set of all sequences (x1 , . . . , x n ), where xi is either Out or a position of candidate i (a number) for i = 1, . . . , n. Player function P(∅) = 1, P(x1 ) = 2 for all x1 , P(x1 , x2 ) = 3 for all (x1 , x2 ), . . . , P(x1 , . . . , xn−1 ) = n for all (x1 , . . . , x n−1 ). Preferences Each candidate’s preferences are represented by a payoff function that assigns n to every terminal history in which she wins, k to every terminal history in which she ties for first place with n − k other candidates, for 1 ≤ k ≤ n − 1, 0 to every terminal history in which she stays out, and −1 to every terminal history in which she loses, where positions attract votes as in Hotelling’s model of electoral competition (Section 3.3). When there are two candidates the analysis of the subgame perfect equilibria is similar to that in the previous exercise. In every subgame perfect equilibrium candidate 1’s strategy is m; candidate 2’s strategy chooses m after the history m, 96 Chapter 6. Extensive Games with Perfect Information: Illustrations some position between x1 and 2m − x1 after the history x1 for any position x1 , and any position after the history Out. Now consider the case of three candidates when the voters’ favorite positions are distributed uniformly from 0 to 1. I claim that every subgame perfect equilibrium results in the first candidate’s entering at 12 , the second candidate’s staying out, and the third candidate’s entering at 12 . To show this, first consider the best response of candidate 3 to each possible pair of actions of candidates 1 and 2. Figure 96.1 illustrates these optimal actions in every case that candidate 1 enters. (If candidate 1 does not enter then the subgame is exactly the two-candidate game.) 1 tie 3 1 d an tie 2 wi ns In (near 12 ); 3 wins 1 2 wi ns d an 2 wins tie tie 1 wins 1, 2, ins 1w In (e.g. at 12 ) 3 wins 2 1 3 3 2 wins 2 2 3 In (e.g. at 12 ) 3 wins an d an d 1, 2, 1 wins ns 1 wi ↑ x2 In (e.g. at z) 3 wins 0 Out 1 3 In; 3 wins 1 2 In; 1 and 3 tie In (e.g. at z) 3 wins 2 3 x1 → 1 In; 3 wins Figure 96.1 The outcome of a best response of candidate 3 to each pair of actions by candidates 1 and 2. The best response for any point in the gray shaded area (including the black boundaries of this area, but excluding the other boundaries) is Out. The outcome at each of the four small disks at the outer corners of the shaded area is that all three candidates tie. The value of z is 1 − 12 (x1 + x2 ). Now consider the optimal action of candidate 2, given x1 and the outcome of candidate 3’s best response, as given in Figure 96.1. In the figure, take a value of x1 and look at the outcomes as x2 varies; find the value of x2 that induces the best outcome for candidate 2. For example, for x1 = 0 the only value of x2 for Chapter 6. Extensive Games with Perfect Information: Illustrations 97 which candidate 2 does not lose is 23 , at which point she ties with the other two candidates. Thus when candidate 1’s strategy is x1 = 0, candidate 2’s best action, given candidate 3’s best response, is x2 = 23 , which leads to a three-way tie. We find that the outcome of the optimal value of x2 , for each value of x1 , is given as follows. 1, 2, and 3 tie (x2 = 23 ) if x 1 = 0 if 0 < x1 < 12 2 wins 1 and 3 tie (2 stays out) if x1 = 12 if 12 < x1 < 1 2 wins 1 1, 2, and 3 tie (x2 = 3 ) if x 1 = 1. Finally, consider candidate 1’s best strategy, given the responses of candidates 2 and 3. If she stays out then candidates 2 and 3 enter at m and tie. If she enters then the best position at which to do so is x1 = 12 , where she ties with candidate 3. (For every other position she either loses or ties with both of the other candidates.) We conclude that in every subgame perfect equilibrium the outcome is that candidate 1 enters at 12 , candidate 2 stays out, and candidate 3 enters at 12 . (There are many subgame perfect equilibria, because after many histories candidate 3’s optimal action is not unique.) (If you’re interested in what may happen when there are many potential candidates, look at http://www.economics.utoronto.ca/osborne/research/CONJECT. HTM.) 195.1 The race G1 (2, 2) The consequences of player 1’s actions at the start of the game are as follows. Take two steps: Player 1 wins. Take one step: Go to the game G2 (1, 2), in which player 2 initially takes two steps and wins. Do not move: If player 2 does not move, the game ends. If she takes one step we go to the game G1 (2, 1), in which player 1 takes two steps and wins. If she takes two steps, she wins. Thus in a subgame perfect equilibrium player 2 takes two steps, and wins. We conclude that in a subgame perfect equilibrium of G1 (2, 2) player 1 initially takes two steps, and wins. 198.1 A race in which the players’ valuations of the prize diﬀer By the arguments in the text for the case in which both players’ valuations of the prize are between 6 and 7, the subgame perfect equilibrium outcomes of all games in which k 1 ≤ 2 or k 2 ≤ 3 are the same as they are when both players’ valuations of the prize are between 6 and 7. If k 2 ≥ 5 then player 1 is the winner in all subgame 98 Chapter 6. Extensive Games with Perfect Information: Illustrations perfect equilibria, because even if player 2 reaches the finish line after taking one step at a time, her payoff is negative. The games Gi (3, 4), Gi (4, 4), Gi (5, 4), and Gi (6, 4) remain. If, in the games G2 (3, 4) and G2 (4, 4), player 2 takes a single step then play moves to a game that player 1 wins. Thus player 2 is better off not moving; the subgame perfect equilibrium outcome is that player 1 takes one step at a time, and wins. In the game Gi (5, 4), the player who moves first can, by taking a single step, reach a game in which she wins regardless of the identity of the first-mover. Thus in this game the winner is the first-mover. Finally, in the game G1 (6, 4) it is not worth player 1’s while taking two steps, to reach a game in which she wins, because her payoff would ultimately be negative. And if she takes one step, play moves to a game in which player 2 is the first-mover, and wins. Thus in this game player 2 wins. Figure 98.1 shows the subgame perfect equilibrium outcomes. ↑ k2 4 2 Finish line 3 1 1 1 1 1 f 2 1 1 f f 2 2 f f 2 2 2 2 f f 2 2 2 2 1 2 5 6 k1 → Finish line 3 4 Figure 98.1 The subgame perfect equilibrium outcomes for the race in Exercise 198.1. Player 1 moves to the left, and player 2 moves down. The labels on the values of (k 1 , k 2 ) indicate the subgame perfect equilibrium outcomes, as in the text. 198.2 Removing stones For n = 1 the game has a unique subgame perfect equilibrium, in which player 1 takes one stone. The outcome is that player 1 wins. For n = 2 the game has a unique subgame perfect equilibrium in which • player 1 takes two stones • after a history in which player 1 takes one stone, player 2 takes one stone. The outcome is that player 1 wins. For n = 3, the subgame following the history in which player 1 takes one stone is the game for n = 2 in which player 2 is the first mover, so player 2 wins. The subgame following the history in which player 1 takes two stones is the game for n = 1 in which player 2 is the first mover, so player 2 wins. Thus there is a subgame Chapter 6. Extensive Games with Perfect Information: Illustrations 99 perfect equilibrium in which player 1 takes one stone initially, and one in which she takes two stones initially. In both subgame perfect equilibria player 2 wins. For n = 4, the subgame following the history in which player 1 takes one stone is the game for n = 3 in which player 2 is the first-mover, so player 1 wins. The subgame following the history in which player 1 takes two stones is the game for n = 2 in which player 2 is the first-mover, so player 2 wins. Thus in every subgame perfect equilibrium player 1 takes one stone initially, and wins. Continuing this argument for larger values of n, we see that if n is a multiple of 3 then in every subgame perfect equilibrium player 2 wins, while if n is not a multiple of 3 then in every subgame perfect equilibrium player 1 wins. We can prove this claim by induction on n. The claim is correct for n = 1, 2, and 3, by the arguments above. Now suppose it is correct for all integers through n − 1. I will argue that it is correct for n. First suppose that n is divisible by 3. The subgames following player 1’s removal of one or two stones are the games for n − 1 and n − 2 in which player 2 is the first-mover. Neither n − 1 nor n − 2 is divisible by 3, so by hypothesis player 2 is the winner in every subgame perfect equilibrium of both of these subgames. Thus player 2 is the winner in every subgame perfect equilibrium of the whole game. Now suppose that n is not divisible by 3. As before, the subgames following player 1’s removal of one or two stones are the games for n − 1 and n − 2 in which player 2 is the first-mover. Either n − 1 or n − 2 is divisible by 3, so in one of these subgames player 1 is the winner in every subgame perfect equilibrium. Thus player 1 is the winner in every subgame perfect equilibrium of the whole game. 199.1 Hungry lions Denote by G(n) the game in which there are n lions. The game G(1) has a unique subgame perfect equilibrium, in which the single lion eats the prey. Consider the game G(2). If lion 1 does not eat, it remains hungry. If it eats, we reach a subgame identical to G(1), which we know has a unique subgame perfect equilibrium, in which lion 2 eats lion 1. Thus G(2) has a unique subgame perfect equilibrium, in which lion 1 does not eat the prey. In G(3), lion 1’s eating the prey leads to G(2), in which we have just concluded that the first mover (lion 2) does not eat the prey (lion 1). Thus G(3) has a unique subgame perfect equilibrium, in which lion 1 eats the prey. For an arbitrary value of n, lion 1’s eating the prey in G(n) leads to G(n − 1). If G(n − 1) has a unique subgame perfect equilibrium, in which the prey is eaten, then G(n) has a unique subgame perfect equilibrium, in which the prey is not eaten; if G(n − 1) has a unique subgame perfect equilibrium, in which the prey is not eaten, then G(n) has a unique subgame perfect equilibrium, in which the prey is eaten. Given that G(1) has a unique subgame perfect equilibrium, in which the 100 Chapter 6. Extensive Games with Perfect Information: Illustrations prey is eaten, we conclude that if n is odd then G(n) has a unique subgame perfect equilibrium, in which lion 1 eats the prey, and if n is even it has a unique subgame perfect equilibrium, in which lion 1 does not eat the prey. 200.1 A race with a liquidity constraint In the absence of the constraint, player 1 initially takes one step. Suppose she does so in the game with the constraint. Consider player 2’s options after player 1’s move. Player 2 takes two steps: Because of the liquidity constraint, player 1 can take at most one step. If she takes one step, player 2’s optimal action is to take one step, and win. Thus player 1’s best action is not to move; player 2’s payoff exceeds 1 (her steps cost 5, and the prize is worth more than 6). Player 2 moves one step: Again because of the liquidity constraint, player 1 can take at most one step. If she takes one step, player 2 can take two steps and win, obtaining a payoff of more than 1 (as in the previous case). Player 2 does not move: Player 1, as before, can take one step on each turn, and win; player 2’s payoff is 0. We conclude that after player 1 moves one step, player 2 should take either one or two steps, and ultimately win; player 1’s payoff is −1. A better option for player 1 is not to move, in which case player 2 can move one step at a time, and win; player 1’s payoff is zero. Thus the subgame perfect equilibrium outcome is that player 1 does not move, and player 2 takes one step at a time and wins. Draft of solutions to exercises in chapter of An introduction to game theory by Martin J. Osborne

[email protected]; www.chass.utoronto.ca/~osborne/index.html Version: 00/11/6. c 1995–2000 by Martin J. Osborne. All rights reserved. No part of this book may be reCopyright produced in any form by any electronic or mechanical means (including photocopying, recording, or information storage and retrieval) without permission in writing from Martin J. Osborne. On request, permission to make one copy for each student will be granted to instructors who wish to use the book in a course, on condition that copies be sold at a price not more than the cost of duplication. 7 Extensive Games with Perfect Information: Extensions and Discussion 206.2 Extensive game with simultaneous moves The game is shown in Figure 101.1. 1 A C D C 4, 2 0, 0 D 0, 0 2, 4 B E F E 3, 1 0, 0 F 0, 0 1, 3 Figure 101.1 The game in Exercise 206.2. The subgame following player 1’s choice of A has two Nash equilibria, (C, C) and (D, D); the subgame following player 1’s choice of B also has two Nash equilibria, (E, E) and (F, F). If the equilibrium reached after player 1 chooses A is (C, C), then regardless of the equilibrium reached after she chooses (E, E), she chooses A at the beginning of the game. If the equilibrium reached after player 1 chooses A is (D, D) and the equilibrium reached after she chooses B is (F, F), she chooses A at the beginning of the game. If the equilibrium reached after player 1 chooses A is (D, D) and the equilibrium reached after she chooses B is (E, E), she chooses B at the beginning of the game. Thus the game has four subgame perfect equilibria: (ACE, CE), (ACF, CF), (ADF, DF), and (BDE, DE) (where the first component of player 1’s strategy is her choice at the start of the game, the second component is her action after she chooses A, and the third component is her action after she chooses B, and the first component of player 2’s strategy is her action after player 1 chooses A at the start of the game and the second component is her action after player 1 chooses B at the start of the game). In the first two equilibria the outcome is that player 1 chooses A and then both players choose C, in the third equilibrium the outcome is that player 1 chooses A and then both players choose D, and in the last equilibrium the outcome is that player 1 chooses B and then both players choose E. 101 102 Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion 206.3 Two-period Prisoner’s Dilemma The extensive game is specified as follows. Players The two people. Terminal histories The set of pairs ((W, X), (Y, Z)), where each component is either Q or F. Player function P(∅) = {1, 2} and P(W, X) = {1, 2} for any pair (W, X) in which both W and X are either Q or F. Actions The set Ai (∅) of player i’s actions at the initial history is {Q, F}, for i = 1, 2; the set Ai (W, X) of player i’s actions after any history (W, X) in which both W and X are either Q or F is {Q, F}, for i = 1, 2. Preferences Each player’s preferences are represented by the payoffs described in the problem. Consider the subgame following some history (W, X) (where W and X are both either Q or F). In this subgame each player chooses either Q or F, and her payoff to each resulting terminal history is the sum of her payoff to (W, X) in the Prisoner’s Dilemma given in Figure 13.1 and her payoff to the pair of actions chosen in the subgame, again as in the Prisoner’s Dilemma. Thus the subgame differs from the Prisoner’s Dilemma given in Figure 13.1 only in that every payoff to a given player is increased by her payoff to the pair of actions (W, X). Thus the subgame has a unique Nash equilibrium, in which both players choose F. Now consider the whole game. Regardless of the actions chosen at the start of the game, the outcome in the second period is (F, F). Thus the payoffs to the pairs of actions chosen in the first period are the payoffs in the Prisoner’s Dilemma plus the payoff to (F, F). We conclude that the game has a unique subgame perfect equilibrium, in which each player chooses F after every history. 207.1 Timing claims on an investment The following extensive game models the situation. Players The two people. Terminal histories The sequences of the form ((N, N), (N, N), . . . , (N, N), xt ), where 1 ≤ t ≤ T, xt is (C, C), (C, N), or (N, C) if t ≤ T − 1 and (C, C), (C, N), (N, C), or (N, N) if t = T, C means “claim”, and N means “do not claim”. Player function The set of players assigned to every nonterminal history is {1, 2} (the two people). Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion 103 Actions The set of actions of each player after every nonterminal history is {C, N}. Preferences Each player’s preferences are represented by a payoff equal to the amount of money she obtains. The consequences of the players’ actions in period T are given in Figure 103.1. We see that the subgame starting in period T has a unique Nash equilibrium, (C, C), in which each player’s payoff is T. C N C T, T 0, 2T N 2T, 0 T, T Figure 103.1 The consequences of the players’ actions in period T of the game in Exercise 207.1. Thus if T = 1 the game has a unique subgame perfect equilibrium, in which both players claim. Now suppose that T ≥ 2, and consider period T − 1. The consequences of the players’ actions in this period, given the equilibrium in the subgame starting in period T, are shown in Figure 103.2. (The entry in the bottom right box, (T, T), is the pair of equilibrium payoffs in the subgame in period T.) If T > 2 then 2(T − 1) > T, so that the subgame starting in period T − 1 has a unique subgame perfect equilibrium, (C, C), in which each player’s payoff is T − 1. If T = 2 then the whole game has two subgame perfect equilibria, in one of which both players claim in both periods, and another in which neither claims in period 1 and both claim in period 2. C N C T − 1, T − 1 0, 2(T − 1) N 2(T − 1), 0 T, T Figure 103.2 The consequences of the players’ actions in period T − 1 of the game in Exercise 207.1, given the equilibrium actions in period T. For T > 2, working back to period 1 we see that the game has two subgame perfect equilibria: one in which each player claims in every period, and one in which neither player claims in period 1 but both players claim in every subsequent period. 207.2 A market game The following extensive game models the situation. Players The seller and m buyers. 104 Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion Terminal histories The set of sequences of the form ((p1 , . . . , pm ), j), where each pi is a price (nonnegative number) and j is either 0 or one of the sellers (an integer from 1 to m), with the interpretation that pi is the offer of buyer i, j = 0 means that the seller accepts no offer, and j ≥ 1 means that the seller accepts buyer j’s offer. Player function P(∅) is the set of buyers and P(p1 , . . . , pm ) is the seller for every history (p1 , . . . , pm ). Actions The set Ai (∅) of actions of buyer i at the start of the game is the set of prices (nonnegative numbers). The set As (p1 , . . . , pm ) of actions of the seller after the buyers have made offers is the set of integers from 0 to m. Preferences Each player’s preferences are represented by the payoffs given in the question. To find the subgame perfect equilibria of the game, first consider the subgame following a history (p1 , . . . , pm ) of offers. The seller’s best action is to accept the highest price, or one of the highest prices in the case of a tie. I claim that a strategy profile is a subgame perfect equilibrium of the whole game if and only if the seller’s strategy is the one just described, and among the buyers’ strategies (p1 , . . . , pm ), every offer pi is at most v and at least two offers are equal to v. Such a strategy profile is a subgame perfect equilibrium by the following argument. If the buyer with whom the seller trades raises her offer then her payoff becomes negative, while if she lowers her offer she no longer trades and her payoff remains zero. If any other buyer raises her offer then either she still does not trade, or she trades at a price greater than v and hence receives a negative payoff. No other profile of actions for the buyers at the start of the game is part of a subgame perfect equilibrium by the following argument. • If some offer exceeds v then the buyer who submits the highest offer can induce a better outcome by reducing her offer to a value below v, so that either the seller does not trade with her, or, if the seller does trade with her, she trades at a lower price. • If all offers are at most v and only one is equal to v, the buyer who offers v can increase her payoff by reducing her offer a little. • If all offers are less than v then one of the buyers whose offer is not accepted can increase her offer to some value between the winning offer and v, induce the seller to trade with her, and obtain a positive payoff. In any equilibrium the buyer who trades with the seller does so at the price v. Thus her payoff is zero. The other buyers do not trade, and hence also obtain the payoff of zero. Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion 105 208.1 Price competition The following game models the situation. Players The two sellers and the two buyers. Terminal histories All sequences ((p1 , p2 ), (x1 , x2 )) where pi (for i = 1, 2) is the price posted by seller i and xi (for i = 1, 2) is the seller chosen by buyer i (either seller 1 or seller 2). Player function P(∅) is the set consisting of the two sellers; P(p1 , p2 ) for any pair (p1 , p2 ) of prices is the set consisting of the two buyers. Actions The set of actions of each seller at the start of the game is the set of prices (nonnegative numbers), and the set of actions of each buyer after any history (p1 , p2 ) is the set consisting of seller 1 and seller 2. Preferences Each seller’s preferences on lotteries over the terminal histories are represented by the expected value of a Bernoulli payoff function that assigns the payoff p to a sale at the price p. Each buyers’ preferences on lotteries over the terminal histories are represented by the expected value of a Bernoulli payoff function that assigns the payoff 1 − p to a purchase at the price p. The payoff of a player who does not trade is 0. In any subgame perfect equilibrium, the buyers’ strategies in the subgame following any history (p 1 , p2 ) must be a Nash equilibrium of the game in Exercise 125.2. This game has a unique Nash equilibrium unless 12 (1 + p 1 ) ≤ p 2 ≤ 2p1 − 1. If 12 (1 + p1 ) < p2 < 2p1 − 1 the game has three Nash equilibria, two pure and one mixed. I claim that for any price p ≥ 12 the extensive game in this exercise has a subgame perfect equilibrium in which if 12 (1 + p 1 ) < p2 < 2p1 − 1 then if either p1 ≤ p or p2 ≤ p, the equilibrium in the subgame is the pure Nash equilibrium in which buyer 1 approaches seller 1 and buyer 2 approaches seller 2, while if p1 > p and p2 > p, the equilibrium in the subgame is the mixed strategy equilibrium. Precisely, I claim that for any p ≥ 12 the following strategy pair is a subgame perfect equilibrium of the game. Sellers’ strategies Each seller announces the price p. Buyers’ strategies • After a history (p1 , p2 ) in which 2p1 − 1 < p2 < 12 (1 + p 1 ) and either p1 ≤ p or p2 ≤ p (or both), buyer 1 approaches seller 1 and buyer 2 approaches seller 2. • After a history (p1 , p2 ) in which 2p1 − 1 < p2 < 12 (1 + p1 ), p1 > p, and p2 > p, each buyer approaches seller 1 with probability (1 − 2p1 + p2 )/(2 − p 1 − p2 ). 106 Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion • After a history (p1 , p2 ) in which p2 ≤ 2p 1 − 1, both buyers approach seller 2. • After a history (p1 , p2 ) in which p 2 ≥ 12 (1 + p1 ), both buyers approach seller 1. By Exercise 125.2, the buyers’ strategy pair is a Nash equilibrium in every subgame. The sellers’ payoffs in the pure equilibrium in which one buyer approaches each seller are (p1 , p2 ); their payoffs in the pure equilibrium in which both buyers approach seller 1 is (p1 , 0); and their payoffs in the pure equilibrium in which both buyers approach seller 1 is (0, p2 ). Their payoffs in the mixed strategy equilibrium are more difficult to calculate. They are (π1∗ (p1 , p2 ), π2∗ (p1 , p2 )) = ((1 − (1 − π)2 )p1 , (1 − π 2 )p2 ), where π = (1 − 2p1 + p2 )/(2 − p 1 − p2 ). After some algebra we obtain (π1∗ (p1 , p2 ), π2∗ (p1 , p2 )) = 3p1 (1 − p 2 )(1 − 2p1 + p2 ) 3p2 (1 − p1 )(1 − 2p2 + p 1 ) , (2 − p 1 − p2 )2 (2 − p1 − p2 )2 These equilibrium payoffs are illustrated in Figure 106.1. 1 ↑ p2 (p1 , 0) = 1 2 p 1) (π1∗ (p1 , p2 ), π2∗ (p1 , p2 )) −1 p2 + (1 p2 = 2p 1 p (0, p2 ) (p1 , p2 ) 0 p p1 → 1 Figure 106.1 The sellers’ payoffs in the game in Exercise 208.1 as a function of their prices, given the buyers’ equilibrium strategies. Now consider the sellers’ choices of prices. Given that p2 = p ≥ 12 and the buyers’ strategies are those defined above, seller 1’s payoff when she sets the price . Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion p1 is 107 if p1 ≤ p p1 π1∗ (p1 , p) if p < p1 ≤ 12 (1 + p) 0 if p > 12 (1 + p). By the claim in the question (verified at the end of this solution), π1∗ (p1 , p2 ) is decreasing in p1 for p1 ≥ p 2 , so that seller 1’s best response to p is p. An analogous argument shows that seller 2’s best response to p is p. We conclude that the strategy pair defined above is a subgame perfect equilibrium. The verification of the last claim of the question (not required as part of an answer) follows. We have π1∗ (p1 , p2 ) = 3p1 (1 − p2 )(1 − 2p1 + p2 ) . (2 − p1 − p2 )2 The derivative of this function with respect to p1 is 3(1 − p2 ) (2 − p1 − p 2 )2 (1 − 2p 1 + p2 − 2p1 ) + 2(2 − p1 − p2 )p1 (1 − 2p 1 + p2 ) (2 − p1 − p2 )4 or 3(1 − p2 )(2 − p1 − p2 ) [(2 − p1 − p 2 )(1 − 4p1 + p2 ) + 2p1 (1 − 2p 1 + p2 )] . (2 − p 1 − p2 )4 This expression is negative if (2 − p1 − p2 )(1 − 4p1 + p2 ) + 2p1 (1 − 2p 1 + p 2 ) < 0, or p1 > (2 − p2 )(1 + p 2 ) . 7 − 5p2 The right-hand side is less than p2 if (2p 2 − 1)(p2 − 1) < 0, which is true if 12 < p 2 < 1, so that seller 1’s equilibrium payoff is decreasing in p1 whenever p1 > p2 > 12 . 210.1 Bertrand’s duopoly game with entry The unique Nash equilibrium of the subgame that follows the challenger’s entry is (c, c), as we found in Section 3.2.2. The challenger’s profit is − f < 0 in this equilibrium. By choosing to stay out the challenger obtains the profit of 0, so in any subgame perfect equilibrium the challenger stays out. After the history in which the challenger stays out, the incumbent chooses its price p1 to maximize its profit (p1 − c)(α − p 1 ). Thus for any value of f > 0 the whole game has a unique subgame perfect equilibrium, in which the strategies are: 108 Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion Challenger • at the start of the game: stay out • after the history in which the challenger enters: choose the price c Incumbent • after the history in which the challenger enters: choose the price c • after the history in which the challenger stays out: choose the price p 1 that maximizes (p1 − c)(α − p 1 ). 212.1 Electoral competition with strategic voters Consider the strategy profile in which each candidate chooses the median m of the citizens’ favorite positions and the citizens’ strategies are defined as follows. • After a history in which every candidate chooses m, each citizen i votes for candidate j, where j is the smallest integer greater than or equal to in/q. (That is, the citizens split their votes equally among the n candidates. If there are 3 candidates and 15 citizens, for example, citizens 1 through 5 vote for candidate 1, citizens 6 through 10 vote for candidate 2, and citizens 11 through 15 vote for candidate 3.) • After a history in which all candidates enter and every candidate but j chooses m, each citizen votes for candidate j if her favorite position is closer to j’s position than it is to m, and for some candidate whose position is m otherwise. (All citizens who do not vote for j vote for the same candidate .) • After any other history, the citizens’ action profile is any Nash equilibrium of the voting subgame in which no citizen’s action is weakly dominated. The outcome induced by this strategy profile is that all candidates enter and choose the median of the citizens’ favorite positions, and tie for first place. After every history of one of the first two types, every citizen votes for one of the candidates who is closest to her favorite position, so no citizen’s strategy is weakly dominated. After a history of the third type, no citizen’s strategy is weakly dominated by construction. The strategy profile is a subgame perfect equilibrium by the following argument. In each voting subgame the citizens’ strategy profile is a Nash equilibrium: • after the history in which the candidates’ positions are the same, equal to m, no citizen’s vote affects the outcome • after a history in which all candidates enter and every candidate but j chooses m, a change in any citizen’s vote either has no effect on the outcome or makes it worse for her Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion 109 • after any other history the citizens’ strategy profile is a Nash equilibrium by construction. Now consider the candidates’ choices at the start of the game. If any candidate deviates by choosing a position different from that of the other candidates, she loses, rather than tying for first place. If any candidate deviates by staying out of the race, the outcome is worse for her than adhering to the equilibrium, and tying for first place. Thus each candidate’s strategy is optimal given the other players’ strategies. [The claim that every voting subgame has a (pure) Nash equilibrium in which no citizen’s action is weakly dominated, which you are not asked to prove, may be demonstrated as follows. Given the candidates’ positions, choose the candidate, say j, ranked last by the smallest number of citizens. Suppose that all citizens except those who rank j last vote for j; distribute the votes of the citizens who rank j last as equally as possible among the other candidates. Each citizen’s action is not weakly dominated (no citizen votes for the candidate she ranks last) and, given q ≥ 2n, no change in any citizen’s vote affects the outcome, so that the list of citizens’ actions is a Nash equilibrium of the voting subgame.] 213.1 Electoral competition with strategic voters I first argue that in any equilibrium each candidate that enters is in the set of winners. If some candidate that enters is not a winner, she can increase her payoff by deviating to Out. Now consider the voting subgame in which there are more than two candidates and not all candidates’ positions are the same. Suppose that the citizens’ votes are equally divided among the candidates. I argue that this list of citizens’ strategies is not a Nash equilibrium of the voting subgame. For either the citizen whose favorite position is 0 or the citizen whose favorite position is 1 (or both), at least two candidates’ positions are better than the position of the candidate furthest from the citizen’s favorite position. Denote a citizen for whom this condition holds by i. (The claim that citizen i exists is immediate if the candidates occupy at least three distinct positions, or they occupy two distinct positions and at least two candidates occupy each position. If the candidates occupy only two positions and one position is occupied by a single candidate, then take the citizen whose favorite position is 0 if the lone candidate’s position exceeds the other candidates’ position; otherwise take the citizen whose favorite position is 1.) Now, given that each candidate obtains the same number of votes, if citizen i switches her vote to one of the candidates whose position is better for her than that of the candidate whose position is furthest from her favorite position, then this candidate wins outright. (If citizen i originally votes for one of these superior candidates, she can switch her vote to the other superior candidate; if she originally votes for neither of the superior candidates, she can switch her vote to either one 110 Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion of them.) Citizen i’s payoff increases when she thus switches her vote, so that the list of citizens’ strategies is not a Nash equilibrium of the voting subgame. We conclude that in every Nash equilibrium of every voting subgame in which there are more than two candidates and not all candidates’ positions are the same at least one candidate loses. Because no candidate loses in a subgame perfect equilibrium (by the first argument in the proof), in any subgame perfect equilibrium either only two candidates enter, or all candidates’ positions are the same. If only two candidates enter, then by the argument in the text for the case n = 2, each candidate’s position is m (the median of the citizens’ favorite positions). Now suppose that more than two candidates enter, and their common position is not equal to m. If a candidate deviates to m then in the resulting voting subgame only two positions are occupied, so that for every citizen, any strategy that is not weakly dominated votes for a candidate at the position closest to her favorite position. Thus a candidate who deviates to m wins outright. We conclude that in any subgame perfect equilibrium in which more than two candidates enter, they all choose the position m. 216.1 Top cycle set a. The top cycle set is the set {x, y, z} of all three alternatives because x beats y beats z beats x. b. The top cycle set is the set {w, x, y, z} of all four alternatives. As in the previous case, x beats y beats z beats x; also y beats w. 217.1 Designing agendas We have: x beats y beats z beats x; x, y, and z all beat v; v beats w; and w does not beat any alternative. Thus the top cycle set is {x, y, z}. An agenda that yields x is shown in Figure 111.1. A similar agenda, with y and x interchanged, yields y, and one with x and z interchanged yields z. No binary agenda yields w because for every other alternative a, a majority of committee members prefer a to w. No binary agenda yields v because the only alternative that v beats is w, which itself is beaten by every other alternative. 217.2 An agenda that yields an undesirable outcome An agenda for which the outcome of sophisticated voting is z is given in Figure 111.2. 220.1 Exit from a declining industry Period t1 is the largest value of t for which Pt (k 1 ) ≥ c, or 60 − t ≥ 10. Thus t1 = 50. Similarly, t2 = 70. Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion 111 vote vote vote vote x v y z w Figure 111.1 A binary agenda for which the alternative x is the outcome of sophisticated voting for the committee in Exercise 217.1. vote vote z vote x y w Figure 111.2 A binary agenda for which the alternative z is the outcome of sophisticated voting for the committee in Exercise 217.2. If both firms are active in period t1 , then firm 2’s profit in this period is (100 − t1 − c − k 1 − k 2 )k 2 = (−20)(20) = −400. Its profit in any period t in which it is alone in the market is (100 − t − c − k 2 )k 2 = (70 − t)(20). Thus its profit from period t1 + 1 through period t2 is (19 + 18 + . . . + 1)(20) = 3800. Hence firm 2’s loss in period t1 when both firms are active is (much) less than the sum of its profits in periods t1 + 1 through t2 when it alone is active. 221.1 Eﬀect of borrowing constraint in declining industry Period t0 is the largest value of t for which Pt (k 1 + k 2 ) ≥ c, or 100 − t − 60 ≥ 10, or t ≤ 30. Thus t0 = 30. From Exercise 220.1 we have t1 = 50 and t2 = 70. Suppose that firm 2 stays in the market for k periods after t0 , then exits in period t0 + k + 1. Firm 1’s total profit from period t0 + 1 on if it stays until period t1 is (Pt0 +1 (k 1 + k 2 ) − c)k 1 + . . . + (Pt0 +k (k 1 + k 2 ) − c)k 1 + 112 Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion (Pt0 +k+1 (k 1 ) − c)k 1 + . . . + (Pt1 (k 1 ) − c)k 1 , or 40[(100 − 30 − 1 − 60 − 10) + . . . + (100 − 30 − k − 60 − 10) + (100 − 30 − k − 1 − 40 − 10) + . . . + (100 − 50 − 40 − 10)], or 40[−1 − . . . − k + (19 − k) + . . . + 0], or 40[− 12 k(k + 1) + 12 (19 − k)(20 − k)] (using the fact that the sum of the first n positive integers is 12 n(n + 1)), or 20(380 − 40k). In order that this profit be nonpositive we need 40k ≥ 380, or k ≥ 9.5. Thus firm 2 needs to survive until at least period 40 (30 + 10) in order to make firm 1’s exit in period t0 + 1 optimal. Firm 2’s total loss from period 31 through period 40 when both firms are in the market is (P31 (k 1 + k 2 ) − c)k 2 + . . . + (P40 (k 1 + k 2 ) − c)k 2 , or 20[(100 − 31 − 60 − 10) + . . . + (100 − 40 − 60 − 10)], or 20(−1 + . . . + −10), or 1100. Thus firm 2 needs to be able to bear a debt of at least 1100 in order for there to be a subgame perfect equilibrium in which firm 1 exits in period t0 + 1. 222.2 Variant of ultimatum game with equity-conscious players The game is defined as follows. Players The two people. Terminal histories The set of sequences (x, β 2 , Z), where x is a number with 0 ≤ x ≤ c (the amount of money that person 1 offers to person 2), β 2 is 0 or 1 (the value of β 2 selected by chance), and Z is either Y (“yes, I accept”) or N (“no, I reject”). Player function β2. P(∅) = 1, P(x) = c for all x, and P(x, β 2 ) = 2 for all x and all Chance probabilities For every history x, chance chooses 0 with probability p and 1 with probability 1 − p. Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion 113 Preferences Each person’s preferences are represented by the expected value of a payoff equal to the amount of money she receives. For any terminal history (x, β 2 , Y) person 1 receives c − x and person 2 receives x; for any terminal history (x, β 2 , N) each person receives 0. Given the result from Exercise 181.1 given in the question, if person 1’s offer x satisfies 0 < x < 13 then the offer is rejected with probability 1 − p, so that person 1’s expected payoff is p(1 − x), while if x > 13 the offer is certainly accepted, independent of the type of person 2. Thus person 1’s optimal offer is 1 3 0 if p < 23 if p > 23 ; if p = 23 then both offers are optimal. If p > 23 we see that in a subgame perfect equilibrium person 1’s offers are rejected by every person 2 with whom she is matched for whom β 2 = 1 (that is, with probability 1 − p). 223.1 Sequential duel The following game models the situation. Players The two people. Terminal histories All sequences of the form (X1 , X2 , . . . , Xk , S, H), where each Xi is either N (“don’t shoot”) or (S, M) (“shoot”, “miss”), and H means “hit”, together with the infinite sequence (S, M, S, M, S, M, . . .). Player function P(h) = 1 for any history h in which the total number of S’s and N’s is even and P(h) = 2 for any history h in which the total number of S’s and N’s is odd. Chance probabilities Whenever chance moves after a move of player 1 it chooses H with probability p 1 and M with probability 1 − p1; whenever chance moves after a move of player 2 it chooses H with probability p2 and M with probability 1 − p2 ; Preferences Each player’s preferences are represented by the expected value of a Bernoulli payoff function that assigns 1 to any history in which she survives and 0 to any history in which she is killed. If neither player ever shoots, both players survive. No outcome is better for either player, so in particular neither player has a strategy that leads to a better outcome for her, given the other player’s strategy. Now suppose that player 2 shoots whenever it is her turn to move. I claim that a best response for player 1 is to shoot whenever it is her turn to move. Denote player 1’s probability of survival when she follows this strategy by π1 . 114 Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion Suppose that player 1 deviates to not shooting at the start of the game (but does not change the remainder of her strategy). If player 2 hits her in the next round, she does not survive. If player 2 misses her, an event with probability 1 − p2 , then we reach a subgame identical to the whole game in which both players always shoot, so that in this subgame player 1’s survival probability is π1 . Thus if player 1 deviates to not shooting at the start of the game her survival probability is (1 − p2 )π1 . We conclude that player 1 is not better off (and worse off if p2 > 0) by deviating at the start of the game. The same argument shows that, given player 2’s strategy, player 1 is not better off deviating after any history that ends with player 2’s shooting and missing, or after any collection of such histories. A change in player 1’s strategy after a history that ends with player 2’s not shooting has no effect on the outcome (because player 2’s is to shoot whenever it is her turn to move). Thus no change in player 1’s strategy increases her expected payoff. A symmetric argument shows that player 2 is not better off changing her strategy. Thus the strategy pair in which each player always shoots is a subgame perfect equilibrium. 223.2 Sequential truel The games are shown in Figure 115.1. (The action marked “0” is that of shooting into the air, which is available only in the second version of the game.) To find the subgame perfect equilibria, first consider the subgame Γ in Figure 115.1. Whomever player C aims at, if she misses then she survives in the company of both A and B. If she aims at B and hits her, then she survives in the company of A; if she aims at A and hits her then she survives in the company of B. Thus C aims at B if p A < p B and at A if p A > p B . Now consider the subgame Γ. Whomever B aims at, the outcome is the same if she misses (because Γ has a unique subgame perfect equilibrium). If B aims at A and hits her, then she survives with probability 1 − pC ; if she aims at C and hits her, then she survives with probability 1. Thus (given pC > 0), the subgame Γ thus has a unique subgame perfect equilibrium, in which B aims at C. Finally, consider the whole game. Whomever A aims at, the outcome is the same if she misses (because Γ has a unique subgame perfect equilibrium). If she aims at B and hits her, then she survives with probability 1 − pC ; if she aims at C and hits her, then she survives with probability 1 − p B. Thus A aims at C if p B < pC and at B if p B > pC . In summary, the game in which no player has the option of shooting into the air has the following unique subgame perfect equilibrium. • At the start of the game, A aims at C if p B < pC and at B if p B > pC . • After a history in which A misses, B aims at C. Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion 115 A B 1 − pA pA C 0 c c C A 1 − pA pA Γ B Γ c pC 1 − pC {C} {A, C} Γ A c 1 − pB pB {B} {A, B} where the game Γ is B A 1 − pB pB C 0 c c 1 − pB pB Γ C B Γ Γ {A, B} c pC 1 − pC {C} {B, C} and the game Γ is C A c pC {B, C} 0 1 − pC {A, B, C} {A, B, C} B c pC {A, C} 1 − pC {A, B, C} Figure 115.1 The games in Exercise 223.2. Only the actions indicated by black lines are available when players do not have the option of shooting into the air (the action “0”). The labels beside the actions of chance are the probabilities with which the actions are chosen; in each case the left action is “hit” and the right action is “miss”. • After a history in which both A and B miss, C aims at B if p A < p B and at A if p A > p B . Player A aims the player who is her more dangerous opponent; she is better off if she eliminates this opponent than if she eliminates her weaker opponent. Player C’s survival probability is (1 − p A )(1 − p B ) = 1 − p A − p B (1 − p A ) if 116 Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion pC > p B , and 1 − p B (1 − p A ) if pC < p B . Thus she is better off if pC < p B than if pC > p B . Now consider the game in which each player has the option of shooting into the air. In the subgame Γ , player C’s best action is to aim at B (given p A < p B ). (If she shoots into the air then the set of survivors is {A, B, C}; if she aims at B she has some chance of eliminating her.) In the subgame Γ we know that if B shoots, her target should be C. If she does so her probability of survival is 1 − (1 − p B )pC . If she shoots into the air her probability of survival is 1 − pC . The former exceeds the latter, so in the subgame Γ player B aims at C. Finally, given the equilibrium actions in the subgames, at the start of the game we know that if A fires she aims at C if p B < pC and at B if p B > pC . Given p A < p B , her shooting into the air results in her certain survival, while her aiming at B or C results in her surviving with probability less than 1. Thus she shoots into the air. We conclude that if p A < p B then the game in which each player has the option of shooting into the air has a unique subgame perfect equilibrium, which differs from the subgame perfect equilibrium in which this option is absent only in that A shoots into the air at the beginning of the game. Player A fires into the air because when she does so B and C fight between themselves; if she shoots at one of them she may eliminate her from the game, giving the remaining player an incentive to shoot at her. 224.1 Cohesion in legislatures Let the initial governing coalition consist of legislators 1 and 2. The US game is defined as follows. Players The three legislators. Terminal histories All sequences (i, (x1 , x2 , x3 ), (A, B, C), j, (y1 , y2 , y3 ), (A , B , C )), where i and j are members of the governing coalition (possibly i = j), (x1 , x2 , x3 ) and (y1 , y2 , y3 ) are partitions of one unit of payoff (x1 + x2 + x3 = y1 + y 2 + y3 = 1, xi ≥ 0, and yi ≥ 0 for i = 1, 2, 3), and A, B, C, A , B , and C are either yes (vote for bill) or no (vote against bill). Player function • P(∅) = c (chance) • P(i) = i • P(i, (x1 , x2 , x3 )) = {1, 2, 3} • P(i, (x1 , x2 , x3 ), (A, B, C)) = c • P(i, (x1 , x2 , x3 ), (A, B, C), j) = j • P(i, (x1 , x2 , x3 ), (A, B, C), j, (y1 , y2 , y3 )) = {1, 2, 3}. Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion Chance probabilities Chance assigns probability whenever it moves. 1 2 to 1 and probability 117 1 2 to 2 Actions • A(∅) = {1, 2} • A(i) = {(x1 , x2 , x3 ) : x1 + x2 + x 3 = 1, xi ≥ 0 for all i} for i = 1, 2 • Ak (i, (x1 , x2 , x3 )) = {yes, no} for all k, i = 1, 2, and all (x1 , x2 , x3 ) • A(i, (x1 , x2 , x3 ), (A, B, C)) = {1, 2} for all i, all (x1 , x2 , x3 ), and all triples (A, B, C) in which A, B, and C are all either yes or no • A(i, (x1 , x2 , x3 ), (A, B, C), j) = {(x1 , x2 , x3 ) : x1 + x2 + x3 = 1, x i ≥ 0 for all i} for i = 1, 2, all (x1 , x2 , x3 ), all triples (A, B, C) in which A, B, and C are all either yes or no, and j = 1, 2 • Ak (i, (x1 , x2 , x3 ), (A, B, C), j, (y1 , y2 , y3 )) = {yes, no} for all k, i = 1, 2, all (x1 , x2 , x3 ), all triples (A, B, C) in which A, B, and C are all either yes or no, j = 1, 2, and all (y1 , y2 , y3 ). Preferences Each legislator i ranks the terminal histories by the amount of money she receives: xi + y i if both bills are passed, xi + d2i if only the first bill is passed, d1i + yi if only the second bill is passed, and d1i + d2i if neither bill is passed. We find a subgame perfect equilibrium as follows. Refer to dti as legislator i’s reservation value in period t. In the second period, denote by k the legislator whose reservation value is lower between the two who do not propose a bill. Each legislator i gets dti if a bill does not pass, and hence votes for a bill only if it gives her a payoff of at least dti . The proposer needs one vote in addition to her own to pass a bill, and can obtain it most cheaply by proposing a bill that gives k the payoff d2k and gives herself the remaining payoff 1 − d2k (which exceeds her reservation value, because all reservation values are less than 12 ). Legislator k and the proposer vote for the bill, which thus passes. (Legislator k is indifferent between voting for or against the bill, but there is no subgame perfect equilibrium in which she votes against the bill, because relative if she uses such a strategy the proposer can increase her offer to k a little, leading k to strictly prefer voting for the bill.) The third player may vote for or against the bill (her vote has no effect on the outcome). In the first period, the pattern of behavior is the same: the bill proposed gives the non-proposer with the lower reservation value that value. In summary, in every subgame perfect equilibrium of the US game the strategy of each member i of the governing coalition has the following properties: • after the move of chance in either period, propose the bill that gives the legislator with the smallest reservation value in the that period her reservation value and gives i the remaining payoff 118 Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion • after a bill is proposed in either period, vote for the bill if it assigns i a positive amount. The equilibrium strategy of the other legislator j satisfies the condition: • after a bill is proposed in either period, vote for the bill if it assigns j a positive amount. (Each legislator’s equilibrium strategy may either vote for or vote against a bill that gives her a payoff of zero.) Thus in the US game there is no cohesion: the supporters of a bill may change from period to period, depending on the values of the reservation values. The UK game is defined as follows. Players The three legislators. Terminal histories All sequences (i, (x1 , x2 , x3 ), (A, B, C), j, (y1 , y2 , y3 ), (A , B , C )), where i is a member of the governing coalition and j is any legislator, (x1 , x2 , x3 ) and (y1 , y2 , y3 ) are partitions of one unit of payoff (x1 + x2 + x3 = y1 + y 2 + y3 = 1, xi ≥ 0, and yi ≥ 0 for i = 1, 2, 3), and A, B, C, A , B , and C are either yes (vote for bill) or no (vote against bill). Player function • P(∅) = c (chance) • P(i) = i • P(i, (x1 , x2 , x3 )) = {1, 2, 3} • P(i, (x1 , x2 , x3 ), (A, B, C)) = c • P(i, (x1 , x2 , x3 ), (A, B, C), j) = j • P(i, (x1 , x2 , x3 ), (A, B, C), j, (y1 , y2 , y3 )) = {1, 2, 3}. Chance probabilities Chance assigns probability 12 to 1 and probability 12 to 2 at the start of the game and after a history (i, (x1 , x2 , x3 ), (A, B, C)) in which at least two of the votes A, B, and C are yes. Chance assigns probability 13 to each legislator after a history (i, (x1 , x2 , x3 ), (A, B, C)) in which at least two of the votes A, B, and C are no. Actions • A(∅) = {1, 2} • A(i) = {(x1 , x2 , x3 ) : x1 + x2 + x 3 = 1, xi ≥ 0 for all i} for i = 1, 2 • Ak (i, (x1 , x2 , x3 )) = {yes, no} for all k, i = 1, 2, and all (x1 , x2 , x3 ) • A(i, (x1 , x2 , x3 ), (A, B, C)) = {1, 2} for all i, all (x1 , x2 , x3 ), and all triples (A, B, C) in which A, B, and C are all either yes or no and at least two are yes, and A(i, (x1 , x2 , x3 ), (A, B, C)) = {1, 2, 3} for all i, all (x1 , x2 , x3 ), and all triples (A, B, C) in which A, B, and C are all either yes or no and at most one is yes Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion 119 • A(i, (x1 , x2 , x3 ), (A, B, C), j) = {(x1 , x2 , x3 ) : x1 + x2 + x3 = 1, x i ≥ 0 for all i} for i = 1, 2, all (x1 , x2 , x3 ), all triples (A, B, C) in which A, B, and C are all either yes or no, and j = 1, 2, 3 • Ak (i, (x1 , x2 , x3 ), (A, B, C), j, (y1 , y2 , y3 )) = {yes, no} for all k, i = 1, 2, all (x1 , x2 , x3 ), all triples (A, B, C) in which A, B, and C are all either yes or no, j = 1, 2, 3, and all (y1 , y2 , y3 ). Preferences Each legislator i ranks the terminal histories by the amount of money she receives: xi + y i if both bills are passed, xi if only the first bill is passed, yi if only the second bill is passed, and 0 if neither bill is passed. To find the subgame perfect equilibria, start with the second period. The defeat of a bill leads each legislator to obtain the payoff of 0, so each legislator optimally votes for every bill. Thus in any subgame perfect equilibrium the proposer’s bill gives the proposer all the pie, and at least one of the other legislators votes for the bill. (As before, each of the other legislators is indifferent between voting for and voting against the bill, but there is no subgame perfect equilibrium in which the bill is voted down.) In the first period, the same argument shows that the proposer’s bill gives the proposer all the pie and that this bill passes. Further, in this period the other member of the governing coalition definitely votes for the bill. The reason is that if she does so, then her chance of being the proposer in the next period is 12 , so that her expected payoff is 12 . If she votes against, then the bill fails, so that she obtains a payoff of 0 in the first period and has a probability of 23 of being in the governing coalition in the second period, so that her expected payoff is 13 . Thus she is better off voting for her comrade’s bill than against it. In summary, in every subgame perfect equilibrium of the UK game the strategy of each legislator i has the following properties: • after the move of chance in either period, propose the bill that gives legislator i the payoff 1 • after a bill is proposed in the first period, vote for the bill if i is a member of the governing coalition. Thus in the UK game the governing coalition is entirely cohesive. 226.1 Nash equilibria when players may make mistakes The players’ best response functions are indicated in Figure 120.1. We see that the game has two Nash equilibria, (A, A, A) and (B, A, A). The action A is not weakly dominated for any player. For player 1, A is better than B if players 2 and 3 both choose B; for players 2 and 3, A is better than B for all actions of the other players. If players 2 and 3 choose A in the modified game, player 1’s expected payoffs to A and B are 120 Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion A B A 1∗ , 1∗ , 1∗ 1∗ , 1∗ , 1∗ B 0, 0, 1∗ 1∗ , 0, 1∗ A B A 0, 1∗ , 0 1∗ , 1∗ , 0 A B 1∗ , 0, 0 0, 0, 0 B Figure 120.1 The player’s best response functions in the game in Exercise 226.1. A: (1 − p2 )(1 − p3 ) + p1 p2 (1 − p 3 ) + p1 (1 − p2 )p3 + (1 − p1 )p2 p3 B: (1 − p 2 )(1 − p3 ) + (1 − p1 )p2 (1 − p3 ) + (1 − p1 )(1 − p2 )p3 + p 1 p2 p3 . The difference between the expected payoff to B and the expected payoff to A is (1 − 2p 1 )[p2 + p 3 − 3p 2 p3 ]. If 0 < pi < 12 for i = 1, 2, 3, this difference is positive, so that (A, A, A) is not a Nash equilibrium of the modified game. 228.1 Nash equilibria of the chain-store game Any terminal history in which the event in each period is either Out or (In, A) is the outcome of a Nash equilibrium. In any period in which challenger chooses Out, the strategy of the chain-store specifies that it choose F in the event that the challenger chooses In. 229.1 Subgame perfect equilibrium of the chain-store game The outcome of the strategy pair is that the only the last 10 challengers enter, and the chain-store acquiesces to their entry. The payoff of each of the first 90 challengers is 1 and the payoff to the remaining 10 is 2. The chain-store’s payoff is 90 × 2 + 10 × 1 = 190. No challenger can profitably deviate in any subgame (if one of the first 90 enters it is fought). However, I claim that the chain-store can increase its payoff by deviating after a history in which the first 89 challengers enter and are fought, and then challenger 90 enters. The chain-store’s strategy calls for it to fight challenger 90 and then subsequently acquiesce to any entry, and the remaining challengers’ strategies call for them to enter. But if instead the chain-store acquiesces to challenger 90, keeping the rest of its strategy the same, it increases its payoff by 1. (Note that the chain-store cannot profitably deviate after a history in which fewer than 89 challengers enter and each of them is fought. Suppose, for example, that each of the first 88 challengers enters and is fought, and then challenger 89 enters. The chain-store’s strategy calls for it to fight challenger 89, which induces challenger 90 to stay out; the remaining challengers enter, and the chain-store acquiesces. Its best deviation is to acquiesce to challenger 89’s entry and that of Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion 121 all subsequent entrants, in which case all remaining challengers, including challenger 90, enter. The outcomes of the two strategies differ in periods 89 and 90. If the challenger sticks to its original strategy it obtains 0 in period 89 and 2 in period 90; if it deviates it obtains 1 in each period.) 229.3 Nash equilibria of the centipede game Consider a strategy pair that results in an outcome in which player 1 stops the game in period k ≥ 2. (That is, each player chooses C through period k − 1 and the player who moves in period k chooses S.) Such a pair is not a Nash equilibrium because the player who moves in period k − 1 can do better (in the whole game, not only the subgame) by choosing S rather than C, given the other player’s strategy. Similarly the strategy pair in which each player always chooses C is not a Nash equilibrium. Thus in every Nash equilibrium player 1 chooses S at the start of the game. Draft of solutions to exercises in chapter of An introduction to game theory by Martin J. Osborne

[email protected]; www.chass.utoronto.ca/~osborne/index.html Version: 00/11/6. c 1995–2000 by Martin J. Osborne. All rights reserved. No part of this book may be reCopyright produced in any form by any electronic or mechanical means (including photocopying, recording, or information storage and retrieval) without permission in writing from Martin J. Osborne. On request, permission to make one copy for each student will be granted to instructors who wish to use the book in a course, on condition that copies be sold at a price not more than the cost of duplication. 8 Coalitional Games and the Core 241.1 Three-player majority game Let (x1 , x2 , x3 ) be an action of the grand coalition. Every coalition consisting of two players can obtain one unit of output, so for (x1 , x2 , x3 ) to be in the core we need x1 + x2 ≥ 1 x1 + x3 ≥ 1 x2 + x3 ≥ 1 x1 + x2 + x3 = 1. Adding the first three conditions we conclude that 2x1 + 2x 2 + 2x 3 ≥ 3, or x1 + x2 + x3 ≥ 32 , contradicting the last condition. Thus no action of the grand coalition satisfies all the conditions, so that the core of the game is empty. In the variant in which player 1 has three votes, a coalition can obtain one unit of output if and only if it contains player 1. (Note that players 2 and 3 together do not have a majority of the votes.) Thus for (x1 , x2 , x3 ) to be in the core we need x1 ≥ 1 x1 + x2 ≥ 1 x1 + x3 ≥ 1 x1 + x2 + x3 = 1. The first and last conditions (and the restriction that amounts of output must be nonnegative) imply that (x1 , x2 , x3 ) = (1, 0, 0), which satisfies the other two conditions. Thus the core consists of the single action (1, 0, 0) in which player 1 obtains all the output. 242.1 Market with one owner and two heterogeneous buyers By the arguments in Example 241.2, in any action in the core the owner does not keep the good, the buyer who obtains the good pays at most her valuation, and 123 124 Chapter 8. Coalitional Games and the Core the other buyer makes no payment. Let a N be an action of the grand coalition in which buyer 2 obtains the good and pays the owner p, and buyer 1 makes no payment. Then p ≤ v < 1, so that the coalition consisting of the owner and buyer 1 can improve upon a N : if the owner transfers the good to buyer 1 in exchange for 1 2 (1 + p) units of money, both the owner and buyer 1 are better off than they are in a N . Thus in any action in the core, buyer 1 obtains the good. The price she pays is at least v (otherwise the coalition consisting of the owner and buyer 2 can improve upon the action). No coalition can improve upon any action in which buyer 1 obtains the good and pays the owner at least v and at most 1 (and buyer 2 makes no payment), so the core consists of all such actions. 242.2 Vote trading a. The core consists of the single action in which all three bills pass, yielding each legislator a payoff of 2. This action cannot be improved upon by any coalition because no single bill or pair of bills gives every member of any majority coalition a payoff of more than 2. No other action is in the core, by the following argument. • The action in which no bill passes (so that each legislator’s payoff is 0) can be improved upon by the coalition of all three legislators, which by passing all three bills raises the payoff of each legislator to 2. • The action in which only A passes can be improved upon by the coalition of legislators 2 and 3, who by passing bills A and B raise both of their payoffs. • Similarly the action in which only B passes can be improved upon by the coalition of legislators 1 and 3, and the action in which only C passes can be improved upon by the coalition of legislators 1 and 2. • The action in which bills A and B pass can be improved upon by the coalition of legislators 1 and 3, who by passing all three bills raise both their payoffs. • Similarly the action in which bills A and C pass can be improved upon by the coalition of legislators 2 and 3, and the action in which bills B and C pass can be improved upon by the coalition of legislators 1 and 2. b. The core consists of two actions: all three bills pass, and bills A and B pass. As in part a, the action in which all three bills pass cannot be improved upon by any coalition. The action in which bills A and B cannot be improved upon either: for no other set of bills are at least two legislators better off. No other action is in the core, by the following argument. • The action in which A passes can be improved upon by the coalition consisting of legislators 2 and 3, who can pass B instead. Chapter 8. Coalitional Games and the Core 125 • The action in which B passes can be improved upon by the coalition consisting of legislators 1 and 2, who can pass A and B instead. • The action in which C passes can be improved upon by the coalition consisting of legislators 2 and 3, who can pass B instead. • The action in which A and C pass can be improved upon by the coalition consisting of legislators 2 and 3, who can pass A and B instead. • The action in which B and C pass can be improved upon by the coalition consisting of legislators 1 and 2, who can pass A and B instead. c. The core is empty. • The action in which no bill passes can be improved upon by the coalition consisting of legislators 1 and 2, who can pass A and B instead. • The action in which any single bill passes can be improved upon by the coalition consisting of the two legislators whose payoffs are −1 if this bill passes; this coalition can do better by passing the other two bills. • The action in which bills A and B pass can be improved upon by the coalition consisting of legislators 2 and 3, who can pass B instead. • Similarly the action in which A and C pass can be improved upon by the coalition consisting of legislators 1 and 2, who can pass A instead, and the action in which B and C pass can be improved upon by the coalition consisting of legislators 1 and 2, who can pass B instead. • The action in which all three bills pass can be improved upon by the coalition consisting of legislators 1 and 2, who can pass A and B instead. 244.1 Core of landowner–worker game Let a N be an action of the grand coalition in which the output received by each worker is at most f (n) − f (n − 1). No coalition consisting solely of workers can obtain any output, so no such coalition can improve upon a N . Let S be a coalition of the landowner and k − 1 workers. The total output received by the members of S in a N is at least f (n) − (n − k)( f (n) − f (n − 1)) (because the total output is f (n), and every other worker receives at most f (n) − f (n − 1)). Now, the output that S can obtain is f (k), so for S to improve upon a N we need f (k) > f (n) − (n − k)( f (n) − f (n − 1)), which contradicts the inequality given in the exercise. 126 Chapter 8. Coalitional Games and the Core 244.2 Unionized workers in landowner–worker game The following game models the situation. Players The landowner and the workers. Actions The set of actions of the grand coalition is the set of all allocations of the output f (n). Every other coalition has a single action, which yields the output 0. Preferences Each player’s preferences are represented by the amount of output she obtains. The core of this game consists of every allocation of the output f (n) among the players. The grand coalition cannot improve upon any allocation x because for every other allocation x there is at least one player whose payoff is lower in x than it is in x. No other coalition can improve upon any allocation because no other coalition can obtain any output. 245.1 Landowner–worker game with increasing marginal products We need to show that no coalition can improve upon the action a N of the grand coalition in which every player receives the output f (n)/n. No coalition of workers can obtain any output, so we need to consider only coalitions containing the landowner. Consider a coalition consisting of the landowner and k workers, which can obtain f (k + 1) units of output by itself. Under a N this coalition obtains the output (k + 1) f (n)/n, and we have f (k + 1)/(k + 1) < f (n)/n because k < n. Thus no coalition can improve upon a N . 250.1 Range of prices in horse market The equality of the number of owners who sell their horses and the number of nonowners who buy horses implies that the common trading price p∗ • is not less than σk∗ , otherwise at most k∗ − 1 owners’ valuations would be less than p∗ and at least k∗ nonowners’ valuations would be greater than p∗ , so that the number of buyers would exceed the number of sellers • is not less than β k∗ +1 , otherwise at most k∗ owners’ valuations would be less than p∗ and at least k∗ + 1 nonowners’ valuations would be greater than p∗ , so that the number of buyers would exceed the number of sellers • is not greater than β k∗ , otherwise at least k∗ owners’ valuations would be less than p∗ and at most k∗ − 1 nonowners’ valuations would be greater than p∗ , so that the number of sellers would exceed the number of buyers Chapter 8. Coalitional Games and the Core 127 • is not greater than σk∗ +1 , otherwise at least k∗ + 1 owners’ valuations would be less than p∗ and at most k∗ nonowners’ valuations would be greater than p∗ , so that the number of sellers would exceed the number of buyers. That is, p∗ ≥ max{σk∗ , β k∗ +1 } and p ∗ ≤ min{β k∗ , σk∗ +1 }. 251.1 Horse trading game with single seller The core consists of the set of actions of the grand coalition in which the owner sells her horse to the nonowner with the highest valuation (nonowner 1) at a price p∗ for which max{β 2 , σ1 } ≤ p∗ ≤ β 1 . (The coalition consisting of the owner and nonwoner 2 can improve any action in which the price is less than β 2 , the owner alone can improve upon any action in which the price is less than σ1 , and nonowner 1 alone can improve upon any action in which the price is greater than β 1 .) 251.2 Horse trading game with large seller In every action in the core, the owner sells one horse to buyer 1 and one horse to buyer 2. The prices at which the trades occur are not necessarily the same. The price p1 paid by buyer 1 satisfies max{β 3 , σ1 } ≤ p 1 ≤ β 1 and the price p2 paid by buyer 2 satisfies max{β 3 , σ1 } ≤ p1 ≤ β 2 . 254.1 House assignment with identical preferences Because the players rank the houses in the same way, we can refer to the “best house”, the “second best house”, and so on. In any assignment in the core, the player who owns the best house is assigned this house (because she has the option of keeping it). Among the remaining players, the one who owns the second best house must be assigned this house (again, because she has the option of keeping it). Continuing to argue in the same way, we see that there is a single assignment in the core, in which every player is assigned the house she owns initially. 255.1 Emptiness of the strong core when preferences are not strict Of the six possible assignments, h1 h2 h3 (i.e. every player keeps the house she owns) and h3 h2 h1 can both be improved upon by {1, 2} (and by {2, 3}). All four of the other assignments are in the core. None of the assignments in the core is in the strong core. The assignments h 1 h3 h2 and h3 h1 h2 can both be weakly improved upon by {1, 2}, and h2 h1 h3 and h2 h3 h1 can both be weakly improved upon by {2, 3}. 128 Chapter 8. Coalitional Games and the Core 257.1 Median voter theorem Denote the median favorite position by m. If x < m then every player whose favorite position is m or greater—a majority of the players—prefers m to x. Similarly, if x > m then every player whose favorite position is m or less—a majority of the players—prefers m to x. 258.1 Cores of q-rule games a. Denote the favorite policy of player i by xi∗ and number the players so that x1∗ ≤ · · · ≤ xn∗ . The q-core is the set of all policies x for which ∗ ≤ x ≤ xq∗ . xn−q+1 Any such policy x is in the core because every coalition of q players contains at least one player whose favorite position is less than x and at least one player whose favorite position is greater than x, so that there is no position y = x that all members of the coalition prefer to x. ∗ is not in the core because the coalition of players Any policy x < xn−q+1 n − q + 1 through n can improve upon x: this coalition contains q players, all ∗ to x. Similarly, no policy greater than xq∗ is in the core. of whom prefer xn−q+1 b. The core is the set of policies in the triangle defined by x1∗ , x2∗ , and x3∗ . Every policy x in this set is in the core because for every other policy y = x at least one player is worse off than she is at x. No policy x outside the set is in the core because the policy y = x closest to x in the set is preferred by all three players. 262.1 Deferred acceptance procedure with proposals by Y’s For the preferences given in Figure 260.1, the progress of the procedure when proposals are made by Y’s is given in Figure 128.1. The matching produced is the same as that produced by the procedure when proposals are made by X’s, namely (x1 , y1 ), (x2 , y2 ), x3 (alone), and y3 (alone). Stage 1 y1 : → x1 y2 : → x2 y3 : → x1 reject Stage 2 → x3 reject Stage 3 → x2 reject Figure 128.1 The progress of the deferred acceptance procedure with proposals by Y’s when the players’ preferences are those given in Figure 260.1. Each row gives the proposals of one X. Chapter 8. Coalitional Games and the Core 129 262.2 Example of deferred acceptance procedure For the preferences in Figure 262.1, the procedure when proposals are made by X’s yields the matching (x1 , y1 ), (x2 , y2 ), (x3 , y3 ); the procedure when proposals are made by Y’s yields the matching (x1 , y1 ), (x2 , y3 ), (x3 , y2 ). In any matching in the core, x1 and y1 are matched, because each is the other’s top-ranked partner. Thus the only two possible matchings are those generated by the two procedures. Player x2 prefers y2 to y3 and player x3 prefers y3 to y2 , so the matching generated by the procedure when proposals are made by X’s yields each X a better partner than does the matching generated by the procedure when proposals are made by Y’s. Similarly, player y2 prefers x3 to x 2 and player y3 prefers x2 to x3 , so the matching generated by the procedure when proposals are made by Y’s yields each Y a better partner than does the matching generated by the procedure when proposals are made by X’s. 263.1 Strategic behavior under the deferred acceptance procedure The matching produced by the deferred acceptance procedure with proposals by X’s is (x 1 , y2 ), (x2 , y3 ), (x3 , y1 ). The matching produced by the deferred acceptance procedure with proposals by Y’s is (x1 , y1 ), (x2 , y3 ), (x3 , y2 ). Of the four other matchings, the coalition {x3 , y2 } can improve upon (x1 , y1 ), (x2 , y2 ), (x3 , y3 ) and (x1 , y2 ), (x2 , y1 ), (x3 , y3 ), and the coalition {x1 , y1 } can improve upon (x1 , y3 ), (x2 , y1 ), (x3 , y2 ) and (x1 , y3 ), (x2 , y2 ), (x3 , y1 ). Thus the core consists of the two matchings produced by the deferred acceptance procedures. If y1 names the ranking (x1 , x2 , x3 ) and every other player names her true ranking, the deferred acceptance procedure with proposals by X’s yields the matching (x1 , y1 ), (x2 , y3 ), (x3 , y2 ), as illustrated in Figure 129.1. Players y1 and y 2 are matched with their favorite partners, so cannot profitably deviate by submitting any other ranking. Player y3 ’s ranking does not affect the outcome of the procedure. Thus, given that submitting her true ranking is a dominant strategy for every X, the game thus has a Nash equilibrium in which player y1 submits the ranking (x1 , x2 , x3 ) and every other player submits her true ranking. Stage 1 x1 : → y2 x2 : → y1 x3 : → y1 Stage 2 reject Stage 3 → y1 reject reject Stage 4 → y3 → y2 Figure 129.1 The progress of the deferred acceptance procedure with proposals by X’s when the players’ preferences differ from those in Exercise 263.1 only in that y1 ’s ranking is (x1 , x2 , x3 ). Each row gives the proposals of one X. 130 Chapter 8. Coalitional Games and the Core 263.2 Empty core in roommate problem Notice that is at the bottom of each of the other players’ preferences. Suppose that she is matched with i. Then j and k are matched, and {i, k} can improve upon the matching. Similarly, if is matched with j then {i, j} can improve upon the matching, and if is matched with k then {j, k} can improve upon the matching. Thus the core is empty ( has to be matched with someone!). 264.1 Spatial preferences in roommate problem The core consists of the single matching µ∗ defined as follows. First match the pair of players whose characteristics are closest. Then match the pair of players in the remaining set whose characteristics are closest. Continue until all players are matched. Number the matches in the order they are made according to this procedure. If a coalition can improve upon µ∗ , then a coalition consisting of two players can do so. Now, neither member of match k is better off being matched with a member of match for any > k, so no two-player coalition can improve upon the matching. Thus µ∗ is in the core. For any other matching µ , at least one of the members of some match k defined by the procedure is matched with a different partner. If she is matched with a member of some match < k then the coalition consisting of the two members of match can improve µ ; if she is matched with a member of some match > k then the coalition consisting of the two member of match k can improve upon µ . Thus no matching µ = µ∗ is in the core. Draft of solutions to exercises in chapter of An introduction to game theory by Martin J. Osborne

[email protected]; www.chass.utoronto.ca/~osborne/index.html Version: 00/11/6. c 1995–2000 by Martin J. Osborne. All rights reserved. No part of this book may be reCopyright produced in any form by any electronic or mechanical means (including photocopying, recording, or information storage and retrieval) without permission in writing from Martin J. Osborne. On request, permission to make one copy for each student will be granted to instructors who wish to use the book in a course, on condition that copies be sold at a price not more than the cost of duplication. 9 Bayesian games 274.1 Equilibria of a variant of BoS with imperfect information If player 1 chooses S then type 1 of player 2 chooses S and type 2 chooses B. But if the two types of player 2 make these choices then player 1 is better off choosing B (which yields her an expected payoff of 1) than choosing S (which yields her an expected payoff of 12 ). Thus there is no Nash equilibrium in which player 1 chooses S. Now consider the mixed strategy Nash equilibria. If both types of player 2 use a pure strategy then player 1’s two actions yield her different payoffs. Thus there is no equilibrium in which both types of player 2 use pure strategies and player 1 randomizes. Now consider an equilibrium in which type 1 of player 2 randomizes. Denote by p the probability that player 1’s mixed strategy assigns to B. In order for type 1 of player 2 to obtain the same expected payoff to B and S we need p = 23 . For this value of p the best action of type 2 of player 2 is S. Denote by q the probability that type 1 of player 2 assigns to B. Given these strategies for the two types of player 2, player 1’s expected payoff if she chooses B is 1 2 · 2q = q and her expected payoff if she chooses S is 1 2 · (1 − q) + 1 2 · 1 = 1 − 12 q. These expected payoffs are equal if and only if q = 23 . Thus the game has a mixed strategy equilibrium in which the mixed strategy of player 1 is ( 23 , 13 ), that of type 1 of player 2 is ( 23 , 13 ), and that of type 2 of player 2 is (0, 1) (that is, type 2 of player 2 uses the pure strategy that assigns probability 1 to S). Similarly the game has a mixed strategy equilibrium in which the strategy of player 1 is ( 13 , 23 ), that of type 1 of player 2 is (0, 1), and that of type 2 of player 2 is ( 23 , 13 ). For no mixed strategy of player 1 are both types of player 2 indifferent between their two actions, so there is no equilibrium in which both types randomize. 275.1 Expected payoﬀs in a variant of BoS with imperfect information The expected payoffs are given in Figure 132.1. 131 132 Chapter 9. Bayesian games (B, B) (B, S) (S, B) (S, S) 0 1 1 2 1 1 2 1 2 0 B S Type n1 of player 1 (B, B) (B, S) (S, B) (S, S) B 1 0 1 3 4 3 0 S 2 3 2 3 2 Type y2 of player 2 (B, B) (B, S) (S, B) (S, S) B 0 2 2 3 2 3 1 S 1 3 4 3 0 Type n2 of player 2 Figure 132.1 The expected payoffs of type n 1 of player 1 and types y2 and n 2 of player 2 in Example 274.2. 280.2 Fighting an opponent of unknown strength The following Bayesian game models the situation. Players The two people. States The set of states is {strong, weak}. Actions The set of actions of each player is {fight, yield}. Signals Player 1 receives the same signal in each state, whereas player 2 receives different signals in the two states. Beliefs The single type of player 1 assigns probability α to the state strong and probability 1 − α to the state weak. Each type of player 2 assigns probability 1 to the single state consistent with her signal. Payoﬀs The players’ Bernoulli payoffs are shown in Figure 133.1. The best responses of each type of player 2 are indicated by asterisks in Figure 133.1. Thus if α < 12 then player 1’s best action is fight, whereas if α > 12 her best action is yield. Thus for α < 12 the game has a unique Nash equilibrium, in which player 1 chooses fight and player 2 chooses fight if she is strong and yield if she is weak, and if α > 12 the game has a unique Nash equilibrium, in which player 1 chooses yield and player 2 chooses fight whether she is strong or weak. Chapter 9. Bayesian games F Y F −1, 1 ∗ 0, 1∗ 133 Y 1, 0 0, 0 F Y State: strong F 1, −1 0, 1∗ Y 1, 0∗ 0, 0 State: weak Figure 133.1 The player’s Bernoulli payoff functions in Exercise 280.2. The asterisks indicate the best responses of each type of player 2. 280.3 An exchange game The following Bayesian game models the situation. Players The two individuals. States The set of all pairs (s1 , s2 ), where si is the number on player i’s ticket (an integer from 1 to m). Actions The set of actions of each player is {Exchange, Don’t exchange}. Signals The signal function of each player i is defined by τi (s1 , s2 ) = si (each player observes her own ticket, but not that of the other player) Beliefs Type si of player i assigns the probability Pr j (s j ) to the state (s1 , s2 ), where j is the other player and Pr j (s j ) is the probability with which player j receives a ticket with the prize s j on it. Payoﬀs Player i’s Bernoulli payoff function is given by ui ((X, Y), ω) = ω j if X = Y = Exchange and ui ((X, Y), ω) = ωi otherwise. Let Mi be the highest type of player i that chooses Exchange. If Mi > 1 then type 1 of player j optimally chooses Exchange: by exchanging her ticket, she cannot obtain a smaller prize, and may receive a bigger one. Thus if Mi ≥ M j and Mi > 1, type Mi of player i optimally chooses Don’t exchange, because the expected value of the prizes of the types of player j that choose Exchange is less than Mi . Thus in any possible Nash equilibrium Mi = M j = 1: the only prizes that may be exchanged are the smallest. 280.4 Adverse selection The game is defined as follows. Players Firms A and T. States The set of possible values of firm T (the integers from 0 to 100). Actions Firm A’s set of actions is its set of possible bids (nonnegative numbers), and firm T’s set of actions is the set of possible cutoffs (nonnegative numbers) above which it will accept A’s offer. 134 Chapter 9. Bayesian games Signals Firm A receives the same signal in every state; firm T receives a different signal in every state. Beliefs The single type of firm A assigns an equal probability to each state; each type of firm T assigns probability 1 to the single state consistent with its signal. Payoﬀ functions If firm A bids y, firm T’s cutoff is at most y, and the state is x, then A’s payoff is 32 x − y and T’s payoff is y. If firm A bids y, firm T’s cutoff is greater than y, and the state is x, then A’s payoff is 0 and T’s payoff is x. To find the Nash equilibria of this game, first consider the behavior of each type of firm T. Type x is at least as well off accepting the offer y than it is rejecting it if and only if y ≥ x. Thus type x’s optimal cutoff for accepting offers is x, regardless of firm A’s action. Now consider firm A. If it bids y then each type x of T with x < y accepts its offer, and each type x of T with x > y rejects the offer. Thus the expected value of the type that accepts an offer y ≤ 100 is 12 y, and the expected value of the type that accepts an offer y > 100 is 50. If the offer y is accepted then A’s payoff is 32 x − y, so that its expected payoff is 32 ( 12 y) − y = − 14 y if y ≤ 100 and 32 (50) − y = 75 − y if y > 100. Thus firm A’s optimal bid is 0! We conclude that the game has a unique Nash equilibrium, in which firm A bids 0 and the cutoff for accepting an offer for each type x of firm T is x. Even though firm A can increase firm T’s value, it is not willing to make a positive bid in equilibrium because firm T’s interest is in accepting only offers that exceed its value, so that the average type that accepts an offer has a value of only half the offer. As A decreases its offer, the value of the average firm that accepts the offer decreases: the selection of firms that accept the offer is adverse to A’s interest. 282.1 Infection argument In any Nash equilibrium, the action of player 1 when she receives the signal τ1 (α) is R, because R strictly dominates L. Now suppose that player 2’s signal is τ2 (α) = τ2 (β). I claim that her best action is R, regardless of player 1’s action in state β. If player 1 chooses L in state β then player 2’s expected payoff to L is 34 · 0 + 14 · 2 = 12 , and her expected payoff to R is 3 1 3 4 · 1 + 4 · 0 = 4 . If player 1 chooses R in state β then player 2’s expected payoff to L is 0, and her expected payoff to R is 1. Thus in any Nash equilibrium player 2’s action when her signal is τ2 (α) = τ2 (β) is R. Now suppose that player 1’s signal is τ1 (β) = τ1 (γ). By the same argument as in the previous paragraph, player 1’s best action is R, regardless of player 2’s action in state γ. Thus in any Nash equilibrium player 1’s action in this case is R. Finally, given that player 1’s action in state γ is R, player 2’s best action in this state is also R. Chapter 9. Bayesian games 135 285.1 Cournot’s duopoly game with imperfect information We have b1 (q L , q H ) = 1 2 (α − c − (θq L 0 + (1 − θ)q H )) if θq L + (1 − θ)q H ≤ α − c otherwise. The best response function of each type of player 2 is similar: 1 b I (q1 ) = 2 (α − c I − q1 ) if q1 ≤ α − c I 0 otherwise for I = L, H. The three equations that define a Nash equilibrium are q∗1 = b1 (q∗L , q∗H ), q∗L = b L (q∗1 ), and q∗H = b H (q∗1 ). Solving these equations under the assumption that they have a solution in which all three outputs are positive, we obtain q∗1 = q∗L q∗H = = 1 3 (α 1 3 (α 1 3 (α − 2c + θc L + (1 − θ)c H ) − 2c L + c) − 16 (1 − θ)(c H − c L ) − 2c H + c) + 16 θ(c H − c L ) If both firms know that the unit costs of the two firms are c1 and c2 then in a Nash equilibrium the output of firm i is 13 (α − 2ci + c j ) (see Exercise 57.1). In the case of imperfect information considered here, firm 2’s output is less than 1 1 3 (α − 2c L + c) if its cost is c L and is greater than 3 (α − 2c H + c) if its cost is c H . Intuitively, the reason is as follows. If firm 1 knew that firm 2’s cost were high then it would produce a relatively large output; if it knew this cost were low then it would produce a relatively small output. Given that it does not know whether the cost is high or low it produces a moderate output, less than it would if it knew firm 2’s cost were high. Thus if firm 2’s cost is in fact high, firm 2 benefits from firm 1’s lack of knowledge and optimally produces more than it would if firm 1 knew its cost. 286.1 Cournot’s duopoly game with imperfect information The best response b0 (q L , q H ) of type 0 of firm 1 is the solution of max[θ(P(q0 + q L ) − c)q0 + (1 − θ)(P(q0 + q H ) − c)q0 ]. q0 The best response b (q L , q H ) of type of firm 1 is the solution of max(P(q + q L ) − c)q q 136 Chapter 9. Bayesian games and the best response bh (q L , q H ) of type h of firm 1 is the solution of max(P(qh + q H ) − c)qh . qh The best response b L (q0 , q , qh ) of type L of firm 2 is the solution of max[(1 − π)(P(q0 + q L ) − c L )q L + π(P(q + q L ) − c L )q L ] qL and the best response b H (q0 , q , qh ) of type H of firm 2 is the solution of max[(1 − π)(P(q0 + q H ) − c H )q H + π(P(qh + q H ) − c H )q H ]. qH A Nash equilibrium is a profile (q∗0 , q∗ , q∗h , q∗L , q∗H ) for which q∗0 , q∗ , and q∗h are best responses to q∗L and q∗H , and q∗L and q∗H are best responses to q∗0 , q∗ , and q∗h . When P(Q) = α − Q for Q ≤ α and P(Q) = 0 for Q > α we find, after some exciting algebra, that q∗0 = q∗ = q∗H = q∗L = q∗H = 1 3 1 3 1 3 1 3 1 3 (α − 2c + c H − θ (c H − c L )) (1 − θ)(1 − π)(c H − c L ) α − 2c + c L + 4−π θ(1 − π)(c H − c L ) α − 2c + c H − 4−π 2(1 − θ)(1 − π)(c H − c L ) α − 2c L + c − 4−π 2θ(1 − π)(c H − c L ) α − 2c H + c + . 4−π When π = 0 we have q∗0 = q∗ = q∗H = q∗L = q∗H = 1 3 1 3 1 3 1 3 1 3 (α − 2c + c H − θ (c H − c L )) (1 − θ)(c H − c L ) α − 2c + c L + 4 θ(c H − c L ) α − 2c + c H − 4 (1 − θ)(c H − c L ) α − 2c L + c − 2 θ(c H − c L ) α − 2c H + c + , 2 so that q∗0 is equal to the equilibrium output of firm 1 in Exercise 285.1, and q∗L and q∗H are the same as the equilibrium outputs of the two types of firm 2 in that exercise. Chapter 9. Bayesian games 137 When π = 1 we have q∗0 = q∗ = q∗H = q∗L = q∗H = 1 3 1 3 1 3 1 3 1 3 (α − 2c + c H − θ (c H − c L )) (α − 2c + c L ) (α − 2c + c H ) (α − 2c L + c) (α − 2c H + c) , so that q∗ and q∗L are the same as the equilibrium outputs when there is perfect information and the costs are c and c L (see Exercise 57.1), and q∗h and q∗H are the same as the equilibrium outputs when there is perfect information and the costs are c and c H . Now, for an arbitrary value of π we have 1 2(1 − θ)(1 − π)(c H − c L ) ∗ qL = α − 2c L + c − 3 4−π 1 2θ(1 − π)(c H − cL ) ∗ qH = α − 2c H + c + . 3 4−π To show that for 0 < π < 1 the values of these variables lie between their values when π = 0 and when π = 1, we need to show that 0≤ (1 − θ)(c L − c H ) 2(1 − θ)(1 − π)(c H − c L ) ≤ 4−π 2 and 0≤ θ(c L − c H ) 2θ(1 − π)(c H − c L ) ≤ . 4−π 2 These inequalities follow from c H ≥ c L , θ ≥ 0, and 0 ≤ π ≤ 1. 288.1 Nash equilibria of game of contributing to a public good Any type v j of any player j with v j < c obtains a negative payoff if she contributes and 0 if she does not. Thus she optimally does not contribute. Any type vi ≥ c of player i obtains the payoff vi − c ≥ 0 if she contributes, and the payoff 0 if she does not, so she optimally contributes. Any type v j ≥ c of any player j = i obtains the payoff v j − c if she contributes, and the payoff (1 − F(c))v j if she does not. (If she does not contribute, the probability that player i does so is 1 − F(c), the probability that player i’s valuation is at least c.) Thus she optimally does not contribute if (1 − F(c))v j ≥ v j − c, or F(c) ≤ c/v j . This condition must hold for all types of every player j = i, so we need F(c) ≤ c/v for the strategy profile to be a Nash equilibrium. 138 Chapter 9. Bayesian games 290.1 Reporting a crime with an unknown number of witnesses A Bayesian game that models the situation is given in Figure 138.1. π 2 C N N v−c 0 State 1 1−π 1−π 1 C N C v − c, v − c v, v − c 1 N v − c, v 0, 0 State 12 π 2 N C v−c N 0 State 2 Figure 138.1 A Bayesian game that models the situation in Exercise 290.1. The action Call is denoted C, and the action Don’t call is denoted N. In state 1, only player 1 is active, in state 2, only player 2 is active, and in state 12, both players are active. In states in which only one players is active, only that player’s payoff is given. A player obtains the payoff v − c if she chooses C and the payoff (1 − π)v if she chooses N. Thus the game has a pure strategy Nash equilibrium in which each player chooses C if and only if v − c ≥ (1 − π)v, or π ≥ c/v. For a mixed strategy Nash equilibrium in which each player chooses C (if she is active) with probability p, where 0 < p < 1, we need each player’s expected payoffs to C and N to be the same, given that the other player chooses C with probability p. Thus we need v − c = (1 − π)pv, or p= v−c . (1 − π)v If π < c/v, this number is less than 1, so that the game indeed has a mixed strategy Nash equilibrium in which each player calls with probability p. When π = 0 we have p = 1 − c/v, as found in Section 4.8. 292.1 Weak domination in second-price sealed-bid action Fix player i, and choose a bid for every type of every other player. Player i, who does not know the other players’ types, is uncertain of the highest bid of the other players. Denote by b this highest bid. Consider a bid bi of type vi of player i for which bi < vi . The dependence of the payoff of type vi of player i on b is shown in Figure 139.1. Player i’s expected payoffs to the bids bi and vi are weighted averages of the payoffs in the columns; each value of b gets the same weight when calculating the expected payoff to bi as it does when calculating the expected payoff to vi . The payoffs in the two rows are the same except when bi ≤ b < vi , in which case vi yields a payoff higher than does bi . Thus the expected payoff to vi is at least as high as the expected payoff to bi , and is greater than the expected payoff to bi unless the other players’ bids lead this range of values of b to get probability 0. Chapter 9. Bayesian games 139 b < bi i’s bid bi < vi vi vi − b vi − b Highest of other players’ bids bi = b bi < b < v i (m-way tie) (v i − b)/m 0 vi − b vi − b b ≥ vi 0 0 Figure 139.1 Player i’s payoffs to her bids bi < vi and vi in a second-price sealed-bid auction as a function of the highest of the other player’s bids, denoted b. Now consider a bid bi of type vi of player i for which bi > vi . The dependence of the payoff of type vi of player i on b is shown in Figure 139.2. b ≤ vi i’s bid vi vi − b bi > v i vi − b Highest of other players’ bids bi = b v i < b < bi (m-way tie) 0 0 vi − b (vi − b)/m b > bi 0 0 Figure 139.2 Player i’s payoffs to her bids vi and bi > vi in a second-price sealed-bid auction as a function of the highest of the other player’s bids, denoted b. As before, player i’s expected payoffs to the bids bi and vi are weighted averages of the payoffs in the columns; each value of b gets the same weight when calculating the expected payoff to vi as it does when calculating the expected payoff to bi . The payoffs in the two rows are the same except when vi < b ≤ bi , in which case vi yields a payoff higher than does bi . (Note that vi − b < 0 for b in this range.) Thus the expected payoff to vi is at least as high as the expected payoff to bi , and is greater than the expected payoff to bi unless the other players’ bids lead this range of values of b to get probability 0. We conclude that for type vi of player i, every bid bi = vi is weakly dominated by the bid vi . 292.2 Nash equilibria of a second-price sealed-bid auction For any player i, the game has a Nash equilibrium in which player i bids v (the highest possible valuation) regardless of her valuation and every other player bids v regardless of her valuation. The outcome is that player i wins and pays v. Player i can do no better by bidding less; no other player can do better by bidding more, because unless she bids at least v she does not win, and if she makes such a bid her payoff is at best zero. (It is zero if her valuation is v, negative otherwise.) 140 Chapter 9. Bayesian games 295.1 Auctions with risk-averse bidders Consider player i. Suppose that the bid of each type v j of player j is given by β j (v j ) = (1 − 1/[m(n − 1) + 1])v j . Then as far as player i is concerned, the bids of every other player are distributed uniformly between 0 and 1 − 1/[m(n − 1) + 1]. Thus for 0 ≤ x ≤ 1 − 1/[m(n − 1) + 1], the probability that any given player’s bid is less than x is (1 + 1/[m(n + 1)])x (1 + 1/[m(n + 1)] being the reciprocal of 1 − 1/[m(n − 1) + 1]), and hence the probability that all the bids of the other n − 1 players are less than x is [(1 + 1/[m(n + 1)])x]n−1 . Consequently, if player i bids more than 1 − 1/[m(n − 1) + 1] then she surely wins, whereas if she bids bi ≤ 1 − 1/[m(n − 1) + 1] she wins with probability [(1 + 1/[m(n + 1)])bi ]n−1 . Thus player i’s payoff as a function of her bid bi is n−1 1 1 1/m (vi − bi ) if 0 ≤ bi ≤ 1 − 1+ bi m(n + 1) m(n − 1) + 1 (140.1) 1 1/m (v i − bi ) . if bi > 1 − m(n − 1) + 1 Now, the value of bi that maximizes the function n−1 1 (vi − bi )1/m 1+ bi m(n + 1) is the same as the value of bi that maximizes the function (v i − bi )1/m (bi )n−1 , which is (n − 1)v i /(n − 1 + 1/m) (by the mathematical fact stated in the exercise), or 1 1− vi . m(n − 1) + 1 We have 1− 1 m(n − 1) + 1 vi ≤ 1 − 1 m(n − 1) + 1 (because vi ≤ 1), and the function in (140.1) is decreasing in bi for bi > 1 − 1/[m(n − 1) + 1], so 1 − 1/[m(n − 1) + 1] is the bid that maximizes player i’s expected payoff, given that the bid of each type v j of player j is (1 − 1/[m(n − 1) + 1])v j . We conclude that, as claimed, the game has a Nash equilibrium in which each type vi of each player i bids (1 − 1/[m(n − 1) + 1])v i . In this equilibrium, the price paid by a bidder with valuation v who wins is (1 − 1/[m(n − 1) + 1])v (the amount she bids). The expected price paid by a bidder in a second-price auction does not depend on the players’ payoff functions. Thus this payoff is equal, by the revenue equivalence result, to the expected price paid by a bidder with valuation v who wins in a first-price auction in which each bidder is risk-neutral, namely (1 − 1/n)v. We have 1 (m − 1)(n − 1) 1 , − 1− = 1− m(n − 1) + 1 n n(m(n − 1) + 1) Chapter 9. Bayesian games 141 which is positive because m > 1. Thus the expected price paid by a bidder with valuation v who wins is greater in a first-price auction than it is in a second-price auction. The probability that a bidder with any given valuation wins is the same in both auctions, so the auctioneer’s expected revenue is greater in a first-price auction than it is in a second-price auction. 297.1 Asymmetric Nash equilibria of second-price sealed-bid common value auctions Suppose that each type t2 of player 2 bids (1 + 1/λ)t2 and that type t1 of player 1 bids b1 . Then by the calculations in the text, with α = 1 and γ = 1/λ, • a bid of b1 by player 1 wins with probability b1 /(1 + 1/λ) • the expected value of player 2’s bid, given that it is less than b1 , is 12 b1 • the expected value of signals that yield a bid of less than b1 is 12 b1 /(1 + 1/λ) (because of the uniformity of the distribution of t2 ). Thus player 1’s expected payoff if she bids b1 is (t1 + 12 b1 /(1 + 1/λ) − 12 b1 )b1 /(1 + 1/λ), or λ · (2(1 + λ)t1 − b1 )b1 . 2(1 + λ)2 This function is maximized at b1 = (1 + λ)t1 . That is, if each type t2 of player 2 bids (1 + 1/λ)t2 , any type t1 of player 1 optimally bids (1 + λ)t1 . Symmetrically, if each type t1 of player 1 bids (1 + λ)t1 , any type t2 of player 2 optimally bids (1 + 1/λ)t2 . Hence the game has the claimed Nash equilibrium. 297.2 First-price sealed-bid auction with common valuations Suppose that each type t2 of player 2 bids 12 (α + γ)t2 and type t1 of player 1 bids b1 . To determine the expected payoff of type t1 of player 1, we need to find the probability with which she wins, and the expected value of player 2’s signal if player 1 wins. (The price she pays is her bid, b1 .) Probability of player 1’s winning: Given that player 2’s bidding function is 1 (α + γ)t2 , player 1’s bid of b1 wins only if b1 ≥ 12 (α + γ)t2 , or if t2 ≤ 2b1 /(α + γ). 2 Now, t2 is distributed uniformly from 0 to 1, so the probability that it is at most 2b1 /(α + γ) is 2b1 /(α + γ). Thus a bid of b1 by player 1 wins with probability 2b1 /(α + γ). Expected value of player 2’s signal if player 1 wins: Player 2’s bid, given her signal t2 , is 12 (α + γ)t2 , so that the expected value of signals that yield a bid of less than b1 is b1 /(α + γ) (because of the uniformity of the distribution of t2 ). Thus player 1’s expected payoff if she bids b1 is 2(αt1 + γb1 /(α + γ) − b1 )b1 /(α + γ), or 2α ((α + γ)t1 − b1 )b1 . (α + γ)2 142 Chapter 9. Bayesian games This function is maximized at b1 = 12 (α + γ)t1 . That is, if each type t2 of player 2 bids 12 (α + γ)t2 , any type t1 of player 1 optimally bids 12 (α + γ)t1 . Hence, as claimed, the game has a Nash equilibrium in which each type ti of player i bids 1 2 (α + γ)t i . 304.1 Signal-independent equilibria in a model of a jury If every juror votes for acquittal regardless of her signal then the action of any single juror has no effect on the outcome. Thus the strategy profile in which every juror votes for acquittal regardless of her signal is always a Nash equilibrium. Now consider the possibility of a Nash equilibrium in which every juror votes for conviction regaredless of her signal. Suppose that every juror other than juror 1 votes for conviction independently of her signal. Then juror 1’s vote determines the outcome, exactly as in the case in which there is a single juror. Thus from the calculations in Section 9.8.2, type b of juror 1 optimally votes for conviction if and only if (1 − p)π z≤ (1 − p)π + q(1 − π) and type g of juror 1 optimally votes for conviction if and only if z≤ pπ . pπ + (1 − q)(1 − π) The assumption that p > 1 − q implies that the term on the right side of the second inequality is greater than the term on the right side of the first inequality, so that we conclude that there is a Nash equilibrium in which every juror votes for conviction regardless of her signal if and only if pπ (1 − p)π ≤z≤ . (1 − p)π + q(1 − π) pπ + (1 − q)(1 − π) 305.1 Swing voter’s curse a. The Bayesian game is defined as follows. Players Citizens 1 and 2. States {A, B}. Actions The set of actions of each player is {0, 1, 2} (where 0 means do not vote). Signals Citizen 1 receives different signals in states A and B, whereas citizen 2 receives the same signal in both states. Beliefs Each type of citizen 1 assigns probability 1 to the single state consistent with her signal. The single type of citizen 2 assigns probability 0.9 to state A and probability 0.1 to state B. Chapter 9. Bayesian games 143 Payoﬀs Both citizens’ Bernoulli payoffs are 1 if either the state is A and candidate 1 receives the most votes or the state is B and candidate 2 receives the most votes; their payoffs are 0 if either the state is B and candidate 1 receives the most votes or the state is A and candidate 2 receives the most votes; and otherwise their payoffs are 12 . (These payoffs are shown in Figure 143.1.) 2 0 1 1, 1 0, 0 1 1 2, 2 0, 0 1, 1 1, 1 1, 1 1 1 2, 2 0, 0 0, 0 1 1 2, 2 0, 0 1 1 2, 2 1, 1 1 1 2, 2 1, 1 0 0 1 2 1 1 2, 2 1 State A 0, 0 0 1 2 2 State B Figure 143.1 The payoffs in the Bayesian game for Exercise 305.1. b. Type A of player 1’s best action depends only on the action of player 2; it is to vote for 1 if player 2 votes for 2 or does not vote, and either to vote for 1 or not vote if player 2 votes for 1. Similarly, type B of player 1’s best action is to vote for 2 if player 2 votes for 1 or does not vote, and either to vote for 2 or not vote if player 2 votes for 2. Player 2’s best action is to vote for 1 if type A of player 1 either does not vote or votes for 2 (regardless of how type B of player 1 votes), not to vote if type A of player 1 votes for 1 and type B of player 1 either votes for 2 or does not vote, and either to vote for 1 or not to vote if both types of player 1 vote for 1. Given the best responses of the two types of player 1, their only possible equilibrium actions are (0, 0) (i.e. both do not vote), (0, 2), (1, 0), and (1, 2). Checking player 2’s best responses we see that the only equilibria are • (0, 2, 1) (player 1 does not vote in state A and votes for 2 in state B; player 2 votes for 1) • (1, 2, 0) (player 1 votes for 1 in state A and for 2 in state B; player 2 does not vote). c. In the equilibrium (0, 2, 1), type A of player 1’s action is weakly dominated by the action of voting for 1: voting for 1 instead of not voting never makes her worse off, and makes her better off in the event that player 2 does not vote. d. In the equilibrium (1, 2, 0), player 2 does not vote because if she does then in the only case in which her vote affects the outcome (i.e. the only case in which she is a “swing voter”), it affects it adversely: if she votes for 1 then her vote makes no difference in state A, whereas it causes a tie, instead of a 144 Chapter 9. Bayesian games win for candidate 2 in state B, and if she votes for 2 then her vote causes a tie, instead of a win for candidate 1 in state A, and makes no difference in state B. 307.2 Properties of the bidding function in a ﬁrst-price sealed-bid auction We have ∗ β (v) = 1 − = 1− = (F(v))n−1 (F(v))n−1 − (n − 1)(F(v))n−2 F (v) (F(v))n − (n − 1)F (v) (n − 1)F (v) > 0 v v v v (F(x))n−1 dx (F(v))2n−2 v n−1 dx v (F(x)) (F(v))n (F(x))n−1 dx (F(v))n if v > v because F (v) > 0 (F is increasing). (The first line uses the quotient rule for derivav f (x)dx with respect to v is f (v) for any tives and the fact that the derivative of function f .) If v > v then the integral in (307.1) is positive, so that β∗ (v) < v. If v = v then both the numerator and denominator of the quotient in (307.1) are zero, so we may use L’Hopital’s ˆ rule to find the value of the quotient as v → v. Taking the derivatives of the numerator and denominator we obtain F(v) (F(v))n−1 , = (n − 1)F (v) (n − 1)(F(v))n−2 F (v) the numerator of which is zero and the denominator of which is positive. Thus the quotient in (307.1) is zero, and hence β∗ (v) = v. 307.3 Example of Nash equilibrium in a ﬁrst-price auction From (307.1) we have ∗ v x n−1 dx n−1 v v n−1 x dx = v − 0 n−1 v = v − v/n = (n − 1)v/n. β (v) = v − 0 Draft of solutions to exercises in chapter of An introduction to game theory by Martin J. Osborne

[email protected]; www.chass.utoronto.ca/~osborne/index.html Version: 00/11/6. c 1995–2000 by Martin J. Osborne. All rights reserved. No part of this book may be reCopyright produced in any form by any electronic or mechanical means (including photocopying, recording, or information storage and retrieval) without permission in writing from Martin J. Osborne. On request, permission to make one copy for each student will be granted to instructors who wish to use the book in a course, on condition that copies be sold at a price not more than the cost of duplication. 11 Strictly competitive games and maxminimization 338.2 Nash equilibrium payoﬀs and maxminimized payoﬀs In the game in Figure 147.1 each player’s maxminimized payoff is 1, while her payoff in the unique Nash equilibrium is 2. T B L 2, 2 0, 1 R 1, 0 0, 0 Figure 147.1 A game in which each player’s Nash equilibrium payoff exceeds her maxminimized payoff. 340.1 Strictly competitive games Left-hand game: Strictly competitive both in pure and in mixed strategies. (Player 2’s preferences are represented by the vNM payoff function −u1 since −u1 (a) = − 12 + 1 2 u2 (a) for every pure outcome a.) Right-hand game: Strictly competitive in pure strategies (since player 1’s ranking of the four outcomes is the reverse of player 2’s ranking). Not strictly competitive in mixed strategies (there exist no values of α and β > 0 such that −u1 (a) = α + βu2 (a) for every outcome a; or, alternatively, player 1 is indifferent between (D, L) and the lottery that yields (U, L) with probability 12 and (U, R) with probability 12 , while player 2 is not indifferent between these two outcomes). 343.2 Maxminimizing in BoS The maxminimizer of player 1 is ( 13 , 23 ) while that of player 2 is ( 23 , 13 ). It is clear that neither of the pure equilibrium strategies of either player guarantees her equilibrium payoff. In the mixed strategy equilibrium player 1’s expected payoff is 23 ; but if, for example, player 2 choose S instead of her equilibrium strategy, then player 1’s expected payoff is 13 . Similarly for player 2. 343.3 Changing payoﬀs in strictly competitive game 147 148 Chapter 11. Strictly competitive games and maxminimization a. Let ui be player i’s payoff function in the game G, let wi be his payoff function in G , and let (x ∗ , y∗ ) be a Nash equilibrium of G . Then, using part (a) of Proposition 341.1, we have w1 (x ∗ , y∗ ) = min y max x w1 (x, y) ≥ min y max x u1 (x, y), which is the value of G. b. This follows from part (a) of Proposition 341.1 and the fact that for any function f we have max x∈X f (x) ≥ max x∈Y f (x) if Y ⊆ X. c. In the unique equilibrium of the game on the left of Figure 148.1 player 1 receives a payoff of 3, while in the unique equilibrium of she receives a payoff of 2. If she is prohibited from using her second action in this second game then she obtains an equilibrium payoff of 3, however. 3, 3 1, 0 1, 1 0, 1 3, 3 4, 0 1, 1 2, 1 Figure 148.1 The games for part c of Exercise 343.3. 344.1 Equilibrium payoﬀ in strictly competitive game The claim is false. In the strictly competitive game in Figure 148.2 the action pair (T, L) is a Nash equilibrium, so that player 1’s unique equilibrium payoff in the game is 0; but (B, R), which also yields player 1 a payoff of 0, is not a Nash equilibrium. T B L 0, 0 −1, 1 R 1, −1 0, 0 Figure 148.2 The game in Exercise 344.1. 344.2 Guessing Morra In the strategic game there are two players, each of whom has four (relevant) actions, S1G2, S1G3, S2G3, and S2G4, where SiGj denotes the strategy (Show i, Guess j). The payoffs in the game are shown in Figure 148.3. S1G2 S1G3 S2G3 S2G4 S1G2 0, 0 −2, 2 3, −3 0, 0 S1G3 2, −2 0, 0 0, 0 −3, 3 S2G3 −3, 3 0, 0 0, 0 4, −4 S2G4 0, 0 3, −3 −4, 4 0, 0 Figure 148.3 The game in Exercise 344.2. Chapter 11. Strictly competitive games and maxminimization 149 Now, if there is a Nash equilibrium in which player 1’s payoff is v then, given the symmetry of the game, there is a Nash equilibrium in which player 2’s payoff is v, so that player 1’s payoff is −v. Since the equilibrium payoff in a strictly competitive game is unique, we have v = 0. Let (p 1 , p2 , p3 , p4 ) be the probabilities that player 1 assigns to her four actions. In order that she obtain a payoff of at least 0 if player 2 uses any of her pure strategies, we need − 2p 2 + 3p 3 ≥ 0 2p1 − 3p4 ≥ 0 −3p1 + 4p4 ≥ 0 3p2 − 4p 3 ≥ 0. The second and third inequalities imply that p1 ≥ 32 p4 and p1 ≤ 43 p4 , so that p1 = p4 = 0, so that p3 = 1 − p2 . The first and fourth inequalities imply that p2 ≤ 32 p3 and p2 ≥ 43 p3 , or p 2 ≤ 35 and p 2 ≥ 47 . We conclude that any pair of mixed strategies ((0, p2 , 1 − p2, 0), (0, q2 , 1 − q2 , 0)) with 47 ≤ p 2 ≤ 35 and 47 ≤ q2 ≤ 35 is an equilibrium. 344.3 Equilibria of a 4 × 4 game a. Denote the probability with which player 1 chooses each of her actions 1, 2, and 3, by p and the probability with which player 2 chooses each of these actions by q. Then all four of player 1’s actions yield the same expected payoff if and only if 4q − 1 = 1 − 6q, or q = 15 , and similarly all four of player 2’s actions yield the same expected payoff if and only if p = 15 . Thus (( 15 , 15 , 15 , 25 ), ( 15 , 15 , 15 , 25 )) is a Nash equilibrium of the game. The players’ payoffs in this equilibrium are (− 15 , 15 ). b. Let (p1 , p2 , p3 , p4 ) be an equilibrium strategy of player 1. In order that it guarantee her the payoff of − 15 , we need −p1 + p2 + p 3 − p4 ≥ − 15 p1 − p2 + p 3 − p4 ≥ − 15 p1 + p2 − p 3 − p4 ≥ − 15 −p1 − p2 − p 3 + p4 ≥ − 15 . Adding these four inequalities, we deduce that p4 ≤ 25 . Adding each pair of the first three inequalities, we deduce that p1 ≤ 15 , p2 ≤ 15 , and p3 ≤ 15 . Since p1 + p2 + p3 + p4 = 1, we deduce that (p1 , p2 , p3 , p4 ) = ( 15 , 15 , 15 , 25 ). A similar analysis of the conditions for player 2’s strategy to guarantee her the payoff of 15 leads to the conclusion that (q1 , q2 , q3 , q4 ) = ( 15 , 15 , 15 , 25 ). Draft of solutions to exercises in chapter of An introduction to game theory by Martin J. Osborne

[email protected]; www.chass.utoronto.ca/~osborne/index.html Version: 00/11/6. c 1995–2000 by Martin J. Osborne. All rights reserved. No part of this book may be reCopyright produced in any form by any electronic or mechanical means (including photocopying, recording, or information storage and retrieval) without permission in writing from Martin J. Osborne. On request, permission to make one copy for each student will be granted to instructors who wish to use the book in a course, on condition that copies be sold at a price not more than the cost of duplication. 12 Rationalizability 354.3 Mixed strategy equilibria of game There is no equilibrium in which player 2 assigns positive probability only to L and C, since if she does so then only M and B are possible best responses for player 1, but if player 1 assigns positive probability only to these actions then L is not optimal for player 2. By a similar argument there is no equilibrium in which player 2 assigns positive probability only to C and R. Assume that player 2 assigns positive probability only to L and R. There are no probabilities for L and R under which player 1 is indifferent between all three of her actions, so player 1 must assign positive probability to at most two actions. If these two actions are T and M then player 2 prefers L to R, while if the two actions are M and B then player 2 prefers R to L. The only possibility is thus that the two actions are T and B. In this case we need player 2 to assign probability 12 to L and R (in order that player 1 be indifferent between T and B); but then M is better for player 1. Thus there is no equilibrium in which player 2 assigns positive probability only to L and R. Finally, if player 2 assigns positive probability to all three of her actions then player 1’s mixed strategy must be such that each of these three actions yields the same payoff. A calculation shows that there is no mixed strategy of player 1 with this property. We conclude that the game has no mixed strategy equilibrium in which either player assigns positive probability to more than one action. 358.1 Example of rationalizable actions I claim that the action R of player 2 is strictly dominated by some mixed strategies that assign positive probability to L and C. Consider such a mixed strategy that assigns probability p to L. In order for this mixed strategy to strictly dominate R we need p + 4(1 − p) > 3 and 8p + 2(1 − p) > 3, or 16 < p < 13 . That is, any such value of p is associated with a mixed strategy that strictly dominated R. In the reduced game (i.e. after R is eliminated), B is dominated by T. Finally, L is dominated by C. Hence the only rationalizable action of player 1 is T and the only rationalizable action of player 2 is C. 151 152 Chapter 12. Rationalizability 358.2 Guessing Morra Take Zi to be all the actions of player i, for i = 1, 2. Then (Z1 , Z2 ) satisfies Definition 354.1. (The action S1G2 is a best response to a belief that assigns probability 1 to S1G3, the action S1G3 is a best response to the belief that assigns probability one to S2G4, the action S2G3 is a best response to the belief that assigns probability one to S1G2, and the action S2G4 is a best response to the belief that assigns probability one to S2G3.) 358.3 Contributing to a public good a. The derivative to player i’s payoff with respect to ci is −2ci − ∑ c j + wi , j = i which, for every possible value of ∑ j=i c j , is negative if ci > 12 wi . Thus the contribution wi /2 yields player i a payoff higher than does any larger contribution, regardless of the other players’ contributions. (Note that this result depends on the sum of the other players’ contributions being nonnegative.) b. The best response function of player i is given by max{0, 12 (w − ∑ c j )}. j = i Let c ≤ w/2 and suppose that each of the other players contributes 12 w − c (which is nonnegative). Then the other players’ total contribution is w − 2c, so that player i’s best response is to contribute c. That is, any contribution c of at most w/2 is a best response to the belief that assigns probability one to each of the other player’s contributing 12 w − c ≤ 12 w. Thus if we set Zi = [0, w/2] for all i in Definition 354.1 we see that any action of player i in [0, w/2] is rationalizable for player i. [Note: This argument does not show that actions outside [0, w/2] are not rationalizable.] c. Denote w1 = w2 = w. First eliminate contributions of more than wi /2 by each player i. In the reduced game the most that players 1 and 2 together contribute is w (since each contributes at most w/2). Now consider player 3. Given the derivative of her payoff function found in part a, her payoff is increasing in her contribution for every remaining possible value of c1 + c2 so long as c3 < 12 (w3 − (c1 + c2 )). Since c1 + c2 ≤ w, player 3’s payoff is thus definitely increasing for c3 < 12 (w3 − w). But w3 ≥ 3w, so player 3’s payoff is definitely increasing for c3 < w. We conclude that in the reduced game every contribution of player 3 of less than w is strictly dominated. Eliminate all such actions of player 3. Chapter 12. Rationalizability 153 In the newly reduced game every contribution of player 3 is in the interval [w, w3 /2]. Now consider player 1. Her payoff is decreasing in her contribution if c1 > 12 (w − (c2 + c3 )). We have c2 ≥ 0 and c3 ≥ w, so player 1’s payoff is decreasing if c1 > 0. Thus every action of player 1 is strictly dominated by a contribution of 0. The same analysis applies to player 2. Eliminate all such actions of player 1 and player 2. Finally, in the game we now have, players 1 and 2 both contribute 0; it follows that all actions of player 3 are dominated except for a contribution of w3 /2, which is her best response to a total contribution of 0 by players 1 and 2. We conclude that the unique action profile that survives iterated elimination of strictly dominated actions is (0, 0, w3 /2). 358.4 Iterated elimination in location game In the first round Out is strictly dominated by the position 12 (since the position 12 guarantees at least a draw, which each player prefers to staying out of the competition). In the next round the positions 0 and 1 are strictly dominated by the position 12 : a player who chooses 12 rather than either 0 or 1 ties rather than loses if her opponent also chooses 12 , and wins outright rather than ties or loses if her opponent chooses any other position. In every subsequent round the two remaining extreme positions are strictly dominated by 12 . The only action that remains is 12 . [Note that under the procedure of iterated elimination of weakly dominated actions, discussed in the next section of the text, there is only one round of elimination: all actions other than 12 are weakly dominated by 12 . (In particular, the game is dominance solvable.)] 361.1 Example of dominance solvability The Nash equilibria of the game are (T, L), any ((0, 0, 1), (0, q, 1 − q)) with 0 ≤ q ≤ 1, and any ((0, p, 1 − p), (0, 0, 1)) with 0 ≤ p ≤ 1. The game is dominance solvable, because T and L are the only weakly dominated actions, and in they are eliminated the only weakly dominated actions are M and C, leaving (B, R), with payoffs (0, 0). If T is eliminated, then L and C, no remaining action is weakly dominated; (M, R) and (B, R) both remain. 361.2 Dominance solvability in demand game In the first round the demands 0, 1, and 2 are eliminated for each player and in the second round the demand 4 is eliminated, leaving the outcome in which each player demands 3 (and receives 2). 154 Chapter 12. Rationalizability 361.3 Dominance solvability in Bertrand’s duopoly game In the first round every price in excess of the monopoly price is weakly dominated by the monopoly price and every price equal to at most c is weakly dominated by the price c + 0.01. At each subsequent round the highest remaining price is weakly dominated by the next highest price. (Note that for any p ≥ c + 0.01 it is better to obtain all the demand at the price p than obtain half of the demand at the price p + 0.01.) The pair of prices that remains is (c + 0.01, c + 0.01). Draft of solutions to exercises in chapter of An introduction to game theory by Martin J. Osborne

[email protected]; www.chass.utoronto.ca/~osborne/index.html Version: 00/11/6. c 1995–2000 by Martin J. Osborne. All rights reserved. No part of this book may be reCopyright produced in any form by any electronic or mechanical means (including photocopying, recording, or information storage and retrieval) without permission in writing from Martin J. Osborne. On request, permission to make one copy for each student will be granted to instructors who wish to use the book in a course, on condition that copies be sold at a price not more than the cost of duplication. 13 Evolutionary equilibrium 370.1 ESSs and weakly dominated actions The ESS a∗ does not necessarily weakly dominate every other action in the game. For example, in the game in Figure 155.1, a∗ is an ESS but does not weakly dominate b. a∗ b a∗ 1, 1 0, 0 b 0, 0 2, 2 Figure 155.1 A game in which an ESS (a∗ ) does not weakly dominate another action. No action can weakly dominate an ESS. To see why, let a∗ be an ESS and let b be another action. Since a∗ is an ESS, (a∗ , a∗ ) is a Nash equilibrium, so that u(b, a∗ ) ≤ u(a∗ , a∗ ). Now, if u(b, a∗ ) < u(a∗ , a∗ ), certainly b does not weakly dominate a∗ , so suppose that u(b, a∗ ) = u(a∗ , a∗ ). Then by the second condition for an ESS we have u(b, b) < u(a∗ , b). We conclude that b does not weakly dominate a∗ . 370.2 Pure ESSs The payoff matrix of the game is given in Figure 155.2. The pure strategy symmet- 1 2 3 1 1, 1 2δ, 2 3δ, 3 2 2, 2δ 2, 2 3δ, 3 3 3, 3δ 3, 3δ 3, 3 Figure 155.2 The game in Exercise 370.2. ric Nash equilibria are (1, 1), (2, 2), and (3, 3). The only pure evolutionarily stable strategy is 1, by the following argument. The action 1 is evolutionarily stable since (1, 1) is a strict Nash equilibrium. The action 2 is not evolutionarily stable, since 1 is a best response to 2 and u(1, 1) = 1 > 2δ = u(2, 1). 155 156 Chapter 13. Evolutionary equilibrium The action 3 is not evolutionarily stable, since 2 is a best response to 3 and u(2, 2) = 2 > 3δ = u(3, 2). In the case that each player has n actions, every pair (i, i) is a Nash equilibrium; only the action 1 is an ESS. 375.1 Hawk–Dove–Retaliator First suppose that v ≥ c. In this case the game has two pure symmetric Nash equilibria, (A, A) and (R, R). However, A is not an ESS, since R is a best response to A and u(R, R) > u(A, R). Since (R, R) is a strict equilibrium, R is an ESS. Now consider the possibility that the game has a mixed strategy equilibrium (α, α). If α assigns positive probability to either P or R (or both) then R yields a payoff higher than does P, so only A and R may be assigned positive probability in a mixed strategy equilibrium. But if a strategy α assigns positive probability to A and R and probability 0 to P, then R yields a payoff higher than does A against an opponent who uses α. Thus the game has no symmetric mixed strategy equilibrium in this case. Now suppose that v < c. Then the only symmetric pure strategy equilibrium is (R, R). This equilibrium is strict, so that R is an ESS. Now consider the possibility that the game has a mixed strategy equilibrium (α, α). If α assigns probability 0 to A then R yields a payoff higher than does P against an opponent who uses α; if α assigns probability 0 to P then R yields a payoff higher than does A against an opponent who uses α. Thus in any mixed strategy equilibrium (α, α), the strategy α must assign positive probability to both A and P. If α assigns probability 0 to R then we need α = (v/c, 1 − v/c) (the calculation is the same as for Hawk–Dove). Since R yields a lower payoff against this strategy than do A and P, and since the strategy is an ESS in Hawk–Dove, it is an ESS in the present game. The remaining possibility is that the game has a mixed strategy equilibrium (α, α) in which α assigns positive probability to all three actions. If so, then the expected payoff to this strategy is less than 12 v, since the pure strategy P yields an expected payoff less than 12 v against any such strategy. But then U(R, R) = 12 v > U(α, R), violating the second condition in the definition of an ESS. In summary: • If v ≥ c then R is the unique ESS of the game. • If v < c then both R and the mixed strategy that assigns probability v/c to A and 1 − v/c to P are ESSs. 375.2 Example of pure and mixed ESSs Since (C, C) is a strict Nash equilibrium, C is an ESS. Chapter 13. Evolutionary equilibrium 157 The game also has a symmetric mixed strategy equilibrium in which each player’s mixed strategy is α∗ = ( 34 , 14 , 0). Every mixed strategy β = (p, 1 − p, 0) is a best response to α∗ , so in order that α∗ is an ESS we need U(β, β) < U(α∗ , β). We have U(β, β) = 4p(1 − p) and U(α∗ , β) = 94 (1 − p) + 14 p, so the inequality is equivalent to (p − 34 )2 > 0, which is true for all p = 34 . Thus α∗ is an ESS. The only other symmetric mixed strategy equilibrium is one in which each player’s strategy is α∗∗ = ( 37 , 17 , 37 ). This strategy is not an ESS, since u(C, C) = 1 while u(α∗∗ , C) = 37 < 1. 375.3 Bargaining The game is given in Figure 157.1. Let α be a mixed strategy that assigns positive 0 2 4 6 8 10 0 5, 5 6, 4 7, 3 8, 2 9, 1 10, 0 2 4, 6 5, 5 6, 4 7, 3 8, 2 0, 0 4 3, 7 4, 6 5, 5 6, 4 0, 0 0, 0 6 2, 8 3, 7 4, 6 0, 0 0, 0 0, 0 8 1, 9 2, 8 0, 0 0, 0 0, 0 0, 0 10 0, 10 0, 0 0, 0 0, 0 0, 0 0, 0 Figure 157.1 A bargaining game. probability only to the demands 2 and 8. For (α, α) to be a Nash equilibrium we need α = ( 25 , 35 ). Each player’s payoff at this strategy pair (α∗ , α∗ ) is 16 5 . Thus the only actions a that are best responses to α∗ are 2 and 8, so that the only mixed strategies that are best responses assign positive probability only to the actions 2 and 8. Let β be the mixed strategy that assigns probability p to 2 and probability 1 − p to 8. We have U(β, β) = 5p(2 − p) and U(α∗ , β) = 6p + 45 . We find that U(α∗ , β) − U(β, β) = 5(p − 25 )2 , which is positive if p = 25 . Hence α∗ is an ESS. Now let α be a mixed strategy that assigns positive probability only to the demands 4 and 6. For (α, α) to be a Nash equilibrium we need α = 45 . Each player’s payoff at this strategy pair (α∗ , α∗ ) is 24 5 . Thus the only actions a that are best re∗ sponses to α are 4 and 6, so that the only mixed strategies that are best responses 158 Chapter 13. Evolutionary equilibrium assign positive probability only to the actions 4 and 6. Let β be the mixed strategy that assigns probability p to 4 and probability 1 − p to 6. We have U(β, β) = 5p(2 − p) and U(α∗ , β) = 2p + 16 5 . We find that U(α∗ , β) − U(β, β) = 5(p − 45 )2 , which is positive if p = 45 . Hence α∗ is an ESS. 379.1 Mixed strategies in an asymmetric Hawk–Dove Let p be the probability that β assigns to AA. In order that AA and DD yield a player the same expected payoff when her opponent uses β, we need p(V + v − 2c) + (1 − p)(2V + 2v) = (1 − p)(V + v), or V+v . 2c Now, if player 2 uses the strategy β then the difference between player 1’s expected payoff to AA and her expected payoff to AP is p= p(v − c) + (1 − p)v = v − pc = 12 (v − V) < 0. Thus the strategy pair (β, β) is not a Nash equilibrium. 379.2 Mixed strategy ESSs Let β be an ESS that assigns positive probability to every action in A∗ . Then (β, β) is a Nash equilibrium (since β is an ESS), so that every mixed strategy that assigns positive probability only to actions in A∗ is a best response to β. In particular, α∗ is a best response to β. Thus if β = α∗ then the second condition in the definition of an ESS, when applied to β, requires that U(α∗ , α∗ ) < U(β, α∗ ). But this inequality contradicts the fact that (α∗ , α∗ ) is a Nash equilibrium. Hence β = α∗ . 380.1 Asymmetric ESSs of BoS The game is shown in Figure 159.1. The strategy pairs (LD,LD) and (DL,DL) are strict symmetric Nash equilibria. Thus both LD and DL are ESSs. By the same argument as in the analysis of Hawk–Dove in the text, the only possible mixed ESS Chapter 13. Evolutionary equilibrium 159 LL 0, 0 1 2, 1 1 2, 1 1, 2 LL LD DL DD LD 1, 12 3 3 2, 2 0, 0 1 2, 1 DL 1, 12 0, 0 3 3 2, 2 1 2, 1 DD 2, 1 1, 12 1, 12 0, 0 Figure 159.1 The game BoS when the players’ roles may differ. assigns positive probability only to LL and DD. Let β be such a strategy; let p be the probability that it assigns to LL. Then for (β, β) to be a Nash equilibrium we need 2(1 − p) = p, or p = 23 . If one of the players uses such a strategy then the other player obtains the same expected payoff to all her four actions, namely 23 . Thus (β, β) is a Nash equilibrium. However, since u(LD, LD) = 3 2 > 5 6 = u(β, LD), the strategy β is not an ESS. Thus the game has two ESSs, each of which is a pure strategy: LD and DL. 385.1 A coordination game between siblings The games with payoff functions v and w are shown in Figure 159.2. If x < 2 then X Y X x, x 1 1 2, 2x Y 1 1 2 x, 2 1, 1 v X Y X x, x 1 1 5, 5x Y 1 1 5 x, 5 1, 1 w Figure 159.2 The games with payoff functions v and w derived from the game in Exercise 385.1. (Y, Y) is a strict Nash equilibrium of both games, so Y is an evolutionarily stable action in the game between siblings. If x > 2 then the only (pure) Nash equilibrium of the game is (X, X), and this equilibrium is strict. Thus the range of values of x for which the only evolutionarily stable action is X is x > 2. 387.1 Darwin’s theory of the sex ratio A normal organism produces pn female offspring and (1 − p)n male offspring (ignoring the small probability that the partner of a normal organism is a mutant). Thus it has pn · n + (1 − p)n · (p/(1 − p))n = 2pn2 grandchildren. A mutant has 12 n female offspring and 12 n male offspring, and hence has 12 n · n + 12 n · (p/(1 − p))n = 12 n2 /(1 − p) grandchildren. 160 Chapter 13. Evolutionary equilibrium Thus the difference between the number of grandchildren produced by normal and mutant organisms is 2 2 2 1 2 n /(1 − p) − 2pn = n (p − 12 )2 , 2 1−p which is positive if p = 12 . (The point is that a higher fraction of the mutant’s offspring are female, which each bear more offspring than each male.) Thus the mutant invades the population; only p = 12 is evolutionarily stable. Draft of solutions to exercises in chapter of An introduction to game theory by Martin J. Osborne

[email protected]; www.chass.utoronto.ca/~osborne/index.html Version: 00/11/6. c 1995–2000 by Martin J. Osborne. All rights reserved. No part of this book may be reCopyright produced in any form by any electronic or mechanical means (including photocopying, recording, or information storage and retrieval) without permission in writing from Martin J. Osborne. On request, permission to make one copy for each student will be granted to instructors who wish to use the book in a course, on condition that copies be sold at a price not more than the cost of duplication. 14 Repeated games: The Prisoner’s Dilemma 395.1 Strategies for an inﬁnitely repeated Prisoner’s Dilemma a. The strategy is shown in Figure 161.1. P0 : C (·, D) ✲ P1 : C ✲ D: D all outcomes Figure 161.1 The strategy in Exercise 395.1a. b. The strategy is shown in Figure 161.2. P0 : C (·, D) ✲ P1 : C (·, D) ✲ D: D Figure 161.2 The strategy in Exercise 395.1b. c. The strategy is shown in Figure 161.3. ☛ ❄ C: C ✟ (C, C) or (D, D) ✲ D: D (D, C) or (C, D) Figure 161.3 The strategy in Exercise 395.1c. 398.1 Nash equilibria of the inﬁnitely repeated Prisoner’s Dilemma a. A player who adheres to the strategy obtains the discounted average payoff of 2. A player who deviates obtains the stream of payoffs (3, 3, 1, 1, . . .), with a discounted average of (1 − δ)(3 + 3δ) √ + δ2 . Thus for an equilibrium we 1 2 require (1 − δ)(3 + 3δ) + δ ≤ 2, or δ ≥ 2 2. b. A player who adheres to the strategy obtains the payoff of 2 in every period. A player who chooses D in the first period and C in every subsequent period obtains the stream of payoffs (3, 2, 2, . . .). Thus for any value of δ a player can 161 162 Chapter 14. Repeated games: The Prisoner’s Dilemma increase her payoff by deviating, so that the strategy pair is not a Nash equilibrium. Further, whatever the one-shot payoffs, a player can increase her payoff by deviating to D in a single period, so that for no payoffs is there any δ such that the strategy pair is a Nash equilibrium of the infinitely repeated game. c. A player who adheres to the strategy obtains the discounted average payoff of 2 (the outcome is (C, C) in every period). If player 1 deviates to D in every period then she induces the outcome to alternate between (D, C) and (D, D), yielding her a discounted average payoff of (1 − δ) · (3 + 3δ2 + 3δ4 + . . .) + (1 − δ)(δ + δ3 + δ5 + . . .) = (1 − δ)[3/(1 − δ2 ) + δ/(1 − δ2 )] = (3 + δ)/(1 + δ). For all δ < 1 this payoff exceeds 2, so that the strategy pair is not a Nash equilibrium of the infinitely repeated game. However, for different payoffs for the one-shot Prisoner’s Dilemma, the strategy pair is a Nash equilibrium of the infinitely repeated game. The point is that the best deviation leads to the sequence of outcomes that alternates between (C, D) and (D, D). If the average payoff of player 2 in these two outcomes is less than her payoff to the outcome (C, C) then the strategy pair is a Nash equilibrium for some values of δ. (For the payoffs in Figure 389.1 the average payoff of the two outcomes (C, D) and (D, D) is exactly equal to the payoff to (C, C).) Consider the general payoffs in Figure 162.1. The dis- C D C x, x y, 0 D 0, y 1, 1 Figure 162.1 A Prisoner’s Dilemma. counted average payoff of the sequence of outcomes that alternates between (C, D) and (D, D) is (y + δ)/(1 + δ), while the discounted average of the constant sequence containing only (C, C) is x. Thus in order for the strategy pair to be a Nash equilibrium we need y+δ ≤ x, 1+δ or δ≥ y−x , x−1 an inequality that is compatible with δ < 1 if x > exceeds the average of 1 and y. 1 2 (y + 1)—that is, if x 406.1 Diﬀerent punishment lengths in the inﬁnitely repeated Prisoner’s Dilemma Yes, there are such subgame perfect equilibria. The only subtlety is that the number of periods for which a player chooses D after a history in which not all the Chapter 14. Repeated games: The Prisoner’s Dilemma 163 outcomes were (C, C) depends on who first deviated. If, for example, player 1 punishes for two periods while player 2 punishes for three periods, then the outcome (C, D) induces player 1 to choose D for two periods (to punish player 2 for her deviation) while the outcome (D, C) induces her to choose D for three periods (while she is being punished by player 2). The strategy of each player in this case is shown in Figure 163.1. ✟ ☛ all outcomes ✲ P2 : D ✯ P1 : D ✟ (·, D) ✟ all ❄ ✟✟ outcomes P0 : C ✟ ❍❍ ✻ (D, ·)❍❍ ❥ P : D ❍ ✲ P : D 1 2 ✡ all outcomes ✲ P : D 3 all outcomes all outcomes ✠ Figure 163.1 A strategy in an infinitely repeated Prisoner’s Dilemma that punishes deviations for two periods and reacts to punishment by choosing D for three periods. 407.1 Tit-for-tat in the inﬁnitely repeated Prisoner’s Dilemma Suppose that player 2 adheres to tit-for-tat. Consider player 1’s behavior in subgames following histories that end in each of the following outcomes. (C, C) If player 1 adheres to tit-for-tat the outcome is (C, C) in every period, so that her discounted average payoff in the subgame is x. If she chooses D, then adheres to tit-for-tat, the outcome alternates between (D, C) and (C, D), and player 1’s discounted average payoff is y/(1 + δ). Thus we need x ≥ y/(1 + δ), or δ ≥ (y − x)/x, in order that tit-for-tat be optimal for player 1. (C, D) If player 1 adheres to tit-for-tat the outcome alternates between (D, C) and (C, D), so that her discounted average payoff is y/(1 + δ). If she deviates to C, then adheres to tit-for-tat, the outcome is (C, C) in every period, and her discounted average payoff is x. Thus we need y/(1 + δ) ≥ x, or δ ≤ (y − x)/x, in order that tit-for-tat be optimal for player 1. (D, C) If player 1 adheres to tit-for-tat the outcome alternates between (C, D) and (D, C), so that her discounted average payoff is δy/(1 + δ). If she deviates to D, then adheres to tit-for-tat, the outcome is (D, D) in every period, and her discounted average payoff is 1. Thus we need δy/(1 + δ) ≥ 1, or δ ≥ 1/(y − 1), in order that tit-for-tat be optimal for player 1. (D, D) If player 1 adheres to tit-for-tat the outcome is (D, D) in every period, so that her discounted average payoff is 1. If she deviates to C, then adheres to tit-for-tat, the outcome alternates between (C, D) and (D, C), and her discounted average payoff is δy/(1 + δ). Thus we need 1 ≥ δy/(1 + δ), or δ ≤ 1/(y − 1), in order that tit-for-tat be optimal for player 1. 164 Chapter 14. Repeated games: The Prisoner’s Dilemma We conclude that for (tit-for-tat,tit-for-tat) to be a subgame perfect equilibrium we need δ = (y − x)/x and δ = 1/(y − 1). Thus only if (y − x)/x = 1/(y − 1), or y − x = 1, is the strategy pair a subgame perfect equilibrium. Given that a subgame perfect equilibrium satisfies the one-deviation property, the strategy pair is indeed a subgame perfect equilibrium in this case when δ = 1/x. Draft of solutions to exercises in chapter of An introduction to game theory by Martin J. Osborne

[email protected]; www.chass.utoronto.ca/~osborne/index.html Version: 00/11/6. c 1995–2000 by Martin J. Osborne. All rights reserved. No part of this book may be reCopyright produced in any form by any electronic or mechanical means (including photocopying, recording, or information storage and retrieval) without permission in writing from Martin J. Osborne. On request, permission to make one copy for each student will be granted to instructors who wish to use the book in a course, on condition that copies be sold at a price not more than the cost of duplication. 17 Mathematical appendix 446.1 Maximizer of quadratic function We can write the function as −x(x − α). Thus r1 = 0 and r2 = α, and hence the maximizer is α/2. 449.3 Sums of sequences In the first case set r = δ2 to transform the sum into 1 + r + r 2 + · · ·, which is equal to 1/(1 − r) = 1/(1 − δ2 ). In the second case split the sum into (1 + δ2 + δ4 + · · ·) + (2δ + 2δ3 + 2δ5 + · · ·); the first part is equal to 1/(1 − δ2 ) and the second part is equal to 2δ(1 + δ2 + δ4 + · · ·), or 2δ/(1 − δ2 ). Thus the complete sum is 1 + 2δ . 1 − δ2 454.1 Bayes’ law Your posterior probability of carrying X given that you test positive is Pr(positive test|X) Pr(X) Pr(positive test|X) Pr(X) + Pr(positive test|¬X) Pr(¬X) where ¬X means “not X”. This probability is equal to 0.9p/(0.9p + 0.2(1 − p)) = 0.9p/(0.2 + 0.7p), which is increasing in p (i.e. a smaller value of p gives a smaller value of the probability). If p = 0.001 then the probability is approximately 0.004. (That is, if 1 in 1,000 people carry the gene then if you test positive on a test that is 90% accurate for people who carry the gene and 80% accurate for people who do not carry the gene, then you should assign probability 0.004 to your carrying the gene.) If the test is 99% accurate in both cases then the posterior probability is (0.99 · 0.001)/[0.99 · 0.001 + 0.01 · 0.999] ≈ 0.09. 171 Corrections and updates for first printing of Osborne’s “An Introduction to Game Theory” (Oxford University Press, 2003) 2004/5/4 I thank the following people for pointing out errors and improvements: T. K. Ahn, Kyung Hwan Baik, Richard Boylan, Hao-Chen Liu, Nathan Nunn, David A. Malueg, Ahmer Tarar, Debraj Ray, Kaouthar Souki. Corrections Page, Line 4 6 31 78 Correction The first letter of the text in Section 1.2 should be upper case. Add a space after the period on the last line. The “A” in the caption of Figure 31.1 should be upright, not italic. In Figure 78.2, replace “B1 ( p2 )” with “B1 (t2 )” and replace “B2 ( p1 )” with “B2 (t1 )”. 83 In the first line of the second paragraph, change “complete” to “perfect” (for consistency with other terminology). 83 The first sentence of the item “Preferences” just below the middle of the page is hard to follow. A better version is: “Denote by bi the bid of player i and by b the highest bid submitted by a player other than i. If either (a) bi > b or (b) bi = b and the number of every other player who bids b is greater than i, then player i’s payoff is vi − b.” 85–87 In the third line of the text on page 85, in the third line of Section 3.5.3 on page 86, and in the fifth line from the bottom of page 87, change “complete” to “perfect” (for consistency with other terminology). 94 The fourth word of the caption of Figure 94.1 should be “shows”. 110 Replace “an” at the end of line 16 with “a”. 143 Replace F(z) on line 21 with Fi (z). 145 Replace “x2 and y2 ” on the line below the display with “x1 and y1 ”. 145 Delete “a1 ” at the end of line −8. 187 Replace “in” with “is” on line 2. 202–203 The term “equilibrium path” is used without explanation. It is synonymous with “equilibrium outcome”. (That is, the equilibrium path is the terminal history generated by the equilibrium strategies.) 216 The word “that” on the fourth line from the bottom of the page should be “than”. 1 In the description of the states above the figure, replace “0 ≤ vi ≤ v” with “v ≤ vi ≤ v”. 291 Replace “a decreasing” with “an increasing” on line 16 and “increases” with “decreases” on line 17. 295 The claim in the last sentence on the page is too strong: the appendix contains only suggestive arguments, not a proof. 303 In the description of the beliefs in the middle of the page, replace the π near the start of the second line with Pr(G | g), the 1 − π near the end of the second line with Pr( I | g), the π near the middle of the fifth line with Pr(G | b), and the expression involving 1 − π near the start of the sixth line with Pr( I | b)(1 − q)k qn−k−1. 307 In part c of Exercise 307.1, replace “one of the player’s actions” with “an action of one of the players”. In part d, replace “second” with “first”. 308–309 To deduce the solution of the differential equation near the bottom of page 308, the initial condition β(v ) = v is needed. Given that this initial condition is needed to find the equilibrium bidding function, the part of Exercise 309.2 asking for a proof that the equilibrium bidding function satisfies the condition should be removed. See the website for the book for a version of Section 9.8.1 that corrects these two points, treats more carefully the boundary cases in which v = v and v = v, and explains the argument more clearly. 310 Two lines below (310.1), replace Pr{X < v} with Pr(X < v). On the following line, delete “= 0”. 319 Change the weak inequality on the next to last line to a strict inequality. 321 In the bottom row of the right-hand table in the bottom panel of Figure 321.1, interchange the entries in the columns headed XY and YX, so that 1/(2 − 4e) is in the column headed XY and 0 is in the column headed YX. 330 In the 7th line of Example 330.1, replace “the history is Acquiesce” with “the history is Unready”. 331 Add a period to the end of the caption of Figure 331.2. 331 Replace Exercise 331.2 (which is incorrect) with the following exercise. E XERCISE 331.2 (Weak sequential equilibrium and Nash equilibrium in subgames) Consider the variant of the game in Figure 331.1 shown in Figure 332.1, in which the challenger’s initial move is broken into two steps. Show that this game has a weak sequential equilibrium in which the players’ actions in the subgame following the history In do not constitute a Nash equilibrium of the subgame. 289 2 332–333 Replace the last word on page 332 and the first word on page 333 with “a weak”, and replace the penultimate word of the sentence with “strong”. 344 Replace each of the seven occurrences of the string t − b with t + b. 389 On line 11, the outcomes that survive are (T, L) and (T, C) (not (T, L) and (T, R)). 415 Add a period to the end of the caption of Figure 415.1. 457 Change k − ` to k − ` + 1 on line −2. Updates Dhillon and Lockwood (2003) is now Dhillon, Amrita, and Ben Lockwood (2004), “When are plurality rule voting games dominance-solvable?”, Games and Economic Behavior 46, 55–75. 3 Publicly-available solutions for AN INTRODUCTION TO GAME THEORY Publicly-available solutions for AN INTRODUCTION TO GAME THEORY M ARTIN J. O SBORNE University of Toronto Copyright © 2004 by Martin J. Osborne All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior permission of Martin J. Osborne. This manual was typeset by the author, who is greatly indebted to Donald Knuth (TEX), Leslie Lamport (LATEX), Diego Puga (mathpazo), Christian Schenk (MiKTEX), Ed Sznyter (ppctr), Timothy van Zandt (PSTricks), and others, for generously making superlative software freely available. The main font is 10pt Palatino. Version 2: 2004-4-27 Contents Preface 1 xi Introduction 1 Exercise 5.3 (Altruistic preferences) 1 Exercise 6.1 (Alternative representations of preferences) 1 2 Nash Equilibrium 3 Exercise 16.1 (Working on a joint project) 3 Exercise 17.1 (Games equivalent to the Prisoner’s Dilemma) 3 Exercise 20.1 (Games without conflict) 3 Exercise 31.1 (Extension of the Stag Hunt) 4 Exercise 34.1 (Guessing two-thirds of the average) 4 Exercise 34.3 (Choosing a route) 5 Exercise 37.1 (Finding Nash equilibria using best response functions) 6 Exercise 38.1 (Constructing best response functions) 6 Exercise 38.2 (Dividing money) 7 Exercise 41.1 (Strict and nonstrict Nash equilibria) 7 Exercise 47.1 (Strict equilibria and dominated actions) 8 Exercise 47.2 (Nash equilibrium and weakly dominated actions) 8 Exercise 50.1 (Other Nash equilibria of the game modeling collective decision-making) 8 Exercise 51.2 (Symmetric strategic games) 9 Exercise 52.2 (Equilibrium for pairwise interactions in a single population) 9 3 Nash Equilibrium: Illustrations 11 Exercise 58.1 (Cournot’s duopoly game with linear inverse demand and different unit costs) 11 Exercise 60.2 (Nash equilibrium of Cournot’s duopoly game and the collusive outcome) 12 Exercise 63.1 (Interaction among resource-users) 12 Exercise 67.1 (Bertrand’s duopoly game with constant unit cost) 13 Exercise 68.1 (Bertrand’s oligopoly game) 13 Exercise 68.2 (Bertrand’s duopoly game with different unit costs) 13 Exercise 73.1 (Electoral competition with asymmetric voters’ preferences) 14 Exercise 75.3 (Electoral competition for more general preferences) 14 Exercise 76.1 (Competition in product characteristics) 15 Exercise 79.1 (Direct argument for Nash equilibria of War of Attrition) 15 Exercise 85.1 (Second-price sealed-bid auction with two bidders) 16 v vi Contents Exercise 86.2 (Nash equilibrium of first-price sealed-bid auction) 17 Exercise 87.1 (First-price sealed-bid auction) 17 Exercise 89.1 (All-pay auctions) 18 Exercise 90.1 (Multiunit auctions) 18 Exercise 90.3 (Internet pricing) 19 Exercise 96.2 (Alternative standards of care under negligence with contributory negligence) 19 4 Mixed Strategy Equilibrium 23 Exercise 101.1 (Variant of Matching Pennies) 23 Exercise 106.2 (Extensions of BoS with vNM preferences) 23 Exercise 110.1 (Expected payoffs) 24 Exercise 111.1 (Examples of best responses) 24 Exercise 114.1 (Mixed strategy equilibrium of Hawk–Dove) 25 Exercise 117.2 (Choosing numbers) 26 Exercise 120.2 (Strictly dominating mixed strategies) 26 Exercise 120.3 (Strict domination for mixed strategies) 26 Exercise 127.1 (Equilibrium in the expert diagnosis game) 27 Exercise 130.3 (Bargaining) 27 Exercise 132.2 (Reporting a crime when the witnesses are heterogeneous) 28 Exercise 136.1 (Best response dynamics in Cournot’s duopoly game) 29 Exercise 139.1 (Finding all mixed strategy equilibria of two-player games) 29 Exercise 145.1 (All-pay auction with many bidders) 30 Exercise 147.2 (Preferences over lotteries) 31 Exercise 149.2 (Normalized vNM payoff functions) 31 5 Extensive Games with Perfect Information: Theory 33 Exercise 163.1 (Nash equilibria of extensive games) 33 Exercise 164.2 (Subgames) 33 Exercise 168.1 (Checking for subgame perfect equilibria) 33 Exercise 174.1 (Sharing heterogeneous objects) 34 Exercise 177.3 (Comparing simultaneous and sequential games) 34 Exercise 179.3 (Three Men’s Morris, or Mill) 35 6 Extensive Games with Perfect Information: Illustrations 37 Exercise 183.1 (Nash equilibria of the ultimatum game) 37 Exercise 183.2 (Subgame perfect equilibria of the ultimatum game with indivisible units) 37 Exercise 186.1 (Holdup game) 37 Exercise 189.1 (Stackelberg’s duopoly game with quadratic costs) 38 Exercise 196.4 (Sequential positioning by three political candidates) 38 Exercise 198.1 (The race G1 (2, 2)) 40 Exercise 203.1 (A race with a liquidity constraint) 40 Contents vii 7 Extensive Games with Perfect Information: Extensions and Discussion 43 Exercise 210.2 (Extensive game with simultaneous moves) 43 Exercise 217.1 (Electoral competition with strategic voters) 43 Exercise 220.1 (Top cycle set) 44 Exercise 224.1 (Exit from a declining industry) 45 Exercise 227.1 (Variant of ultimatum game with equity-conscious players) 45 Exercise 230.1 (Nash equilibria when players may make mistakes) 46 Exercise 233.1 (Nash equilibria of the chain-store game) 46 8 Coalitional Games and the Core 47 Exercise 245.1 (Three-player majority game) 47 Exercise 248.1 (Core of landowner–worker game) 47 Exercise 249.1 (Unionized workers in landowner–worker game) 47 Exercise 249.2 (Landowner–worker game with increasing marginal products) 48 Exercise 254.1 (Range of prices in horse market) 48 Exercise 258.1 (House assignment with identical preferences) 49 Exercise 261.1 (Median voter theorem) 49 Exercise 267.2 (Empty core in roommate problem) 49 9 Bayesian Games 51 Exercise 276.1 (Equilibria of a variant of BoS with imperfect information) 51 Exercise 277.1 (Expected payoffs in a variant of BoS with imperfect information) 51 Exercise 282.2 (An exchange game) 52 Exercise 287.1 (Cournot’s duopoly game with imperfect information) 53 Exercise 288.1 (Cournot’s duopoly game with imperfect information) 53 Exercise 290.1 (Nash equilibria of game of contributing to a public good) 55 Exercise 294.1 (Weak domination in second-price sealed-bid action) 56 Exercise 299.1 (Asymmetric Nash equilibria of second-price sealed-bid common value auctions) 57 Exercise 299.2 (First-price sealed-bid auction with common valuations) 57 Exercise 309.2 (Properties of the bidding function in a first-price sealed-bid auction) 58 Exercise 309.3 (Example of Nash equilibrium in a first-price auction) 58 10 Extensive Games with Imperfect Information 59 Exercise 316.1 (Variant of card game) 59 Exercise 318.2 (Strategies in variants of card game and entry game) 59 Exercise 331.2 (Weak sequential equilibrium and Nash equilibrium in subgames) 60 Exercise 340.1 (Pooling equilibria of game in which expenditure signals quality) 60 Exercise 346.1 (Comparing the receiver’s expected payoff in two equilibria) 61 Exercise 350.1 (Variant of model with piecewise linear payoff functions) 61 viii Contents 11 Strictly Competitive Games and Maxminimization 63 Exercise 363.1 (Maxminimizers in a bargaining game) 63 Exercise 363.3 (Finding a maxminimizer) 63 Exercise 366.2 (Determining strictly competitiveness) 64 Exercise 370.2 (Maxminimizing in BoS) 64 Exercise 372.2 (Equilibrium in strictly competitive game) 64 Exercise 372.4 (O’Neill’s game) 64 12 Rationalizability 67 Exercise 379.2 (Best responses to beliefs) 67 Exercise 384.1 (Mixed strategy equilibria of game in Figure 384.1) 67 Exercise 387.2 (Finding rationalizable actions) 68 Exercise 387.5 (Hotelling’s model of electoral competition) 68 Exercise 388.2 (Cournot’s duopoly game) 68 Exercise 391.1 (Example of dominance-solvable game) 69 Exercise 391.2 (Dividing money) 69 Exercise 392.2 (Strictly competitive extensive games with perfect information) 69 13 Evolutionary Equilibrium 71 Exercise 400.1 (Evolutionary stability and weak domination) 71 Exercise 405.1 (Hawk–Dove–Retaliator) 71 Exercise 405.3 (Bargaining) 72 Exercise 408.1 (Equilibria of C and of G) 72 Exercise 414.1 (A coordination game between siblings) 73 Exercise 414.2 (Assortative mating) 73 Exercise 416.1 (Darwin’s theory of the sex ratio) 74 14 Repeated Games: The Prisoner’s Dilemma 75 Exercise 423.1 (Equivalence of payoff functions) 75 Exercise 426.1 (Subgame perfect equilibrium of finitely repeated Prisoner’s Dilemma) 75 Exercise 428.1 (Strategies in an infinitely repeated Prisoner’s Dilemma) 76 Exercise 439.1 (Finitely repeated Prisoner’s Dilemma with switching cost) 76 Exercise 442.1 (Deviations from grim trigger strategy) 78 Exercise 443.2 (Different punishment lengths in subgame perfect equilibrium) 78 Exercise 445.1 (Tit-for-tat as a subgame perfect equilibrium) 79 15 Repeated Games: General Results 81 Exercise 454.3 (Repeated Bertrand duopoly) 81 Exercise 459.2 (Detection lags) 82 16 Bargaining 83 Exercise 468.1 (Two-period bargaining with constant cost of delay) 83 Exercise 468.2 (Three-period bargaining with constant cost of delay) 83 Contents ix 17 Appendix: Mathematics 85 Exercise 497.1 (Maximizer of quadratic function) 85 Exercise 499.3 (Sums of sequences) 85 Exercise 504.2 (Bayes’ law) 85 References 87 Preface This manual contains all publicly-available solutions to exercises in my book An Introduction to Game Theory (Oxford University Press, 2004). The sources of the problems are given in the section entitled “Notes” at the end of each chapter of the book. Please alert me to errors. M ARTIN J. O SBORNE

[email protected] Department of Economics, 150 St. George Street, University of Toronto, Toronto, Canada M5S 3G7 xi 1 Introduction 5.3 Altruistic preferences Person 1 is indifferent between (1, 4) and (3, 0), and prefers both of these to (2, 1). The payoff function u defined by u( x, y) = x + 21 y, where x is person 1’s income and y is person 2’s, represents person 1’s preferences. Any function that is an increasing function of u also represents her preferences. For example, the functions k( x + 12 y) for any positive number k, and ( x + 12 y)2 , do so. 6.1 Alternative representations of preferences The function v represents the same preferences as does u (because u( a) < u(b) < u(c) and v( a) < v(b) < v(c)), but the function w does not represent the same preferences, because w( a) = w(b) while u( a) < u(b). 1 2 Nash Equilibrium 16.1 Working on a joint project The game in Figure 3.1 models this situation (as does any other game with the same players and actions in which the ordering of the payoffs is the same as the ordering in Figure 3.1). Work hard 3, 3 2, 0 Work hard Goof off Goof off 0, 2 1, 1 Figure 3.1 Working on a joint project (alternative version). 17.1 Games equivalent to the Prisoner’s Dilemma The game in the left panel differs from the Prisoner’s Dilemma in both players’ preferences. Player 1 prefers (Y, X ) to ( X, X ) to ( X, Y ) to (Y, Y ), for example, which differs from her preference in the Prisoner’s Dilemma, which is ( F, Q) to ( Q, Q) to ( F, F ) to ( Q, F ), whether we let X = F or X = Q. The game in the right panel is equivalent to the Prisoner’s Dilemma. If we let X = Q and Y = F then player 1 prefers ( F, Q) to ( Q, Q) to ( F, F ) to ( Q, F ) and player 2 prefers ( Q, F ) to ( Q, Q) to ( F, F ) to ( F, Q), as in the Prisoner’s Dilemma. 20.1 Games without conflict Any two-player game in which each player has two actions and the players have the same preferences may be represented by a table of the form given in Figure 3.2, where a, b, c, and d are any numbers. T B L a, a c, c R b, b d, d Figure 3.2 A strategic game in which conflict is absent. 3 4 Chapter 2. Nash Equilibrium 31.1 Extension of the Stag Hunt Every profile (e, . . . , e), where e is an integer from 0 to K, is a Nash equilibrium. In the equilibrium (e, . . . , e), each player’s payoff is e. The profile (e, . . . , e) is a Nash equilibrium since if player i chooses ei < e then her payoff is 2ei − ei = ei < e, and if she chooses ei > e then her payoff is 2e − ei < e. Consider an action profile (e1 , . . . , en ) in which not all effort levels are the same. Suppose that ei is the minimum. Consider some player j whose effort level exceeds ei . Her payoff is 2ei − e j < ei , while if she deviates to the effort level ei her payoff is 2ei − ei = ei . Thus she can increase her payoff by deviating, so that (e1 , . . . , en ) is not a Nash equilibrium. (This game is studied experimentally by van Huyck, Battalio, and Beil (1990). See also Ochs (1995, 209–233).) 34.1 Guessing two-thirds of the average If all three players announce the same integer k ≥ 2 then any one of them can deviate to k − 1 and obtain $1 (since her number is now closer to 32 of the average than the other two) rather than $ 13 . Thus no such action profile is a Nash equilibrium. If all three players announce 1, then no player can deviate and increase her payoff; thus (1, 1, 1) is a Nash equilibrium. Now consider an action profile in which not all three integers are the same; denote the highest by k∗ . • Suppose only one player names k∗ ; denote the other integers named by k1 and k2 , with k1 ≥ k2 . The average of the three integers is 31 (k∗ + k1 + k2 ), so that 32 of the average is 92 (k∗ + k1 + k2 ). If k1 ≥ 92 (k∗ + k1 + k2 ) then k∗ is further from 32 of the average than is k1 , and hence does not win. If k1 < 92 (k∗ + k1 + k2 ) then the difference between k∗ and 32 of the average is k∗ − 29 (k∗ + k1 + k2 ) = 97 k∗ − 29 k1 − 29 k2 , while the difference between k1 and 2 2 ∗ 2 ∗ 7 2 3 of the average is 9 ( k + k 1 + k 2 ) − k 1 = 9 k − 9 k 1 + 9 k 2 . The difference between the former and the latter is 95 k∗ + 59 k1 − 94 k2 > 0, so k1 is closer to 32 of the average than is k∗ . Hence the player who names k∗ does not win, and is better off naming k2 , in which case she obtains a share of the prize. Thus no such action profile is a Nash equilibrium. • Suppose two players name k∗ , and the third player names k < k∗ . The average of the three integers is then 31 (2k∗ + k), so that 32 of the average is 4 ∗ 2 4 ∗ 2 1 ∗ 4 1 2 1 9 k + 9 k. We have 9 k + 9 k < 2 ( k + k ) (since 9 < 2 and 9 < 2 ), so that the player who names k is the sole winner. Thus either of the other players can switch to naming k and obtain a share of the prize rather obtaining nothing. Thus no such action profile is a Nash equilibrium. We conclude that there is only one Nash equilibrium of this game, in which all three players announce the number 1. (This game is studied experimentally by Nagel (1995).) Chapter 2. Nash Equilibrium 5 34.3 Choosing a route A strategic game that models this situation is: Players The four people. Actions The set of actions of each person is {X, Y} (the route via X and the route via Y). Preferences Each player’s payoff is the negative of her travel time. In every Nash equilibrium, two people take each route. (In any other case, a person taking the more popular route is better off switching to the other route.) For any such action profile, each person’s travel time is either 29.9 or 30 minutes (depending on the route they take). If a person taking the route via X switches to the route via Y her travel time becomes 12 + 21.8 = 33.8 minutes; if a person taking the route via Y switches to the route via X her travel time becomes 22 + 12 = 34 minutes. For any other allocation of people to routes, at least one person can decrease her travel time by switching routes. Thus the set of Nash equilibria is the set of action profiles in which two people take the route via X and two people take the route via Y. Now consider the situation after the road from X to Y is built. There is no equilibrium in which the new road is not used, by the following argument. Because the only equilibrium before the new road is built has two people taking each route, the only possibility for an equilibrium in which no one uses the new road is for two people to take the route A–X–B and two to take A–Y–B, resulting in a total travel time for each person of either 29.9 or 30 minutes. However, if a person taking A– X–B switches to the new road at X and then takes Y–B her total travel time becomes 9 + 7 + 12 = 28 minutes. I claim that in any Nash equilibrium, one person takes A–X–B, two people take A–X–Y–B, and one person takes A–Y–B. For this assignment, each person’s travel time is 32 minutes. No person can change her route and decrease her travel time, by the following argument. • If the person taking A–X–B switches to A–X–Y–B, her travel time increases to 12 + 9 + 15 = 36 minutes; if she switches to A–Y–B her travel time increases to 21 + 15 = 36 minutes. • If one of the people taking A–X–Y–B switches to A–X–B, her travel time increases to 12 + 20.9 = 32.9 minutes; if she switches to A–Y–B her travel time increases to 21 + 12 = 33 minutes. • If the person taking A–Y–B switches to A–X–B, her travel time increases to 15 + 20.9 = 35.9 minutes; if she switches to A–X–Y–B, her travel time increases to 15 + 9 + 12 = 36 minutes. For every other allocation of people to routes at least one person can switch routes and reduce her travel time. For example, if one person takes A–X–B, one 6 Chapter 2. Nash Equilibrium person takes A–X–Y–B, and two people take A–Y–B, then the travel time of those taking A–Y–B is 21 + 12 = 33 minutes; if one of them switches to A–X–B then her travel time falls to 12 + 20.9 = 32.9 minutes. Or if one person takes A–Y–B, one person takes A–X–Y–B, and two people take A–X–B, then the travel time of those taking A–X–B is 12 + 20.9 = 32.9 minutes; if one of them switches to A–X–Y–B then her travel time falls to 12 + 8 + 12 = 32 minutes. Thus in the equilibrium with the new road every person’s travel time increases, from either 29.9 or 30 minutes to 32 minutes. 37.1 Finding Nash equilibria using best response functions a. The Prisoner’s Dilemma and BoS are shown in Figure 6.1; Matching Pennies and the two-player Stag Hunt are shown in Figure 6.2. Quiet 2 ,2 3∗ , 0 Quiet Fink Fink 0 , 3∗ 1∗ , 1∗ Bach 2∗ , 1∗ 0 ,0 Bach Stravinsky Prisoner’s Dilemma Stravinsky 0 ,0 1∗ , 2∗ BoS Figure 6.1 The best response functions in the Prisoner’s Dilemma (left) and in BoS (right). Head Tail Head 1 ∗ , −1 −1 , 1 ∗ Tail −1 , 1 ∗ 1 ∗ , −1 Stag Hare Matching Pennies Stag 2∗ , 2∗ 1 ,0 Hare 0 ,1 1∗ , 1∗ Stag Hunt Figure 6.2 The best response functions in Matching Pennies (left) and the Stag Hunt (right). b. The best response functions are indicated in Figure 6.3. The Nash equilibria are ( T, C ), ( M, L), and ( B, R). T M B L 2 ,2 3∗ , 1∗ 1 , 0∗ C 1∗ , 3∗ 0 ,0 0 , 0∗ R 0∗ , 1 0∗ , 0 0∗ , 0∗ Figure 6.3 The game in Exercise 37.1. 38.1 Constructing best response functions The analogue of Figure 38.2 in the book is given in Figure 7.1. Chapter 2. Nash Equilibrium 7 R A2 C L |T M {z A1 B} Figure 7.1 The players’ best response functions for the game in Exercise 38.1b. Player 1’s best responses are indicated by circles, and player 2’s by dots. The action pairs for which there is both a circle and a dot are the Nash equilibria. 38.2 Dividing money For each amount named by one of the players, the other player’s best responses are given in the following table. Other player’s action 0 1 2 3 4 5 6 7 8 9 10 Sets of best responses {10} {9, 10} {8, 9, 10} {7, 8, 9, 10} {6, 7, 8, 9, 10} {5, 6, 7, 8, 9, 10} {5, 6} {6 } {7 } {8 } {9 } The best response functions are illustrated in Figure 8.1 (circles for player 1, dots for player 2). From this figure we see that the game has four Nash equilibria: (5, 5), (5, 6), (6, 5), and (6, 6). 41.1 Strict and nonstrict Nash equilibria Only the Nash equilibrium ( a1∗ , a2∗ ) is strict. For each of the other equilibria, player 2’s action a2 satisfies a2∗∗∗ ≤ a2 ≤ a2∗∗ , and for each such action player 1 has multiple best responses, so that her payoff is the same for a range of actions, only one of which is such that ( a1 , a2 ) is a Nash equilibrium. 8 Chapter 2. Nash Equilibrium 10 9 8 7 6 A2 5 4 3 2 1 0 5 6 7 8 9 10} |0 1 2 3 4 {z A1 Figure 8.1 The players’ best response functions for the game in Exercise 38.2. 47.1 Strict equilibria and dominated actions For player 1, T is weakly dominated by M, and strictly dominated by B. For player 2, no action is weakly or strictly dominated. The game has a unique Nash equilibrium, ( M, L). This equilibrium is not strict. (When player 2 choose L, B yields player 1 the same payoff as does M.) 47.2 Nash equilibrium and weakly dominated actions The only Nash equilibrium of the game in Figure 8.2 is ( T, L). The action T is weakly dominated by M and the action L is weakly dominated by C. (There are of course many other games that satisfy the conditions.) T M B L 1, 1 1, 0 0, 0 C 0, 1 2, 1 1, 1 R 0, 0 1, 2 2, 0 Figure 8.2 A game with a unique Nash equilibrium, in which both players’ equilibrium actions are weakly dominated. (The unique Nash equilibrium is ( T, L ).) 50.1 Other Nash equilibria of the game modeling collective decision-making Denote by i the player whose favorite policy is the median favorite policy. The set of Nash equilibria includes every action profile in which (i) i’s action is her favorite policy xi∗ , (ii) every player whose favorite policy is less than x i∗ names a Chapter 2. Nash Equilibrium 9 policy equal to at most xi∗ , and (iii) every player whose favorite policy is greater than xi∗ names a policy equal to at least xi∗ . To show this, first note that the outcome is xi∗ , so player i cannot induce a better outcome for herself by changing her action. Now, if a player whose favorite position is less than xi∗ changes her action to some x < xi∗ , the outcome does not change; if such a player changes her action to some x > x i∗ then the outcome either remains the same (if some player whose favorite position exceeds xi∗ names xi∗ ) or increases, so that the player is not better off. A similar argument applies to a player whose favorite position is greater than xi∗ . The set of Nash equilibria also includes, for any positive integer k ≤ n, every action profile in which k players name the median favorite policy x i∗ , at most 21 (n − 3) players name policies less than xi∗ , and at most 21 (n − 3) players name policies greater than xi∗ . (In these equilibria, the favorite policy of a player who names a policy less than xi∗ may be greater than xi∗ , and vice versa. The conditions on the numbers of players who name policies less than xi∗ and greater than xi∗ ensure that no such player can, by naming instead her favorite policy, move the median policy closer to her favorite policy.) Any action profile in which all players name the same, arbitrary, policy is also a Nash equilibrium; the outcome is the common policy named. More generally, any profile in which at least three players name the same, arbitrary, policy x, at most (n − 3)/2 players name a policy less than x, and at most (n − 3)/2 players name a policy greater than x is a Nash equilibrium. (In both cases, no change in any player’s action has any effect on the outcome.) 51.2 Symmetric strategic games The games in Exercise 31.2, Example 39.1, and Figure 47.2 (both games) are symmetric. The game in Exercise 42.1 is not symmetric. The game in Section 2.8.4 is symmetric if and only if u1 = u2 . 52.2 Equilibrium for pairwise interactions in a single population The Nash equilibria are ( A, A), ( A, C ), and (C, A). Only the equilibrium ( A, A) is relevant if the game is played between the members of a single population—this equilibrium is the only symmetric equilibrium. 3 Nash Equilibrium: Illustrations 58.1 Cournot’s duopoly game with linear inverse demand and different unit costs Following the analysis in the text, the best response function of firm 1 is 1 b1 (q2 ) = 2 (α − c1 − q2 ) if q2 ≤ α − c1 0 otherwise while that of firm 2 is b2 (q1 ) = 1 2 ( α − c2 0 − q1 ) if q1 ≤ α − c2 otherwise. To find the Nash equilibrium, first plot these two functions. Each function has the same general form as the best response function of either firm in the case studied in the text. However, the fact that c1 6= c2 leads to two qualitatively different cases when we combine the two functions to find a Nash equilibrium. If c1 and c2 do not differ very much then the functions in the analogue of Figure 59.1 intersect at a pair of outputs that are both positive. If c1 and c2 differ a lot, however, the functions intersect at a pair of outputs in which q1 = 0. Precisely, if c1 ≤ 21 (α + c2 ) then the downward-sloping parts of the best response functions intersect (as in Figure 59.1), and the game has a unique Nash equilibrium, given by the solution of the two equations q1 = q2 = 1 2 ( α − c1 1 2 ( α − c2 − q2 ) − q1 ) . This solution is (q1∗ , q2∗ ) = 1 3 ( α − 2c1 + c2 ), 31 (α − 2c2 + c1 ) . If c1 > 21 (α + c2 ) then the downward-sloping part of firm 1’s best response function lies below the downward-sloping part of firm 2’s best response function (as in Figure 12.1), and the game has a unique Nash equilibrium, (q1∗ , q2∗ ) = (0, 21 (α − c2 )). In summary, the game always has a unique Nash equilibrium, defined as follows: 1 (α − 2c1 + c2 ), 1 (α − 2c2 + c1 ) if c1 ≤ 21 (α + c2 ) 3 3 0, 1 (α − c2 ) if c1 > 21 (α + c2 ). 2 The output of firm 2 exceeds that of firm 1 in every equilibrium. 11 12 Chapter 3. Nash Equilibrium: Illustrations ↑ q2 α− c2 2 (q1∗ , q2∗ ) α − c1 b2 (q1 ) b1 (q2 ) α− c1 2 0 α − c2 q1 → Figure 12.1 The best response functions in Cournot’s duopoly game under the assumptions of Exercise 58.1 when α − c1 < 12 ( α − c2 ). The unique Nash equilibrium in this case is ( q1∗ , q2∗ ) = (0, 12 ( α − c2 )). If c2 decreases then firm 2’s output increases and firm 1’s output either falls, if c1 ≤ 12 (α + c2 ), or remains equal to 0, if c1 > 21 (α + c2 ). The total output increases and the price falls. 60.2 Nash equilibrium of Cournot’s duopoly game and the collusive outcome The firms’ total profit is (q1 + q2 )(α − c − q1 − q2 ), or Q(α − c − Q), where Q denotes total output. This function is a quadratic in Q that is zero when Q = 0 and when Q = α − c, so that its maximizer is Q∗ = 21 (α − c). If each firm produces 14 (α − c) then its profit is 18 (α − c)2 . This profit exceeds its Nash equilibrium profit of 91 (α − c)2 . If one firm produces Q∗ /2, the other firm’s best response is bi ( Q∗ /2) = 21 (α − c − 14 (α − c)) = 83 (α − c). That is, if one firm produces Q∗ /2, the other firm wants to produce more than Q∗ /2. 63.1 Interaction among resource-users The game is given as follows. Players The firms. Actions Each firm’s set of actions is the set of all nonnegative numbers (representing the amount of input it uses). Preferences The payoff of each firm i is xi (1 − ( x1 + · · · + xn )) 0 if x1 + · · · + xn ≤ 1 if x1 + · · · + xn > 1. Chapter 3. Nash Equilibrium: Illustrations 13 This game is the same as that in Exercise 61.1 for c = 0 and α = 1. Thus it has a unique Nash equilibrium, ( x1 , . . . , xn ) = (1/(n + 1), . . . , 1/(n + 1)). In this Nash equilibrium, each firm’s output is (1/(n + 1))(1 − n/(n + 1)) = 1/(n + 1)2 . If xi = 1/(2n) for i = 1, . . . , n then each firm’s output is 1/(4n), which exceeds 1/(n + 1)2 for n ≥ 2. (We have 1/(4n) − 1/(n + 1)2 = (n − 1)2 /(4n(n + 1)2 ) > 0 for n ≥ 2.) 67.1 Bertrand’s duopoly game with constant unit cost The pair (c, c) of prices remains a Nash equilibrium; the argument is the same as before. Further, as before, there is no other Nash equilibrium. The argument needs only very minor modification. For an arbitrary function D there may exist no monopoly price pm ; in this case, if pi > c, p j > c, pi ≥ p j , and D ( p j ) = 0 then firm i can increase its profit by reducing its price slightly below p (for example). 68.1 Bertrand’s oligopoly game Consider a profile ( p1, . . . , pn ) of prices in which pi ≥ c for all i and at least two prices are equal to c. Every firm’s profit is zero. If any firm raises its price its profit remains zero. If a firm charging more than c lowers its price, but not below c, its profit also remains zero. If a firm lowers its price below c then its profit is negative. Thus any such profile is a Nash equilibrium. To show that no other profile is a Nash equilibrium, we can argue as follows. • If some price is less than c then the firm charging the lowest price can increase its profit (to zero) by increasing its price to c. • If exactly one firm’s price is equal to c then that firm can increase its profit by raising its price a little (keeping it less than the next highest price). • If all firms’ prices exceed c then the firm charging the highest price can increase its profit by lowering its price to some price between c and the lowest price being charged. 68.2 Bertrand’s duopoly game with different unit costs a. If all consumers buy from firm 1 when both firms charge the price c2 , then ( p1, p2 ) = (c2 , c2 ) is a Nash equilibrium by the following argument. Firm 1’s profit is positive, while firm 2’s profit is zero (since it serves no customers). • If firm 1 increases its price, its profit falls to zero. • If firm 1 reduces its price, say to p, then its profit changes from (c2 − c1 )(α − c2 ) to ( p − c1 )(α − p). Since c2 is less than the maximizer of ( p − c1 )(α − p), firm 1’s profit falls. 14 Chapter 3. Nash Equilibrium: Illustrations • If firm 2 increases its price, its profit remains zero. • If firm 2 decreases its price, its profit becomes negative (since its price is less than its unit cost). Under this rule no other pair of prices is a Nash equilibrium, by the following argument. • If pi < c1 for i = 1, 2 then the firm with the lower price (or either firm, if the prices are the same) can increase its profit (to zero) by raising its price above that of the other firm. • If p1 > p2 ≥ c2 then firm 2 can increase its profit by raising its price a little. • If p2 > p1 ≥ c1 then firm 1 can increase its profit by raising its price a little. • If p2 ≤ p1 and p2 < c2 then firm 2’s profit is negative, so that it can increase its profit by raising its price. • If p1 = p2 > c2 then at least one of the firms is not receiving all of the demand, and that firm can increase its profit by lowering its price a little. b. Now suppose that the rule for splitting up the customers when the prices are equal specifies that firm 2 receives some customers when both prices are c2 . By the argument for part a, the only possible Nash equilibrium is ( p1 , p2 ) = (c2 , c2 ). (The argument in part a that every other pair of prices is not a Nash equilibrium does not use the fact that customers are split equally when ( p1 , p2 ) = (c2 , c2 ).) But if ( p1, p2 ) = (c2 , c2 ) and firm 2 receives some customers, firm 1 can increase its profit by reducing its price a little and capturing the entire market. 73.1 Electoral competition with asymmetric voters’ preferences The unique Nash equilibrium remains (m, m); the direct argument is exactly the same as before. (The dividing line between the supporters of two candidates with different positions changes. If xi < x j , for example, the dividing line is 31 xi + 23 x j rather than 12 ( xi + x j ). The resulting change in the best response functions does not affect the Nash equilibrium.) 75.3 Electoral competition for more general preferences a. If x ∗ is a Condorcet winner then for any y 6= x ∗ a majority of voters prefer x ∗ to y, so y is not a Condorcet winner. Thus there is no more than one Condorcet winner. b. Suppose that one of the remaining voters prefers y to z to x, and the other prefers z to x to y. For each position there is another position preferred by a majority of voters, so no position is a Condorcet winner. Chapter 3. Nash Equilibrium: Illustrations 15 c. Now suppose that x ∗ is a Condorcet winner. Then the strategic game described the exercise has a unique Nash equilibrium in which both candidates choose x ∗ . This pair of actions is a Nash equilibrium because if either candidate chooses a different position she loses. For any other pair of actions either one candidate loses, in which case that candidate can deviate to the position x ∗ and at least tie, or the candidates tie at a position different from x ∗ , in which case either of them can deviate to x ∗ and win. If there is no Condorcet winner then for every position there is another position preferred by a majority of voters. Thus for every pair of distinct positions the loser can deviate and win, and for every pair of identical positions either candidate can deviate and win. Thus there is no Nash equilibrium. 76.1 Competition in product characteristics Suppose there are two firms. If the products are different, then either firm increases its market share by making its product more similar to that of its rival. Thus in every possible equilibrium the products are the same. But if x1 = x2 6= m then each firm’s market share is 50%, while if it changes its product to be closer to m then its market share rises above 50%. Thus the only possible equilibrium is ( x1 , x2 ) = (m, m). This pair of positions is an equilibrium, since each firm’s market share is 50%, and if either firm changes its product its market share falls below 50%. Now suppose there are three firms. If all firms’ products are the same, each obtains one-third of the market. If x1 = x2 = x3 = m then any firm, by changing its product a little, can obtain close to one-half of the market. If x1 = x2 = x3 6= m then any firm, by changing its product a little, can obtain more than one-half of the market. If the firms’ products are not all the same, then at least one of the extreme products is different from the other two products, and the firm that produces it can increase its market share by making it more similar to the other products. Thus when there are three firms there is no Nash equilibrium. 79.1 Direct argument for Nash equilibria of War of Attrition • If t1 = t2 then either player can increase her payoff by conceding slightly later (in which case she obtains the object for sure, rather than getting it with probability 21 ). • If 0 < ti < t j then player i can increase her payoff by conceding at 0. • If 0 = ti < t j < vi then player i can increase her payoff (from 0 to almost vi − t j > 0) by conceding slightly after t j . Thus there is no Nash equilibrium in which t1 = t2 , 0 < ti < t j , or 0 = ti < t j < vi (for i = 1 and j = 2, or i = 2 and j = 1). The remaining possibility is that 0 = ti < t j and t j ≥ vi for i = 1 and j = 2, or i = 2 and j = 1. In this case player i’s 16 Chapter 3. Nash Equilibrium: Illustrations payoff is 0, while if she concedes later her payoff is negative; player j’s payoff is v j , her highest possible payoff in the game. 85.1 Second-price sealed-bid auction with two bidders If player 2’s bid b2 is less than v1 then any bid of b2 or more is a best response of player 1 (she wins and pays the price b2 ). If player 2’s bid is equal to v1 then every bid of player 1 yields her the payoff zero (either she wins and pays v1 , or she loses), so every bid is a best response. If player 2’s bid b2 exceeds v1 then any bid of less than b2 is a best response of player 1. (If she bids b2 or more she wins, but pays the price b2 > v1 , and hence obtains a negative payoff.) In summary, player 1’s best response function is if b2 < v1 {b1 : b1 ≥ b2 } B1 (b2 ) = {b1 : b1 ≥ 0} if b2 = v1 {b1 : 0 ≤ b1 < b2 } if b2 > v1 . By similar arguments, player 2’s best response function is if b1 < v2 {b2 : b2 > b1 } B2 (b1 ) = {b2 : b2 ≥ 0} if b1 = v2 . {b2 : 0 ≤ b2 ≤ b1 } if b1 > v2 . These best response functions are shown in Figure 16.1. ↑ b2 ↑ b2 B1 (b2 ) v1 v1 v2 v2 0 v2 v1 B2 (b1 ) b1 → v2 v1 b1 → Figure 16.1 The players’ best response functions in a two-player second-price sealed-bid auction (Exercise 85.1). Player 1’s best response function is in the left panel; player 2’s is in the right panel. (Only the edges marked by a black line are included.) Superimposing the best response functions, we see that the set of Nash equilibria is the shaded set in Figure 17.1, namely the set of pairs (b1 , b2 ) such that either b1 ≤ v2 and b2 ≥ v1 or b1 ≥ v2 , b1 ≥ b2 , and b2 ≤ v1 . Chapter 3. Nash Equilibrium: Illustrations 17 ↑ b2 v1 v2 0 v2 v1 b1 → Figure 17.1 The set of Nash equilibria of a two-player second-price sealed-bid auction (Exercise 85.1). 86.2 Nash equilibrium of first-price sealed-bid auction The profile (b1 , . . . , bn ) = (v2 , v2 , v3 , . . . , vn ) is a Nash equilibrium by the following argument. • If player 1 raises her bid she still wins, but pays a higher price and hence obtains a lower payoff. If player 1 lowers her bid then she loses, and obtains the payoff of 0. • If any other player changes her bid to any price at most equal to v2 the outcome does not change. If she raises her bid above v2 she wins, but obtains a negative payoff. 87.1 First-price sealed-bid auction A profile of bids in which the two highest bids are not the same is not a Nash equilibrium because the player naming the highest bid can reduce her bid slightly, continue to win, and pay a lower price. By the argument in the text, in any equilibrium player 1 wins the object. Thus she submits one of the highest bids. If the highest bid is less than v2 , then player 2 can increase her bid to a value between the highest bid and v2 , win, and obtain a positive payoff. Thus in an equilibrium the highest bid is at least v2 . If the highest bid exceeds v1 , player 1’s payoff is negative, and she can increase this payoff by reducing her bid. Thus in an equilibrium the highest bid is at most v1 . Finally, any profile (b1 , . . . , bn ) of bids that satisfies the conditions in the exercise is a Nash equilibrium by the following argument. 18 Chapter 3. Nash Equilibrium: Illustrations • If player 1 increases her bid she continues to win, and reduces her payoff. If player 1 decreases her bid she loses and obtains the payoff 0, which is at most her payoff at (b1 , . . . , bn ). • If any other player increases her bid she either does not affect the outcome, or wins and obtains a negative payoff. If any other player decreases her bid she does not affect the outcome. 89.1 All-pay auctions Second-price all-pay auction with two bidders: The payoff function of bidder i is ui (b1 , b2 ) = − bi vi − bj if bi < b j if bi > b j , with u1 (b, b) = v1 − b and u2 (b, b) = −b for all b. This payoff function differs from that of player i in the War of Attrition only in the payoffs when the bids are equal. The set of Nash equilibria of the game is the same as that for the War of Attrition: the set of all pairs (0, b2 ) where b2 ≥ v1 and (b1 , 0) where b1 ≥ v2 . (The pair (b, b) of actions is not a Nash equilibrium for any value of b because player 2 can increase her payoff by either increasing her bid slightly or by reducing it to 0.) First-price all-pay auction with two bidders: In any Nash equilibrium the two highest bids are equal, otherwise the player with the higher bid can increase her payoff by reducing her bid a little (keeping it larger than the other player’s bid). But no profile of bids in which the two highest bids are equal is a Nash equilibrium, because the player with the higher index who submits this bid can increase her payoff by slightly increasing her bid, so that she wins rather than loses. 90.1 Multiunit auctions Discriminatory auction To show that the action of bidding vi and wi is not dominant for player i, we need only find actions for the other players and alternative bids for player i such that player i’s payoff is higher under the alternative bids than it is under the vi and wi , given the other players’ actions. Suppose that each of the other players submits two bids of 0. Then if player i submits one bid between 0 and vi and one bid between 0 and wi she still wins two units, and pays less than when she bids v i and wi . Uniform-price auction Suppose that some bidder other than i submits one bid between wi and vi and one bid of 0, and all the remaining bidders submit two bids of 0. Then bidder i wins one unit, and pays the price wi . If she replaces her bid of wi with a bid between 0 and wi then she pays a lower price, and hence is better off. Chapter 3. Nash Equilibrium: Illustrations 19 Vickrey auction Suppose that player i bids v i and wi . Consider separately the cases in which the bids of the players other than i are such that player i wins 0, 1, and 2 units. Player i wins 0 units: In this case the second highest of the other players’ bids is at least vi , so that if player i changes her bids so that she wins one or more units, for any unit she wins she pays at least v i . Thus no change in her bids increases her payoff from its current value of 0 (and some changes lower her payoff). Player i wins 1 unit: If player i raises her bid of v i then she still wins one unit and the price remains the same. If she lowers this bid then either she still wins and pays the same price, or she does not win any units. If she raises her bid of wi then either the outcome does not change, or she wins a second unit. In the latter case the price she pays is the previously-winning bid she beat, which is at least wi , so that her payoff either remains zero or becomes negative. Player i wins 2 units: Player i’s raising either of her bids has no effect on the outcome; her lowering a bid either has no effect on the outcome or leads her to lose rather than to win, leading her to obtain the payoff of zero. 90.3 Internet pricing The situation may be modeled as a multiunit auction in which k units are available, and each player attaches a positive value to only one unit and submits a bid for only one unit. The k highest bids win, and each winner pays the (k + 1)st highest bid. By a variant of the argument for a second-price auction, in which “highest of the other players’ bids” is replaced by “highest rejected bid”, each player’s action of bidding her value is weakly dominates all her other actions. 96.2 Alternative standards of care under negligence with contributory negligence First consider the case in which X1 = aˆ 1 and X2 ≤ aˆ2 . The pair ( aˆ 1 , aˆ 2 ) is a Nash equilibrium by the following argument. If a2 = aˆ 2 then the victim’s level of care is sufficient (at least X2 ), so that the injurer’s payoff is given by (94.1) in the text. Thus the argument that the injurer’s action aˆ 1 is a best response to aˆ 2 is exactly the same as the argument for the case X2 = aˆ 2 in the text. Since X1 is the same as before, the victim’s payoff is the same also, so that by the argument in the text the victim’s best response to aˆ1 is aˆ 2 . Thus ( aˆ 1 , aˆ 2 ) is a Nash equilibrium. To show that ( aˆ 1 , aˆ 2 ) is the only Nash equilibrium of the game, we study the players’ best response functions. First consider the injurer’s best response function. As in the text, we split the analysis into three cases. 20 Chapter 3. Nash Equilibrium: Illustrations a2 < X2 : In this case the injurer does not have to pay any compensation, regardless of her level of care; her payoff is − a1 , so that her best response is a1 = 0. a2 = X2 : In this case the injurer’s best response is aˆ 1 , as argued when showing that ( aˆ 1 , aˆ 2 ) is a Nash equilibrium. a2 > X2 : In this case the injurer’s best response is at most aˆ1 , since her payoff is equal to − a1 for larger values of a1 . Thus the injurer’s best response takes a form like that shown in the left panel of Figure 20.1. (In fact, b1 ( a2 ) = aˆ1 for X2 ≤ a2 ≤ aˆ2 , but the analysis depends only on the fact that b1 ( a2 ) ≤ aˆ1 for a2 > X2 .) ↑ a2 b1 ( a2 ) aˆ 2 X2 0 ↑ a2 X2 aˆ 1 a1 → 0 b2 ( a1 ) ? aˆ 1 a1 → Figure 20.1 The players’ best response functions under the rule of negligence with contributory negligence when X1 = aˆ1 and X2 = aˆ2 . Left panel: the injurer’s best response function b1 . Right panel: the victim’s best response function b2 . (The position of the victim’s best response function for a1 > aˆ 1 is not significant, and is not determined in the solution.) is Now consider the victim’s best response function. The victim’s payoff function − a2 if a1 < aˆ 1 and a2 ≥ X2 u2 ( a 1 , a 2 ) = − a2 − L( a1 , a2 ) if a1 ≥ aˆ 1 or a2 < X2 . As before, for a1 < aˆ 1 we have − a2 − L( a1 , a2 ) < − aˆ 2 for all a2 , so that the victim’s best response is X2 . As in the text, the nature of the victim’s best responses to levels of care a1 for which a1 > aˆ 1 are not significant. Combining the two best response functions we see that ( aˆ 1 , aˆ 2 ) is the unique Nash equilibrium of the game. Now consider the case in which X1 = M and a2 = aˆ2 , where M ≥ aˆ 1 . The injurer’s payoff is − a1 − L( a1 , a2 ) if a1 < M and a2 ≥ aˆ 2 u1 ( a 1 , a 2 ) = − a1 if a1 ≥ M or a2 < aˆ 2 . Now, the maximizer of − a1 − L( a1 , aˆ 2 ) is aˆ 1 (see the argument following (94.1) in the text), so that if M is large enough then the injurer’s best response to aˆ 2 is aˆ 1 . As before, if a2 < aˆ2 then the injurer’s best response is 0, and if a2 > aˆ2 then the Chapter 3. Nash Equilibrium: Illustrations ↑ a2 ↑ a2 b1 ( a2 ) aˆ 2 0 21 aˆ 2 aˆ 1 M a1 → 0 b2 ( a1 ) ? aˆ1 M a1 → Figure 21.1 The players’ best response functions under the rule of negligence with contributory negligence when ( X1 , X2 ) = ( M, aˆ 2 ), with M ≥ aˆ1 . Left panel: the injurer’s best response function b1 . Right panel: the victim’s best response function b2 . (The position of the victim’s best response function for a1 > M is not significant, and is not determined in the text.) injurer’s payoff decreases for a1 > M, so that her best response is less than M. The injurer’s best response function is shown in the left panel of Figure 21.1. The victim’s payoff is − a2 if a1 < M and a2 ≥ aˆ 2 u2 ( a 1 , a 2 ) = − a2 − L( a1 , a2 ) if a1 ≥ M or a2 < aˆ 2 . If a1 ≤ aˆ1 then the victim’s best response is aˆ 2 by the same argument as the one in the text. If a1 is such that aˆ1 < a1 < M then the victim’s best response is at most aˆ 2 (since her payoff is decreasing for larger values of a2 ). This information about the victim’s best response function is recorded in the right panel of Figure 21.1; it is sufficient to deduce that ( aˆ 1 , aˆ 2 ) is the unique Nash equilibrium of the game. 4 Mixed Strategy Equilibrium 101.1 Variant of Matching Pennies The analysis is the same as for Matching Pennies. There is a unique steady state, in which each player chooses each action with probability 21 . 106.2 Extensions of BoS with vNM preferences In the first case, when player 1 is indifferent between going to her less preferred concert in the company of player 2 and the lottery in which with probability 21 she and player 2 go to different concerts and with probability 12 they both go to her more preferred concert, the Bernoulli payoffs that represent her preferences satisfy the condition u1 (S, S) = 12 u1 (S, B) + 12 u1 ( B, B). If we choose u1 (S, B) = 0 and u1 ( B, B) = 2, then u1 (S, S) = 1. Similarly, for player 2 we can set u2 ( B, S) = 0, u2 (S, S) = 2, and u2 ( B, B) = 1. Thus the Bernoulli payoffs in the left panel of Figure 23.1 are consistent with the players’ preferences. In the second case, when player 1 is indifferent between going to her less preferred concert in the company of player 2 and the lottery in which with probability 34 she and player 2 go to different concerts and with probability 41 they both go to her more preferred concert, the Bernoulli payoffs that represent her preferences satisfy the condition u1 (S, S) = 34 u1 (S, B) + 14 u1 ( B, B). If we choose u1 (S, B) = 0 and u1 ( B, B) = 2 (as before), then u1 (S, S) = 12 . Similarly, for player 2 we can set u2 ( B, S) = 0, u2 (S, S) = 2, and u2 ( B, B) = 12 . Thus the Bernoulli payoffs in the right panel of Figure 23.1 are consistent with the players’ preferences. Bach Stravinsky Bach 2, 1 0, 0 Stravinsky 0, 0 1, 2 Bach Stravinsky Bach 2, 12 0, 0 Figure 23.1 The Bernoulli payoffs for two extensions of BoS. 23 Stravinsky 0, 0 1 2, 2 24 Chapter 4. Mixed Strategy Equilibrium ↑ Player 1’s expected payoff 2 q=1 1 1 q= 1 2 1 2 q=0 p→ 1 0 Figure 24.1 Player 1’s expected payoff as a function of the probability p that she assigns to B in BoS, when the probability q that player 2 assigns to B is 0, 12 , and 1. 110.1 Expected payoffs For BoS, player 1’s expected payoff is shown in Figure 24.1. For the game in the right panel of Figure 21.1 in the book, player 1’s expected payoff is shown in Figure 24.2. ↑ Player 1’s expected payoff 3 q=1 2 3 2 q= 1 1 2 q=0 0 p→ 1 Figure 24.2 Player 1’s expected payoff as a function of the probability p that she assigns to Refrain in the game in the right panel of Figure 21.1 in the book, when the probability q that player 2 assigns to Refrain is 0, 12 , and 1. 111.1 Examples of best responses For BoS: for q = 0 player 1’s unique best response is p = 0 and for q = 21 and q = 1 her unique best response is p = 1. For the game in the right panel of Figure 21.1: for q = 0 player 1’s unique best response is p = 0, for q = 21 her set of best responses is the set of all her mixed strategies (all values of p), and for q = 1 her unique best response is p = 1. Chapter 4. Mixed Strategy Equilibrium 25 114.1 Mixed strategy equilibrium of Hawk–Dove Denote by ui a payoff function whose expected value represents player i’s preferences. The conditions in the problem imply that for player 1 we have u1 (Passive, Passive) = 21 u1 (Aggressive, Aggressive) + 12 u1 (Aggressive, Passive) and u1 (Passive, Aggressive) = 23 u1 (Aggressive, Aggressive) + 13 u1 (Passive, Passive). Given u1 (Aggressive, Aggressive) = 0 and u1 (Passive, Aggressive = 1, we have u1 (Passive, Passive) = 12 u1 (Aggressive, Passive) and 1 = 13 u1 (Passive, Passive), so that u1 (Passive, Passive) = 3 and u1 (Aggressive, Passive) = 6. Similarly, u2 (Passive, Passive) = 3 and u2 (Passive, Aggressive) = 6. Thus the game is given in the left panel of Figure 25.1. The players’ best response functions are shown in the right panel. The game has three mixed strategy Nash equilibria: ((0, 1), (1, 0)), (( 34 , 14 ), ( 34 , 14 )), and ((1, 0), (0, 1)). ↑ 1 q 3 4 Aggressive Passive Aggressive 0, 0 1, 6 B1 Passive 6, 1 3, 3 B2 0 3 4 1 p→ Figure 25.1 An extension of Hawk–Dove (left panel) and the players’ best response functions when randomization is allowed in this game (right panel). The probability that player 1 assigns to Aggressive is p and the probability that player 2 assigns to Aggressive is q. The disks indicate the Nash equilibria (two pure, one mixed). 26 Chapter 4. Mixed Strategy Equilibrium 117.2 Choosing numbers a. To show that the pair of mixed strategies in the question is a mixed strategy equilibrium, it suffices to verify the conditions in Proposition 116.2. Thus, given that each player’s strategy specifies a positive probability for every action, it suffices to show that each action of each player yields the same expected payoff. Player 1’s expected payoff to each pure strategy is 1/K, because with probability 1/K player 2 chooses the same number, and with probability 1 − 1/K player 2 chooses a different number. Similarly, player 2’s expected payoff to each pure strategy is −1/K, because with probability 1/K player 1 chooses the same number, and with probability 1 − 1/K player 2 chooses a different number. Thus the pair of strategies is a mixed strategy Nash equilibrium. b. Let ( p∗ , q∗ ) be a mixed strategy equilibrium, where p∗ and q∗ are vectors, the jth components of which are the probabilities assigned to the integer j by each player. Given that player 2 uses the mixed strategy q∗ , player 1’s expected payoff if she chooses the number k is q∗k . Hence if p∗k > 0 then (by the first condition in Proposition 116.2) we need q∗k ≥ q∗j for all j, so that, in particular, q∗k > 0 (q∗j cannot be zero for all j!). But player 2’s expected payoff if she chooses the number k is − p∗k , so given q∗k > 0 we need p∗k ≤ p∗j for all j (again by the first condition in Proposition 116.2), and, in particular, p ∗k ≤ 1/K (p∗j cannot exceed 1/K for all j!). We conclude that any probability p ∗k that is positive must be at most 1/K. The only possibility is that p ∗k = 1/K for all k. A similar argument implies that q∗k = 1/K for all k. 120.2 Strictly dominating mixed strategies Denote the probability that player 1 assigns to T by p and the probability she assigns to M by r (so that the probability she assigns to B is 1 − p − r). A mixed strategy of player 1 strictly dominates T if and only if p + 4r > 1 and p + 3(1 − p − r ) > 1, or if and only if 1 − 4r < p < 1 − 23 r. For example, the mixed strategies ( 41 , 14 , 12 ) and (0, 41 , 34 ) both strictly dominate T. 120.3 Strict domination for mixed strategies (a) True. Suppose that the mixed strategy α0i assigns positive probability to the action a0i , which is strictly dominated by the action ai . Then ui ( ai , a−i ) > ui ( a0i , a−i ) for all a−i . Let αi be the mixed strategy that differs from α0i only in the weight that α0i assigns to a0i is transferred to ai . That is, αi is defined by αi ( a0i ) = 0, αi ( ai ) = α0i ( a0i ) + α0i ( ai ), and αi (bi ) = α0i (bi ) for every other action bi . Then αi strictly dominates α0i : for any a−i we have U (αi , a−i ) − U (α0i , a−i ) = α0i ( a0i )(u( ai , a−i ) − ui ( a0i , a−i )) > 0. Chapter 4. Mixed Strategy Equilibrium 27 (b) False. Consider a variant of the game in Figure 120.1 in the text in which player 1’s payoffs to ( T, L) and to ( T, R) are both 52 instead of 1. Then player 1’s mixed strategy that assigns probability 21 to M and probability 12 to B is strictly dominated by T, even though neither M nor B is strictly dominated. 127.1 Equilibrium in the expert diagnosis game When E = rE0 + (1 − r ) I 0 the consumer is indifferent between her two actions when p = 0, so that her best response function has a vertical segment at p = 0. Referring to Figure 126.1 in the text, we see that the set of mixed strategy Nash equilibria correspond to p = 0 and π/π 0 ≤ q ≤ 1. 130.3 Bargaining The game is given in Figure 27.1. 0 2 4 6 8 10 0 5, 5 6, 4 7, 3 8, 2 9, 1 10, 0 2 4, 6 5, 5 6, 4 7, 3 8, 2 0, 0 4 3, 7 4, 6 5, 5 6, 4 0, 0 0, 0 6 2, 8 3, 7 4, 6 0, 0 0, 0 0, 0 8 1, 9 2, 8 0, 0 0, 0 0, 0 0, 0 10 0, 10 0, 0 0, 0 0, 0 0, 0 0, 0 Figure 27.1 A bargaining game. By inspection it has a single symmetric pure strategy Nash equilibrium, (10, 10). Now consider situations in which the common mixed strategy assigns positive probability to two actions. Suppose that player 2 assigns positive probability only to 0 and 2. Then player 1’s payoff to her action 4 exceeds her payoff to either 0 or 2. Thus there is no symmetric equilibrium in which the actions assigned positive probability are 0 and 2. By a similar argument we can rule out equilibria in which the actions assigned positive probability are any pair except 2 and 8, or 4 and 6. If the actions to which player 2 assigns positive probability are 2 and 8 then player 1’s expected payoffs to 2 and 8 are the same if the probability player 2 assigns to 2 is 25 (and the probability she assigns to 8 is 35 ). Given these probabilities, player 1’s expected payoff to her actions 2 and 8 is 16 5 , and her expected payoff to every other action is less than 16 . Thus the pair of mixed strategies in which every 5 player assigns probability 25 to 2 and 35 to 8 is a symmetric mixed strategy Nash equilibrium. Similarly, the game has a symmetric mixed strategy equilibrium (α∗ , α∗ ) in which α∗ assigns probability 54 to the demand of 4 and probability 51 to the demand of 6. 28 Chapter 4. Mixed Strategy Equilibrium In summary, the game has three symmetric mixed strategy Nash equilibria in which each player’s strategy assigns positive probability to at most two actions: one in which probability 1 is assigned to 10, one in which probability 52 is assigned to 2 and probability 35 is assigned to 8, and one in which probability 54 is assigned to 4 and probability 15 is assigned to 6. 132.2 Reporting a crime when the witnesses are heterogeneous Denote by pi the probability with which each witness with cost ci reports the crime, for i = 1, 2. For each witness with cost c1 to report with positive probability less than one, we need or v − c1 = v · Pr{at least one other person calls} = v 1 − ( 1 − p 1 ) n1 −1 ( 1 − p 2 ) n2 , c 1 = v ( 1 − p 1 ) n1 −1 ( 1 − p 2 ) n2 . (28.1) Similarly, for each witness with cost c2 to report with positive probability less than one, we need or v − c2 = v · Pr{at least one other person calls} = v 1 − ( 1 − p 1 ) n1 ( 1 − p 2 ) n2 −1 , c 2 = v ( 1 − p 1 ) n1 ( 1 − p 2 ) n2 −1 . (28.2) Dividing (28.1) by (28.2) we obtain 1 − p2 = c1 (1 − p1 )/c2 . Substituting this expression for 1 − p2 into (28.1) we get p1 = 1 − c1 · v c2 c1 n2 1/(n−1) . p2 = 1 − c2 · v c1 c2 n1 1/(n−1) . Similarly, For these two numbers to be probabilities, we need each of them to be nonnegative and at most one, which requires c2n2 v !1/(n2 −1) 1/n1 < c1 < vc2n1 −1 . Chapter 4. Mixed Strategy Equilibrium 29 136.1 Best response dynamics in Cournot’s duopoly game The best response functions of both firms are the same, so if the firms’ outputs are initially the same, they are the same in every period: q1t = q2t for every t. For each period t, we thus have qti = 21 (α − c − qti ). Given that q1i = 0 for i = 1, 2, solving this first-order difference equation we have qti = 31 (α − c)[1 − (− 21 )t−1 ] for each period t. When t is large, qti is close to 13 (α − c), a firm’s equilibrium output. 5 In the first few periods, these outputs are 0, 12 (α − c), 14 (α − c), 38 (α − c), 16 (α − c ). 139.1 Finding all mixed strategy equilibria of two-player games Left game: • There is no equilibrium in which each player’s mixed strategy assigns positive probability to a single action (i.e. there is no pure equilibrium). • Consider the possibility of an equilibrium in which one player assigns probability 1 to a single action while the other player assigns positive probability to both her actions. For neither action of player 1 is player 2’s payoff the same for both her actions, and for neither action of player 2 is player 1’s payoff the same for both her actions, so there is no mixed strategy equilibrium of this type. • Consider the possibility of a mixed strategy equilibrium in which each player assigns positive probability to both her actions. Denote by p the probability player 1 assigns to T and by q the probability player 2 assigns to L. For player 1’s expected payoff to her two actions to be the same we need 6q = 3q + 6(1 − q), or q = 23 . For player 2’s expected payoff to her two actions to be the same we need 2(1 − p) = 6p, or p = 14 . We conclude that the game has a unique mixed strategy equilibrium, (( 14 , 43 ), ( 23 , 13 )). Right game: • By inspection, ( T, R) and ( B, L) are the pure strategy equilibria. 30 Chapter 4. Mixed Strategy Equilibrium • Consider the possibility of a mixed strategy equilibrium in which one player assigns probability 1 to a single action while the other player assigns positive probability to both her actions. ◦ { T } for player 1, { L, R} for player 2: no equilibrium, because player 2’s payoffs to ( T, L) and ( T, R) are not the same. ◦ { B} for player 1, { L, R} for player 2: no equilibrium, because player 2’s payoffs to ( B, L) and ( B, R) are not the same. ◦ { T, B} for player 1, { L} for player 2: no equilibrium, because player 1’s payoffs to ( T, L) and ( B, L) are not the same. ◦ { T, B} for player 1, { R} for player 2: player 1’s payoffs to ( T, R) and ( B, R) are the same, so there is an equilibrium in which player 1 uses T with probability p if player 2’s expected payoff to R, which is 2p + 1 − p, is at least her expected payoff to L, which is p + 2(1 − p). That is, the game has equilibria in which player 1’s mixed strategy is ( p, 1 − p), with p ≥ 12 , and player 2 uses R with probability 1. • Consider the possibility of an equilibrium in which both players assign positive probability to both their actions. Denote by q the probability that player 2 assigns to L. For player 1’s expected payoffs to T and B to be the same we need 0 = 2q, or q = 0, so there is no equilibrium in which both players assign positive probability to both their actions. In summary, the mixed strategy equilibria of the game are ((0, 1), (1, 0)) (i.e. the pure equilibrium ( B, L)) and (( p, 1 − p), (0, 1)) for 12 ≤ p ≤ 1 (of which one equilibrium is the pure equilibrium ( T, R)). 145.1 All-pay auction with many bidders Denote the common mixed strategy by F. Look for an equilibrium in which the largest value of z for which F (z) = 0 is 0 and the smallest value of z for which F (z) = 1 is z = K. A player who bids ai wins if and only if the other n − 1 players all bid less than she does, an event with probability ( F ( ai ))n−1 . Thus, given that the probability that she ties for the highest bid is zero, her expected payoff is (K − ai )( F ( ai ))n−1 + (− ai )(1 − ( F ( ai ))n−1 ). Given the form of F, for an equilibrium this expected payoff must be constant for all values of ai with 0 ≤ ai ≤ K. That is, for some value of c we have K ( F ( ai ))n−1 − ai = c for all 0 ≤ ai ≤ K. For F (0) = 0 we need c = 0, so that F ( ai ) = ( ai /K )1/(n−1) is the only candidate for an equilibrium strategy. Chapter 4. Mixed Strategy Equilibrium 31 The function F is a cumulative probability distribution on the interval from 0 to K because F (0) = 0, F (K ) = 1, and F is increasing. Thus F is indeed an equilibrium strategy. We conclude that the game has a mixed strategy Nash equilibrium in which each player randomizes over all her actions according to the probability distribution F ( ai ) = ( ai /K )1/(n−1); each player’s equilibrium expected payoff is 0. Each player’s mean bid is K/n. 147.2 Preferences over lotteries The first piece of information about the decision-maker’s preferences among lotteries is consistent with her preferences being represented by the expected value of a payoff function: set u( a1 ) = 0, u( a2 ) equal to any number between 21 and 14 , and u( a3 ) = 1. The second piece of information about the decision-maker’s preferences is not consistent with these preferences being represented by the expected value of a payoff function, by the following argument. For consistency with the information about the decision-maker’s preferences among the four lotteries, we need 0.4u( a1 ) + 0.6u( a3 ) > 0.5u( a2 ) + 0.5u( a3 ) > 0.3u( a1 ) + 0.2u( a2 ) + 0.5u( a3 ) > 0.45u( a1 ) + 0.55u( a3 ). The first inequality implies u( a2 ) < 0.8u( a1 ) + 0.2u( a3 ) and the last inequality implies u( a2 ) > 0.75u( a1 ) + 0.25u( a3 ). Because u( a1 ) < u( a3 ), we have 0.75u( a1 ) + 0.25u( a3 ) > 0.8u( a1 ) + 0.2u( a3 ), so that the two inequalities are incompatible. 149.2 Normalized vNM payoff functions Let a be the best outcome according to her preferences and let a be the worse outcome. Let η = −u( a)/(u( a) − u( a)) and θ = 1/(u( a) − u( a)) > 0. Lemma 148.1 implies that the function v defined by v( x ) = η + θu( x ) represents the same preferences as does u; we have v( a) = 0 and v( a) = 1. 5 Extensive Games with Perfect Information: Theory 163.1 Nash equilibria of extensive games The strategic form of the game in Exercise 156.2a is given in Figure 33.1. C D EG 1, 0 2, 3 EH 1, 0 0, 1 FG 3, 2 2, 3 FH 3, 2 0, 1 Figure 33.1 The strategic form of the game in Exercise 156.2a. The Nash equilibria of the game are (C, FG), (C, FH), and ( D, EG). The strategic form of the game in Figure 160.1 is given in Figure 33.2. CG CH DG DH E 1, 2 0, 0 2, 0 2, 0 F 3, 1 3, 1 2, 0 2, 0 Figure 33.2 The strategic form of the game in Figure 160.1. The Nash equilibria of the game are (CH, F ), (DG, E), and (DH, E). 164.2 Subgames The subgames of the game in Exercise 156.2c are the whole game and the six games in Figure 34.1. 168.1 Checking for subgame perfect equilibria The Nash equilibria (CH, F ) and (DH, E) are not subgame perfect equilibria: in the subgame following the history (C, E), player 1’s strategies CH and DH induce the strategy H, which is not optimal. The Nash equilibrium (DG, E) is a subgame perfect equilibrium: (a) it is a Nash equilibrium, so player 1’s strategy is optimal at the start of the game, given player 2’s strategy, (b) in the subgame following the history C, player 2’s strategy E induces the strategy E, which is optimal given player 1’s strategy, and (c) in the subgame following the history (C, E), player 1’s strategy DG induces the strategy G, which is optimal. 33 34 Chapter 5. Extensive Games with Perfect Information: Theory R B E H E E B H 0, 0, 0 1, 2, 1 B E B 1, 2, 1 2, 1, 2 0, 0, 0 B 0, 0, 0 H B R B 0, 0, 0 1, 2, 1 E H H R H 0, 0, 0 B 0, 0, 0 R H 2, 1, 2 B 1, 2, 1 H 2, 1, 2 R H 0, 0, 0 B 0, 0, 0 H 2, 1, 2 Figure 34.1 The proper subgames of the game in Exercise 156.2c. 174.1 Sharing heterogeneous objects Let n = 2 and k = 3, and call the objects a, b, and c. Suppose that the values person 1 attaches to the objects are 3, 2, and 1 respectively, while the values player 2 attaches are 1, 3, 2. If player 1 chooses a on the first round, then in any subgame perfect equilibrium player 2 chooses b, leaving player 1 with c on the second round. If instead player 1 chooses b on the first round, in any subgame perfect equilibrium player 2 chooses c, leaving player 1 with a on the second round. Thus in every subgame perfect equilibrium player 1 chooses b on the first round (though she values a more highly.) Now I argue that for any preferences of the players, G (2, 3) has a subgame perfect equilibrium of the type described in the exercise. For any object chosen by player 1 in round 1, in any subgame perfect equilibrium player 2 chooses her favorite among the two objects remaining in round 2. Thus player 2 never obtains the object she least prefers; in any subgame perfect equilibrium, player 1 obtains that object. Player 1 can ensure she obtains her more preferred object of the two remaining by choosing that object on the first round. That is, there is a subgame perfect equilibrium in which on the first round player 1 chooses her more preferred object out of the set of objects excluding the object player 2 least prefers, and on the last round she obtains x3 . In this equilibrium, player 2 obtains the object less preferred by player 1 out of the set of objects excluding the object player 2 least prefers. That is, player 2 obtains x2 . (Depending on the players’ preferences, the game also may have a subgame perfect equilibrium in which player 1 chooses x3 on the first round.) 177.3 Comparing simultaneous and sequential games a. Denote by ( a1∗ , a2∗ ) a Nash equilibrium of the strategic game in which player 1’s payoff is maximal in the set of Nash equilibria. Because ( a1∗ , a2∗ ) is a Nash equilibrium, a2∗ is a best response to a1∗ . By assumption, it is the only best Chapter 5. Extensive Games with Perfect Information: Theory 35 response to a1∗ . Thus if player 1 chooses a1∗ in the extensive game, player 2 must choose a2∗ in any subgame perfect equilibrium of the extensive game. That is, by choosing a1∗ , player 1 is assured of a payoff of at least u1 ( a1∗ , a2∗ ). Thus in any subgame perfect equilibrium player 1’s payoff must be at least u1 ( a1∗ , a2∗ ). b. Suppose that A1 = { T, B}, A2 = { L, R}, and the payoffs are those given in Figure 35.1. The strategic game has a unique Nash equilibrium, ( T, L), in which player 2’s payoff is 1. The extensive game has a unique subgame perfect equilibrium, ( B, LR) (where the first component of player 2’s strategy is her action after the history T and the second component is her action after the history B). In this subgame perfect equilibrium player 2’s payoff is 2. T B L 1, 1 0, 0 R 3, 0 2, 2 Figure 35.1 The payoffs for the example in Exercise 177.3b. c. Suppose that A1 = { T, B}, A2 = { L, R}, and the payoffs are those given in Figure 35.2. The strategic game has a unique Nash equilibrium, ( T, L), in which player 2’s payoff is 2. A subgame perfect equilibrium of the extensive game is ( B, RL) (where the first component of player 2’s strategy is her action after the history T and the second component is her action after the history B). In this subgame perfect equilibrium player 1’s payoff is 1. (If you read Chapter 4, you can find the mixed strategy Nash equilibria of the strategic game; in all these equilibria, as in the pure strategy Nash equilibrium, player 1’s expected payoff exceeds 1.) T B L 2, 2 1, 1 R 0, 2 3, 0 Figure 35.2 The payoffs for the example in Exercise 177.3c. 179.3 Three Men’s Morris, or Mill Number the squares 1 through 9, starting at the top left, working across each row. The following strategy of player 1 guarantees she wins, so that the subgame perfect equilibrium outcome is that she wins. First player 1 chooses the central square (5). • Suppose player 2 then chooses a corner; take it to be square 1. Then player 1 chooses square 6. Now player 2 must choose square 4 to avoid defeat; player 1 must choose square 7 to avoid defeat; and then player 2 must choose square 36 Chapter 5. Extensive Games with Perfect Information: Theory 3 to avoid defeat (otherwise player 1 can move from square 6 to square 3 on her next turn). If player 1 now moves from square 6 to square 9, then whatever player 2 does she can subsequently move her counter from square 5 to square 8 and win. • Suppose player 2 then chooses a noncorner; take it to be square 2. Then player 1 chooses square 7. Now player 2 must choose square 3 to avoid defeat; player 1 must choose square 1 to avoid defeat; and then player 2 must choose square 4 to avoid defeat (otherwise player 1 can move from square 5 to square 4 on her next turn). If player 1 now moves from square 7 to square 8, then whatever player 2 does she can subsequently move from square 8 to square 9 and win. 6 Extensive Games with Perfect Information: Illustrations 183.1 Nash equilibria of the ultimatum game For every amount x there are Nash equilibria in which person 1 offers x. For example, for any value of x there is a Nash equilibrium in which person 1’s strategy is to offer x and person 2’s strategy is to accept x and any offer more favorable, and reject every other offer. (Given person 2’s strategy, person 1 can do no better than offer x. Given person 1’s strategy, person 2 should accept x; whether person 2 accepts or rejects any other offer makes no difference to her payoff, so that rejecting all less favorable offers is, in particular, optimal.) 183.2 Subgame perfect equilibria of the ultimatum game with indivisible units In this case each player has finitely many actions, and for both possible subgame perfect equilibrium strategies of player 2 there is an optimal strategy for player 1. If player 2 accepts all offers then player 1’s best strategy is to offer 0, as before. If player 2 accepts all offers except 0 then player 1’s best strategy is to offer one cent (which player 2 accepts). Thus the game has two subgame perfect equilibria: one in which player 1 offers 0 and player 2 accepts all offers, and one in which player 1 offers one cent and player 2 accepts all offers except 0. 186.1 Holdup game The game is defined as follows. Players Two people, person 1 and person 2. Terminal histories The set of all sequences (low, x, Z ), where x is a number with 0 ≤ x ≤ c L (the amount of money that person 1 offers to person 2 when the pie is small), and (high, x, Z ), where x is a number with 0 ≤ x ≤ c H (the amount of money that person 1 offers to person 2 when the pie is large) and Z is either Y (“yes, I accept”) or N (“no, I reject”). Player function P(∅) = 2, P(low) = P(high) = 1, and P(low, x ) = P(high, x ) = 2 for all x. Preferences Person 1’s preferences are represented by payoffs equal to the amounts of money she receives, equal to c L − x for any terminal history (low, x, Y ) with 0 ≤ x ≤ c L , equal to c H − x for any terminal history 37 38 Chapter 6. Extensive Games with Perfect Information: Illustrations (high, x, Y ) with 0 ≤ x ≤ c H , and equal to 0 for any terminal history (low, x, N ) with 0 ≤ x ≤ c L and for any terminal history (high, x, N ) with 0 ≤ x ≤ c H . Person 2’s preferences are represented by payoffs equal to x − L for the terminal history (low, x, Y ), x − H for the terminal history (high, x, Y ), − L for the terminal history (low, x, N ), and − H for the terminal history (high, x, N ). 189.1 Stackelberg’s duopoly game with quadratic costs From Exercise 59.1, the best response function of firm 2 is the function b2 defined by 1 (α − q1 ) if q1 ≤ α b2 (q1 ) = 4 0 if q1 > α. Firm 1’s subgame perfect equilibrium strategy is the value of q1 that maximizes q1 (α − q1 − b2 (q1 )) − q21 , or q1 (α − q1 − 41 (α − q1 )) − q21 , or 41 q1 (3α − 7q1 ). The 3 maximizer is q1 = 14 α. We conclude that the game has a unique subgame perfect equilibrium, in which 3 firm 1’s strategy is the output 14 α and firm 2’s strategy is its best response function b2 . The outcome of the subgame perfect equilibrium is that firm 1 produces q1∗ = 3 3 11 ∗ 14 α units of output and firm 2 produces q2 = b2 ( 14 α ) = 56 α units. In a Nash equilibrium of Cournot’s (simultaneous-move) game each firm produces 51 α (see Exercise 59.1). Thus firm 1 produces more in the subgame perfect equilibrium of the sequential game than it does in the Nash equilibrium of Cournot’s game, and firm 2 produces less. 196.4 Sequential positioning by three political candidates The following extensive game models the situation. Players The candidates. Terminal histories The set of all sequences ( x1 , . . . , xn ), where xi is either Out or a position of candidate i (a number) for i = 1, . . . , n. Player function P(∅) = 1, P( x1 ) = 2 for all x1 , P( x1, x2 ) = 3 for all ( x1 , x2 ), . . . , P( x1, . . . , xn−1 ) = n for all ( x1 , . . . , xn−1 ). Preferences Each candidate’s preferences are represented by a payoff function that assigns n to every terminal history in which she wins, k to every terminal history in which she ties for first place with n − k other candidates, for 1 ≤ k ≤ n − 1, 0 to every terminal history in which she stays out, and −1 to every terminal history in which she loses, where positions attract votes as in Hotelling’s model of electoral competition (Section 3.3). Chapter 6. Extensive Games with Perfect Information: Illustrations 39 When there are two candidates the analysis of the subgame perfect equilibria is similar to that in the previous exercise. In every subgame perfect equilibrium candidate 1’s strategy is m; candidate 2’s strategy chooses m after the history m, some position between x1 and 2m − x1 after the history x1 for any position x1 , and any position after the history Out. Now consider the case of three candidates when the voters’ favorite positions are distributed uniformly from 0 to 1. I claim that every subgame perfect equilibrium results in the first candidate’s entering at 12 , the second candidate’s staying out, and the third candidate’s entering at 12 . To show this, first consider the best response of candidate 3 to each possible pair of actions of candidates 1 and 2. Figure 39.1 illustrates these optimal actions in every case that candidate 1 enters. (If candidate 1 does not enter then the subgame is exactly the two-candidate game.) 1 tie 3 an d 1 wins 1, 1 2 2 wins d an 2 3 In (e.g. at 12 ) 3 wins ins 1w 2, ↑ x2 In (e.g. at z) 3 wins tie 2 wi ns In (near 12 ); 3 wins 2 wi ns 1 d an tie tie 1 wins 3 ns 1 wi 1, 2, 3 wins 2 wins an d In (e.g. at 12 ) 2 1 3 0 Out 1 3 In; 3 wins 1 2 In; 1 and 3 tie In (e.g. at z) 3 wins 2 3 x1 → 1 In; 3 wins Figure 39.1 The outcome of a best response of candidate 3 to each pair of actions by candidates 1 and 2. The best response for any point in the gray shaded area (including the black boundaries of this area, but excluding the other boundaries) is Out. The outcome at each of the four small disks at the outer corners of the shaded area is that all three candidates tie. The value of z is 1 − 12 ( x1 + x2 ). 40 Chapter 6. Extensive Games with Perfect Information: Illustrations Now consider the optimal action of candidate 2, given x1 and the outcome of candidate 3’s best response, as given in Figure 39.1. In the figure, take a value of x1 and look at the outcomes as x2 varies; find the value of x2 that induces the best outcome for candidate 2. For example, for x1 = 0 the only value of x2 for which candidate 2 does not lose is 32 , at which point she ties with the other two candidates. Thus when candidate 1’s strategy is x1 = 0, candidate 2’s best action, given candidate 3’s best response, is x2 = 32 , which leads to a three-way tie. We find that the outcome of the optimal value of x2 , for each value of x1 , is given as follows. 1, 2, and 3 tie ( x2 = 32 ) if x1 = 0 if 0 < x1 < 21 2 wins 1 and 3 tie (2 stays out) if x1 = 21 if 12 < x1 < 1 2 wins 1 1, 2, and 3 tie ( x2 = 3 ) if x1 = 1. Finally, consider candidate 1’s best strategy, given the responses of candidates 2 and 3. If she stays out then candidates 2 and 3 enter at m and tie. If she enters then the best position at which to do so is x1 = 21 , where she ties with candidate 3. (For every other position she either loses or ties with both of the other candidates.) We conclude that in every subgame perfect equilibrium the outcome is that candidate 1 enters at 12 , candidate 2 stays out, and candidate 3 enters at 12 . (There are many subgame perfect equilibria, because after many histories candidate 3’s optimal action is not unique.) (The case in which there are many potential candidates, is discussed on the page http://www.economics.utoronto.ca/osborne/research/CONJECT.HTM.) 198.1 The race G1 (2, 2) The consequences of player 1’s actions at the start of the game are as follows. Take two steps: Player 1 wins. Take one step: Go to the game G2 (1, 2), in which player 2 initially takes two steps and wins. Do not move: If player 2 does not move, the game ends. If she takes one step we go to the game G1 (2, 1), in which player 1 takes two steps and wins. If she takes two steps, she wins. Thus in a subgame perfect equilibrium player 2 takes two steps, and wins. We conclude that in a subgame perfect equilibrium of G1 (2, 2) player 1 initially takes two steps, and wins. 203.1 A race with a liquidity constraint In the absence of the constraint, player 1 initially takes one step. Suppose she does so in the game with the constraint. Consider player 2’s options after player 1’s move. Chapter 6. Extensive Games with Perfect Information: Illustrations 41 Player 2 takes two steps: Because of the liquidity constraint, player 1 can take at most one step. If she takes one step, player 2’s optimal action is to take one step, and win. Thus player 1’s best action is not to move; player 2’s payoff exceeds 1 (her steps cost 5, and the prize is worth more than 6). Player 2 moves one step: Again because of the liquidity constraint, player 1 can take at most one step. If she takes one step, player 2 can take two steps and win, obtaining a payoff of more than 1 (as in the previous case). Player 2 does not move: Player 1, as before, can take one step on each turn, and win; player 2’s payoff is 0. We conclude that after player 1 moves one step, player 2 should take either one or two steps, and ultimately win; player 1’s payoff is −1. A better option for player 1 is not to move, in which case player 2 can move one step at a time, and win; player 1’s payoff is zero. Thus the subgame perfect equilibrium outcome is that player 1 does not move, and player 2 takes one step at a time and wins. 7 Extensive Games with Perfect Information: Extensions and Discussion 210.2 Extensive game with simultaneous moves The game is shown in Figure 43.1. 1 A C D C 4, 2 0, 0 D 0, 0 2, 4 B E F E 3, 1 0, 0 F 0, 0 1, 3 Figure 43.1 The game in Exercise 210.2. The subgame following player 1’s choice of A has two Nash equilibria, (C, C ) and ( D, D ); the subgame following player 1’s choice of B also has two Nash equilibria, ( E, E) and ( F, F ). If the equilibrium reached after player 1 chooses A is (C, C ), then regardless of the equilibrium reached after she chooses ( E, E), she chooses A at the beginning of the game. If the equilibrium reached after player 1 chooses A is ( D, D ) and the equilibrium reached after she chooses B is ( F, F ), she chooses A at the beginning of the game. If the equilibrium reached after player 1 chooses A is ( D, D ) and the equilibrium reached after she chooses B is ( E, E), she chooses B at the beginning of the game. Thus the game has four subgame perfect equilibria: ( ACE, CE), ( ACF, CF ), ( ADF, DF ), and ( BDE, DE) (where the first component of player 1’s strategy is her choice at the start of the game, the second component is her action after she chooses A, and the third component is her action after she chooses B, and the first component of player 2’s strategy is her action after player 1 chooses A at the start of the game and the second component is her action after player 1 chooses B at the start of the game). In the first two equilibria the outcome is that player 1 chooses A and then both players choose C, in the third equilibrium the outcome is that player 1 chooses A and then both players choose D, and in the last equilibrium the outcome is that player 1 chooses B and then both players choose E. 217.1 Electoral competition with strategic voters I first argue that in any equilibrium each candidate that enters is in the set of winners. If some candidate that enters is not a winner, she can increase her payoff by deviating to Out. 43 44 Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion Now consider the voting subgame in which there are more than two candidates and not all candidates’ positions are the same. Suppose that the citizens’ votes are equally divided among the candidates. I argue that this list of citizens’ strategies is not a Nash equilibrium of the voting subgame. For either the citizen whose favorite position is 0 or the citizen whose favorite position is 1 (or both), at least two candidates’ positions are better than the position of the candidate furthest from the citizen’s favorite position. Denote a citizen for whom this condition holds by i. (The claim that citizen i exists is immediate if the candidates occupy at least three distinct positions, or they occupy two distinct positions and at least two candidates occupy each position. If the candidates occupy only two positions and one position is occupied by a single candidate, then take the citizen whose favorite position is 0 if the lone candidate’s position exceeds the other candidates’ position; otherwise take the citizen whose favorite position is 1.) Now, given that each candidate obtains the same number of votes, if citizen i switches her vote to one of the candidates whose position is better for her than that of the candidate whose position is furthest from her favorite position, then this candidate wins outright. (If citizen i originally votes for one of these superior candidates, she can switch her vote to the other superior candidate; if she originally votes for neither of the superior candidates, she can switch her vote to either one of them.) Citizen i’s payoff increases when she thus switches her vote, so that the list of citizens’ strategies is not a Nash equilibrium of the voting subgame. We conclude that in every Nash equilibrium of every voting subgame in which there are more than two candidates and not all candidates’ positions are the same at least one candidate loses. Because no candidate loses in a subgame perfect equilibrium (by the first argument in the proof), in any subgame perfect equilibrium either only two candidates enter, or all candidates’ positions are the same. If only two candidates enter, then by the argument in the text for the case n = 2, each candidate’s position is m (the median of the citizens’ favorite positions). Now suppose that more than two candidates enter, and their common position is not equal to m. If a candidate deviates to m then in the resulting voting subgame only two positions are occupied, so that for every citizen, any strategy that is not weakly dominated votes for a candidate at the position closest to her favorite position. Thus a candidate who deviates to m wins outright. We conclude that in any subgame perfect equilibrium in which more than two candidates enter, they all choose the position m. 220.1 Top cycle set a. The top cycle set is the set { x, y, z} of all three alternatives because x beats y beats z beats x. b. The top cycle set is the set {w, x, y, z} of all four alternatives. As in the previous case, x beats y beats z beats x; also y beats w. Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion 45 224.1 Exit from a declining industry Period t1 is the largest value of t for which Pt (k1 ) ≥ c, or 60 − t ≥ 10. Thus t1 = 50. Similarly, t2 = 70. If both firms are active in period t1 , then firm 2’s profit in this period is (100 − t1 − c − k1 − k2 )k2 = (−20)(20) = −400. Its profit in any period t in which it is alone in the market is (100 − t − c − k2 )k2 = (70 − t)(20). Thus its profit from period t1 + 1 through period t2 is (19 + 18 + . . . + 1)(20) = 3800. Hence firm 2’s loss in period t1 when both firms are active is (much) less than the sum of its profits in periods t1 + 1 through t2 when it alone is active. 227.1 Variant of ultimatum game with equity-conscious players The game is defined as follows. Players The two people. Terminal histories The set of sequences ( x, β 2 , Z ), where x is a number with 0 ≤ x ≤ c (the amount of money that person 1 offers to person 2), β 2 is 0 or 1 (the value of β 2 selected by chance), and Z is either Y (“yes, I accept”) or N (“no, I reject”). Player function P(∅) = 1, P( x ) = c for all x, and P( x, β 2 ) = 2 for all x and all β2. Chance probabilities For every history x, chance chooses 0 with probability p and 1 with probability 1 − p. Preferences Each person’s preferences are represented by the expected value of a payoff equal to the amount of money she receives. For any terminal history ( x, β 2, Y ) person 1 receives c − x and person 2 receives x; for any terminal history ( x, β 2 , N ) each person receives 0. Given the result from Exercise 183.4 given in the question, if person 1’s offer x satisfies 0 < x < 13 then the offer is rejected with probability 1 − p, so that person 1’s expected payoff is p(1 − x ), while if x > 31 the offer is certainly accepted, independent of the type of person 2. Thus person 1’s optimal offer is 1 2 3 if p < 3 0 if p > 23 ; if p = 32 then both offers are optimal. If p > 23 we see that in a subgame perfect equilibrium person 1’s offers are rejected by every person 2 with whom she is matched for whom β 2 = 1 (that is, with probability 1 − p). 46 Chapter 7. Extensive Games with Perfect Information: Extensions and Discussion 230.1 Nash equilibria when players may make mistakes The players’ best response functions are indicated in Figure 46.1. We see that the game has two Nash equilibria, ( A, A, A) and ( B, A, A). A B A 1∗ , 1∗ , 1∗ 1∗ , 1∗ , 1∗ B 0, 0, 1∗ 1∗ , 0, 1∗ A B A 0, 1∗ , 0 1∗ , 1∗ , 0 A B 1∗ , 0, 0 0, 0, 0 B Figure 46.1 The player’s best response functions in the game in Exercise 230.1. The action A is not weakly dominated for any player. For player 1, A is better than B if players 2 and 3 both choose B; for players 2 and 3, A is better than B for all actions of the other players. If players 2 and 3 choose A in the modified game, player 1’s expected payoffs to A and B are A: (1 − p2 )(1 − p3 ) + p1 p2 (1 − p3 ) + p1 (1 − p2 ) p3 + (1 − p1 ) p2 p3 B: (1 − p2 )(1 − p3 ) + (1 − p1 ) p2 (1 − p3 ) + (1 − p1 )(1 − p2 ) p3 + p1 p2 p3 . The difference between the expected payoff to B and the expected payoff to A is (1 − 2p1 )[ p2 + p3 − 3p2 p3 ]. If 0 < pi < 21 for i = 1, 2, 3, this difference is positive, so that ( A, A, A) is not a Nash equilibrium of the modified game. 233.1 Nash equilibria of the chain-store game Any terminal history in which the event in each period is either Out or (In, A) is the outcome of a Nash equilibrium. In any period in which challenger chooses Out, the strategy of the chain-store specifies that it choose F in the event that the challenger chooses In. 8 Coalitional Games and the Core 245.1 Three-player majority game Let ( x1 , x2 , x3 ) be an action of the grand coalition. Every coalition consisting of two players can obtain one unit of output, so for ( x1 , x2 , x3 ) to be in the core we need x1 + x2 ≥ 1 x1 + x3 ≥ 1 x2 + x3 ≥ 1 x1 + x2 + x3 = 1. Adding the first three conditions we conclude that 2x1 + 2x2 + 2x3 ≥ 3, or x1 + x2 + x3 ≥ 23 , contradicting the last condition. Thus no action of the grand coalition satisfies all the conditions, so that the core of the game is empty. 248.1 Core of landowner–worker game Let a N be an action of the grand coalition in which the output received by each worker is at most f (n) − f (n − 1). No coalition consisting solely of workers can obtain any output, so no such coalition can improve upon a N . Let S be a coalition of the landowner and k − 1 workers. The total output received by the members of S in a N is at least f (n) − (n − k)( f (n) − f (n − 1)) (because the total output is f (n), and every other worker receives at most f (n) − f (n − 1)). Now, the output that S can obtain is f (k), so for S to improve upon a N we need f (k) > f (n) − (n − k)( f (n) − f (n − 1)), which contradicts the inequality given in the exercise. 249.1 Unionized workers in landowner–worker game The following game models the situation. Players The landowner and the workers. 47 48 Chapter 8. Coalitional Games and the Core Actions The set of actions of the grand coalition is the set of all allocations of the output f (n). Every other coalition has a single action, which yields the output 0. Preferences Each player’s preferences are represented by the amount of output she obtains. The core of this game consists of every allocation of the output f (n) among the players. The grand coalition cannot improve upon any allocation x because for every other allocation x 0 there is at least one player whose payoff is lower in x 0 than it is in x. No other coalition can improve upon any allocation because no other coalition can obtain any output. 249.2 Landowner–worker game with increasing marginal products We need to show that no coalition can improve upon the action a N of the grand coalition in which every player receives the output f (n)/n. No coalition of workers can obtain any output, so we need to consider only coalitions containing the landowner. Consider a coalition consisting of the landowner and k workers, which can obtain f (k + 1) units of output by itself. Under a N this coalition obtains the output (k + 1) f (n)/n, and we have f (k + 1)/(k + 1) < f (n)/n because k < n. Thus no coalition can improve upon a N . 254.1 Range of prices in horse market The equality of the number of owners who sell their horses and the number of nonowners who buy horses implies that the common trading price p∗ • is not less than σk∗ , otherwise at most k∗ − 1 owners’ valuations would be less than p∗ and at least k∗ nonowners’ valuations would be greater than p∗ , so that the number of buyers would exceed the number of sellers • is not less than β k∗ +1 , otherwise at most k∗ owners’ valuations would be less than p∗ and at least k∗ + 1 nonowners’ valuations would be greater than p∗ , so that the number of buyers would exceed the number of sellers • is not greater than β k∗ , otherwise at least k∗ owners’ valuations would be less than p∗ and at most k∗ − 1 nonowners’ valuations would be greater than p∗ , so that the number of sellers would exceed the number of buyers • is not greater than σk∗ +1 , otherwise at least k∗ + 1 owners’ valuations would be less than p∗ and at most k∗ nonowners’ valuations would be greater than p∗ , so that the number of sellers would exceed the number of buyers. That is, p∗ ≥ max{σk∗ , β k∗ +1 } and p∗ ≤ min{ β k∗ , σk∗ +1 }. Chapter 8. Coalitional Games and the Core 49 258.1 House assignment with identical preferences Because the players rank the houses in the same way, we can refer to the “best house”, the “second best house”, and so on. In any assignment in the core, the player who owns the best house is assigned this house (because she has the option of keeping it). Among the remaining players, the one who owns the second best house must be assigned this house (again, because she has the option of keeping it). Continuing to argue in the same way, we see that there is a single assignment in the core, in which every player is assigned the house she owns initially. 261.1 Median voter theorem Denote the median favorite position by m. If x < m then every player whose favorite position is m or greater—a majority of the players—prefers m to x. Similarly, if x > m then every player whose favorite position is m or less—a majority of the players—prefers m to x. 267.2 Empty core in roommate problem Notice that ` is at the bottom of each of the other players’ preferences. Suppose that she is matched with i. Then j and k are matched, and {i, k} can improve upon the matching. Similarly, if ` is matched with j then {i, j} can improve upon the matching, and if ` is matched with k then { j, k} can improve upon the matching. Thus the core is empty (` has to be matched with someone!). 9 Bayesian Games 276.1 Equilibria of a variant of BoS with imperfect information If player 1 chooses S then type 1 of player 2 chooses S and type 2 chooses B. But if the two types of player 2 make these choices then player 1 is better off choosing B (which yields her an expected payoff of 1) than choosing S (which yields her an expected payoff of 12 ). Thus there is no Nash equilibrium in which player 1 chooses S. Now consider the mixed strategy Nash equilibria. If both types of player 2 use a pure strategy then player 1’s two actions yield her different payoffs. Thus there is no equilibrium in which both types of player 2 use pure strategies and player 1 randomizes. Now consider an equilibrium in which type 1 of player 2 randomizes. Denote by p the probability that player 1’s mixed strategy assigns to B. In order for type 1 of player 2 to obtain the same expected payoff to B and S we need p = 32 . For this value of p the best action of type 2 of player 2 is S. Denote by q the probability that type 1 of player 2 assigns to B. Given these strategies for the two types of player 2, player 1’s expected payoff if she chooses B is 1 2 · 2q = q and her expected payoff if she chooses S is 1 2 · (1 − q) + 21 · 1 = 1 − 21 q. These expected payoffs are equal if and only if q = 32 . Thus the game has a mixed strategy equilibrium in which the mixed strategy of player 1 is ( 32 , 13 ), that of type 1 of player 2 is ( 23 , 13 ), and that of type 2 of player 2 is (0, 1) (that is, type 2 of player 2 uses the pure strategy that assigns probability 1 to S). Similarly the game has a mixed strategy equilibrium in which the strategy of player 1 is ( 13 , 23 ), that of type 1 of player 2 is (0, 1), and that of type 2 of player 2 is ( 23 , 13 ). For no mixed strategy of player 1 are both types of player 2 indifferent between their two actions, so there is no equilibrium in which both types randomize. 277.1 Expected payoffs in a variant of BoS with imperfect information The expected payoffs are given in Figure 52.1. 51 52 Chapter 9. Bayesian Games B S ( B, B) ( B, S) (S, B) (S, S) 0 1 1 2 1 1 2 1 2 0 Type n1 of player 1 B ( B, B) 1 S 0 ( B, S) (S, B) 2 3 2 3 1 3 4 3 (S, S) 0 2 Type y2 of player 2 B ( B, B) 0 S 2 ( B, S) (S, B) 1 3 4 3 2 3 2 3 (S, S) 1 0 Type n2 of player 2 Figure 52.1 The expected payoffs of type n1 of player 1 and types y2 and n2 of player 2 in Example 276.2. 282.2 An exchange game The following Bayesian game models the situation. Players The two individuals. States The set of all pairs (s1 , s2 ), where si is the number on player i’s ticket (an integer from 1 to m). Actions The set of actions of each player is {Exchange, Don’t exchange}. Signals The signal function of each player i is defined by τi (s1 , s2 ) = si (each player observes her own ticket, but not that of the other player) Beliefs Type si of player i assigns the probability Prj (s j ) to the state (s1 , s2 ), where j is the other player and Prj (s j ) is the probability with which player j receives a ticket with the prize s j on it. Payoffs Player i’s Bernoulli payoff function is given by ui (( X, Y ), ω ) = ω j if X = Y = Exchange and ui (( X, Y ), ω ) = ωi otherwise. Let Mi be the highest type of player i that chooses Exchange. If Mi > 1 then type 1 of player j optimally chooses Exchange: by exchanging her ticket, she cannot obtain a smaller prize, and may receive a bigger one. Thus if Mi ≥ M j and Mi > 1, type Mi of player i optimally chooses Don’t exchange, because the expected value of the prizes of the types of player j that choose Exchange is less than Mi . Thus in any possible Nash equilibrium Mi = M j = 1: the only prizes that may be exchanged are the smallest. Chapter 9. Bayesian Games 53 287.1 Cournot’s duopoly game with imperfect information We have b1 (q L , q H ) = 1 2 ( α − c − ( θq L 0 + (1 − θ )q H )) if θq L + (1 − θ )q H ≤ α − c otherwise. The best response function of each type of player 2 is similar: b I ( q1 ) = 1 2 (α − c I 0 − q1 ) if q1 ≤ α − c I otherwise for I = L, H. The three equations that define a Nash equilibrium are q1∗ = b1 (q∗L , q∗H ), q∗L = b L (q1∗ ), and q∗H = b H (q1∗ ). Solving these equations under the assumption that they have a solution in which all three outputs are positive, we obtain q1∗ = q∗L q∗H = = 1 3 ( α − 2c + θc L + (1 − θ ) c H ) 1 1 3 ( α − 2c L + c ) − 6 (1 − θ )( c H − c L ) 1 1 3 ( α − 2c H + c ) + 6 θ ( c H − c L ) If both firms know that the unit costs of the two firms are c1 and c2 then in a Nash equilibrium the output of firm i is 31 (α − 2ci + c j ) (see Exercise 58.1). In the case of imperfect information considered here, firm 2’s output is less than 1 1 3 ( α − 2c L + c ) if its cost is c L and is greater than 3 ( α − 2c H + c ) if its cost is c H . Intuitively, the reason is as follows. If firm 1 knew that firm 2’s cost were high then it would produce a relatively large output; if it knew this cost were low then it would produce a relatively small output. Given that it does not know whether the cost is high or low it produces a moderate output, less than it would if it knew firm 2’s cost were high. Thus if firm 2’s cost is in fact high, firm 2 benefits from firm 1’s lack of knowledge and optimally produces more than it would if firm 1 knew its cost. 288.1 Cournot’s duopoly game with imperfect information The best response b0 (q L , q H ) of type 0 of firm 1 is the solution of max[θ ( P(q0 + q L ) − c)q0 + (1 − θ )( P(q0 + q H ) − c)q0 ]. q0 The best response b` (q L , q H ) of type ` of firm 1 is the solution of max( P(q` + q L ) − c)q` q` 54 Chapter 9. Bayesian Games and the best response bh (q L , q H ) of type h of firm 1 is the solution of max( P(qh + q H ) − c)qh . qh The best response b L (q0 , q` , qh ) of type L of firm 2 is the solution of max[(1 − π )( P(q0 + q L ) − c L )q L + π ( P(q` + q L ) − c L )q L ] qL and the best response b H (q0 , q` , qh ) of type H of firm 2 is the solution of max[(1 − π )( P(q0 + q H ) − c H )q H + π ( P(qh + q H ) − c H )q H ]. qH A Nash equilibrium is a profile (q0∗ , q∗` , q∗h , q∗L , q∗H ) for which q0∗ , q∗` , and q∗h are best responses to q∗L and q∗H , and q∗L and q∗H are best responses to q0∗ , q∗` , and q∗h . When P( Q) = α − Q for Q ≤ α and P( Q) = 0 for Q > α we find, after some exciting algebra, that q0∗ = q∗` = q∗H = q∗L = q∗H = 1 (α − 2c + c H − θ (c H − c L )) 3 1 (1 − θ )(1 − π )(c H − c L ) α − 2c + c L + 3 4−π 1 θ (1 − π )(c H − c L ) α − 2c + c H − 3 4−π 1 2(1 − θ )(1 − π )(c H − c L ) α − 2c L + c − 3 4−π 1 2θ (1 − π )(c H − c L ) α − 2c H + c + . 3 4−π When π = 0 we have q0∗ = q∗` = q∗H = q∗L = q∗H = 1 (α − 2c + c H − θ (c H − c L )) 3 1 (1 − θ )(c H − c L ) α − 2c + c L + 3 4 1 θ (c H − c L ) α − 2c + c H − 3 4 1 (1 − θ )(c H − c L ) α − 2c L + c − 3 2 1 θ (c H − c L ) α − 2c H + c + , 3 2 so that q0∗ is equal to the equilibrium output of firm 1 in Exercise 287.1, and q∗L and q∗H are the same as the equilibrium outputs of the two types of firm 2 in that exercise. Chapter 9. Bayesian Games 55 When π = 1 we have q0∗ = q∗` = q∗H = q∗L = q∗H = 1 (α − 2c + c H − θ (c H − c L )) 3 1 (α − 2c + c L ) 3 1 (α − 2c + c H ) 3 1 (α − 2c L + c) 3 1 (α − 2c H + c) , 3 so that q∗` and q∗L are the same as the equilibrium outputs when there is perfect information and the costs are c and c L (see Exercise 58.1), and q∗h and q∗H are the same as the equilibrium outputs when there is perfect information and the costs are c and c H . Now, for an arbitrary value of π we have 1 2(1 − θ )(1 − π )(c H − c L ) ∗ qL = α − 2c L + c − 3 4−π 1 2θ ( 1 − π )( cH − cL) ∗ qH = α − 2c H + c + . 3 4−π To show that for 0 < π < 1 the values of these variables lie between their values when π = 0 and when π = 1, we need to show that 0≤ 2(1 − θ )(1 − π )(c H − c L ) (1 − θ )(c L − c H ) ≤ 4−π 2 and 0≤ 2θ (1 − π )(c H − c L ) θ (c L − c H ) ≤ . 4−π 2 These inequalities follow from c H ≥ c L , θ ≥ 0, and 0 ≤ π ≤ 1. 290.1 Nash equilibria of game of contributing to a public good Any type v j of any player j with v j < c obtains a negative payoff if she contributes and 0 if she does not. Thus she optimally does not contribute. Any type vi ≥ c of player i obtains the payoff vi − c ≥ 0 if she contributes, and the payoff 0 if she does not, so she optimally contributes. Any type v j ≥ c of any player j 6= i obtains the payoff v j − c if she contributes, and the payoff (1 − F (c))v j if she does not. (If she does not contribute, the probability that player i does so is 1 − F (c), the probability that player i’s valuation is at least c.) Thus she optimally does not contribute if (1 − F (c))v j ≥ v j − c, or F (c) ≤ c/v j . This condition must hold for all types of every player j 6= i, so we need F (c) ≤ c/v for the strategy profile to be a Nash equilibrium. 56 Chapter 9. Bayesian Games 294.1 Weak domination in second-price sealed-bid action Fix player i, and choose a bid for every type of every other player. Player i, who does not know the other players’ types, is uncertain of the highest bid of the other players. Denote by b this highest bid. Consider a bid bi of type vi of player i for which bi < vi . The dependence of the payoff of type vi of player i on b is shown in Figure 56.1. b < bi i’s bid bi < v i vi − b vi vi − b Highest of other players’ bids bi = b bi < b < v i (m-way tie) (vi − b)/m 0 vi − b vi − b b ≥ vi 0 0 Figure 56.1 Player i’s payoffs to her bids bi < vi and vi in a second-price sealed-bid auction as a function of the highest of the other player’s bids, denoted b. Player i’s expected payoffs to the bids bi and vi are weighted averages of the payoffs in the columns; each value of b gets the same weight when calculating the expected payoff to bi as it does when calculating the expected payoff to vi . The payoffs in the two rows are the same except when bi ≤ b < vi , in which case vi yields a payoff higher than does bi . Thus the expected payoff to vi is at least as high as the expected payoff to bi , and is greater than the expected payoff to bi unless the other players’ bids lead this range of values of b to get probability 0. Now consider a bid bi of type vi of player i for which bi > vi . The dependence of the payoff of type vi of player i on b is shown in Figure 56.2. b ≤ vi i’s bid vi vi − b bi > v i vi − b Highest of other players’ bids bi = b v i < b < bi (m-way tie) 0 0 vi − b (vi − b)/m b > bi 0 0 Figure 56.2 Player i’s payoffs to her bids vi and bi > vi in a second-price sealed-bid auction as a function of the highest of the other player’s bids, denoted b. As before, player i’s expected payoffs to the bids bi and vi are weighted averages of the payoffs in the columns; each value of b gets the same weight when calculating the expected payoff to vi as it does when calculating the expected payoff to bi . The payoffs in the two rows are the same except when v i < b ≤ bi , in which case vi yields a payoff higher than does bi . (Note that vi − b < 0 for b in this range.) Thus the expected payoff to vi is at least as high as the expected payoff to bi , and is greater than the expected payoff to bi unless the other players’ bids lead this range of values of b to get probability 0. We conclude that for type vi of player i, every bid bi 6= vi is weakly dominated by the bid vi . Chapter 9. Bayesian Games 57 299.1 Asymmetric Nash equilibria of second-price sealed-bid common value auctions Suppose that each type t2 of player 2 bids (1 + 1/λ)t2 and that type t1 of player 1 bids b1 . Then by the calculations in the text, with α = 1 and γ = 1/λ, • a bid of b1 by player 1 wins with probability b1 /(1 + 1/λ) • the expected value of player 2’s bid, given that it is less than b1 , is 21 b1 • the expected value of signals that yield a bid of less than b1 is 21 b1 /(1 + 1/λ) (because of the uniformity of the distribution of t2 ). Thus player 1’s expected payoff if she bids b1 is (t1 + 21 b1 /(1 + 1/λ) − 12 b1 ) · or b1 , 1 + 1/λ λ · (2(1 + λ)t1 − b1 )b1 . 2(1 + λ )2 This function is maximized at b1 = (1 + λ)t1 . That is, if each type t2 of player 2 bids (1 + 1/λ)t2 , any type t1 of player 1 optimally bids (1 + λ)t1 . Symmetrically, if each type t1 of player 1 bids (1 + λ)t1 , any type t2 of player 2 optimally bids (1 + 1/λ)t2 . Hence the game has the claimed Nash equilibrium. 299.2 First-price sealed-bid auction with common valuations Suppose that each type t2 of player 2 bids 21 (α + γ)t2 and type t1 of player 1 bids b1 . To determine the expected payoff of type t1 of player 1, we need to find the probability with which she wins, and the expected value of player 2’s signal if player 1 wins. (The price she pays is her bid, b1 .) Probability of player 1’s winning: Given that player 2’s bidding function is 1 1 2 ( α + γ ) t2 , player 1’s bid of b1 wins only if b1 ≥ 2 ( α + γ ) t2 , or if t2 ≤ 2b1 /( α + γ ). Now, t2 is distributed uniformly from 0 to 1, so the probability that it is at most 2b1 /(α + γ) is 2b1 /(α + γ). Thus a bid of b1 by player 1 wins with probability 2b1 /(α + γ). Expected value of player 2’s signal if player 1 wins: Player 2’s bid, given her signal t2 , is 12 (α + γ)t2, so that the expected value of signals that yield a bid of less than b1 is b1 /(α + γ) (because of the uniformity of the distribution of t2 ). Thus player 1’s expected payoff if she bids b1 is 2(αt1 + γb1 /(α + γ) − b1 ) · or b1 , α+γ 2α ((α + γ)t1 − b1 )b1 . ( α + γ )2 58 Chapter 9. Bayesian Games This function is maximized at b1 = 21 (α + γ)t1 . That is, if each type t2 of player 2 bids 21 (α + γ)t2 , any type t1 of player 1 optimally bids 21 (α + γ)t1 . Hence, as claimed, the game has a Nash equilibrium in which each type ti of player i bids 1 2 (α + γ)ti . 309.2 Properties of the bidding function in a first-price sealed-bid auction We have ∗0 β (v) = 1 − = 1− = ( F (v))n−1 ( F (v))n−1 − (n − 1)( F (v))n−2 F 0 (v) ( F (v))2n−2 R v ( F (v))n − (n − 1) F 0 (v) v ( F ( x ))n−1 dx Rv v ( F ( x ))n−1 dx ( F (v))n ( n − 1) F 0 ( v ) Rv v ( F ( x ))n−1 dx ( F (v))n > 0 if v > v because F 0 (v) > 0 (F is increasing). (TheRfirst line uses the quotient rule for derivav tives and the fact that the derivative of f ( x )dx with respect to v is f (v) for any function f .) If v > v then the integral in (309.1) is positive, so that β∗ (v) < v. If v = v then both the numerator and denominator of the quotient in (309.1) are zero, so we may use L’Hôpital’s rule to find the value of the quotient as v → v. Taking the derivatives of the numerator and denominator we obtain ( F (v))n−1 F(v) , = n − 2 0 ( n − 1) F 0 ( v ) (n − 1)( F (v)) F (v) the numerator of which is zero and the denominator of which is positive. Thus the quotient in (309.1) is zero, and hence β∗ (v) = v. 309.3 Example of Nash equilibrium in a first-price auction From (309.1) we have ∗ Rv x n−1 dx n −1 R v v n −1 x dx = v − 0 n −1 v = v − v/n = (n − 1)v/n. β (v) = v − 0 10 Extensive Games with Imperfect Information 316.1 Variant of card game An extensive game that models the game is shown in Figure 59.1. 0, 0 1, −1 0, 0 Pass Meet 1, −1 HH LH HL ( 14 ) Raise Pass 1, −1 See 0, 0 Raise 2 1 + k, −1 − k −1, 1 1 LL ( 14 ) Meet See ( 14 ) Chance See Meet Raise ( 14 ) 1 1, −1 Pass 2 Raise See −1 − k, 1 + k Pass Meet 1, −1 0, 0 Figure 59.1 An extensive game that models the situation in Exercise 316.1. 318.2 Strategies in variants of card game and entry game Card game: Each player has two information sets, and has two actions at each information set. Thus each player has four strategies: SS, SR, RS, and RR for player 1 (where S stands for See and R for Raise, the first letter of each strategy is player 1’s action if her card is High, and the second letter if her action is her card is Low), and PP, PM, MP, and MM for player 2 (where P stands for Pass and M for Meet). Entry game: The challenger has a single information set (the empty history) and has three actions after this history, so it has three strategies—Ready, Unready, and Out. The incumbent also has a single information set, at which two actions are available, so it has two strategies—Acquiesce and Fight. 59 60 Chapter 10. Extensive Games with Imperfect Information 331.2 Weak sequential equilibrium and Nash equilibrium in subgames Consider the assessment in which the Challenger’s strategy is (Out, R), the Incumbent’s strategy is F, and the Incumbent’s belief assigns probability 1 to the history (In, U ) at her information set. Each player’s strategy is sequentially rational. The Incumbent’s belief satisfies the condition of weak consistency because her information set is not reached when the Challenger follows her strategy. Thus the assessment is a weak sequential equilibrium. The players’ actions in the subgame following the history In do not constitute a Nash equilibrium of the subgame because the Incumbent’s action F is not optimal when the Challenger chooses R. (The Incumbent’s action F is optimal given her belief that the history is (In, U ), as it is in the weak sequential equilibrium. In a Nash equilibrium she acts as if she has a belief that coincides with the Challenger’s action in the subgame.) 340.1 Pooling equilibria of game in which expenditure signals quality We know that in the second period the high-quality firm charges the price H and the low-quality firm charges any nonnegative price, and the consumer buys the good from a high-quality firm, does not buy the good from a low-quality firm that charges a positive price, and may or may not buy from a low-quality firm that charges a price of 0. Consider an assessment in which each type of firm chooses ( p ∗ , E∗ ) in the first period, the consumer believes the firm is high-quality with probability π if it observes ( p∗ , E∗ ) and low quality if it observes any other (price, expenditure) pair, and buys the good if and only if it observes ( p ∗ , E∗ ). The payoff of a high-quality firm under this assessment is p ∗ + H − E∗ − 2c H , that of a low-quality firm is p∗ − E∗ , and that of the consumer is π ( H − p ∗ ) + (1 − π )(− p∗ ) = πH − p∗ . This assessment is consistent—the only first-period action of the firm observed in equilibrium is ( p∗ , E∗ ), and after observing this pair the consumer believes, correctly, that the firm is high-quality with probability π. Under what conditions is the assessment sequentially rational? Firm If the firm chooses a (price, expenditure) pair different from ( p ∗ , E∗ ) then the consumer does not buy the good, and the firm’s profit is 0. Thus for the assessment to be an equilibrium we need p∗ + H − E∗ − 2c H ≥ 0 (for the high-quality firm) and p∗ − E∗ ≥ 0 (for the low-quality firm). Consumer If the consumer does not buy the good after observing ( p ∗ , E∗ ) then its payoff is 0, so for the assessment to be an equilibrium we need πH − p ∗ ≥ 0. In summary, the assessment is a weak sequential equilibrium if and only if max{ E∗ , E∗ − H + 2c H } ≤ p∗ ≤ πH. Chapter 10. Extensive Games with Imperfect Information 61 346.1 Comparing the receiver’s expected payoff in two equilibria The receiver’s payoff as a function of the state t in each equilibrium is shown in Figure 61.1. The area above the black curve is smaller than the area above the gray curve: if you shift the black curve 12 t1 to the left and move the section from 0 to 12 t1 to the interval from 1 − 12 t1 to 1 then the area above the black curve is a subset of the area above the gray curve. 0 1 2 t1 t1 1 2 1 2 ( t1 + 1) −( 12 t1 − t)2 t→ 1 −( 21 (t1 + 1) − t)2 −( 12 − t)2 Figure 61.1 The gray curve gives the receiver’s payoff in each state in the equilibrium in which no information is transferred. The black curve gives her payoff in each state in the two-report equilibrium. 350.1 Variant of model with piecewise linear payoff functions The equilibria of the variant are exactly the same as the equilibria of the original model. 11 Strictly Competitive Games and Maxminimization 363.1 Maxminimizers in a bargaining game If a player demands any amount x up to $5 then her payoff is x regardless of the other player’s action. If she demands $6 then she may get as little as $5 (if the other player demands $5 or $6). If she demands x ≥ $7 then she may get as little as $(11 − x ) (if the other player demands x − 1). For each amount that a player demands, the smallest amount that you may get is given in Figure 63.1. We see that each player’s maxminimizing pure strategies are $5 and $6 (for both of which the worst possible outcome is that the player receives $5). Amount demanded 0 Smallest amount obtained 0 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 5 4 3 2 1 Figure 63.1 The lowest payoffs that a player receives in the game in Exercise 38.2 for each of her possible actions, as the other player’s action varies. 363.3 Finding a maxminimizer The analog of Figure 364.1 in the text is Figure 63.2. From this figure we see that the maxminimizer for player 2 is the strategy that assigns probability 53 to L. Player 2’s maxminimized payoff is − 51 . ↑ payoff of 1 player 1 − 15 2 B 3 5 0 T q→ 1 −1 −2 Figure 63.2 The expected payoff of player 2 in the game in Figure 363.1 for each of player 1’s actions, as a function of the probability q that player 2 assigns to L. 63 64 Chapter 11. Strictly Competitive Games and Maxminimization 366.2 Determining strictly competitiveness Game in Exercise 365.1: Strictly competitive in pure strategies (because player 1’s ranking of the four outcomes is the reverse of player 2’s ranking). Not strictly competitive in mixed strategies (there exist no values of π and θ > 0 such that −u1 ( a) = π + θu2 ( a) for every outcome a; or, alternatively, player 1 is indifferent between ( B, L) and the lottery that yields ( T, L) with probability 12 and ( T, R) with probability 12 , whereas player 2 is not indifferent between these two outcomes). Game in Figure 367.1: Strictly competitive both in pure and in mixed strategies. (Player 2’s preferences are represented by the expected value of the Bernoulli payoff function −u1 because −u1 ( a) = − 12 + 12 u2 ( a) for every pure outcome a.) 370.2 Maxminimizing in BoS Player 1’s maxminimizer is ( 31 , 23 ) while player 2’s is ( 32 , 13 ). Clearly neither pure equilibrium strategy of either player guarantees her equilibrium payoff. In the mixed strategy equilibrium, player 1’s expected payoff is 23 . But if, for example, player 2 choose S instead of her equilibrium strategy, then player 1’s expected payoff is 13 . Similarly for player 2. 372.2 Equilibrium in strictly competitive game The claim is false. In the strictly competitive game in Figure 64.1 the action pair ( T, L) is a Nash equilibrium, so that player 1’s unique equilibrium payoff in the game is 0. But ( B, R), which also yields player 1 a payoff of 0, is not a Nash equilibrium. T B L 0, 0 −1, 1 R 1, −1 0, 0 Figure 64.1 The game in Exercise 372.2. 372.4 O’Neill’s game a. Denote the probability with which player 1 chooses each of her actions 1, 2, and 3, by p, and the probability with which player 2 chooses each of these actions by q. Then all four of player 1’s actions yield the same expected payoff if and only if 4q − 1 = 1 − 6q, or q = 51 , and similarly all four of player 2’s actions yield the same expected payoff if and only if p = 51 . Thus (( 51 , 15 , 15 , 52 ), ( 15 , 15 , 51 , 25 )) is a Nash equilibrium of the game. The players’ payoffs in this equilibrium are (− 51 , 15 ). Chapter 11. Strictly Competitive Games and Maxminimization 65 b. Let ( p1, p2, p3 , p4 ) be an equilibrium strategy of player 1. In order that it guarantee her the payoff of − 51 , we need − p1 + p2 + p3 − p4 ≥ − 51 p1 − p2 + p3 − p4 ≥ − 51 p1 + p2 − p3 − p4 ≥ − 51 − p1 − p2 − p3 + p4 ≥ − 51 . Adding these four inequalities, we deduce that p4 ≤ 52 . Adding each pair of the first three inequalities, we deduce that p1 ≤ 51 , p2 ≤ 15 , and p3 ≤ 15 . We have p1 + p2 + p3 + p4 = 1, so we deduce that ( p1 , p2, p3, p4 ) = ( 51 , 15 , 51 , 25 ). A similar analysis of the conditions for player 2’s strategy to guarantee her the payoff of 15 leads to the conclusion that (q1 , q2 , q3 , q4 ) = ( 51 , 15 , 15 , 25 ). 12 Rationalizability 379.2 Best responses to beliefs Consider a two-player game in which player 1’s payoffs are given in Figure 67.1. The action B of player 1 is a best response to the belief that assigns probability 21 to both L and R, but is not a best response to any belief that assigns probability 1 to either action. T M B L 3 0 2 R 0 3 2 Figure 67.1 The action B is a best response to a belief that assigns probability a best response to any belief that assigns probability 1 to either L or R. 1 2 to L and to R, but is not 384.1 Mixed strategy equilibria of game in Figure 384.1 The game has no equilibrium in which player 2 assigns positive probability only to L and C, because if she does so then only M and B are possible best responses for player 1, but if player 1 assigns positive probability only to these actions then L is not optimal for player 2. Similarly, the game has no equilibrium in which player 2 assigns positive probability only to C and R, because if she does so then only T and M are possible best responses for player 1, but if player 1 assigns positive probability only to these actions then R is not optimal for player 2. Now assume that player 2 assigns positive probability only to L and R. There are no probabilities for L and R under which player 1 is indifferent between all three of her actions, so player 1 must assign positive probability to at most two actions. If these two actions are T and M then player 2 prefers L to R, while if the two actions are M and B then player 2 prefers R to L. The only possibility is thus that the two actions are T and B. In this case we need player 2 to assign probability 12 to L and R (in order that player 1 be indifferent between T and B); but then M is better for player 1. Thus there is no equilibrium in which player 2 assigns positive probability only to L and R. Finally, if player 2 assigns positive probability to all three of her actions then player 1’s mixed strategy must be such that each of these three actions yields the 67 68 Chapter 12. Rationalizability same payoff. A calculation shows that there is no mixed strategy of player 1 with this property. We conclude that the game has no mixed strategy equilibrium in which either player assigns positive probability to more than one action. 387.2 Finding rationalizable actions I claim that the action R of player 2 is strictly dominated. Consider a mixed strategy of player 2 that assigns probability p to L and probability 1 − p to C. Such a mixed strategy strictly dominates R if p + 4(1 − p) > 3 and 8p + 2(1 − p) > 3, or if 1 1 6 < p < 3 . Now eliminate R from the game. In the reduced game, B is dominated by T. In the game obtained by eliminating B, L is dominated by C. Thus the only rationalizable action of player 1 is T and the only rationalizable action of player 2 is C. 387.5 Hotelling’s model of electoral competition The positions 0 and ` are strictly dominated by the position m: • if her opponent chooses m, a player who chooses m ties whereas a player who chooses 0 loses • if her opponent chooses 0 or `, a player who chooses m wins whereas a player who chooses 0 or ` either loses or ties • if her opponent chooses any other position, a player who chooses m wins whereas a player who chooses 0 or ` loses. In the game obtained by eliminating the two positions 0 and `, the positions 1 and ` − 1 are similarly strictly dominated. Continuing in the same way, we are left with the position m. 388.2 Cournot’s duopoly game From Figure 58.1 we see that firm 1’s payoff to any output greater than 12 (α − c) is less than its payoff to the output 12 (α − c) for any output q2 of firm 2. Thus any output greater than 12 (α − c) is strictly dominated by the output 12 (α − c) for firm 1; the same argument applies to firm 2. Now eliminate all outputs greater than 12 (α − c) for each firm. The maximizer of firm 1’s payoff function for q2 = 12 (α − c) is 41 (α − c), so from Figure 58.1 we see that firm 1’s payoff to any output less than 14 (α − c) is less than its payoff to the output 14 (α − c) for any output q2 ≤ 12 (α − c) of firm 2. Thus any output less than 1 1 4 ( α − c ) is strictly dominated by the output 4 ( α − c ) for firm 1; the same argument applies to firm 2. Chapter 12. Rationalizability 69 Now eliminate all outputs less than 14 (α − c) for each firm. Then by another similar argument, any output greater than 38 (α − c) is strictly dominated by 38 (α − c). Continuing in this way, we see from Figure 59.1 that in a finite number of rounds (given the finite number of possible outputs for each firm) we reach the Nash equilibrium output 13 (α − c). 391.1 Example of dominance-solvable game The Nash equilibria of the game are ( T, L), any ((0, 0, 1), (0, q, 1 − q)) with 0 ≤ q ≤ 1, and any ((0, p, 1 − p), (0, 0, 1)) with 0 ≤ p ≤ 1. The game is dominance solvable, because T and L are the only weakly dominated actions, and when they are eliminated the only weakly dominated actions are M and C, leaving ( B, R), with payoffs (0, 0). If T is eliminated, then L and C, no remaining action is weakly dominated; ( M, R) and ( B, R) both remain. 391.2 Dividing money In the first round every action ai ≤ 5 of each player i is weakly dominated by 6. No other action is weakly dominated, because 100 is a strict best response to 0 and every other action ai ≥ 6 is a strict best response to ai + 1. In the second round, 10 is weakly dominated by 6 for each player, and each other remaining action ai of player i is a strict best response to a1 + 1, so no other action is weakly dominated. Similarly, in the third round, 9 is weakly dominated by 6, and no other action is weakly dominated. In the fourth and fifth rounds 8 and 7 are eliminated, leaving the single action pair (6, 6), with payoffs (5, 5). 392.2 Strictly competitive extensive games with perfect information Every finite extensive game with perfect information has a (pure strategy) subgame perfect equilibrium (Proposition 173.1). This equilibrium is a pure strategy Nash equilibrium of the strategic form of the game. Because the game has only two possible outcomes, one of the players prefers the Nash equilibrium outcome to the other possible outcome. By Proposition 368.1, this player’s equilibrium strategy guarantees her equilibrium payoff, so this strategy weakly dominates all her nonequilibrium strategies. After all dominated strategies are eliminated, every remaining pair of strategies generates the same outcome. 13 Evolutionary Equilibrium 400.1 Evolutionary stability and weak domination The ESS a∗ does not necessarily weakly dominate every other action in the game. For example, in the game in Figure 395.1 of the text, X is an ESS but does not weakly dominate Y. No action can weakly dominate an ESS. To see why, let a∗ be an ESS and let b be another action. Because a∗ is an ESS, ( a∗ , a∗ ) is a Nash equilibrium, so that u(b, a∗ ) ≤ u( a∗ , a∗ ). Now, if u(b, a∗ ) < u( a∗ , a∗ ), certainly b does not weakly dominate a∗ , so suppose that u(b, a∗ ) = u( a∗ , a∗ ). Then by the second condition for an ESS we have u(b, b) < u( a∗ , b). We conclude that b does not weakly dominate a∗ . 405.1 Hawk–Dove–Retaliator First suppose that v ≥ c. In this case the game has two pure symmetric Nash equilibria, ( A, A) and ( R, R). However, A is not an ESS, because R is a best response to A and u( R, R) > u( A, R). The action pair ( R, R) is a strict equilibrium, so R is an ESS. Now consider the possibility that the game has a mixed strategy equilibrium (α, α). If α assigns positive probability to either P or R (or both) then R yields a payoff higher than does P, so only A and R may be assigned positive probability in a mixed strategy equilibrium. But if a strategy α assigns positive probability to A and R and probability 0 to P, then R yields a payoff higher than does A against an opponent who uses α. Thus the game has no symmetric mixed strategy equilibrium in this case. Now suppose that v < c. Then the only symmetric pure strategy equilibrium is ( R, R). This equilibrium is strict, so that R is an ESS. Now consider the possibility that the game has a mixed strategy equilibrium (α, α). If α assigns probability 0 to A then R yields a payoff higher than does P against an opponent who uses α; if α assigns probability 0 to P then R yields a payoff higher than does A against an opponent who uses α. Thus in any mixed strategy equilibrium (α, α), the strategy α must assign positive probability to both A and P. If α assigns probability 0 to R then we need α = (v/c, 1 − v/c) (the calculation is the same as for Hawk–Dove). Because R yields a lower payoff against this strategy than do A and P, and the strategy is an ESS in Hawk–Dove, it is an ESS in the present game. The remaining possibility is that the game has a mixed strategy equilibrium (α, α) in which α assigns positive probability to all three actions. If so, then the expected payoff to this strategy is less than 12 v, because the pure strategy P yields an expected payoff 71 72 Chapter 13. Evolutionary Equilibrium less than 12 v against any such strategy. But then U ( R, R) = 21 v > U (α, R), violating the second condition in the definition of an ESS. In summary: • If v ≥ c then R is the unique ESS of the game. • If v < c then both R and the mixed strategy that assigns probability v/c to A and 1 − v/c to P are ESSs. 405.3 Bargaining The game is given in Figure 27.1. The pure strategy of demanding 10 is not an ESS because 2 is a best response to 10 and u(2, 2) > u(10, 2). Now let α be the mixed strategy that assigns probability 25 to 2 and 35 to 8. Each player’s payoff at the strategy pair (α, α) is 16 5 . Thus the only actions a that are best responses to α are 2 and 8, so that the only mixed strategies that are best responses to α assign positive probability only to the actions 2 and 8. Let β be the mixed strategy that assigns probability p to 2 and probability 1 − p to 8. We have U ( β, β) = 5p(2 − p) and U (α, β) = 6p + 45 . We find that U (α, β) − U ( β, β) = 5( p − 25 )2 , which is positive if p 6= 52 . Hence α is an ESS. Finally let α be the mixed strategy that assigns probability 45 to 4 and 15 to 6. Each player’s payoff at the strategy pair (α, α) is 24 5 . Thus the only actions a that are best responses to α are 4 and 6, so that the only mixed strategies that are best responses assign positive probability only to the actions 4 and 6. Let β be the mixed strategy that assigns probability p to 4 and probability 1 − p to 6. We have U ( β, β) = 5p(2 − p) and U (α∗ , β) = 2p + 16 5 . We find that U (α, β) − U ( β, β) = 5( p − 45 )2 , which is positive if p 6= 54 . Hence α∗ is an ESS. 408.1 Equilibria of C and of G First suppose that (α1 , α2 ) is a mixed strategy Nash equilibrium of C. Then for all mixed strategies β 1 of player 1 and all mixed strategies β 2 of player 2 we have U1 (α1 , α2 ) ≥ U1 ( β 1 , α2 ) and U2 (α1 , α2 ) ≥ U2 (α1 , β 2 ). Chapter 13. Evolutionary Equilibrium 73 Thus u((α1 , α2 ), (α1 , α2 )) = 12 U1 (α1 , α2 ) + 12 U2 (α1 , α2 ) ≥ 12 U1 ( β 1 , α2 ) + 12 U2 (α1 , β 2 ) = u(( β 1 , β 2 ), (α1, α2 )), so that ((α1 , α2 ), (α1 , α2 )) is a Nash equilibrium of G. If (α1 , α2 ) is a strict Nash equilibrium of C then the inequalities are strict, and ((α1 , α2 ), (α1 , α2 )) is a strict Nash equilibrium of G. Now assume that ((α1 , α2 ), (α1 , α2 )) is a Nash equilibrium of G. Then u((α1 , α2 ), (α1 , α2 )) ≥ u(( β 1 , β 2 ), (α1, α2 )), or 1 1 2 U1 ( α1 , α2 ) + 2 U2 ( α1 , α2 ) ≥ 12 U1 ( β 1 , α2 ) + 12 U2 (α1 , β 2 ), for all conditional strategies ( β 1 , β 2 ). Taking β 2 = α2 we see that α1 is a best response to α2 in C, and taking β 1 = α1 we see that α2 is a best response to α1 in C. Thus (α1 , α2 ) is a Nash equilibrium of G. 414.1 A coordination game between siblings The game with payoff function v is shown in Figure 73.1. If x < 2 then (Y, Y ) is a strict Nash equilibrium of the games, so Y is an evolutionarily stable action in the game between siblings. If x > 2 then the only Nash equilibrium of the game is ( X, X ), and this equilibrium is strict. Thus the range of values of x for which the only evolutionarily stable action is X is x > 2. X Y X x, x 1 1 2 x, 2 Y 1 1 2, 2x 1, 1 v Figure 73.1 The game with payoff function v derived from the game in Exercise 414.1. 414.2 Assortative mating Under assortative mating, all siblings take the same action, so the analysis is the same as that for asexual reproduction. (A difficulty with the assumption of assortative mating is that a rare mutant will have to go to great lengths to find a mate that is also a mutant.) 74 Chapter 13. Evolutionary Equilibrium 416.1 Darwin’s theory of the sex ratio A normal organism produces pn male offspring and (1 − p)n female offspring (ignoring the small probability that the partner of a normal organism is a mutant). Thus it has pn · ((1 − p)/p)n + (1 − p)n · n = 2(1 − p)n2 grandchildren. A mutant has 12 n male offspring and 12 n female offspring, and hence 12 n · ((1 − p)/p)n + 12 n · n = 21 n2 /p grandchildren. Thus the difference between the number of grandchildren produced by mutant and normal organisms is 1 2 2 1 2 n /p − 2 ( 1 − p ) n = n (1 − 2p)2, 2 2p which is positive if p 6= 12 . (The point is that if p > 12 then the fraction of a mutant’s offspring that are males is higher than the fraction of a normal organism’s offspring that are males, and males each bear more offspring than females. Similarly, if p < 21 then the fraction of a mutant’s offspring that are females is higher than the fraction of a normal organism’s offspring that are females, and females each bear more offspring than males.) Thus any mutant with p 6= 21 invades the population; only p = 21 is evolutionarily stable. 14 Repeated Games: The Prisoner’s Dilemma 423.1 Equivalence of payoff functions Suppose that a person’s preferences are represented by the discounted sum of payoffs with payoff function u and discount factor δ. Then if the two sequences of outcomes ( x1 , x2 , . . .) and (y1 , y2 , . . .) are indifferent, we have ∞ ∞ t =0 t =0 ∑ δ t −1 u ( x t ) = ∑ δ t −1 u ( y t ) . Now let v( x ) = α + βu( x ) for all x, with β > 0. Then ∞ ∞ ∞ ∞ t =0 t =0 t =0 t =0 ∞ ∑ δt−1 v(x t ) = ∑ δt−1 [α + βu(x t )] = ∑ δt−1 α + β ∑ δt−1 u(x t ) and similarly ∑ δ t −1 v ( y t ) = ∑ δt−1 [α + βu(yt )] = ∞ ∑ δ t −1 α + β ∑ δ t −1 u ( y t ) , ∞ t =0 t =0 t =0 t =0 ∞ so that ∞ ∞ ∑ δ t −1 v ( x t ) = ∑ δ t −1 v ( y t ) . t =0 t =0 Thus the person’s preferences are represented also by the discounted sum of payoffs with payoff function v and discount factor δ. 426.1 Subgame perfect equilibrium of finitely repeated Prisoner’s Dilemma Use backward induction. In the last period, the action C is strictly dominated for each player, so each player chooses D, regardless of history. Now consider period T − 1. Each player’s action in this period affects only the outcome in this period—it has no effect on the outcome in period T, which is ( D, D ). Thus in choosing her action in period T − 1, a player considers only her payoff in that period. As in period T, her action D strictly dominates her action C, so that in any subgame perfect equilibrium she chooses D. A similar argument applies to all previous periods, leading to the conclusion that in every subgame perfect equilibrium each player chooses D in every period, regardless of history. 75 76 Chapter 14. Repeated Games: The Prisoner’s Dilemma P0 : C - P1 : C (·, D ) - D: D all outcomes Figure 76.1 The strategy in Exercise 428.1a. 428.1 Strategies in an infinitely repeated Prisoner’s Dilemma a. The strategy is shown in Figure 76.1. b. The strategy is shown in Figure 76.2. P0 : C - P1 : C (·, D ) (·, D ) - D: D Figure 76.2 The strategy in Exercise 428.1b. c. The strategy is shown in Figure 76.3. ? C: C (C, C ) or ( D, D ) - D: D ( D, C ) or (C, D ) Figure 76.3 The strategy in Exercise 428.1c. 439.1 Finitely repeated Prisoner’s Dilemma with switching cost a. Consider deviations by player 1, given that player 2 adheres to her strategy, in the subgames following histories that end in each of the four outcomes of the game. (C, C ): If player 1 adheres to her strategy, her payoff is 3 in every period. If she deviates in the first period of the subgame, but otherwise follows her strategy, her payoff is 4 − e in the first period of the subgame, and 2 in every subsequent period. Given e > 1, player 1’s deviation is not profitable, even if it occurs in the last period of the game. ( D, C ) or ( D, D ): If player 1 adheres to her strategy, her payoff is 2 in every period. If she deviates in the first period of the subgame, but otherwise follows her strategy, her payoff is −e in the first period of the subgame, 2 − e in the next period, and 2 subsequently. Thus adhering to her strategy is optimal for player 1. (C, D ): If player 1 adheres to her strategy, her payoff is 2 − e in the first period of the subgame, and 2 subsequently. If she deviates in the first period of the subgame, but otherwise follows her strategy, her payoff Chapter 14. Repeated Games: The Prisoner’s Dilemma 77 is 0 in the first period of the subgame, 2 − e in the next period, and 2 subsequently. Given e < 2, player 1’s deviation is not optimal even if it occurs in the last period of the game. b. Given e > 2, a player does not gain from deviating from (C, C ) in the nextto-last or last periods, even if she is not punished, and does not optimally punish such a deviation by her opponent. Consider the strategy that chooses C at the start of the game and after any history that ends with (C, C ), chooses D after any other history that has length at most T − 2, and chooses the action it chose in period T − 1 after any history of length T − 1 (where T is the length of the game). I claim that the strategy pair in which both players use this strategy is a subgame perfect equilibrium. Consider deviations by player 1, given that player 2 adheres to her strategy, in the subgames following the various possible histories. History ending in (C, C ), length ≤ T − 3: If player 1 adheres to her strategy, her payoff is 3 in every period of the subgame. If she deviates in the first period of the subgame, but otherwise follows her strategy, her payoff is 4 − e in the first period of the subgame, and 2 in every subsequent period (her opponent switches to D). Given e > 1, player 1’s deviation is not profitable. History ending in (C, C ), length ≥ T − 2: If player 1 adheres to her strategy, her payoff is 3 in each period of the subgame. If she deviates to D in the first period of the subgame, her payoff is 4 − e in that period, and 4 subsequently (her deviation is not punished). The length of the subgame is at most 2, so given e > 2, her deviation is not profitable. History ending in ( D, C ) or ( D, D ): If player 1 adheres to her strategy, her payoff is 2 in every period. If she deviates in the first period of the subgame, but otherwise follows her strategy, her payoff is −e in the first period of the subgame, 2 − e in the next period, and 2 subsequently. Thus adhering to her strategy is optimal for player 1. History ending in (C, D ), length ≤ T − 2: If player 1 adheres to her strategy, her payoff is 2 − e in the first period of the subgame (she switches to D), and 2 subsequently. If she deviates in the first period of the subgame, but otherwise follows her strategy, her payoff is 0 in the first period of the subgame, 2 − e in the next period, and 2 subsequently. History ending in (C, D ), length T − 1: If player 1 adheres to her strategy, her payoff is 0 in period T (the outcome is (C, D )). If she deviates to D, her payoff is 2 − e in period T. Given e > 2, adhering to her strategy is thus optimal. 78 Chapter 14. Repeated Games: The Prisoner’s Dilemma 442.1 Deviations from grim trigger strategy • If player 1 adheres to the strategy, she subsequently chooses D (because player 2 chose D in the first period). Player 2 chooses C in the first period of the subgame (player 1 chose C in the first period of the game), and then chooses D (because player 1 chooses D in the first period of the subgame). Thus the sequence of outcomes in the subgame is (( D, C ), ( D, D ), ( D, D), . . .), yielding player 1 a discounted average payoff in the subgame of δ 2 3 = 3 − 2δ. (1 − δ)(3 + δ + δ + δ + · · ·) = (1 − δ) 3 + 1−δ • If player 1 refrains from punishing player 2 for her lapse, and simply chooses C in every subsequent period, then the outcome in period 2 and subsequently is (C, C ), so that the sequence of outcomes in the subgame yields player 1 a discounted average payoff of 2. If δ > 12 then 2 > 3 − 2δ, so that player 1 prefers to ignore player 2’s deviation rather than to adhere to her strategy and punish player 2 by choosing D. (Note that the theory does not consider the possibility that player 1 takes player 2’s play of D as a signal that she is using a strategy different from the grim trigger strategy.) 443.2 Different punishment lengths in subgame perfect equilibrium Yes, an infinitely repeated Prisoner’s Dilemma has such subgame perfect equilibria. As for the modified grim trigger strategy, each player’s strategy has to switch to D not only if the other player chooses D but also if the player herself chooses D. The only subtlety is that the number of periods for which a player chooses D after a history in which not all the outcomes were (C, C ) must depend on the identity of the player who first deviated. If, for example, player 1 punishes for two periods while player 2 punishes for three periods, then the outcome (C, D ) induces player 1 to choose D for two periods (to punish player 2 for her deviation) while the outcome ( D, C ) induces her to choose D for three periods (while she is being punished by player 2). The strategy of each player in this case is shown in Figure 79.1. Viewed as a strategy of player 1, the top part of the figure entails punishment of player 2 and the bottom part entails player 1’s reaction to her own deviation. Similarly, viewed as a strategy of player 2, the bottom part of the figure entails punishment of player 1 and the top part entails player 2’s reaction to her own deviation. To find the values of δ for which the strategy pair in which each player uses the strategy in Figure 79.1 is a subgame perfect equilibrium, consider the result of each player’s deviating at the start of a subgame. First consider player 1. If she deviates when both players are in state P0 , she induces the outcome ( D, C ) followed by three periods of ( D, D ), and then (C, C ) subsequently. This outcome path is worse for her than (C, C ) in every period if Chapter 14. Repeated Games: The Prisoner’s Dilemma 79 all outcomes - P2 : D * P1 : D (·, D ) all ? outcomes P0: C H H 6 ( D, ·)HH j P0 : D H - P0 : D 2 1 all outcomes - P0 : D 3 all outcomes all outcomes Figure 79.1 A strategy in an infinitely repeated Prisoner’s Dilemma that punishes deviations for two periods and reacts to punishment by choosing D for three periods. and only if δ3 − 2δ + 1 ≤ 0, or if and only if δ is at least around 0.62 (as we found in Section 14.7.2). If she deviates when both players are in one of the other states then she is worse off in the period of her deviation and her deviation does not affect the subsequent outcomes. Thus player 1 cannot profitably deviate in the first period of any subgame if δ is at least around 0.62. The same argument applies to player 2, except that a deviation when both players are in state P0 induces (C, D ) followed by three, rather than two periods of ( D, D ). This outcome path is worse for player 2 than (C, C ) in every period if and only if δ4 − 2δ + 1 ≤ 0, or if and only if δ is at least around 0.55 (as we found in Section 14.7.2). We conclude that the strategy pair in which each player uses the strategy in Figure 79.1 is a subgame perfect equilibrium if and only if δ3 − 2δ + 1 ≤ 0, or if and only if δ is at least around 0.62. 445.1 Tit-for-tat as a subgame perfect equilibrium Suppose that player 2 adheres to tit-for-tat. Consider player 1’s behavior in subgames following histories that end in each of the following outcomes. (C, C ) If player 1 adheres to tit-for-tat the outcome is (C, C ) in every period, so that her discounted average payoff in the subgame is x. If she chooses D in the first period of the subgame, then adheres to tit-for-tat, the outcome alternates between ( D, C ) and (C, D ), and her discounted average payoff is y/(1 + δ). Thus we need x ≥ y/(1 + δ), or δ ≥ (y − x )/x, for a one-period deviation from tit-for-tat not to be profitable for player 1. (C, D ) If player 1 adheres to tit-for-tat the outcome alternates between ( D, C ) and (C, D ), so that her discounted average payoff is y/(1 + δ). If she deviates to C in the first period of the subgame, then adheres to tit-for-tat, the outcome is (C, C ) in every period, and her discounted average payoff is x. Thus we need y/(1 + δ) ≥ x, or δ ≤ (y − x )/x, for a one-period deviation from tit-for-tat not to be profitable for player 1. 80 Chapter 14. Repeated Games: The Prisoner’s Dilemma ( D, C ) If player 1 adheres to tit-for-tat the outcome alternates between (C, D ) and ( D, C ), so that her discounted average payoff is δy/(1 + δ). If she deviates to D in the first period of the subgame, then adheres to tit-for-tat, the outcome is ( D, D ) in every period, and her discounted average payoff is 1. Thus we need δy/(1 + δ) ≥ 1, or δ ≥ 1/(y − 1), for a one-period deviation from tit-for-tat not to be profitable for player 1. ( D, D ) If player 1 adheres to tit-for-tat the outcome is ( D, D ) in every period, so that her discounted average payoff is 1. If she deviates to C in the first period of the subgame, then adheres to tit-for-tat, the outcome alternates between (C, D ) and ( D, C ), and her discounted average payoff is δy/(1 + δ). Thus we need 1 ≥ δy/(1 + δ), or δ ≤ 1/(y − 1), for a one-period deviation from tit-for-tat not to be profitable for player 1. The same arguments apply to deviations by player 2, so we conclude that (tit-for-tat, tit-for-tat) is a subgame perfect equilibrium if and only if δ = (y − x )/x and δ = 1/(y − 1), or y − x = 1 and δ = 1/x. 15 Repeated Games: General Results 454.3 Repeated Bertrand duopoly a. Suppose that firm i uses the strategy si . If the other firm, j, uses s j , then its discounted average payoff is (1 − δ) 12 π ( pm ) + 12 δπ ( pm ) + · · · = 12 π ( pm ). If, on the other hand, firm j deviates to a price p then the closer this price is to pm , the higher is j’s profit, because the punishment does not depend on p. Thus by choosing p close enough to p m the firm can obtain a profit as close as it wishes to π ( pm ) in the period of its deviation. Its profit during its punishment in the following k periods is zero. Once its punishment is complete, it can either revert to pm or deviate once again. If it can profit from deviating initially then it can profit by deviating once its punishment is complete, so its maximal profit from deviating is (1 − δ ) π ( p m ) . (1 − δ) π ( pm ) + δk+1 π ( pm ) + δ2k+2 π ( pm ) + · · · = 1 − δ k +1 Thus for (s1 , s2 ) to be a Nash equilibrium we need 1−δ ≤ 21 , 1 − δ k +1 or δk+1 − 2δ + 1 ≤ 0. (This condition is the same as the one we found for a pair of k-period punishment strategies to be a Nash equilibrium in the Prisoner’s Dilemma (Section 14.7.2).) b. Suppose that firm i uses the strategy si . If the other firm does so then its discounted average payoff is 12 π ( pm ), as in part a. If the other firm deviates to some price p with c < p < pm in the first period, and maintains this price subsequently, then it obtains π ( p) in the first period and shares π ( p) in each subsequent period, so that its discounted average payoff is (1 − δ) π ( p) + 12 δπ ( p) + 21 δ2 π ( p) + · · · = 12 (2 − δ)π ( p). If p is close to pm then π ( p) is close to π ( p m ) (because π is continuous). In fact, for any δ < 1 we have 2 − δ > 1, so that we can find p < p m such that (2 − δ)π ( p) > π ( pm ). Hence the strategy pair is not a Nash equilibrium of the infinitely repeated game for any value of δ. 81 82 Chapter 15. Repeated Games: General Results 459.2 Detection lags a. The best deviations involve prices slightly less than p∗ . Such a deviation by firm i yields a discounted average payoff close to (1 − δ) π ( p∗ ) + δπ ( p∗ ) + · · · + δk i −1 π ( p∗ ) = (1 − δk i )π ( p∗ ), whereas compliance with the strategy yields the discounted average payoff 1 ∗ 2 π ( p ). Thus the strategy pair is a subgame perfect equilibrium for any value of p∗ if δk1 ≥ 21 and δk2 ≥ 21 , and is not a subgame perfect equilibrium for any value of p∗ if δk1 < 12 or δk2 < 12 . That is, the most profitable price for which the strategy pair is a subgame perfect equilibrium is p m if δk1 ≥ 21 and δk2 ≥ 21 and is c if δk1 < 12 or δk2 < 12 . ∗ b. Denote by k∗i the critical value of k i found in part a. (That is, δk i ≥ ∗ δk i +1 < 12 .) 1 2 and If k i > k∗i then no change in k j affects the outcome of the price-setting subgame, so j’s best action at the start of the game is θ, in which case i’s best action is the same. Thus in one subgame perfect equilibrium both firms choose θ at the start of the game, and c regardless of history in the rest of the game. If k i ≤ k∗i then j’s best action is k∗j if the cost of choosing k∗j is at most 12 π ( pm ). Thus if the cost of choosing k∗i is at most 21 π ( pm ) for each firm then the game has another subgame perfect equilibrium, in which each firm i chooses k∗i at the start of the game and the strategy si in the price-setting subgame. A promise by firm i to beat another firm’s price is an inducement for consumers to inform firm i of deviations by other firms, and thus reduce its detection time. To this extent, such a promise tends to promote collusion. 16 Bargaining 468.1 Two-period bargaining with constant cost of delay In the second period, player 1 accepts any proposal that gives a positive amount of the pie. Thus in any subgame perfect equilibrium player 2 proposes (0, 1) in period 2, which player 1 accepts, obtaining the payoff −c1 . Now consider the first period. Given the second period outcome of any subgame perfect equilibrium, player 2 accepts any proposal that gives her more than 1 − c2 and rejects any proposal that gives her less than 1 − c2 . Thus in any subgame perfect equilibrium player 1 proposes (c2 , 1 − c2 ), which player 2 accepts. In summary, the game has a unique subgame perfect equilibrium, in which • player 1 proposes (c2 , 1 − c2 ) in period 1, and accepts all proposals in period 2 • player 2 accepts a proposal in period 1 if and only if it gives her at least 1 − c2 , and proposes (0, 1) in period 2 after any history. The outcome of the equilibrium is that the proposal (c2 , 1 − c2 ) is made by player 1 and immediately accepted by player 2. 468.2 Three-period bargaining with constant cost of delay The subgame following a rejection by player 2 in period 1 is a two-period game in which player 2 makes the first proposal. Thus by the result of Exercise 468.1, the subgame has a unique subgame perfect equilibrium, in which player 2 proposes (1 − c1 , c1 ), which player 1 immediately accepts. Now consider the first period. • If c1 ≥ c2 , player 2 rejects any offer of less than c1 − c2 (which she obtains if she rejects an offer), and accepts any offer of more than c1 − c2 . Thus in an equilibrium player 1 offers her c1 − c2 , which she accepts. • If c1 < c2 , player 2 accepts all offers, so that player 1 proposes (1, 0), which player 2 accepts. In summary, the game has a unique subgame perfect equilibrium, in which • player 1 proposes (1 − (c1 − c2 ), c1 − c2 ) if c1 ≥ c2 and (1, 0) otherwise in period 1, accepts any proposal that gives her at least 1 − c1 in period 2, and proposes (1, 0) in period 3 83 84 Chapter 16. Bargaining • player 2 accepts any proposal that gives her at least c1 − c2 if c1 ≥ c2 and accepts all proposals otherwise in period 1, proposes (1 − c1 , c1 ) in period 2, and accepts all proposals in period 3. 17 Appendix: Mathematics 497.1 Maximizer of quadratic function We can write the function as − x ( x − α). Thus r1 = 0 and r2 = α, and hence the maximizer is α/2. 499.3 Sums of sequences In the first case set r = δ2 to transform the sum into 1 + r + r2 + · · ·, which is equal to 1/(1 − r ) = 1/(1 − δ2 ). In the second case split the sum into (1 + δ2 + δ4 + · · ·) + (2δ + 2δ3 + 2δ5 + · · ·); the first part is equal to 1/(1 − δ2 ) and the second part is equal to 2δ(1 + δ2 + δ4 + · · ·), or 2δ/(1 − δ2 ). Thus the complete sum is 1 + 2δ . 1 − δ2 504.2 Bayes’ law Your posterior probability of carrying X given that you test positive is Pr(positive test| X ) Pr( X ) Pr(positive test| X ) Pr( X ) + Pr(positive test|¬ X ) Pr(¬ X ) where ¬ X means “not X”. This probability is equal to 0.9p/(0.9p + 0.2(1 − p)) = 0.9p/(0.2 + 0.7p), which is increasing in p (i.e. a smaller value of p gives a smaller value of the probability). If p = 0.001 then the probability is approximately 0.004. (That is, if 1 in 1,000 people carry the gene then if you test positive on a test that is 90% accurate for people who carry the gene and 80% accurate for people who do not carry the gene, then you should assign probability 0.004 to your carrying the gene.) If the test is 99% accurate in both cases then the posterior probability is (0.99 · 0.001)/[0.99 · 0.001 + 0.01 · 0.999] ≈ 0.09. 85 References The page numbers on which the references are cited are given in brackets after each item. Nagel, Rosemarie (1995), “Unraveling in guessing games: an experimental study”, American Economic Review 85, 1313–1326. [8] Ochs, Jack (1995), “Coordination problems”, in Handbook of experimental economics (John H. Kagel and Alvin E. Roth, eds.), 195–251. Princeton: Princeton University Press. [6] Shubik, Martin (1982), Game theory in the social sciences. Cambridge, MA: MIT Press. [55] Van Huyck, John B., Raymond C. Battalio, and Richard O. Beil (1990), “Tacit coordination games, strategic uncertainty, and coordination failure”, American Economic Review 80, 234–248. [6] 87