Game theory and exercises

Game Theory and Exercises Game Theory and Exercises introduces the main concepts of game theory, along with interactive exercises to aid readers’ learning and understanding Game theory is used to help players understand decisionmaking, risk-taking and strategy and the impact that the choices they make have on other players; and how the choices of those players, in turn, influence their own behaviour So, it is not surprising that game theory is used in politics, economics, law and management This book covers classic topics of game theory including dominance, Nash equilibrium, backward induction, repeated games, perturbed strategies, beliefs, perfect equilibrium, perfect Bayesian equilibrium and replicator dynamics It also covers recent topics in game theory such as level-k reasoning, best reply matching, regret minimization and quantal responses This textbook provides many economic applications, namely on auctions and negotiations It studies original games that are not usually found in other textbooks, including Nim games and traveller’s dilemma The many exercises and the inserts for students throughout the chapters aid the reader’s understanding of the concepts With more than 20 years’ teaching experience, Umbhauer’s expertise and classroom experience helps students understand what game theory is and how it can be applied to real life examples This textbook is suitable for both undergraduate and postgraduate students who study game theory, behavioural economics and microeconomics Gisèle Umbhauer is Associate Professor of Economics at the University of Strasbourg, France Routledge Advanced Texts in Economics and Finance Financial Econometrics Peijie Wang Macroeconomics for Developing Countries, 2nd edition Raghbendra Jha Advanced Mathematical Economics Rakesh Vohra Advanced Econometric Theory John S Chipman Understanding Macroeconomic Theory John M Barron, Bradley T Ewing and Gerald J Lynch Regional Economics Roberta Capello Mathematical Finance: Core Theory, Problems and Statistical Algorithms Nikolai Dokuchaev Applied Health Economics Andrew M Jones, Nigel Rice, Teresa Bago d’Uva and Silvia Balia â•⁄9 Information Economics Urs Birchler and Monika Bütler 10 Financial Econometrics (Second Edition) Peijie Wang 11 Development Finance Debates, Dogmas and New Directions Stephen Spratt 12 Culture and Economics On Values, Economics and International Business Eelke de Jong 13 Modern Public Economics, Second Edition Raghbendra Jha 14 Introduction to Estimating Economic Models Atsushi Maki 15 Advanced Econometric Theory John Chipman 16 Behavioral Economics Edward Cartwright 17 Essentials of Advanced Macroeconomic Theory Ola Olsson 18 Behavioral Economics and Finance Michelle Baddeley 19 Applied Health Economics, Second Edition Andrew M Jones, Nigel Rice, Teresa Bago d’Uva and Silvia Balia 20 Real Estate Economics A Point to Point Handbook Nicholas G Pirounakis 25 Strategic Entrepreneurial Finance From Value Creation to Realization Darek Klonowski 21 Finance in Asia Institutions, Regulation and Policy Qiao Liu, Paul Lejot and Douglas Arner 26 Computational Economics A concise introduction Oscar Afonso and Paulo Vasconcelos 22 Behavioral Economics, Second Edition Edward Cartwright 27 Regional Economics, Second Edition Roberta Capello 23 Understanding Financial Risk Management Angelo Corelli 28 Game Theory and Exercises Gisèle Umbhauer 24 Empirical Development Economics Måns Söderbom and Francis Teal with Markus Eberhardt, Simon Quinn and Andrew Zeitlin This page intentionally left blank Game Theory and Exercises Gisèle Umbhauer First published 2016 by Routledge Park Square, Milton Park, Abingdon, Oxon OX14 4RN by Routledge 711 Third Avenue, New York, NY 10017 Routledge is an imprint of the Taylor & Francis Group, an informa business © 2016 Gisèle Umbhauer The right of Gisèle Umbhauer to be identified as author of this work has been asserted by her in accordance with the Copyright, Designs and Patent Act 1988 All rights reserved No part of this book may be reprinted or reproduced or utilized in any form or by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying and recording, or in any information storage or retrieval system, without permission in writing from the publishers Trademark notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloging in Publication Data Umbhauer, Gisele Game theory and exercises / Gisele Umbhauer Game theory I Title HB144.U43 2015 519.3 dc23 2015020124 ISBN: 978-0-415-60421-5 (hbk) ISBN: 978-0-415-60422-2 (pbk) ISBN: 978-1-315-66906-9 (ebk) Typeset in Times New Roman and Bell Gothic by Saxon Graphics Ltd, Derby Dedication To my son Victor This page intentionally left blank Contents Acknowledgementsxix Introduction1 HOW TO BUILD A GAME INTRODUCTION5 STRATEGIC OR EXTENSIVE FORM GAMES? 1.1 Strategic/normal form games 1.1.1 Definition 1.1.2 Story strategic/normal form games and behavioural comments 1.1.3 All pay auction 12 1.2 Extensive form games 18 1.2.1 Definition 18 1.2.2 Story extensive form games and behavioural comments 20 1.2.3 Subgames 27 2 STRATEGIES 28 2.1 Strategies in strategic/normal form games 28 2.1.1 Pure strategies 28 2.1.2 Mixed strategies 28 2.2 Strategies in extensive form games 30 2.2.1 Pure strategies: a complete description of the behaviour 30 2.2.2 The Fort Boyard Sticks game and the Envelope game 31 2.2.3 Behavioural strategies 35 2.3 Strategic/normal form games and extensive form games: is there a difference? 36 INFORMATION AND UTILITIES 40 3.1 Perfect/imperfect information 40 3.1.1 A concept linked to the information sets 40 3.1.2 The prisoners’ disks game 41 3.2 Complete/incomplete information 43 3.2.1 Common knowledge: does it make sense? 43 45 3.2.2 Signalling games and screening games 3.3 Utilities 47 3.3.1 Taking risks into account 47 3.3.2 Which is the game you have in mind? 49 3.3.3 Fairness and reciprocity 49 3.3.4 Strategic feelings and paranoia 51 3.3.5 How to bypass utilities 52 CONCLUSION54 ix╇ ■ ANSWERS If x≤P Payoff with x=x+(x+P)(M–x)=A Payoff with x+1=x+1+(x+1+P)(M–x–1)=B B–A=M–P–2x If x=P+1 Payoff with x=x+(x+P)(M–x)=C Payoff with x+1=1+x+1+(x+1+P)(M–x–1)=D D–C=M–P–2x+1=M–3P–1 If x>P+1 Payoff with x=1+…+x–1–P+x+(x+P)(M–x)=E Payoff with x+1=1+…+x–1–P+x–P+x+1+(x+1+P)(M–x–1)=F F–E=M–2P–x So let us consider three cases: Case P0 for x≤P and D–C>0 So the payoff grows till M–2P–x=0, i.e x=M–2P (and M–2P>P because M>3P) So, for P small relative to M, both M–2P and M–2P+1 ensure the highest payoff For example, for M=12 and P=2, M–2P=8, and the highest payoff is obtained for and Case P>(M–1)/3 In that case M–2P–x(P+1), and M–3P–1M/3, the payoff stops growing when M–P–2x=0, hence x=(M–P)/2 which is lower than P So for P large relative to M, if (M–P)/2 is not an integer, the highest payoff is obtained for (M–P+1)/2 If (M–P)/2 is an integer, the highest payoff is obtained for both (M–P)/2 and (M–P)/2+1 Especially, for M=12 and P=5, (M–P)/2=3.5, hence the highest payoff is obtained for x=(M–P+1)/2=4 Case P=(M–1)/3 In that case the payoff is maximal for x=P+1 and x=P+2 What matters is that both M–2P and (M–P)/2 have no link with M/2, the best answer to the mean demand (2+M)/2 For large values of P (P>M/3), we can add that the demand that maximizes level–1 player’s payoff, (M–P)/2 is lower than M/2 By contrast, for P small relative to M (P4P, the demand is higher than M/2 For example, we saw that for M=12 and P=2, a level–1 player plays and (M–2P and M–2P+1) with probability ½, and being higher than (=M/2) Let us make an additional remark: for P3=3) and level–1 players to play and with probabilities ½ because 4=4>0 This leads level–2 players to also play because (4+1)/2>(3+1)/2>(3+0)/2 Level–2 players again play and each with probability ½ So the process ends in the state where all level–k players play (i.e B1), and all level–k players play and (i.e B2 and C2) with probability ½ We are very far from the cycle ((B1½, C1½), (A2½, C2½))(C1, C2)) obtained in Exercise 8! ♣♣ EXERCISE 10 LEVEL–1 AND LEVEL–K REASONING IN THE STUDENTS’ TRAVELLER’S DILEMMA Player 1’s payoffs are given in Matrix A6.8 The last column gives player 1’s mean payoff when her opponent is a level–0 player who asks for each amount with the same probability 1/7 Player P2 2 4 4 4 26/7 3 5 5 29/7 2 6 6 32/7 3 7 35/7 4 4 8 38/7 5 5 41/7 6 6 6 44/7 Matrix A6.8 A level–1 player should play 8, i.e the highest possible amount Once again, the fact that level–0 players play the uniform distribution on [2,M] does not mean that level–1 players have to react to (M+2)/2 A level–1 player does not best respond to 5, the mean value of [2,8], he does not play and 8, each with probability ½, but he plays More generally, let us write the mean payoff (multiplied by M–1) of a level–1 player, who reacts to a uniform distribution on [2,M] for P low, so that P0 The payoff associated with x is increasing in x up to M, if P(M–1)/2 ■╇430 ANSWERS We get again: Player 1’s payoff when she plays x is A=(x–P)(x–2)+x+(x+P)(M–x)=A Her payoff when she plays x+1 is B=(x+1–P)(x–1)+x+1+(x+1+P) (M–x–1)=B B–A=M–1–2P(M–1)/2, leading the player to play As regards my L3 players, let me observe that 37.2% played and that 21.2% played 100 for P=70 and M=100 So 58.4% of them played like level–1 and level–2 players (see Exercise 17 in chapter for more details on the students’ behaviour) Anyway, regardless of the starting point – i.e regardless of the level–1 action, regardless of the values of P and M – we get a cycle which leads from M to max(2, M–2P) and from max(2, M–2P) to M, which is not very intuitive Cognition hierarchy would not lead to a cycle but would also lead to mix behaviour on all values between max(2, M–2P) and M Clearly a level–k reasoning better suits for games that slowly converge to a Nash equilibrium ♣♣ EXERCISE 11 STABLE EQUILIBRIUM, PERFECT EQUILIBRIUM AND PERTURBATIONS Call p1, p2, p3, p4 the probabilities assigned to A1a1, A1b1, B1a1 and B1b1 Player is best off playing A2a2 rather than B2a2 if 2(p1+p2)+p3+p4≥2(p1+p2)+4p3, i.e p4≥3p3 Call ε3 and ε4 the minimal probabilities on B1a1 and B1b1 When ε4≥3ε3, there is no problem to ensure p4≥3p3 But if ε4ε4 So player has to play B1b1 with a probability higher than the minimal one, which is impossible, given that she gets with A1a1 and at most with B1b1 So (A1a1, A2a2) is not a stable equilibrium This is a rather expected result, given that (A1a1, A2a2) is not even subgame perfect Let us set π1x1k(A1)=1–ε1k, π1x1k(B1)=ε1k, π1x3k(a1)=1–e1k, π1x3k(b1)=e1k, π2x2k(A2)=ε2k, π2x2k(B2)=1–ε2k, π2hk(a2)=1–e2k, π2hk(b2)=e2k, where ε1k, ε2k, e1k, e2k are the minimal probabilities assigned to B1, A2, b1 and b2 ε1k, ε2k, e1k, e2k→0 when k→∞ πk→π*, where π* are the behavioural strategies assigned to E*=(A1a1, B2a2) Player at x3: a1 leads to the payoff 4ε1k (1–ε2k)(1–e2k), b1 leads to the payoff ε1k (1–ε2k)e2k So assigning the minimal probability e1k to b1 and the remaining one to a1 is player 1’s best local strategy in the perturbed game Player at x1: A1 leads to the payoff 5, B1 leads to a payoff lower than So player is right assigning the minimal probability ε1k to B1 and the remaining one to A1 Player at h: a2 leads to the payoff 4ε1k (1–ε2k)(1–e1k), b2 leads to the payoff ε1k (1–ε2k)e1k So player is right assigning the minimal probability e2k to b2 and the remaining one to a2 Player at x2: A2 leads to the payoff ε1k, B2 leads to the payoff ε1k [4(1–e1k)(1–e2k)+e1k e2k] which goes to 4ε1k when k→∞ So player is right assigning the minimal probability ε2k to A2 and the remaining one to B2 So πk is a Nash equilibrium of the perturbed game and E* is a perfect equilibrium, without any restriction on the introduced minimal probabilities 431╇ ■ ANSWERS We would like to conclude that it is also a stable equilibrium given that it resists the introduction of any small probabilities Yet, strangely enough, this is not the case To see why, we turn back to the normal form game and observe that player is best off playing B2a2 rather than A2a2 if 2(p1+p2)+p3+p4≤2(p1+p2)+4p3, i.e p4≤3p3 We call again ε3 and ε4 the minimal probabilities on B1a1 and B1b1 When ε4≤3ε3 , there is no problem to ensure p4≤3p3 But if ε4>3ε3 then 3p3≥p4 leads to 3p3≥p4≥ε4>3ε3 So player has to play B1a1 with a probability higher than the minimal one, which is clearly impossible, given that she gets with A1a1 and at most with B1a1 So (A1a1, B2a2) is not a stable equilibrium despite it resists any small perturbations on the extensive form game This astonishing result is due to the fact that perfection introduces perturbations in the extensive form, hence on local strategies, whereas Kohlberg and Mertens’ stability introduces perturbations on the normal form game These two kinds of perturbations are completely different As a matter of fact, to play B1a1 in the extensive form game, player has to make only one error, at x1, because she plays the unexpected action B1 at x1 but the expected action a1 at x3: the probability of this strategy in the perturbed game is ε1k(1–e1k) By contrast, to play B1b1, player has to make two errors, because she plays the unexpected actions B1 at x1 and b1 at x3 The probability of this strategy in the perturbed game is ε1ke1k If we take these probabilities as minimal probabilities in the normal form game, instead of ε3 and ε4, p4≤3p3 is always fulfilled because ε1ke1k