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solutions Publicly-available for
AN INTRODUCTION TO GAME THEORY
Publicly-available solutions for
AN INTRODUCTION TO GAME THEORY
MARTIN J. OSBORNE
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 A (TEX), Leslie Lamport (L TEX), 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
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Introduction 1 Exercise 5.3 (Altruistic preferences) 1 Exercise 6.1 (Alternative representations of preferences)
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Nash Equilibrium 3 Exercise 16.1 (Working on a joint project) 3 Exercise 17.1 (Games equivalent to the Prisoners Dilemma) 3 Exercise 20.1 (Games without conict) 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 Nash Equilibrium: Illustrations 11 Exercise 58.1 (Cournots duopoly game with linear inverse demand and different unit costs) 11 Exercise 60.2 (Nash equilibrium of Cournots duopoly game and the collusive outcome) 12 Exercise 63.1 (Interaction among resource-users) 12 Exercise 67.1 (Bertrands duopoly game with constant unit cost) 13 Exercise 68.1 (Bertrands oligopoly game) 13 Exercise 68.2 (Bertrands 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
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Contents
Exercise 86.2 (Nash equilibrium of rst-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 HawkDove) 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 Cournots 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 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 Mens Morris, or Mill) 35 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 (Stackelbergs 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
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Contents
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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 Coalitional Games and the Core 47 Exercise 245.1 (Three-player majority game) 47 Exercise 248.1 (Core of landownerworker game) 47 Exercise 249.1 (Unionized workers in landownerworker game) 47 Exercise 249.2 (Landownerworker 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 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 (Cournots duopoly game with imperfect information) 53 Exercise 288.1 (Cournots 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 rst-price sealed-bid auction) 58 Exercise 309.3 (Example of Nash equilibrium in a rst-price auction) 58
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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 receivers expected payoff in two equilibria) 61 Exercise 350.1 (Variant of model with piecewise linear payoff functions) 61
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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 (ONeills 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 (Hotellings model of electoral competition) 68 Exercise 388.2 (Cournots 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 (HawkDoveRetaliator) 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 (Darwins theory of the sex ratio) 74 14 Repeated Games: The Prisoners Dilemma 75 Exercise 423.1 (Equivalence of payoff functions) 75 Exercise 426.1 (Subgame perfect equilibrium of nitely repeated Prisoners Dilemma) 75 Exercise 428.1 (Strategies in an innitely repeated Prisoners Dilemma) 76 Exercise 439.1 (Finitely repeated Prisoners 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
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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. MARTIN J. OSBORNE Martin.Osborne@utoronto.ca
Department of Economics, 150 St. George Street, University of Toronto, Toronto, Canada M5S 3G7
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1
Introduction
5.3 Altruistic preferences Person 1 is indifferent between (1, 4) and (3, 0), and prefers both of these to (2, 1). 1 The payoff function u dened by u(x, y) = x + 2 y, where x is person 1s income and y is person 2s, represents person 1s preferences. Any function that is an increasing function of u also represents her preferences. For example, the functions k(x + 1 y) for any positive number k, and (x + 1 y)2 , do so. 2 2 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).
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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 Goof off 0, 2 1, 1
Work hard Goof off
Figure 3.1 Working on a joint project (alternative version).
17.1 Games equivalent to the Prisoners Dilemma The game in the left panel differs from the Prisoners 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 Prisoners 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 Prisoners 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 Prisoners Dilemma.
20.1 Games without conict 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. L a, a c, c R b, b d, d
T B
Figure 3.2 A strategic game in which conict is absent.
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Chapter 2. Nash Equilibrium
31.1 Extension of the Stag Hunt Every prole (e, . . . , e), where e is an integer from 0 to K, is a Nash equilibrium. In the equilibrium (e, . . . , e), each players payoff is e. The prole (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 prole (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, 209233).) 34.1 Guessing two-thirds of the average If all three players announce the same integer k 2 then any one of them can devi2 ate to k 1 and obtain $1 (since her number is now closer to 3 of the average than 1 the other two) rather than $ 3 . Thus no such action prole 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 prole 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 1 and k2 , with k1 k2 . The average of the three integers is 3 (k + k1 + k2 ), 2 2 2 so that 3 of the average is 9 (k + k1 + k2 ). If k1 9 (k + k1 + k2 ) then is further from 2 of the average than is k , and hence does not win. If k 1 3 2 2 k1 < 9 (k + k1 + k2 ) then the difference between k and 3 of the average is 2 (k + k + k ) = 7 k 2 k 2 k , while the difference between k and k 2 1 1 9 9 91 92 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 5 4 2 between the former and the latter is 9 k + 5 k1 9 k2 > 0, so k1 is closer to 3 9 . Hence the player who names k does not win, and of the average than is k is better off naming k2 , in which case she obtains a share of the prize. Thus no such action prole is a Nash equilibrium. Suppose two players name k , and the third player names k < k . The 2 1 average of the three integers is then 3 (2k + k), so that 3 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 prole 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 players 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 prole, each persons 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 proles 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 AXB and two to take AYB, resulting in a total travel time for each person of either 29.9 or 30 minutes. However, if a person taking A XB switches to the new road at X and then takes YB her total travel time becomes 9 + 7 + 12 = 28 minutes. I claim that in any Nash equilibrium, one person takes AXB, two people take AXYB, and one person takes AYB. For this assignment, each persons travel time is 32 minutes. No person can change her route and decrease her travel time, by the following argument. If the person taking AXB switches to AXYB, her travel time increases to 12 + 9 + 15 = 36 minutes; if she switches to AYB her travel time increases to 21 + 15 = 36 minutes. If one of the people taking AXYB switches to AXB, her travel time increases to 12 + 20.9 = 32.9 minutes; if she switches to AYB her travel time increases to 21 + 12 = 33 minutes. If the person taking AYB switches to AXB, her travel time increases to 15 + 20.9 = 35.9 minutes; if she switches to AXYB, 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 AXB, one
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Chapter 2. Nash Equilibrium
person takes AXYB, and two people take AYB, then the travel time of those taking AYB is 21 + 12 = 33 minutes; if one of them switches to AXB then her travel time falls to 12 + 20.9 = 32.9 minutes. Or if one person takes AYB, one person takes AXYB, and two people take AXB, then the travel time of those taking AXB is 12 + 20.9 = 32.9 minutes; if one of them switches to AXYB then her travel time falls to 12 + 8 + 12 = 32 minutes. Thus in the equilibrium with the new road every persons travel time increases, from either 29.9 or 30 minutes to 32 minutes. 37.1 Finding Nash equilibria using best response functions a. The Prisoners 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 Fink 0 , 3 1 , 1 Bach 2 , 1 0 ,0 Stravinsky 0 ,0 1 , 2 BoS
Quiet Fink
Bach Stravinsky
Prisoners Dilemma
Figure 6.1 The best response functions in the Prisoners Dilemma (left) and in BoS (right).
Head Tail
Head 1 , 1 1 , 1
Tail 1 , 1 1 , 1
Stag Hare
Stag 2 , 2 1 ,0
Hare 0 ,1 1 , 1
Matching Pennies
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). L 2 ,2 3 , 1 1 , 0 C 1 , 3 0 ,0 0 , 0 R 0 , 1 0 , 0 0 , 0
T M B
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
A2 C L
R
T
M A1
B
Figure 7.1 The players best response functions for the game in Exercise 38.1b. Player 1s best responses are indicated by circles, and player 2s 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 players best responses are given in the following table. Other players 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 gure 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 a a , and for each such action player 1 has multi2s action a2 satises a2 2 2 ple 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.
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Chapter 2. Nash Equilibrium
10 9 8 7 6 A2 5 4 3 2 1 0
0 1 2 3 4 5 6 7 8 9 10 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.) L 1, 1 1, 0 0, 0 C 0, 1 2, 1 1, 1 R 0, 0 1, 2 2, 0
T M B
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 prole in which (i) is 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, rst 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 1 action prole in which k players name the median favorite policy x i , at most 2 (n 1 3) players name policies less than xi , and at most 2 (n 3) players name policies . (In these equilibria, the favorite policy of a player who names a greater than xi 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 prole in which all players name the same, arbitrary, policy is also a Nash equilibrium; the outcome is the common policy named. More generally, any prole 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 players 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 populationthis equilibrium is the only symmetric equilibrium.
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Nash Equilibrium: Illustrations
58.1 Cournots duopoly game with linear inverse demand and different unit costs Following the analysis in the text, the best response function of rm 1 is b1 (q2 ) = while that of rm 2 is b2 (q1 ) =
1 2 ( c2 1 2 ( c1
0
q2 ) if q2 c1 otherwise
0
q1 ) if q1 c2 otherwise.
To nd the Nash equilibrium, rst plot these two functions. Each function has the same general form as the best response function of either rm 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 nd 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. 1 Precisely, if c1 2 ( + 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 = This solution is
(q1 , q2 ) =
1 3 ( 2c1 1 + c2 ), 3 ( 2c2 + c1 ) . 1 2 ( c1 1 2 ( c2
q2 ) q1 ).
1 If c1 > 2 ( + c2 ) then the downward-sloping part of rm 1s best response function lies below the downward-sloping part of rm 2s best response func tion (as in Figure 12.1), and the game has a unique Nash equilibrium, (q1 , q2 ) = 1 (0, 2 ( c2 )). In summary, the game always has a unique Nash equilibrium, dened as follows: 1 1 ( 2c1 + c2 ), 1 ( 2c2 + c1 ) if c1 2 ( + c2 ) 3 3
The output of rm 2 exceeds that of rm 1 in every equilibrium. 11
0, 1 ( c2 ) 2
1 if c1 > 2 ( + c2 ).
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Chapter 3. Nash Equilibrium: Illustrations
q2
(q1 , q2 )
c1 b2 (q1 ) b1 (q2 )
c2 2
0
c1 2
c2
q1
Figure 12.1 The best response functions in Cournots duopoly game under the assumptions of Exer cise 58.1 when c1 < 1 ( c2 ). The unique Nash equilibrium in this case is (q1 , q2 ) = (0, 1 ( c2 )). 2 2
If c2 decreases then rm 2s output increases and rm 1s output either falls, if 1 c1 1 ( + c2 ), or remains equal to 0, if c1 > 2 ( + c2 ). The total output increases 2 and the price falls. 60.2 Nash equilibrium of Cournots duopoly game and the collusive outcome The rms total prot is (q1 + q2 )( c q1 q2 ), or Q( c Q), where Q denotes total output. This function is a quadratic in Q that is zero when Q = 0 and 1 when Q = c, so that its maximizer is Q = 2 ( c). If each rm produces 1 ( c) then its prot is 1 ( c)2 . This prot exceeds 4 8 1 its Nash equilibrium prot of 9 ( c)2 . 1 If one rm produces Q /2, the other rms best response is bi (Q /2) = 2 ( 1 3 /2, the other rm wants c 4 ( c)) = 8 ( c). That is, if one rm produces Q to produce more than Q /2. 63.1 Interaction among resource-users The game is given as follows. Players The rms. Actions Each rms set of actions is the set of all nonnegative numbers (representing the amount of input it uses). Preferences The payoff of each rm i is xi (1 (x1 + + xn )) 0 if x1 + + xn 1 if x1 + + xn > 1.
Chapter 3. Nash Equilibrium: Illustrations
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This game is the same as that in Exercise 61.1 for c = 0 and = 1. Thus it has a unique Nash equilibrium, (x1 , . . . , xn ) = (1/(n + 1), . . . , 1/(n + 1)). In this Nash equilibrium, each rms output is (1/(n + 1))(1 n/(n + 1)) = 1/(n + 1)2 . If xi = 1/(2n) for i = 1, . . . , n then each rms 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 Bertrands 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 modication. 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 rm i can increase its prot by reducing its price slightly below p (for example). 68.1 Bertrands oligopoly game Consider a prole (p1, . . . , pn ) of prices in which pi c for all i and at least two prices are equal to c. Every rms prot is zero. If any rm raises its price its prot remains zero. If a rm charging more than c lowers its price, but not below c, its prot also remains zero. If a rm lowers its price below c then its prot is negative. Thus any such prole is a Nash equilibrium. To show that no other prole is a Nash equilibrium, we can argue as follows. If some price is less than c then the rm charging the lowest price can increase its prot (to zero) by increasing its price to c. If exactly one rms price is equal to c then that rm can increase its prot by raising its price a little (keeping it less than the next highest price). If all rms prices exceed c then the rm charging the highest price can increase its prot by lowering its price to some price between c and the lowest price being charged. 68.2 Bertrands duopoly game with different unit costs a. If all consumers buy from rm 1 when both rms charge the price c2 , then (p1, p2 ) = (c2 , c2 ) is a Nash equilibrium by the following argument. Firm 1s prot is positive, while rm 2s prot is zero (since it serves no customers). If rm 1 increases its price, its prot falls to zero. If rm 1 reduces its price, say to p, then its prot changes from (c2 c1 )( c2 ) to (p c1 )( p). Since c2 is less than the maximizer of (p c1 )( p), rm 1s prot falls.
14
Chapter 3. Nash Equilibrium: Illustrations
If rm 2 increases its price, its prot remains zero. If rm 2 decreases its price, its prot 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 rm with the lower price (or either rm, if the prices are the same) can increase its prot (to zero) by raising its price above that of the other rm. If p1 > p2 c2 then rm 2 can increase its prot by raising its price a little. If p2 > p1 c1 then rm 1 can increase its prot by raising its price a little. If p2 p1 and p2 < c2 then rm 2s prot is negative, so that it can increase its prot by raising its price. If p1 = p2 > c2 then at least one of the rms is not receiving all of the demand, and that rm can increase its prot by lowering its price a little. b. Now suppose that the rule for splitting up the customers when the prices are equal species that rm 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 rm 2 receives some customers, rm 1 can increase its prot 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 1 different positions changes. If xi < x j , for example, the dividing line is 3 xi + 2 x j 3 1 rather than 2 (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 = 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 rms. If the products are different, then either rm 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 rms 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 rms market share is 50%, and if either rm changes its product its market share falls below 50%. Now suppose there are three rms. If all rms products are the same, each obtains one-third of the market. If x1 = x2 = x3 = m then any rm, by changing its product a little, can obtain close to one-half of the market. If x1 = x2 = x3 = m then any rm, by changing its product a little, can obtain more than one-half of the market. If the rms products are not all the same, then at least one of the extreme products is different from the other two products, and the rm that produces it can increase its market share by making it more similar to the other products. Thus when there are three rms 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 1 probability 2 ). 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 is
16
Chapter 3. Nash Equilibrium: Illustrations
payoff is 0, while if she concedes later her payoff is negative; player js payoff is v j , her highest possible payoff in the game. 85.1 Second-price sealed-bid auction with two bidders If player 2s 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 2s 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 2s 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 1s 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 2s 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
v1 v2
B1 (b2 )
b2
v1 v2
B2 (b1 )
0
v2
v1
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 1s best response function is in the left panel; player 2s 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 rst-price sealed-bid auction The prole (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 prole 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 1s 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 prole (b1 , . . . , bn ) of bids that satises 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 players bid). But no prole 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 nd actions for the other players and alternative bids for player i such that player is 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 is 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 players 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 = a1 and X2 a2 . The pair ( a1 , a2 ) is a Nash equilibrium by the following argument. If a2 = a2 then the victims level of care is sufcient (at least X2 ), so that the injurers payoff is given by (94.1) in the text. Thus the argument that the injurers action a1 is a best response to a2 is exactly the same as the argument for the case X2 = a2 in the text. Since X1 is the same as before, the victims payoff is the same also, so that by the argument in the text the victims best response to a1 is a2 . Thus ( a1 , a2 ) is a Nash equilibrium. To show that ( a1 , a2 ) is the only Nash equilibrium of the game, we study the players best response functions. First consider the injurers 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 injurers best response is a1 , as argued when showing that ( a1 , a2 ) is a Nash equilibrium. a2 > X2 : In this case the injurers best response is at most a1 , since her payoff is equal to a1 for larger values of a1 . Thus the injurers best response takes a form like that shown in the left panel of Figure 20.1. (In fact, b1 (a2 ) = a1 for X2 a2 a2 , but the analysis depends only on the fact that b1 (a2 ) a1 for a2 > X2 .)
a2
a2 X2
b1 (a2 )
a2
X2
b2 (a1 )
? a1
0
a1
a1
0
a1
Figure 20.1 The players best response functions under the rule of negligence with contributory negli gence when X1 = a1 and X2 = a2 . Left panel: the injurers best response function b1 . Right panel: the victims best response function b2 . (The position of the victims best response function for a1 > a1 is not signicant, and is not determined in the solution.)
Now consider the victims best response function. The victims payoff function is u2 (a1 , a2 ) = a2 if a1 < a1 and a2 X2 a2 L(a1 , a2 ) if a1 a1 or a2 < X2 .
As before, for a1 < a1 we have a2 L(a1 , a2 ) < a2 for all a2 , so that the victims best response is X2 . As in the text, the nature of the victims best responses to levels of care a1 for which a1 > a1 are not signicant. Combining the two best response functions we see that ( a1 , a2 ) is the unique Nash equilibrium of the game. Now consider the case in which X1 = M and a2 = a2 , where M a1 . The injurers payoff is u1 (a1 , a2 ) = a1 L(a1 , a2 ) if a1 < M and a2 a2 a1 if a1 M or a2 < a2 .
Now, the maximizer of a1 L(a1 , a2 ) is a1 (see the argument following (94.1) in the text), so that if M is large enough then the injurers best response to a2 is a1 . As before, if a2 < a2 then the injurers best response is 0, and if a2 > a2 then the
Chapter 3. Nash Equilibrium: Illustrations
21
a2
a2
b1 (a2 )
a2
a2 b2 (a1 )
?
0
a1 M
a1
0
a1 M
a1
Figure 21.1 The players best response functions under the rule of negligence with contributory negli gence when (X1 , X2 ) = (M, a2 ), with M a1 . Left panel: the injurers best response function b1 . Right panel: the victims best response function b2 . (The position of the victims best response function for a1 > M is not signicant, and is not determined in the text.)
injurers payoff decreases for a1 > M, so that her best response is less than M. The injurers best response function is shown in the left panel of Figure 21.1. The victims payoff is u2 (a1 , a2 ) = a2 if a1 < M and a2 a2 a2 L(a1 , a2 ) if a1 M or a2 < a2 .
If a1 a1 then the victims best response is a2 by the same argument as the one in the text. If a1 is such that a1 < a1 < M then the victims best response is at most a2 (since her payoff is decreasing for larger values of a2 ). This information about the victims best response function is recorded in the right panel of Figure 21.1; it is sufcient to deduce that ( a1 , a2 ) 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 1 which each player chooses each action with probability 2 . 106.2 Extensions of BoS with vNM preferences In the rst case, when player 1 is indifferent between going to her less preferred 1 concert in the company of player 2 and the lottery in which with probability 2 she and player 2 go to different concerts and with probability 1 they both go to her 2 more preferred concert, the Bernoulli payoffs that represent her preferences satisfy the condition u1 (S, S) = 1 u1 (S, B) + 1 u1 (B, B). 2 2 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 probabil1 ity 3 she and player 2 go to different concerts and with probability 4 they both go 4 to her more preferred concert, the Bernoulli payoffs that represent her preferences satisfy the condition u1 (S, S) = 3 u1 (S, B) + 1 u1 (B, B). 4 4 If we choose u1 (S, B) = 0 and u1 (B, B) = 2 (as before), then u1 (S, S) = 1 . Similarly, 2 for player 2 we can set u2 (B, S) = 0, u2 (S, S) = 2, and u2 (B, B) = 1 . Thus the 2 Bernoulli payoffs in the right panel of Figure 23.1 are consistent with the players preferences. Bach 2, 1 0, 0 Stravinsky 0, 0 1, 2 Bach 2, 1 2 0, 0 Stravinsky 0, 0 1 2, 2
Bach Stravinsky
Bach Stravinsky
Figure 23.1 The Bernoulli payoffs for two extensions of BoS.
23
24
Chapter 4. Mixed Strategy Equilibrium
Player 1s expected payoff
q=1 1 q=
1 2 1 2
2
1
q=0 0 p 1
Figure 24.1 Player 1s 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, 1 , and 1. 2
110.1 Expected payoffs For BoS, player 1s expected payoff is shown in Figure 24.1. For the game in the right panel of Figure 21.1 in the book, player 1s expected payoff is shown in Figure 24.2.
Player 1s expected payoff
q=1 2
3 2
3
1
q=
1 2
q=0 0 p 1
Figure 24.2 Player 1s 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, 1 , and 1. 2
111.1 Examples of best responses
1 For BoS: for q = 0 player 1s unique best response is p = 0 and for q = 2 and q = 1 her unique best response is p = 1. For the game in the right panel of Figure 21.1: 1 for q = 0 player 1s unique best response is p = 0, for q = 2 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 HawkDove Denote by ui a payoff function whose expected value represents player is preferences. The conditions in the problem imply that for player 1 we have
1 u1 (Passive, Passive) = 2 u1 (Aggressive, Aggressive) + 1 u1 (Aggressive, Passive) 2
and u1 (Passive, Aggressive) = 2 u1 (Aggressive, Aggressive) + 1 u1 (Passive, Passive). 3 3 Given u1 (Aggressive, Aggressive) = 0 and u1 (Passive, Aggressive = 1, we have u1 (Passive, Passive) = 1 u1 (Aggressive, Passive) 2 and 1 = 1 u1 (Passive, Passive), 3 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)), (( 3 , 1 ), ( 3 , 1 )), and ((1, 0), (0, 1)). 44 44
1 q
3 4
B1 B2
Aggressive Passive
Aggressive 0, 0 1, 6
Passive 6, 1 3, 3
0
3 4
1 p
Figure 25.1 An extension of HawkDove (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 sufces to verify the conditions in Proposition 116.2. Thus, given that each players strategy species a positive probability for every action, it sufces to show that each action of each player yields the same expected payoff. Player 1s 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 2s 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 1s expected payoff if she chooses the number k is q . Hence if p > 0 then (by k k the rst condition in Proposition 116.2) we need q q for all j, so that, in j k particular, q > 0 (q cannot be zero for all j!). But player 2s expected payoff j k if she chooses the number k is p , so given q > 0 we need p p for all j k k k j (again by the rst condition in Proposition 116.2), and, in particular, p k 1/K (p cannot exceed 1/K for all j!). We conclude that any probability p j k that is positive must be at most 1/K. The only possibility is that p = 1/K k for all k. A similar argument implies that qk = 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,
3 1 or if and only if 1 4r < p < 1 2 r. For example, the mixed strategies ( 4 , 1 , 1 ) 42 13 and (0, 4 , 4 ) both strictly dominate T.
120.3 Strict domination for mixed strategies (a) True. Suppose that the mixed strategy i assigns positive probability to the action ai , which is strictly dominated by the action ai . Then ui (ai , ai ) > ui (ai , ai ) for all ai . Let i be the mixed strategy that differs from i only in the weight that i assigns to ai is transferred to ai . That is, i is dened by i (ai ) = 0, i (ai ) = i (ai ) + i (ai ), and i (bi ) = i (bi ) for every other action bi . Then i strictly dominates i : for any ai we have U(i , ai ) U(i , ai ) = i (ai )(u(ai , ai ) ui (ai , ai )) > 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 1s payoffs to (T, L) and to (T, R) are both 5 instead of 1. Then player 1s 2 1 mixed strategy that assigns probability 2 to M and probability 1 to B is strictly 2 dominated by T, even though neither M nor B is strictly dominated. 127.1 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. Referring to Figure 126.1 in the text, we see that the set of mixed strategy Nash equilibria correspond to p = 0 and / q 1. 130.3 Bargaining The game is given in Figure 27.1. 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
0 2 4 6 8 10
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 1s 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 1s expected payoffs to 2 and 8 are the same if the probability player 2 assigns to 2 is 2 (and the probability she assigns to 8 is 3 ). Given these probabilities, 5 5 player 1s expected payoff to her actions 2 and 8 is 16 , and her expected payoff to 5 every other action is less than 16 . Thus the pair of mixed strategies in which every 5 player assigns probability 2 to 2 and 3 to 8 is a symmetric mixed strategy Nash 5 5 equilibrium. Similarly, the game has a symmetric mixed strategy equilibrium ( , ) in 4 which assigns probability 5 to the demand of 4 and probability 1 to the demand 5 of 6.
28
Chapter 4. Mixed Strategy Equilibrium
In summary, the game has three symmetric mixed strategy Nash equilibria in which each players strategy assigns positive probability to at most two actions: 2 one in which probability 1 is assigned to 10, one in which probability 5 is assigned 4 to 2 and probability 3 is assigned to 8, and one in which probability 5 is assigned 5 1 to 4 and probability 5 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 v c1 = v Pr{at least one other person calls}
= v 1 (1 p1 )n1 1 (1 p2 )n2 ,
or c1 = v(1 p1 )n1 1 (1 p2 )n2 . (28.1) Similarly, for each witness with cost c2 to report with positive probability less than one, we need v c2 = v Pr{at least one other person calls}
= v 1 (1 p1 )n1 (1 p2 )n2 1 ,
or c2 = v(1 p1 )n1 (1 p2 )n2 1 . 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 Similarly, p2 = 1 c2 v c1 c2
n1 1/(n1)
(28.2)
c1 v
c2 c1
n2
1/(n1)
.
.
For these two numbers to be probabilities, we need each of them to be nonnegative and at most one, which requires
n c2 2 v 1/(n2 1) n < c1 < vc2 1 1 1/n1
.
Chapter 4. Mixed Strategy Equilibrium
29
136.1 Best response dynamics in Cournots duopoly game The best response functions of both rms are the same, so if the rms outputs are t t initially the same, they are the same in every period: q1 = q2 for every t. For each period t, we thus have 1 qt = 2 ( c qt ). i i Given that q1 = 0 for i = 1, 2, solving this rst-order difference equation we have i
1 1 qt = 3 ( c)[1 ( 2 )t1 ] i
for each period t. When t is large, qt is close to 1 ( c), a rms equilibrium i 3 output. 5 In the rst few periods, these outputs are 0, 1 ( c), 1 ( c), 3 ( c), 16 ( 2 4 8 c). 139.1 Finding all mixed strategy equilibria of two-player games Left game: There is no equilibrium in which each players 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 2s payoff the same for both her actions, and for neither action of player 2 is player 1s 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 1s expected payoff to her two actions to be the same we need 6q = 3q + 6(1 q), or q = 2 . For player 2s expected payoff to her two actions to be the same we 3 need 2(1 p) = 6p, or p = 1 . We conclude that the game has a unique mixed strategy equilib4 3 rium, (( 1 , 4 ), ( 2 , 1 )). 4 33 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 2s payoffs to (T, L) and (T, R) are not the same. {B} for player 1, {L, R} for player 2: no equilibrium, because player 2s payoffs to (B, L) and (B, R) are not the same. {T, B} for player 1, {L} for player 2: no equilibrium, because player 1s payoffs to (T, L) and (B, L) are not the same. {T, B} for player 1, {R} for player 2: player 1s 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 2s 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 1s mixed strategy is (p, 1 p), with p 1 , and player 2 uses R with probability 1. 2
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 1s 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 1 p 1 (of which one 2 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 ))n1 . Thus, given that the probability that she ties for the highest bid is zero, her expected payoff is
(K ai )(F(ai ))n1 + (ai )(1 (F(ai ))n1 ).
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 ))n1 ai = c for all 0 ai K. For F(0) = 0 we need c = 0, so that F(ai ) = (ai /K)1/(n1) 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/(n1); each players equilibrium expected payoff is 0. Each players mean bid is K/n. 147.2 Preferences over lotteries The rst piece of information about the decision-makers preferences among lotteries is consistent with her preferences being represented by the expected value 1 of a payoff function: set u(a1 ) = 0, u(a2 ) equal to any number between 2 and 1 , 4 and u(a3 ) = 1. The second piece of information about the decision-makers 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-makers 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 rst 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 = u(a)/(u(a) u(a)) and = 1/(u(a) u(a)) > 0. Lemma 148.1 implies that the function v dened by v(x) = + u(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. EG 1, 0 2, 3 EH 1, 0 0, 1 FG 3, 2 2, 3 FH 3, 2 0, 1
C D
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. E 1, 2 0, 0 2, 0 2, 0 F 3, 1 3, 1 2, 0 2, 0
CG CH DG DH
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 1s 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 1s strategy is optimal at the start of the game, given player 2s strategy, (b) in the subgame following the history C, player 2s strategy E induces the strategy E, which is optimal given player 1s strategy, and (c) in the subgame following the history (C, E), player 1s strategy DG induces the strategy G, which is optimal. 33
34
Chapter 5. Extensive Games with Perfect Information: Theory
R B E B 1, 2, 1 E B 1, 2, 1 H 0, 0, 0 B 0, 0, 0 H 0, 0, 0 B 0, 0, 0 E H 2, 1, 2 B 1, 2, 1 H E H 2, 1, 2 B 1, 2, 1 R H 0, 0, 0 R H 0, 0, 0 B
E H R B 0, 0, 0 R B 0, 0, 0 H 2, 1, 2 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 rst 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 rst 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 rst 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 rst round. That is, there is a subgame perfect equilibrium in which on the rst 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 rst round.) 177.3 Comparing simultaneous and sequential games
a. Denote by (a1 , a2 ) a Nash equilibrium of the strategic game in which player 1s payoff is maximal in the set of Nash equilibria. Because (a1 , a2 ) is a Nash is a best response to a . By assumption, it is the only best equilibrium, a2 1
Chapter 5. Extensive Games with Perfect Information: Theory
35
response to a1 . Thus if player 1 chooses a1 in the extensive game, player 2 in any subgame perfect equilibrium of the extensive game. must choose a2 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 1s 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 2s payoff is 1. The extensive game has a unique subgame perfect equilibrium, (B, LR) (where the rst component of player 2s 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 2s payoff is 2. L 1, 1 0, 0 R 3, 0 2, 2
T B
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 2s payoff is 2. A subgame perfect equilibrium of the extensive game is (B, RL) (where the rst component of player 2s 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 1s payoff is 1. (If you read Chapter 4, you can nd the mixed strategy Nash equilibria of the strategic game; in all these equilibria, as in the pure strategy Nash equilibrium, player 1s expected payoff exceeds 1.) L 2, 2 1, 1 R 0, 2 3, 0
T B
Figure 35.2 The payoffs for the example in Exercise 177.3c.
179.3 Three Mens 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 1s strategy is to offer x and person 2s strategy is to accept x and any offer more favorable, and reject every other offer. (Given person 2s strategy, person 1 can do no better than offer x. Given person 1s 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 nitely 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 1s best strategy is to offer 0, as before. If player 2 accepts all offers except 0 then player 1s 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 dened 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 1s 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 2s 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 Stackelbergs duopoly game with quadratic costs From Exercise 59.1, the best response function of rm 2 is the function b2 dened by 1 ( q1 ) if q1 b2 (q1 ) = 4 0 if q1 > . Firm 1s subgame perfect equilibrium strategy is the value of q1 that maximizes 1 1 q1 ( q1 b2 (q1 )) q2 , or q1 ( q1 4 ( q1 )) q2 , or 4 q1 (3 7q1 ). The 1 1 3 maximizer is q1 = 14 . We conclude that the game has a unique subgame perfect equilibrium, in which 3 rm 1s strategy is the output 14 and rm 2s strategy is its best response function b2 . The outcome of the subgame perfect equilibrium is that rm 1 produces q1 = 3 = b ( 3 ) = 11 units. In a Nash 2 14 14 units of output and rm 2 produces q2 56 1 equilibrium of Cournots (simultaneous-move) game each rm produces 5 (see Exercise 59.1). Thus rm 1 produces more in the subgame perfect equilibrium of the sequential game than it does in the Nash equilibrium of Cournots game, and rm 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, . . . , xn1 ) = n for all (x1 , . . . , xn1 ). Preferences Each candidates 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 rst 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 Hotellings 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 1s strategy is m; candidate 2s 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 rst candidates entering at 1 , the second candidates staying 2 out, and the third candidates entering at 1 . 2 To show this, rst 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
x2
an d
3
1 wins
tie
In (e.g. at z) 3 wins
In (e.g. at 1 ) 2 3 wins
2,
ins 1w
1,
1 d an
2 3
2 wins
2 tie
In (near 1 ); 3 wins 2
2 wi ns
2 wi ns
1 d an
1 3
In (e.g. at 1 ) 2 3 wins
2 wins
2 tie
ns 1 wi
1,
0 Out In; wins
2 3 3
2,
In (e.g. at z) 3 wins x1 In; 3 wins 1
1 3 1 2
In; 1 and 3 tie
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 1 (x1 + x2 ). 2
an d
1 wins
3
tie
40
Chapter 6. Extensive Games with Perfect Information: Illustrations
Finally, consider candidate 1s 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 1 the best position at which to do so is x1 = 2 , 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 1 , candidate 2 stays out, and candidate 3 enters at 1 . (There 2 2 are many subgame perfect equilibria, because after many histories candidate 3s 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 1s 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 2s options after player 1s move.
Now consider the optimal action of candidate 2, given x1 and the outcome of candidate 3s best response, as given in Figure 39.1. In the gure, take a value of x1 and look at the outcomes as x2 varies; nd the value of x2 that induces the best outcome for candidate 2. For example, for x1 = 0 the only value of x2 for 2 which candidate 2 does not lose is 3 , at which point she ties with the other two candidates. Thus when candidate 1s strategy is x1 = 0, candidate 2s best action, 2 given candidate 3s best response, is x2 = 3 , which leads to a three-way tie. We nd that the outcome of the optimal value of x2 , for each value of x1 , is given as follows. 2 1, 2, and 3 tie (x2 = 3 ) if x1 = 0 2 wins 1 if 0 < x1 < 2 1 1 and 3 tie (2 stays out) if x1 = 2 2 wins if 1 < x1 < 1 2 1 1, 2, and 3 tie (x2 = 3 ) if x1 = 1.
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 2s optimal action is to take one step, and win. Thus player 1s best action is not to move; player 2s 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 2s 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 1s 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 1s 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 4, 2 0, 0 D 0, 0 2, 4 B E 3, 1 0, 0 F 0, 0 1, 3
C D
E F
Figure 43.1 The game in Exercise 210.2.
The subgame following player 1s choice of A has two Nash equilibria, (C, C) and (D, D); the subgame following player 1s 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 rst component of player 1s 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 rst component of player 2s 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 rst 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 rst 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 citizens 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 candidates 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 is 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 rst 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 candidates 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 rms are active in period t1 , then rm 2s prot in this period is (100 t1 c k1 k2 )k2 = (20)(20) = 400. Its prot in any period t in which it is alone in the market is (100 t c k2 )k2 = (70 t)(20). Thus its prot from period t1 + 1 through period t2 is
(19 + 18 + . . . + 1)(20) = 3800.
Hence rm 2s loss in period t1 when both rms are active is (much) less than the sum of its prots in periods t1 + 1 through t2 when it alone is active. 227.1 Variant of ultimatum game with equity-conscious players The game is dened as follows. Players The two people. Terminal histories The set of sequences (x, 2 , Z), where x is a number with 0 x c (the amount of money that person 1 offers to person 2), 2 is 0 or 1 (the value of 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, 2 ) = 2 for all x and all 2. Chance probabilities For every history x, chance chooses 0 with probability p and 1 with probability 1 p. Preferences Each persons preferences are represented by the expected value of a payoff equal to the amount of money she receives. For any terminal history (x, 2, Y) person 1 receives c x and person 2 receives x; for any terminal history (x, 2 , N) each person receives 0. Given the result from Exercise 183.4 given in the question, if person 1s offer x satises 0 < x < 1 then the offer is rejected with probability 1 p, so that per3 1 son 1s expected payoff is p(1 x), while if x > 3 the offer is certainly accepted, independent of the type of person 2. Thus person 1s optimal offer is
1 3
0
2 if p < 3 2 if p > 3 ;
2 if p = 3 then both offers are optimal. If p > 2 we see that in a subgame perfect equilibrium person 1s offers are 3 rejected by every person 2 with whom she is matched for whom 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 A B 1 , 1 , 1 1 , 1 , 1 A B 0, 0, 1 1 , 0, 1 A 0, 1 , 0 1 , 1 , 0 B B 1 , 0, 0 0, 0, 0
A B
Figure 46.1 The players 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 modied game, player 1s 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 ].
1 If 0 < pi < 2 for i = 1, 2, 3, this difference is positive, so that (A, A, A) is not a Nash equilibrium of the modied 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 species 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 rst three conditions we conclude that 2x1 + 2x2 + 2x3 3,
3 or x1 + x2 + x3 2 , contradicting the last condition. Thus no action of the grand coalition satises all the conditions, so that the core of the game is empty.
248.1 Core of landownerworker 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 landownerworker 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 players 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. 249.2 Landownerworker 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 k , 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 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 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 k +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{k , k +1 } and p min{ k , k +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 greatera majority of the playersprefers m to x. Similarly, if x > m then every player whose favorite position is m or lessa majority of the playersprefers 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 1 ). Thus there is no Nash equilibrium in which player 1 chooses 2 S. Now consider the mixed strategy Nash equilibria. If both types of player 2 use a pure strategy then player 1s 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 1s mixed strategy assigns to B. In order for type 1 2 of player 2 to obtain the same expected payoff to B and S we need p = 3 . 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 1s expected payoff if she chooses B is
1 2
2q = q
and her expected payoff if she chooses S is
1 2 1 1 (1 q) + 2 1 = 1 2 q.
2 These expected payoffs are equal if and only if q = 3 . Thus the game has a mixed 2 strategy equilibrium in which the mixed strategy of player 1 is ( 3 , 1 ), that of type 1 3 21 of player 2 is ( 3 , 3 ), 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 ( 1 , 2 ), that of type 1 of player 2 is (0, 1), and that of type 2 of player 2 is 33 ( 2 , 1 ). 33 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, B)
B S 0 1
(B, S)
1
1 2
(S, B)
1
1 2
(S, S)
2 0
Type n1 of player 1
B S
(B, B) 1
0
(B, S)
2 3 2 3
(S, B)
1 3 4 3
(S, S) 0
2
Type y2 of player 2
B S
(B, B) 0
2
(B, S)
1 3 4 3
(S, B)
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 is ticket (an integer from 1 to m). Actions The set of actions of each player is {Exchange, Dont exchange}. Signals The signal function of each player i is dened by i (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 is Bernoulli payoff function is given by ui ((X, Y), ) = j if X = Y = Exchange and ui ((X, Y), ) = i 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 Dont 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 Cournots duopoly game with imperfect information We have b1 (q L , q H ) =
1 2 ( c (q L
0
+ (1 )q H )) if q L + (1 )q H c otherwise.
The best response function of each type of player 2 is similar: b I (q1 ) =
1 2 ( c I
0
q1 ) if q1 c I otherwise
for I = L, H. The three equations that dene a Nash equilibrium are
q1 = b1 (q , q ), q = b L (q1 ), and q = b H (q1 ). LH L H
Solving these equations under the assumption that they have a solution in which all three outputs are positive, we obtain
q1 =
1 3 ( 2c + c L + (1 )c H ) 1 1 3 ( 2c L + c) 6 (1 )(c H c L ) 1 1 3 ( 2c H + c) + 6 (c H c L )
q = L q = H
If both rms know that the unit costs of the two rms are c1 and c2 then in 1 a Nash equilibrium the output of rm i is 3 ( 2ci + c j ) (see Exercise 58.1). In the case of imperfect information considered here, rm 2s output is less than 1 1 3 ( 2c L + c) if its cost is c L and is greater than 3 ( 2c H + c) if its cost is c H . Intuitively, the reason is as follows. If rm 1 knew that rm 2s 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 rm 2s cost were high. Thus if rm 2s cost is in fact high, rm 2 benets from rm 1s lack of knowledge and optimally produces more than it would if rm 1 knew its cost. 288.1 Cournots duopoly game with imperfect information The best response b0 (q L , q H ) of type 0 of rm 1 is the solution of max[(P(q0 + q L ) c)q0 + (1 )(P(q0 + q H ) c)q0 ].
q0
The best response b (q L , q H ) of type of rm 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 rm 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 rm 2 is the solution of max[(1 )(P(q0 + q L ) c L )q L + (P(q + q L ) c L )q L ]
qL
and the best response b H (q0 , q , qh ) of type H of rm 2 is the solution of max[(1 )(P(q0 + q H ) c H )q H + (P(qh + q H ) c H )q H ].
qH
A Nash equilibrium is a prole (q0 , q , qh , q , q ) for which q0 , q , and q are LH h and q , and q and q are best responses to q , q , and q . best responses to q L H L H 0 h When P(Q) = Q for Q and P(Q) = 0 for Q > we nd, after some exciting algebra, that q0 =
q = q = H q = L q = H
1 ( 2c + c H (c H c L )) 3 1 (1 )(1 )(c H c L ) 2c + c L + 3 4 1 (1 )(c H c L ) 2c + c H 3 4 1 2(1 )(1 )(c H c L ) 2c L + c 3 4 1 2(1 )(c H c L ) 2c H + c + . 3 4
When = 0 we have
q0 =
q = q = H q = L q = H
1 ( 2c + c H (c H c L )) 3 1 (1 )(c H c L ) 2c + c L + 3 4 1 (c H c L ) 2c + c H 3 4 1 (1 )(c H c L ) 2c L + c 3 2 1 (c H c L ) 2c H + c + , 3 2
so that q0 is equal to the equilibrium output of rm 1 in Exercise 287.1, and q L are the same as the equilibrium outputs of the two types of rm 2 in that and q H exercise.
Chapter 9. Bayesian Games
55
When = 1 we have
q0 =
q = q = H q = L q = H
1 ( 2c + c H (c H c L )) 3 1 ( 2c + c L ) 3 1 ( 2c + c H ) 3 1 ( 2c L + c) 3 1 ( 2c H + c) , 3
so that q and q are the same as the equilibrium outputs when there is perfect L information and the costs are c and c L (see Exercise 58.1), and q and q are the H h same as the equilibrium outputs when there is perfect information and the costs are c and c H . Now, for an arbitrary value of we have q = L q H 1 3 1 = 3 2(1 )(1 )(c H c L ) 4 2(1 )(c H c L ) 2c H + c + . 4 2c L + c
To show that for 0 < < 1 the values of these variables lie between their values when = 0 and when = 1, we need to show that 0 and 0 2(1 )(1 )(c H c L ) (1 )(c L c H ) 4 2 2(1 )(c H c L ) (c L c H ) . 4 2
These inequalities follow from c H c L , 0, and 0 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 = 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 is 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 prole 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. Highest of other players bids bi = b bi < b < v i (m-way tie) (vi b)/m 0 vi b vi b
b < bi is bid bi < v i vi vi b vi b
b vi 0 0
Figure 56.1 Player is payoffs to her bids bi < vi and vi in a second-price sealed-bid auction as a function of the highest of the other players bids, denoted b.
Player is 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. Highest of other players bids bi = b v i < b < bi (m-way tie) 0 0 vi b
b vi is bid vi bi > v i vi b vi b
b > bi 0 0
(vi b)/m
Figure 56.2 Player is payoffs to her bids vi and bi > vi in a second-price sealed-bid auction as a function of the highest of the other players bids, denoted b.
As before, player is 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 = 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/)t2 and that type t1 of player 1 bids b1 . Then by the calculations in the text, with = 1 and = 1/, a bid of b1 by player 1 wins with probability b1 /(1 + 1/)
1 the expected value of player 2s bid, given that it is less than b1 , is 2 b1 1 the expected value of signals that yield a bid of less than b1 is 2 b1 /(1 + 1/) (because of the uniformity of the distribution of t2 ).
Thus player 1s expected payoff if she bids b1 is
1 (t1 + 2 b1 /(1 + 1/) 1 b1 ) 2
b1 , 1 + 1/
or
(2(1 + )t1 b1 )b1 . 2(1 + )2
This function is maximized at b1 = (1 + )t1 . That is, if each type t2 of player 2 bids (1 + 1/)t2 , any type t1 of player 1 optimally bids (1 + )t1 . Symmetrically, if each type t1 of player 1 bids (1 + )t1 , any type t2 of player 2 optimally bids (1 + 1/)t2 . Hence the game has the claimed Nash equilibrium. 299.2 First-price sealed-bid auction with common valuations
1 Suppose that each type t2 of player 2 bids 2 ( + )t2 and type t1 of player 1 bids b1 . To determine the expected payoff of type t1 of player 1, we need to nd the probability with which she wins, and the expected value of player 2s signal if player 1 wins. (The price she pays is her bid, b1 .) Probability of player 1s winning: Given that player 2s bidding function is 1 1 ( + )t2 , player 1s bid of b1 wins only if b1 2 ( + )t2 , or if t2 2b1 /( + ). 2 Now, t2 is distributed uniformly from 0 to 1, so the probability that it is at most 2b1 /( + ) is 2b1 /( + ). Thus a bid of b1 by player 1 wins with probability 2b1 /( + ). Expected value of player 2s signal if player 1 wins: Player 2s bid, given her signal t2 , is 1 ( + )t2, so that the expected value of signals that yield a bid of less 2 than b1 is b1 /( + ) (because of the uniformity of the distribution of t2 ). Thus player 1s expected payoff if she bids b1 is
2(t1 + b1 /( + ) b1 ) or
b1 , +
2 (( + )t1 b1 )b1 . ( + )2
58
Chapter 9. Bayesian Games
1 This function is maximized at b1 = 2 ( + )t1 . That is, if each type t2 of player 2 1 1 bids 2 ( + )t2 , any type t1 of player 1 optimally bids 2 ( + )t1 . Hence, as claimed, the game has a Nash equilibrium in which each type ti of player i bids 1 2 ( + )t i .
309.2 Properties of the bidding function in a rst-price sealed-bid auction We have (v) = 1
(F(v))n1 (F(v))n1 (n 1)(F(v))n2 F (v) (F(v))2n2 (F(v))n (n 1)F (v) (F(v))n
v (v) v (F(x))n1 dx (F(v))n v n1 dx v (F(x))
v n1 dx v (F(x))
= 1 =
(n 1)F
> 0 if v > v
because F (v) > 0 (F is increasing). (The rst 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 (v) < v. If v = v then both the numerator and denominator of the quotient in (309.1) are zero, so we may use LHpitals rule to nd the value of the quotient as v v. Taking the derivatives of the numerator and denominator we obtain
(F(v))n1 F(v) , = n2 F (v) (n 1)F (v) (n 1)(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 (v) = v. 309.3 Example of Nash equilibrium in a rst-price auction From (309.1) we have (v) = v x n1 dx vn1 v n1 x dx = v 0 n1 v = v v/n = (n 1)v/n.
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. 1, 1 Pass See 0, 0 Meet Raise HH 1 1, 1 See Raise Pass 1, 1 Meet 1 + k, 1 k HL ( 1 ) 4 2 (1) 4 LH (1) 4 1 See Raise Pass 1, 1 Meet 0, 0 0, 0 2 1, 1 Pass Raise
1 k, 1 + k
Meet See
0, 0
1, 1
Chance LL ( 1 ) 4
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 rst letter of each strategy is player 1s 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 strategiesReady, Unready, and Out. The incumbent also has a single information set, at which two actions are available, so it has two strategiesAcquiesce 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 Challengers strategy is (Out, R), the Incumbents strategy is F, and the Incumbents belief assigns probability 1 to the history (In, U) at her information set. Each players strategy is sequentially rational. The Incumbents belief satises 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 Incumbents action F is not optimal when the Challenger chooses R. (The Incumbents 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 Challengers 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 rm charges the price H and the low-quality rm charges any nonnegative price, and the consumer buys the good from a high-quality rm, does not buy the good from a low-quality rm that charges a positive price, and may or may not buy from a low-quality rm that charges a price of 0. Consider an assessment in which each type of rm chooses (p , E ) in the rst period, the consumer believes the rm is high-quality with probability 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 rm under this assessment is p + H E 2c H , that of a low-quality rm is p E , and that of the consumer is (H p ) + (1 )(p ) = H p . This assessment is consistentthe only rst-period action of the rm observed in equilibrium is (p , E ), and after observing this pair the consumer believes, correctly, that the rm is high-quality with probability . Under what conditions is the assessment sequentially rational? Firm If the rm chooses a (price, expenditure) pair different from (p , E ) then the consumer does not buy the good, and the rms prot is 0. Thus for the assessment to be an equilibrium we need p + H E 2c H 0 (for the high-quality rm) and p E 0 (for the low-quality rm). 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 H p 0. In summary, the assessment is a weak sequential equilibrium if and only if max{E , E H + 2c H } p H.
Chapter 10. Extensive Games with Imperfect Information
61
346.1 Comparing the receivers expected payoff in two equilibria The receivers 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 1 t1 to the left and move the section from 0 to 1 t1 2 2 to the interval from 1 1 t1 to 1 then the area above the black curve is a subset of 2 the area above the gray curve. 0
1 2 t1
t1
1 2
1 2 (t1
+ 1)
t
1
( 1 t1 t)2 2
1 ( 2 (t1 + 1) t)2
( 1 t)2 2
Figure 61.1 The gray curve gives the receivers 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 players 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 players 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 12 12 34 34 56 55 78 43 9 10 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 players action varies.
363.3 Finding a maxminimizer The analog of Figure 364.1 in the text is Figure 63.2. From this gure we see that the 3 maxminimizer for player 2 is the strategy that assigns probability 5 to L. Player 2s 1 maxminimized payoff is 5 .
payoff of 1 player 1 1 5
B 0
3 5
2 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 1s 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 1s ranking of the four outcomes is the reverse of player 2s ranking). Not strictly competitive in mixed strategies (there exist no values of and > 0 such that u1 (a) = + u2 (a) for every outcome a; or, alternatively, player 1 is indifferent between (B, L) and the lottery that yields (T, L) with probability 1 and (T, R) with 2 probability 1 , whereas player 2 is not indifferent between these two outcomes). 2 Game in Figure 367.1: Strictly competitive both in pure and in mixed strategies. (Player 2s preferences are represented by the expected value of the Bernoulli payoff function u1 because u1 (a) = 1 + 1 u2 (a) for every pure outcome a.) 2 2 370.2 Maxminimizing in BoS
2 1 Player 1s maxminimizer is ( 3 , 2 ) while player 2s is ( 3 , 1 ). Clearly neither pure 3 3 equilibrium strategy of either player guarantees her equilibrium payoff. In the mixed strategy equilibrium, player 1s expected payoff is 2 . But if, for example, 3 player 2 choose S instead of her equilibrium strategy, then player 1s expected payoff is 1 . Similarly for player 2. 3
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 1s 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. L 0, 0 1, 1 R 1, 1 0, 0
T B
Figure 64.1 The game in Exercise 372.2.
372.4 ONeills 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 1s actions yield the same ex1 pected payoff if and only if 4q 1 = 1 6q, or q = 5 , and similarly all 1 four of player 2s actions yield the same expected payoff if and only if p = 5 . 1112 1112 Thus (( 5 , 5 , 5 , 5 ), ( 5 , 5 , 5 , 5 )) is a Nash equilibrium of the game. The players 1 payoffs in this equilibrium are ( 5 , 1 ). 5
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 1 guarantee her the payoff of 5 , we need
1 p1 + p2 + p3 p4 5 1 p1 p2 + p3 p4 5 1 p1 + p2 p3 p4 5 1 p1 p2 p3 + p4 5 . 2 Adding these four inequalities, we deduce that p4 5 . Adding each pair of 1 the rst three inequalities, we deduce that p1 5 , p2 1 , and p3 1 . We 5 5 1 1 have p1 + p2 + p3 + p4 = 1, so we deduce that (p1 , p2, p3, p4 ) = ( 5 , 1 , 5 , 2 ). 5 5 A similar analysis of the conditions for player 2s strategy to guarantee her 1 the payoff of 1 leads to the conclusion that (q1 , q2 , q3 , q4 ) = ( 5 , 1 , 1 , 2 ). 5 555
12
Rationalizability
379.2 Best responses to beliefs Consider a two-player game in which player 1s payoffs are given in Figure 67.1. 1 The action B of player 1 is a best response to the belief that assigns probability 2 to both L and R, but is not a best response to any belief that assigns probability 1 to either action. L 3 0 2 R 0 3 2
1 2
T M B
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.
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 1 to L and R (in order that player 1 be indifferent between T and B); 2 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 1s 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 Hotellings 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 ...

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