高级微观经济学博弈论讲义复旦大学CCESYongqinWang1

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1、Static Games of Complete Information-Lecture 1Nash Equilibrium-Pure StrategyYongqin Wang, CCES, Fudan UniversityOutline of Static Games of Complete Information nIntroduction to gamesnNormal-form (or strategic-form) representation nIterated elimination of strictly dominated strategies nNash equilibri

2、umnApplications of Nash equilibrium nMixed strategy Nash equilibriumDec, 2006, Fudan University 2Game Theory-Lecture 1AgendanWhat is game theorynExamplesPrisoners dilemmaThe battle of the sexesMatching penniesnStatic (or simultaneous-move) games of complete informationnNormal-form or strategic-form

3、representationDec, 2006, Fudan University 3Game Theory-Lecture 1What is game theory?nWe focus on games where:There are at least two rational playersEach player has more than one choicesThe outcome depends on the strategies chosen by all players; there is strategic interactionnExample: Six people go

4、to a restaurant.Each person pays his/her own meal a simple decision problemBefore the meal, every person agrees to split the bill evenly among them a gameDec, 2006, Fudan University 4Game Theory-Lecture 1What is game theory?nGame theory is a formal way to analyze strategic interaction among a group

5、of rational players (or agents) nGame theory has applicationsEconomicsPoliticsSociologyLawetc.Dec, 2006, Fudan University 5Game Theory-Lecture 1Classic Example: Prisoners DilemmanTwo suspects held in separate cells are charged with a major crime. However, there is not enough evidence. nBoth suspects

6、 are told the following policy:If neither confesses then both will be convicted of a minor offense and sentenced to one month in jail. If both confess then both will be sentenced to jail for six months. If one confesses but the other does not, then the confessor will be released but the other will b

7、e sentenced to jail for nine months. -1 , -1-9 , 0 0 , -9-6 , -6Prisoner 1Prisoner 2ConfessMumConfessMumDec, 2006, Fudan University 6Game Theory-Lecture 1Example: The battle of the sexesnAt the separate workplaces, Chris and Pat must choose to attend either an opera or a prize fight in the evening.

8、nBoth Chris and Pat know the following:Both would like to spend the evening together. But Chris prefers the opera.Pat prefers the prize fight. 2 , 1 0 , 0 0 , 0 1 , 2ChrisPatPrize FightOperaPrize FightOperaDec, 2006, Fudan University 7Game Theory-Lecture 1Example: Matching penniesnEach of the two pl

9、ayers has a penny. nTwo players must simultaneously choose whether to show the Head or the Tail. nBoth players know the following rules:If two pennies match (both heads or both tails) then player 2 wins player 1s penny. Otherwise, player 1 wins player 2s penny. -1 , 1 1 , -1 1 , -1-1 , 1Player 1Play

10、er 2TailHeadTailHeadDec, 2006, Fudan University 8Game Theory-Lecture 1Static (or simultaneous-move) games of complete informationnA set of players (at least two players)nFor each player, a set of strategies/actionsnPayoffs received by each player for the combinations of the strategies, or for each p

11、layer, preferences over the combinations of the strategiesPlayer 1, Player 2, . Player nS1 S2 . Snui(s1, s2, .sn), for all s1 S1, s2 S2, . sn Sn.A static (or simultaneous-move) game consists of:Dec, 2006, Fudan University 9Game Theory-Lecture 1Static (or simultaneous-move) games of complete informat

12、ionnSimultaneous-moveEach player chooses his/her strategy without knowledge of others choices.nComplete information (on games structure)Each players strategies and payoff function are common knowledge among all the players.nAssumptions on the playersRationalityPlayers aim to maximize their payoffsPl

13、ayers are perfect calculatorsEach player knows that other players are rationalDec, 2006, Fudan University 10Game Theory-Lecture 1Static (or simultaneous-move) games of complete informationnThe players cooperate?No. Only non-cooperative games Methodological individualismnThe timingEach player i choos

14、es his/her strategy si without knowledge of others choices.Then each player i receives his/her payoff ui(s1, s2, ., sn).The game ends.Dec, 2006, Fudan University 11Game Theory-Lecture 1Definition: normal-form or strategic-form representationnThe normal-form (or strategic-form) representation of a ga

15、me G specifies:A finite set of players 1, 2, ., n,players strategy spaces S1 S2 . Sn andtheir payoff functions u1 u2 . un where ui : S1 S2 . SnR.Dec, 2006, Fudan University 12Game Theory-Lecture 1Normal-form representation: 2-player gamenBi-matrix representationn2 players: Player 1 and Player 2nEach

16、 player has a finite number of strategiesnExample:S1=s11, s12, s13 S2=s21, s22Player 2s21s22Player 1s11u1(s11,s21), u2(s11,s21)u1(s11,s22), u2(s11,s22)s12u1(s12,s21), u2(s12,s21)u1(s12,s22), u2(s12,s22)s13u1(s13,s21), u2(s13,s21)u1(s13,s22), u2(s13,s22)Dec, 2006, Fudan University 13Game Theory-Lectu

17、re 1Classic example: Prisoners Dilemma:normal-form representationnSet of players:Prisoner 1, Prisoner 2nSets of strategies: S1 = S2 = Mum, ConfessnPayoff functions: u1(M, M)=-1, u1(M, C)=-9, u1(C, M)=0, u1(C, C)=-6;u2(M, M)=-1, u2(M, C)=0, u2(C, M)=-9, u2(C, C)=-6-1 , -1-9 , 0 0 , -9-6 , -6Prisoner

18、1Prisoner 2ConfessMumConfessMumPlayersStrategiesPayoffsDec, 2006, Fudan University 14Game Theory-Lecture 1Example: The battle of the sexesnNormal (or strategic) form representation:Set of players: Chris, Pat (=Player 1, Player 2)Sets of strategies: S1 = S2 = Opera, Prize FightPayoff functions: u1(O,

19、 O)=2, u1(O, F)=0, u1(F, O)=0, u1(F, O)=1; u2(O, O)=1, u2(O, F)=0, u2(F, O)=0, u2(F, F)=2 2 , 1 0 , 0 0 , 0 1 , 2ChrisPatPrize FightOperaPrize FightOperaDec, 2006, Fudan University 15Game Theory-Lecture 1Example: Matching penniesnNormal (or strategic) form representation:Set of players:Player 1, Pla

20、yer 2Sets of strategies: S1 = S2 = Head, Tail Payoff functions: u1(H, H)=-1, u1(H, T)=1, u1(T, H)=1, u1(H, T)=-1; u2(H, H)=1, u2(H, T)=-1, u2(T, H)=-1, u2(T, T)=1-1 , 1 1 , -1 1 , -1-1 , 1Player 1Player 2TailHeadTailHeadDec, 2006, Fudan University 16Game Theory-Lecture 1Example: Tourists & NativesnO

21、nly two bars (bar 1, bar 2) in a citynCan charge price of $2, $4, or $5n6000 tourists pick a bar randomlyn4000 natives select the lowest price barnExample 1:Both charge $2each gets 5,000 customers and $10,000nExample 2: Bar 1 charges $4, Bar 2 charges $5Bar 1 gets 3000+4000=7,000 customers and $28,0

22、00Bar 2 gets 3000 customers and $15,000Dec, 2006, Fudan University 17Game Theory-Lecture 1Example: Cournot model of duopolynA product is produced by only two firms: firm 1 and firm 2. The quantities are denoted by q1 and q2, respectively. Each firm chooses the quantity without knowing the other firm

23、 has chosen.nThe market price is P(Q)=a-Q, where Q=q1+q2.nThe cost to firm i of producing quantity qi is Ci(qi)=cqi.The normal-form representation:Set of players: Firm 1, Firm 2Sets of strategies: S1=0, +), S2=0, +)Payoff functions: u1(q1, q2)=q1(a-(q1+q2)-c), u2(q1, q2)=q2(a-(q1+q2)-c)Dec, 2006, Fu

24、dan University 18Game Theory-Lecture 1One More ExamplenEach of n players selects a number between 0 and 100 simultaneously. Let xi denote the number selected by player i.nLet y denote the average of these numbersnPlayer is payoff = xi 3y/5nThe normal-form representation:Dec, 2006, Fudan University 1

25、9Game Theory-Lecture 1Solving Prisoners DilemmanConfess always does better whatever the other player choosesnDominated strategyThere exists another strategy which always does better regardless of other players choices-1 , -1-9 , 0 0 , -9-6 , -6Prisoner 1Prisoner 2ConfessMumConfessMumPlayersStrategie

26、sPayoffsDec, 2006, Fudan University 20Game Theory-Lecture 1Definition: strictly dominated strategy-1 , -1-9 , 0 0 , -9-6 , -6Prisoner 1Prisoner 2ConfessMumConfessMumregardless of other players choicessi” is strictly better than siDec, 2006, Fudan University 21Game Theory-Lecture 1ExamplenTwo firms,

27、Reynolds and Philip, share some marketnEach firm earns $60 million from its customers if neither do advertisingnAdvertising costs a firm $20 millionnAdvertising captures $30 million from competitorPhilipNo AdAdReynoldsNo Ad60 , 6030 , 70Ad70 , 3040 , 40Dec, 2006, Fudan University 22Game Theory-Lectu

28、re 12-player game with finite strategiesnS1=s11, s12, s13 S2=s21, s22ns11 is strictly dominated by s12 if u1(s11,s21)u1(s12,s21) and u1(s11,s22)u1(s12,s22).ns21 is strictly dominated by s22 if u2(s1i,s21) u2(s1i,s22), for i = 1, 2, 3Player 2s21s22Player 1s11u1(s11,s21), u2(s11,s21)u1(s11,s22), u2(s1

29、1,s22)s12u1(s12,s21), u2(s12,s21)u1(s12,s22), u2(s12,s22)s13u1(s13,s21), u2(s13,s21)u1(s13,s22), u2(s13,s22)Dec, 2006, Fudan University 23Game Theory-Lecture 1Definition: weakly dominated strategy 1 , 1 2 , 0 0 , 2 2 , 2Player 1Player 2RUBLregardless of other players choicessi” is at least as good a

30、s siDec, 2006, Fudan University 24Game Theory-Lecture 1Strictly and weakly dominated strategynA rational player never chooses a strictly dominated strategy. Hence, any strictly dominated strategy can be eliminated.nA rational player may choose a weakly dominated strategy. nThe order of elimination d

31、oes not matter for strict dominance elimination (pin down the same equilibrium), but does for weak one.Dec, 2006, Fudan University 25Game Theory-Lecture 1Iterated elimination of strictly dominated strategiesnIf a strategy is strictly dominated, eliminate itnThe size and complexity of the game is red

32、ucednEliminate any strictly dominated strategies from the reduced gamenContinue doing so successivelyDec, 2006, Fudan University 26Game Theory-Lecture 1Iterated elimination of strictly dominated strategies: an example 1 , 0 1 , 2 0 , 1 0 , 3 0 , 1 2 , 0Player 1Player 2MiddleUpDownLeft 1 , 0 1 , 2 0

33、, 3 0 , 1Player 1Player 2MiddleUpDownLeftRightDec, 2006, Fudan University 27Game Theory-Lecture 1Example: Tourists & NativesnOnly two bars (bar 1, bar 2) in a citynCan charge price of $2, $4, or $5n6000 tourists pick a bar randomlyn4000 natives select the lowest price barnExample 1:Both charge $2eac

34、h gets 5,000 customers and $10,000nExample 2: Bar 1 charges $4, Bar 2 charges $5Bar 1 gets 3000+4000=7,000 customers and $28,000Bar 2 gets 3000 customers and $15,000Dec, 2006, Fudan University 28Game Theory-Lecture 1Example: Tourists & NativesBar 2$2$4$5Bar 1$210 , 1014 , 1214 , 15$412 , 1420 , 2028

35、 , 15$515 , 1415 , 2825 , 25Payoffs are in thousands of dollarsBar 2$4$5Bar 1$420 , 2028 , 15$515 , 2825 , 25Dec, 2006, Fudan University 29Game Theory-Lecture 1One More ExamplenEach of n players selects a number between 0 and 100 simultaneously. Let xi denote the number selected by player i.nLet y d

36、enote the average of these numbersnPlayer is payoff = xi 3y/5Dec, 2006, Fudan University 30Game Theory-Lecture 1One More ExamplenThe normal-form representation:Players: player 1, player 2, ., player nStrategies: Si =0, 100, for i = 1, 2, ., n.Payoff functions: ui(x1, x2, ., xn) = xi 3y/5nIs there an

37、y dominated strategy?nWhat numbers should be selected?Dec, 2006, Fudan University 31Game Theory-Lecture 1New solution concept: Nash equilibriumPlayer 2LCRPlayer 1T0 , 44 , 05 , 3M4 , 00 , 45 , 3B3 , 53 , 56 , 6The combination of strategies (B, R) has the following property:Player 1 CANNOT do better

38、by choosing a strategy different from B, given that player 2 chooses R.Player 2 CANNOT do better by choosing a strategy different from R, given that player 1 chooses B.Dec, 2006, Fudan University 32Game Theory-Lecture 1New solution concept: Nash equilibriumPlayer 2LCRPlayer 1T0 , 44 , 03 , 3M4 , 00

39、, 43 , 3B3 , 33 , 33.5 , 3.6The combination of strategies (B, R) has the following property:Player 1 CANNOT do better by choosing a strategy different from B, given that player 2 chooses R.Player 2 CANNOT do better by choosing a strategy different from R, given that player 1 chooses B.Dec, 2006, Fud

40、an University 33Game Theory-Lecture 1Nash Equilibrium: ideanNash equilibriumA set of strategies, one for each player, such that each players strategy is best for her, given that all other players are playing their equilibrium strategiesDec, 2006, Fudan University 34Game Theory-Lecture 1Definition: N

41、ash EquilibriumGiven others choices, player i cannot be better-off if she deviates from si* (cf: dominated strategy)Prisoner 2MumConfessPrisoner 1Mum-1 , -1-9 , 0Confess 0 , -9-6 , -6Dec, 2006, Fudan University 35Game Theory-Lecture 12-player game with finite strategiesnS1=s11, s12, s13 S2=s21, s22n

42、(s11, s21)is a Nash equilibrium if u1(s11,s21) u1(s12,s21), u1(s11,s21) u1(s13,s21) andu2(s11,s21) u2(s11,s22).Player 2s21s22Player 1s11u1(s11,s21), u2(s11,s21)u1(s11,s22), u2(s11,s22)s12u1(s12,s21), u2(s12,s21)u1(s12,s22), u2(s12,s22)s13u1(s13,s21), u2(s13,s21)u1(s13,s22), u2(s13,s22)Dec, 2006, Fud

43、an University 36Game Theory-Lecture 1Finding a Nash equilibrium: cell-by-cell inspection 1 , 0 1 , 2 0 , 1 0 , 3 0 , 1 2 , 0Player 1Player 2MiddleUpDownLeft 1 , 0 1 , 2 0 , 3 0 , 1Player 1Player 2MiddleUpDownLeftRightDec, 2006, Fudan University 37Game Theory-Lecture 1Example: Tourists & NativesBar 2

44、$2$4$5Bar 1$210 , 1014 , 1214 , 15$412 , 1420 , 2028 , 15$515 , 1415 , 2825 , 25Payoffs are in thousands of dollarsBar 2$4$5Bar 1$420 , 2028 , 15$515 , 2825 , 25Dec, 2006, Fudan University 38Game Theory-Lecture 1One More ExamplenThe normal-form representation:Players: player 1, player 2, ., player n

45、Strategies: Si =0, 100, for i = 1, 2, ., n.Payoff functions: ui(x1, x2, ., xn) = xi 3y/5nWhat is the Nash equilibrium?Dec, 2006, Fudan University 39Game Theory-Lecture 1Best response function: examplenIf Player 2 chooses L then Player 1s best strategy is MnIf Player 2 chooses C then Player 1s best s

46、trategy is TnIf Player 2 chooses R then Player 1s best strategy is BnIf Player 1 chooses B then Player 2s best strategy is RnBest response: the best strategy one player can play, given the strategies chosen by all other playersPlayer 2LCRPlayer 1T0 , 44 , 03 , 3M4 , 00 , 43 , 3B3 , 33 , 33.5 , 3.6De

47、c, 2006, Fudan University 40Game Theory-Lecture 1Example: Tourists & Nativesnwhat is Bar 1s best response to Bar 2s strategy of $2, $4 or $5?nwhat is Bar 2s best response to Bar 1s strategy of $2, $4 or $5?Bar 2$2$4$5Bar 1$210 , 1014 , 1214 , 15$412 , 1420 , 2028 , 15$515 , 1415 , 2825 , 25Payoffs a

48、re in thousands of dollarsDec, 2006, Fudan University 41Game Theory-Lecture 12-player game with finite strategiesnS1=s11, s12, s13 S2=s21, s22nPlayer 1s strategy s11 is her best response to Player 2s strategy s21 if u1(s11,s21) u1(s12,s21) andu1(s11,s21) u1(s13,s21).Player 2s21s22Player 1s11u1(s11,s

49、21), u2(s11,s21)u1(s11,s22), u2(s11,s22)s12u1(s12,s21), u2(s12,s21)u1(s12,s22), u2(s12,s22)s13u1(s13,s21), u2(s13,s21)u1(s13,s22), u2(s13,s22)Dec, 2006, Fudan University 42Game Theory-Lecture 1Using best response function to find Nash equilibriumnIn a 2-player game, ( s1, s2 ) is a Nash equilibrium

50、if and only if player 1s strategy s1 is her best response to player 2s strategy s2, and player 2s strategy s2 is her best response to player 1s strategy s1. -1 , -1-9 , 0 0 , -9-6 , -6Prisoner 1Prisoner 2ConfessMumConfessMumDec, 2006, Fudan University 43Game Theory-Lecture 1Using best response funct

51、ion to find Nash equilibrium: examplenM is Player 1s best response to Player 2s strategy L nT is Player 1s best response to Player 2s strategy C nB is Player 1s best response to Player 2s strategy R nL is Player 2s best response to Player 1s strategy TnC is Player 2s best response to Player 1s strat

52、egy M nR is Player 2s best response to Player 1s strategy BPlayer 2LCRPlayer 1T0 , 44 , 03 , 3M4 , 00 , 43 , 3B3 , 33 , 33.5 , 3.6Dec, 2006, Fudan University 44Game Theory-Lecture 1Example: Tourists & NativesBar 2$2$4$5Bar 1$210 , 1014 , 1214 , 15$412 , 1420 , 2028 , 15$515 , 1415 , 2825 , 25Payoffs

53、 are in thousands of dollarsUse best response function to find the Nash equilibrium. Dec, 2006, Fudan University 45Game Theory-Lecture 1Example: The battle of the sexesnOpera is Player 1s best response to Player 2s strategy OperanOpera is Player 2s best response to Player 1s strategy OperaHence, (Op

54、era, Opera) is a Nash equilibriumnFight is Player 1s best response to Player 2s strategy FightnFight is Player 2s best response to Player 1s strategy FightHence, (Fight, Fight) is a Nash equilibrium 2 , 1 0 , 0 0 , 0 1 , 2ChrisPatPrize FightOperaPrize FightOperaDec, 2006, Fudan University 46Game The

55、ory-Lecture 1Example: Matching penniesnHead is Player 1s best response to Player 2s strategy TailnTail is Player 2s best response to Player 1s strategy TailnTail is Player 1s best response to Player 2s strategy HeadnHead is Player 2s best response to Player 1s strategy HeadHence, NO Nash equilibrium

56、-1 , 1 1 , -1 1 , -1-1 , 1Player 1Player 2TailHeadTailHeadDec, 2006, Fudan University 47Game Theory-Lecture 1Definition: best response functionPlayer is best responseGiven the strategies chosen by other playersDec, 2006, Fudan University 48Game Theory-Lecture 1Definition: best response functionPlaye

57、r is best response to other players strategies is an optimal solution toDec, 2006, Fudan University 49Game Theory-Lecture 1Using best response function to define Nash equilibriumA set of strategies, one for each player, such that each players strategy is best for her, given that all other players ar

58、e playing their strategies, orA stable situation that no player would like to deviate if others stick to itDec, 2006, Fudan University 50Game Theory-Lecture 1Cournot model of duopolynA product is produced by only two firms: firm 1 and firm 2. The quantities are denoted by q1 and q2, respectively. Ea

59、ch firm chooses the quantity without knowing the other firm has chosen.nThe market priced is P(Q)=a-Q, where a is a constant number and Q=q1+q2.nThe cost to firm i of producing quantity qi is Ci(qi)=cqi.Dec, 2006, Fudan University 51Game Theory-Lecture 1Cournot model of duopolyThe normal-form repres

60、entation:Set of players: Firm 1, Firm 2Sets of strategies: S1=0, +), S2=0, +)Payoff functions: u1(q1, q2)=q1(a-(q1+q2)-c)u2(q1, q2)=q2(a-(q1+q2)-c)Dec, 2006, Fudan University 52Game Theory-Lecture 1Cournot model of duopolynHow to find a Nash equilibriumFind the quantity pair (q1*, q2*) such that q1*

61、 is firm 1s best response to Firm 2s quantity q2* and q2* is firm 2s best response to Firm 1s quantity q1* That is, q1* solves Max u1(q1, q2*)=q1(a-(q1+q2*)-c)subject to 0 q1 +and q2* solvesMax u2(q1*, q2)=q2(a-(q1*+q2)-c)subject to 0 q2 +Dec, 2006, Fudan University 53Game Theory-Lecture 1Cournot mo

62、del of duopolynHow to find a Nash equilibriumSolve Max u1(q1, q2*)=q1(a-(q1+q2*)-c)subject to 0 q1 +FOC: a - 2q1 - q2*- c = 0 q1 = (a - q2*- c)/2Dec, 2006, Fudan University 54Game Theory-Lecture 1Cournot model of duopolynHow to find a Nash equilibriumSolveMax u2(q1*, q2)=q2(a-(q1*+q2)-c)subject to 0

63、 q2 +FOC: a - 2q2 q1* c = 0 q2 = (a q1* c)/2Dec, 2006, Fudan University 55Game Theory-Lecture 1Cournot model of duopolynHow to find a Nash equilibriumThe quantity pair (q1*, q2*) is a Nash equilibrium ifq1* = (a q2* c)/2 q2* = (a q1* c)/2Solving these two equations gives us q1* = q2* = (a c)/3Dec, 2

64、006, Fudan University 56Game Theory-Lecture 1Cournot model of duopolynBest response functionFirm 1s best function to firm 2s quantity q2: R1(q2) = (a q2 c)/2 if q2 a c; 0, othwiseFirm 2s best function to firm 1s quantity q1: R2(q1) = (a q1 c)/2 if q1 a c; 0, othwiseq1q2(a c)/2(a c)/2a ca cNash equil

65、ibriumDec, 2006, Fudan University 57Game Theory-Lecture 1Cournot model of oligopolynA product is produced by only n firms: firm 1 to firm n. Firm is quantity is denoted by qi. Each firm chooses the quantity without knowing the other firms choices.nThe market priced is P(Q)=a-Q, where a is a constant

66、 number and Q=q1+q2+.+qn.nThe cost to firm i of producing quantity qi is Ci(qi)=cqi.Dec, 2006, Fudan University 58Game Theory-Lecture 1Cournot model of oligopolyThe normal-form representation:Set of players: Firm 1, . Firm nSets of strategies: Si=0, +), for i=1, 2, ., nPayoff functions: ui(q1 ,., qn

67、)=qi(a-(q1+q2 +.+qn)-c)for i=1, 2, ., nDec, 2006, Fudan University 59Game Theory-Lecture 1Cournot model of oligopolynHow to find a Nash equilibriumFind the quantities (q1*, . qn*) such that qi* is firm is best response to other firms quantitiesThat is, q1* solves Max u1(q1, q2*, ., qn*)=q1(a-(q1+q2*

68、 +.+qn*)-c)subject to 0 q1 +and q2* solvesMax u2(q1*, q2 , q3*, ., qn*)=q2(a-(q1*+q2+q3*+ .+ qn*)-c)subject to 0 q2 +.Dec, 2006, Fudan University 60Game Theory-Lecture 1Bertrand model of duopoly (differentiated products)nTwo firms: firm 1 and firm 2. nEach firm chooses the price for its product with

69、out knowing the other firm has chosen. The prices are denoted by p1 and p2, respectively.nThe quantity that consumers demand from firm 1: q1(p1, p2) = a p1 + bp2.nThe quantity that consumers demand from firm 2: q2(p1, p2) = a p2 + bp1.nThe cost to firm i of producing quantity qi is Ci(qi)=cqi.Dec, 2

70、006, Fudan University 61Game Theory-Lecture 1Bertrand model of duopoly (differentiated products)The normal-form representation:Set of players: Firm 1, Firm 2Sets of strategies: S1=0, +), S2=0, +)Payoff functions: u1(p1, p2)=(a p1 + bp2 )(p1 c)u2(p1, p2)=(a p2 + bp1 )(p2 c)Dec, 2006, Fudan University

71、 62Game Theory-Lecture 1Bertrand model of duopoly (differentiated products)nHow to find a Nash equilibriumFind the price pair (p1*, p2*) such that p1* is firm 1s best response to Firm 2s price p2* and p2* is firm 2s best response to Firm 1s price p1* That is, p1* solves Max u1(p1, p2*) = (a p1 + bp2

72、* )(p1 c)subject to 0 p1 +and p2* solvesMax u2(p1*, p2) = (a p2 + bp1* )(p2 c)subject to 0 p2 +Dec, 2006, Fudan University 63Game Theory-Lecture 1Bertrand model of duopoly (differentiated products)nHow to find a Nash equilibriumSolve firm 1s maximization problemMax u1(p1, p2*) = (a p1 + bp2* )(p1 c)

73、subject to 0 p1 +FOC: a + c 2p1 + bp2* = 0 p1 = (a + c + bp2*)/2Dec, 2006, Fudan University 64Game Theory-Lecture 1Bertrand model of duopoly (differentiated products)nHow to find a Nash equilibriumSolve firm 2s maximization problem Max u2(p1*, p2)=(a p2 + bp1* )(p2 c)subject to 0 p2 + FOC: a + c 2p2

74、 + bp1* = 0 p2 = (a + c + bp1*)/2Dec, 2006, Fudan University 65Game Theory-Lecture 1Bertrand model of duopoly (differentiated products)nHow to find a Nash equilibriumThe price pair (p1*, p2*) is a Nash equilibrium if p1* = (a + c + bp2*)/2 p2* = (a + c + bp1*)/2Solving these two equations gives us p

75、1* = p2* = (a + c)/(2 b)Dec, 2006, Fudan University 66Game Theory-Lecture 1Bertrand model of duopoly (homogeneous products)nTwo firms: firm 1 and firm 2. nEach firm chooses the price for its product without knowing the other firm has chosen. The prices are denoted by p1 and p2, respectively.nThe qua

76、ntity that consumers demand from firm 1: q1(p1, p2) = a p1 if p1 p2 ; = (a p1)/2 if p1 = p2 ; =0, ow.nThe quantity that consumers demand from firm 2: q2(p1, p2) = a p2 if p2 p1 ; = (a p2)/2 if p1 = p2 ; =0, ow.nThe cost to firm i of producing quantity qi is Ci(qi)=cqi.Dec, 2006, Fudan University 67G

77、ame Theory-Lecture 1Bertrand model of duopoly (homogeneous products)The normal-form representation:Set of players: Firm 1, Firm 2Sets of strategies: S1=0, +), S2=0, +)Payoff functions: Dec, 2006, Fudan University 68Game Theory-Lecture 1Bertrand model of duopoly (homogeneous products)Best response fu

78、nctions: pm =( a + c )/2Dec, 2006, Fudan University 69Game Theory-Lecture 1Bertrand model of duopoly (homogeneous products)Best response functions:p1p2ccpmpmp1p2ccpmpmFirm 1s best response to Firm 2s p2Firm 2s best response to Firm 1s p1Dec, 2006, Fudan University 70Game Theory-Lecture 1Bertrand mod

79、el of duopoly (homogeneous products)Best response functions:p1p2ccpmpmNash Equilibrium( c, c )Dec, 2006, Fudan University 71Game Theory-Lecture 1Contributing to a public goodnTwo persons: person 1 and person 2. person 1 has wealth w1 and person 2 has wealth w2, nEach person chooses how much to contr

80、ibute without knowing the other person has chosen. The amounts are denoted by c1 and c2, respectively.nThe amount of the public good is equal to the sum of contributions.nPerson 1s payoff: u1(c1, c2) = v1(c1 + c2) + w1 c1nPerson 2s payoff: u2(c1, c2) = v2(c1 + c2) + w2 c2nv1(c1 + c2) and v2(c1 + c2)

81、 are both concave functionsDec, 2006, Fudan University 72Game Theory-Lecture 1Contributing to a public goodThe normal-form representation:Set of players: Person 1, Person 2Sets of strategies: S1=0, w1, S2=0, w2Payoff functions: u1(c1, c2) = v1(c1 + c2) + w1 c1 u2(c1, c2) = v2(c1 + c2) + w2 c2Dec, 20

82、06, Fudan University 73Game Theory-Lecture 1Contributing to a public goodnHow to find a Nash equilibriumFind the contribution pair (c1*, c2*) such that c1* is person 1s best response to person 2s contribution c2* and c2* is person 2s best response to person 1s contribution c1* That is, c1* solves Ma

83、x u1(c1, c2*) = v1(c1 + c2*) + w1 c1 subject to 0 c1 w1 and c2* solvesMax u2(c1*, c2) = v2(c1* + c2) + w2 c2 subject to 0 c2 w2Dec, 2006, Fudan University 74Game Theory-Lecture 1Contributing to a public goodnHow to find a Nash equilibriumSolve person 1s maximization problemMax u1(c1, c2*) = v1(c1 +

84、c2*) + w1 c1 subject to 0 c1 w1 Dec, 2006, Fudan University 75Game Theory-Lecture 1Contributing to a public goodnHow to find a Nash equilibriumSolve person 2s maximization problemMax u2(c1*, c2) = v2(c1* + c2) + w2 c2 subject to 0 c2 w2Dec, 2006, Fudan University 76Game Theory-Lecture 1Contributing

85、to a public goodnHow to find a Nash equilibriumThe contribution pair (c1*, c2*) is a Nash equilibrium ifDec, 2006, Fudan University 77Game Theory-Lecture 1Contributing to a public goodnBest response functionPerson 1s best function to person 2s contribution c2: R1(c2) = r1 c2 if c2 r1; =0, if c2 r1 P

86、erson 2s best function to person 1s contribution c1: R2(c1) = r2 c1 if c1 r2Dec, 2006, Fudan University 78Game Theory-Lecture 1The problems of commonsnn farmers in a village. Each summer, all the farmers graze their goats on the village green.nLet gi denote the number of goats owned by farmer i.nThe

87、 cost of buying and caring for a goat is c, independent of how many goats a farmer owns.nThe value of a goat is v(G) per goat, where G = g1 + g2 + . + gn nThere is a maximum number of goats that can be grazed on the green. That is, v(G)0 if G Gmax, and v(G)=0 if G Gmax.nAssumptions on v(G): v(G) 0 a

88、nd v”(G) 0.nEach spring, all the farmers simultaneously choose how many goats to own.Dec, 2006, Fudan University 79Game Theory-Lecture 1The problems of commonsThe normal-form representation:Set of players: Farmer 1, . Farmer nSets of strategies: Si=0, Gmax), for i=1, 2,., nPayoff functions: ui(g1, .

89、, gn)=gi v(g1 + .+ gn) c gi for i = 1, 2, ., n.Dec, 2006, Fudan University 80Game Theory-Lecture 1The problems of commonsnHow to find a Nash equilibriumFind (g1*, g2*, ., gn*) such that gi* is farmer is best response to other farmers choices.That is, g1* solves Max u1(g1, g2*, ., gn*)= g1 v(g1 + g2*

90、 .+ gn*) c g1 subject to 0 g1 Gmaxand g2* solvesMax u2(g1*, g2 , g3*, ., gn*)= g2v(g1*+g2+g3*+ .+ gn*)cg2 subject to 0 g2 Gmax .Dec, 2006, Fudan University 81Game Theory-Lecture 1The problems of commonsnHow to find a Nash equilibriumand gn* solvesMax un(g1*, ., gn-1*, gn)= gnv(g1*+.+ gn-1*+ gn)cgn s

91、ubject to 0 gn Gmax .Dec, 2006, Fudan University 82Game Theory-Lecture 1The problems of commonsnFOCs:Dec, 2006, Fudan University 83Game Theory-Lecture 1The problems of commonsnHow to find a Nash equilibrium(g1*, g2*, ., gn*) is a Nash equilibrium ifDec, 2006, Fudan University 84Game Theory-Lecture 1

92、The problems of commonsnSumming over all n farmers FOCs and then dividing by n yieldsDec, 2006, Fudan University 85Game Theory-Lecture 1The problems of commonsnThe social problemDec, 2006, Fudan University 86Game Theory-Lecture 1The problems of commonsDec, 2006, Fudan University 87Game Theory-Lectur

93、e 1Static (or Simultaneous-Move) Games of Complete Information -Mixed Strategy EquilibriumYongqin Wang,CCES, FudanMatching penniesnHead is Player 1s best response to Player 2s strategy TailnTail is Player 2s best response to Player 1s strategy TailnTail is Player 1s best response to Player 2s strate

94、gy HeadnHead is Player 2s best response to Player 1s strategy HeadHence, NO Nash equilibrium-1 , 1 1 , -1 1 , -1-1 , 1Player 1Player 2TailHeadTailHeadDec, 2006, Fudan University 89Game Theory-Lecture 1Solving matching penniesnRandomize your strategiesPlayer 1 chooses Head and Tail with probabilities

95、 r and 1-r, respectively. Player 2 chooses Head and Tail with probabilities q and 1-q, respectively.nMixed Strategy:Specifies that an actual move be chosen randomly from the set of pure strategies with some specific probabilities.Player 2HeadTailPlayer 1Head-1 , 1 1 , -1Tail 1 , -1-1 , 1q1-qr1-rDec,

96、 2006, Fudan University 90Game Theory-Lecture 1Mixed strategynA mixed strategy of a player is a probability distribution over players (pure) strategies.A mixed strategy for Chris is a probability distribution (p, 1-p), where p is the probability of playing Opera, and 1-p is that probability of playi

97、ng Prize Fight.If p=1 then Chris actually plays Opera. If p=0 then Chris actually plays Prize Fight.Battle of sexesPatOperaPrize FightChrisOpera (p)2 , 10 , 0Prize Fight (1-p)0 , 01 , 2Dec, 2006, Fudan University 91Game Theory-Lecture 1Solving matching penniesnPlayer 1s expected payoffsIf Player 1 c

98、hooses Head, -q+(1-q)=1-2qIf Player 1 chooses Tail, q-(1-q)=2q-1Player 2HeadTailPlayer 1Head-1 , 1 1 , -1Tail 1 , -1-1 , 1q1-q1-2q2q-1Expected payoffsr1-rDec, 2006, Fudan University 92Game Theory-Lecture 11qr11/21/2Solving matching penniesnPlayer 1s best responseB1(q):For q0.5, Tail (r=0)For q=0.5,

99、indifferent (0r1)Player 2HeadTailPlayer 1Head-1 , 1 1 , -1Tail 1 , -1-1 , 1q1-q1-2q2q-1Expected payoffsr1-rDec, 2006, Fudan University 93Game Theory-Lecture 1Solving matching penniesnPlayer 2s expected payoffsIf Player 2 chooses Head, r-(1-r)=2r-1If Player 2 chooses Tail, -r+(1-r)=1-2rPlayer 2HeadTa

100、ilPlayer 1Head-1 , 1 1 , -1Tail 1 , -1-1 , 11-2q2q-1Expected payoffsr1-rq1-qExpected payoffs2r-11-2rDec, 2006, Fudan University 94Game Theory-Lecture 1Solving matching penniesnPlayer 2s best responseB2(r):For r0.5, Head (q=1)For r=0.5, indifferent (0q1)Player 2HeadTailPlayer 1Head-1 , 1 1 , -1Tail 1

101、 , -1-1 , 1q1-q1-2q2q-1Expected payoffsr1-rExpected payoffs2r-11-2r1qr11/21/2Dec, 2006, Fudan University 95Game Theory-Lecture 11qr11/21/2Solving matching penniesnPlayer 1s best responseB1(q):For q0.5, Tail (r=0)For q=0.5, indifferent (0 r 1)nPlayer 2s best responseB2(r):For r0.5, Head (q=1)For r=0.

102、5, indifferent (0 q 1)Check r = 0.5 B1(0.5)q = 0.5 B2(0.5)Player 2HeadTailPlayer 1Head-1 , 1 1 , -1Tail 1 , -1-1 , 1r1-rq1-qMixed strategy Nash equilibriumDec, 2006, Fudan University 96Game Theory-Lecture 1Mixed strategy: examplenMatching penniesnPlayer 1 has two pure strategies: H and T( 1(H)=0.5,

103、1(T)=0.5 ) is a Mixed strategy. That is, player 1 plays H and T with probabilities 0.5 and 0.5, respectively.( 1(H)=0.3, 1(T)=0.7 ) is another Mixed strategy. That is, player 1 plays H and T with probabilities 0.3 and 0.7, respectively.Dec, 2006, Fudan University 97Game Theory-Lecture 1Mixed strateg

104、y: examplenPlayer 1: (3/4, 0, ) is a mixed strategy. That is, 1(T)=3/4, 1(M)=0 and 1(B)=1/4.nPlayer 2: (0, 1/3, 2/3) is a mixed strategy. That is, 2(L)=0, 2(C)=1/3 and 2(R)=2/3.Player 2L (0)C (1/3)R (2/3)Player 1T (3/4)0 , 23 , 31 , 1M (0)4 , 00 , 42 , 3B (1/4)3 , 45 , 10 , 7Dec, 2006, Fudan Univers

105、ity 98Game Theory-Lecture 1Expected payoffs: 2 players each with two pure strategiesnPlayer 1 plays a mixed strategy (r, 1- r ). Player 2 plays a mixed strategy ( q, 1- q ).Player 1s expected payoff of playing s11: EU1(s11, (q, 1-q)=qu1(s11, s21)+(1-q)u1(s11, s22)Player 1s expected payoff of playing

106、 s12: EU1(s12, (q, 1-q)= qu1(s12, s21)+(1-q)u1(s12, s22)nPlayer 1s expected payoff from her mixed strategy:v1(r, 1-r), (q, 1-q)=r EU1(s11, (q, 1-q)+(1-r) EU1(s12, (q, 1-q)Player 2s21 ( q )s22 ( 1- q )Player 1s11 ( r )u1(s11, s21), u2(s11, s21)u1(s11, s22), u2(s11, s22)s12 (1- r )u1(s12, s21), u2(s12

107、, s21)u1(s12, s22), u2(s12, s22)Dec, 2006, Fudan University 99Game Theory-Lecture 1Expected payoffs: 2 players each with two pure strategiesnPlayer 1 plays a mixed strategy (r, 1- r ). Player 2 plays a mixed strategy ( q, 1- q ).Player 2s expected payoff of playing s21: EU2(s21, (r, 1-r)=ru2(s11, s2

108、1)+(1-r)u2(s12, s21)Player 2s expected payoff of playing s22: EU2(s22, (r, 1-r)= ru2(s11, s22)+(1-r)u2(s12, s22)nPlayer 2s expected payoff from her mixed strategy:v2(r, 1-r),(q, 1-q)=q EU2(s21, (r, 1-r)+(1-q) EU2(s22, (r, 1-r)Player 2s21 ( q )s22 ( 1- q )Player 1s11 ( r )u1(s11, s21), u2(s11, s21)u1

109、(s11, s22), u2(s11, s22)s12 (1- r )u1(s12, s21), u2(s12, s21)u1(s12, s22), u2(s12, s22)Dec, 2006, Fudan University 100Game Theory-Lecture 1Expected payoffs: examplenPlayer 1:EU1(H, (0.3, 0.7) = 0.3(-1) + 0.71=0.4EU1(T, (0.3, 0.7) = 0.31 + 0.7(-1)=-0.4v1(0.4, 0.6), (0.3, 0.7)=0.4 0.4+0.6 (-0.4)=-0.08

110、nPlayer 2:EU2(H, (0.4, 0.6) = 0.41+0.6(-1) = -0.2EU2(T, (0.4, 0.6) = 0.4(-1)+0.61 = 0.2v2(0.4, 0.6), (0.3, 0.7)=0.3(-0.2)+0.70.2=0.08Player 2H (0.3)T (0.7)Player 1H (0.4)-1 , 1 1 , -1T (0.6) 1 , -1-1 , 1Dec, 2006, Fudan University 101Game Theory-Lecture 1Expected payoffs: examplenMixed strategies: p

111、1=( 3/4, 0, ); p2=( 0, 1/3, 2/3 ).nPlayer 1: EU1(T, p2)=3(1/3)+1(2/3)=5/3, EU1(M, p2)=0(1/3)+2(2/3)=4/3EU1(B, p2)=5(1/3)+0(2/3)=5/3. v1(p1, p2) = 5/3nPlayer 2: EU2(L, p1)=2(3/4)+4(1/4)=5/2, EU2(C, p1)=3(3/4)+3(1/4)=5/2,EU2(R, p1)=1(3/4)+7(1/4)=5/2. v1(p1, p2) = 5/2Player 2L (0)C (1/3)R (2/3)Player 1

112、T (3/4)0 , 23 , 31 , 1M (0)4 , 00 , 42 , 3B (1/4)3 , 45 , 10 , 7Dec, 2006, Fudan University 102Game Theory-Lecture 1Mixed strategy equilibriumnMixed strategy equilibriumA probability distribution for each playerThe distributions are mutual best responses to one another in the sense of expected payof

113、fsIt is a stochastic steady stateDec, 2006, Fudan University 103Game Theory-Lecture 1Mixed strategy equilibrium: 2-player each with two pure strategiesnMixed strategy Nash equilibrium:nA pair of mixed strategies (r*, 1-r*), (q*, 1-q*)is a Nash equilibrium if (r*,1-r*) is a best response to (q*, 1-q*

114、), and (q*, 1-q*) is a best response to (r*,1-r*). That is,v1(r*, 1-r*), (q*, 1-q*) v1(r, 1-r), (q*, 1-q*), for all 0 r 1v2(r*, 1-r*), (q*, 1-q*) v2(r*, 1-r*), (q, 1-q), for all 0 q 1Player 2s21 ( q )s22 ( 1- q )Player 1s11 ( r )u1(s11, s21), u2(s11, s21)u1(s11, s22), u2(s11, s22)s12 (1- r )u1(s12,

115、s21), u2(s12, s21)u1(s12, s22), u2(s12, s22)Dec, 2006, Fudan University 104Game Theory-Lecture 1Find mixed strategy equilibrium in 2-player each with two pure strategiesnFind the best response correspondence for player 1, given player 2s mixed strategynFind the best response correspondence for playe

116、r 2, given player 1s mixed strategynUse the best response correspondences to determine mixed strategy Nash equilibria.Dec, 2006, Fudan University 105Game Theory-Lecture 1Employee MonitoringnEmployees can work hard or shirknSalary: $100K unless caught shirking nCost of effort: $50KnManagers can monit

117、or or notnValue of employee output: $200KnProfit if employee doesnt work: $0nCost of monitoring: $10KDec, 2006, Fudan University 106Game Theory-Lecture 1nEmployees best response B1(q):Shirk (r=0) if q0.5Any mixed strategy (0 r 1) if q=0.5Employee MonitoringManagerMonitor ( q )Not Monitor (1-q)Employ

118、eeWork ( r )50 , 9050 , 100Shirk (1-r )0 , -10100 , -10050100(1-q)Expected payoffsExpected payoffs100r-10200r-100Dec, 2006, Fudan University 107Game Theory-Lecture 1nManagers best response B2(r):Monitor (q=1) if r0.9 Any mixed strategy (0 q 1) if r=0.9Employee MonitoringManagerMonitor ( q )Not Monit

119、or (1-q)EmployeeWork ( r )50 , 9050 , 100Shirk (1-r )0 , -10100 , -10050100(1-q)Expected payoffsExpected payoffs100r-10200r-100Dec, 2006, Fudan University 108Game Theory-Lecture 11qr10.5nEmployees best response B1(q):Shirk (r=0) if q0.5 Any mixed strategy (0 r 1) if q=0.5nManagers best response B2(r

120、):Monitor (q=1) if r0.9 Any mixed strategy (0 q 1) if r=0.9 Employee Monitoring0.9Mixed strategy Nash equilibrium(0.9,0.1),(0.5,0.5)Dec, 2006, Fudan University 109Game Theory-Lecture 1nChris expected payoff of playing Opera: 2qnChris expected payoff of playing Prize Fight: 1-qnChris best response B1

121、(q):Prize Fight (r=0) if q1/3 Any mixed strategy (0r1) if q=1/3Battle of sexesPatOpera (q)Prize Fight (1-q)ChrisOpera ( r )2 , 10 , 0Prize Fight (1-r)0 , 01 , 2Dec, 2006, Fudan University 110Game Theory-Lecture 1nPats expected payoff of playing Opera: rnPats expected payoff of playing Prize Fight: 2

122、(1-r)nPats best response B2(r):Prize Fight (q=0) if r2/3Any mixed strategy (0q1) if r=2/3, Battle of sexesPatOpera (q)Prize Fight (1-q)ChrisOpera ( r )2 , 10 , 0Prize Fight (1-r)0 , 01 , 2Dec, 2006, Fudan University 111Game Theory-Lecture 11qr1nChris best response B1(q):Prize Fight (r=0) if q1/3 Any

123、 mixed strategy (0r1) if q=1/3nPats best response B2(r):Prize Fight (q=0) if r2/3 Any mixed strategy (0q1) if r=2/3Battle of sexes2/3Three Nash equilibria:(1, 0), (1, 0)(0, 1), (0, 1)(2/3, 1/3), (1/3, 2/3)1/3Dec, 2006, Fudan University 112Game Theory-Lecture 1Battle of sexesnThe limitation of best-r

124、esponse approachnCan only deal with simple gamesnComments of mixed strategiesnRandomizationnvNM expected utilitynChoice under uncertaintyDec, 2006, Fudan University 113Game Theory-Lecture 12-player each with two strategiesnTheorem 1 (property of mixed Nash equilibrium)nA pair of mixed strategies (r*

125、, 1-r*), (q*, 1-q*) is a Nash equilibrium if and only if v1(r*, 1-r*), (q*, 1-q*) EU1(s11, (q*, 1-q*)v1(r*, 1-r*), (q*, 1-q*) EU1(s12, (q*, 1-q*) v2(r*, 1-r*), (q*, 1-q*) EU2(s21, (r*, 1-r*)v2(r*, 1-r*), (q*, 1-q*) EU2(s22, (r*, 1-r*)Player 2s21 ( q )s22 ( 1- q )Player 1s11 ( r )u1(s11, s21), u2(s11

126、, s21)u1(s11, s22), u2(s11, s22)s12 (1- r )u1(s12, s21), u2(s12, s21)u1(s12, s22), u2(s12, s22)Dec, 2006, Fudan University 114Game Theory-Lecture 1Expected payoffs: 2 players each with two pure strategiesnPlayer 1 plays a mixed strategy (r, 1- r ). Player 2 plays a mixed strategy ( q, 1- q ).Player

127、1s expected payoff of playing s11: EU1(s11, (q, 1-q)=qu1(s11, s21)+(1-q)u1(s11, s22)Player 1s expected payoff of playing s12: EU1(s12, (q, 1-q)= qu1(s12, s21)+(1-q)u1(s12, s22)nPlayer 1s expected payoff from her mixed strategy:v1(r, 1-r), (q, 1-q)=r EU1(s11, (q, 1-q)+(1-r) EU1(s12, (q, 1-q)Player 2s

128、21 ( q )s22 ( 1- q )Player 1s11 ( r )u1(s11, s21), u2(s11, s21)u1(s11, s22), u2(s11, s22)s12 (1- r )u1(s12, s21), u2(s12, s21)u1(s12, s22), u2(s12, s22)Dec, 2006, Fudan University 115Game Theory-Lecture 1Expected payoffs: 2 players each with two pure strategiesnPlayer 1 plays a mixed strategy (r, 1-

129、 r ). Player 2 plays a mixed strategy ( q, 1- q ).Player 2s expected payoff of playing s21: EU2(s21, (r, 1-r)=ru2(s11, s21)+(1-r)u2(s12, s21)Player 2s expected payoff of playing s22: EU2(s22, (r, 1-r)= ru2(s11, s22)+(1-r)u2(s12, s22)nPlayer 2s expected payoff from her mixed strategy:v2(r, 1-r),(q, 1

130、-q)=q EU2(s21, (r, 1-r)+(1-q) EU2(s22, (r, 1-r)Player 2s21 ( q )s22 ( 1- q )Player 1s11 ( r )u1(s11, s21), u2(s11, s21)u1(s11, s22), u2(s11, s22)s12 (1- r )u1(s12, s21), u2(s12, s21)u1(s12, s22), u2(s12, s22)Dec, 2006, Fudan University 116Game Theory-Lecture 1Theorem 1: illustrationnPlayer 1:EU1(H,

131、(0.5, 0.5) = 0.5(-1) + 0.51=0EU1(T, (0.5, 0.5) = 0.51 + 0.5(-1)=0v1(0.5, 0.5), (0.5, 0.5)=0.5 0+0.5 0=0nPlayer 2:EU2(H, (0.5, 0.5) = 0.51+0.5(-1) =0EU2(T, (0.5, 0.5) = 0.5(-1)+0.51 = 0v2(0.5, 0.5), (0.5, 0.5)=0.50+0.50=0Matching penniesPlayer 2H (0.5)T (0.5)Player 1H (0.5)-1 , 1 1 , -1T (0.5) 1 , -1

132、-1 , 1Dec, 2006, Fudan University 117Game Theory-Lecture 1Theorem 1: illustrationnPlayer 1:v1(0.5, 0.5), (0.5, 0.5) EU1(H, (0.5, 0.5)v1(0.5, 0.5), (0.5, 0.5) EU1(T, (0.5, 0.5)nPlayer 2:v2(0.5, 0.5), (0.5, 0.5) EU2(H, (0.5, 0.5)v2(0.5, 0.5), (0.5, 0.5) EU2(T, (0.5, 0.5)nHence, (0.5, 0.5), (0.5, 0.5)

133、is a mixed strategy Nash equilibrium by Theorem 1.Matching penniesPlayer 2H (0.5)T (0.5)Player 1H (0.5)-1 , 1 1 , -1T (0.5) 1 , -1-1 , 1Dec, 2006, Fudan University 118Game Theory-Lecture 1nEmployees expected payoff of playing “work”EU1(Work, (0.5, 0.5) = 0.550 + 0.550=50 nEmployees expected payoff o

134、f playing “shirk”EU1(Shirk, (0.5, 0.5) = 0.50 + 0.5100=50nEmployees expected payoff of her mixed strategy v1(0.9, 0.1), (0.5, 0.5)=0.9 50+0.1 50=50Theorem 1: illustrationEmployee MonitoringManagerMonitor (0.5)Not Monitor (0.5)EmployeeWork (0.9)50 , 9050 , 100Shirk (0.1)0 , -10100 , -100Dec, 2006, Fu

135、dan University 119Game Theory-Lecture 1nManagers expected payoff of playing “Monitor”EU2(Monitor, (0.9, 0.1) = 0.990+0.1(-10) =80nManagers expected payoff of playing “Not”EU2(Not, (0.9, 0.1) = 0.9100+0.1(-100) = 80nManagers expected payoff of her mixed strategy v2(0.9, 0.1), (0.5, 0.5)=0.580+0.580=8

136、0Theorem 1: illustrationEmployee MonitoringManagerMonitor (0.5)Not Monitor (0.5)EmployeeWork (0.9)50 , 9050 , 100Shirk (0.1)0 , -10100 , -100Dec, 2006, Fudan University 120Game Theory-Lecture 1nEmployeev1(0.9, 0.1), (0.5, 0.5) EU1(Work, (0.5, 0.5)v1(0.9, 0.1), (0.5, 0.5) EU1(Shirk, (0.5, 0.5)nManage

137、rv2(0.9, 0.1), (0.5, 0.5) EU2(Monitor, (0.9, 0.1)v2(0.9, 0.1), (0.5, 0.5) EU2(Not, (0.9, 0.1)n Hence, (0.9, 0.1), (0.5, 0.5) is a mixed strategy Nash equilibrium by Theorem 1.Theorem 1: illustrationEmployee MonitoringManagerMonitor (0.5)No Monitor (0.5)EmployeeWork (0.9)50 , 9050 , 100Shirk (0.1)0 ,

138、 -10100 , -100Dec, 2006, Fudan University 121Game Theory-Lecture 1nUse Theorem 1 to check whether (2/3, 1/3), (1/3, 2/3) is a mixed strategy Nash equilibrium.Theorem 1: illustrationBattle of sexesPatOpera (1/3)Prize Fight (2/3)ChrisOpera (2/3 )2 , 10 , 0Prize Fight (1/3)0 , 01 , 2Dec, 2006, Fudan Un

139、iversity 122Game Theory-Lecture 1Mixed strategy equilibrium: 2-player each with two strategiesnTheorem 2 Let (r*, 1-r*), (q*, 1-q*) be a pair of mixed strategies, where 0 r*1, 0q*0 and p1n*0 then EU1(s1m, p2*) = EU1(s1n, p2*); if p1m*0 and p1n*=0 then EU1(s1m, p2*) EU1(s1n, p2*)player 2: for any i a

140、nd k, if p2i*0 and p2k*0 then EU2(s2i, p1*) = EU2(s2k, p1*); if p2i*0 and p2k*=0 then EU2(s2i, p1*) EU2(s2k, p1*) Dec, 2006, Fudan University 152Game Theory-Lecture 12-player each with a finite number of pure strategiesnWhat does Theorem 4 tell us? nA pair of mixed strategies (p1*, p2*), wherep1*=(p

141、11*, p12*, ., p1J* ), p2*=(p21*, p22*, ., p2K* ) is a mixed strategy Nash equilibrium if and only if they satisfies the following conditions:nGiven player 2s p2*, player 1s expected payoff of every pure strategy to which she assigns positive probability is the same, and player 1s expected payoff of

142、any pure strategy to which she assigns positive probability is not less than the expected payoff of any pure strategy to which she assigns zero probability.nGiven player 1s p1*, player 2s expected payoff of every pure strategy to which she assigns positive probability is the same, and player 2s expe

143、cted payoff of any pure strategy to which she assigns positive probability is not less than the expected payoff of any pure strategy to which she assigns zero probability.Dec, 2006, Fudan University 153Game Theory-Lecture 12-player each with a finite number of pure strategiesnTheorem 4 implies that

144、we have a mixed strategy Nash equilibrium in the following situation Given player 2s mixed strategy, Player 1 is indifferent among the pure strategies to which she assigns positive probabilities. The expected payoff of any pure strategy she assigns positive probability is not less than the expected

145、payoff of any pure strategy she assigns zero probability.Given player 1s mixed strategy, Player 2 is indifferent among the pure strategies to which she assigns positive probabilities. The expected payoff of any pure strategy she assigns positive probability is not less than the expected payoff of an

146、y pure strategy she assigns zero probability.Dec, 2006, Fudan University 154Game Theory-Lecture 1Theorem 4: illustrationnCheck whether (3/4, 0, 1/4), (0, 1/3, 2/3) is a mixed strategy Nash equilibrium nPlayer 1: EU1(T, p2) = 0 0+3 (1/3)+1 (2/3)=5/3, EU1(M, p2) = 4 0+0 (1/3)+2 (2/3)=4/3EU1(B, p2) = 3

147、 0+5 (1/3)+0 (2/3)=5/3.Hence, EU1(T, p2) = EU1(B, p2) EU1(M, p2)Player 2L (0)C (1/3)R (2/3)Player 1T (3/4)0 , 23 , 31 , 1M (0)4 , 00 , 42 , 3B (1/4)3 , 45 , 10 , 7Dec, 2006, Fudan University 155Game Theory-Lecture 1Theorem 4: illustrationnPlayer 2: EU2(L, p1)=2(3/4) + 00 + 4(1/4)=5/2, EU2(C, p1)=3(3

148、/4) + 40 + 1(1/4)=5/2,EU2(R, p1)=1(3/4) + 30 + 7(1/4)=5/2.Hence, EU2(C, p1)=EU2(R, p1)EU2(L, p1)Therefore, (3/4, 0, 1/4), (0, 1/3, 2/3) is a mixed strategy Nash equilibrium by Theorem 4.Player 2L (0)C (1/3)R (2/3)Player 1T (3/4)0 , 23 , 31 , 1M (0)4 , 00 , 42 , 3B (1/4)3 , 45 , 10 , 7Dec, 2006, Fuda

149、n University 156Game Theory-Lecture 1Example: Rock, paper and scissorsnCheck whether there is a mixed strategy Nash equilibrium in which p110, p120, p130, p210, p220, p230.Player 2Rock (p21)Paper (p22)Scissors (p23)Player 1Rock (p11) 0 , 0-1 , 1 1 , -1Paper (p12) 1 , -1 0 , 0-1 , 1Scissors (p13)-1 ,

150、 1 1 , -1 0 , 0Dec, 2006, Fudan University 157Game Theory-Lecture 1Example: Rock, paper and scissorsnIf each player assigns positive probability to every of her pure strategy, then by Theorem 4, each player is indifferent among her three pure strategies.Player 2Rock (p21)Paper (p22)Scissors (p23)Pla

151、yer 1Rock (p11) 0 , 0-1 , 1 1 , -1Paper (p12) 1 , -1 0 , 0-1 , 1Scissors (p13)-1 , 1 1 , -1 0 , 0Dec, 2006, Fudan University 158Game Theory-Lecture 1Example: Rock, paper and scissorsnPlayer 1 is indifferent among her three pure strategies: EU1(Rock, p2) = 0 p21+(-1) p22+1 p23EU1(Paper, p2) = 1 p21+0

152、 p22+(-1) p23EU1(Scissors, p2) = (-1) p21+1 p22+0 p23nEU1(Rock, p2)= EU1(Paper, p2)= EU1(Scissors, p2)nTogether with p21+ p22+ p23=1, we have three equations and three unknowns. Player 2Rock (p21)Paper (p22)Scissors (p23)Player 1Rock (p11) 0 , 0-1 , 1 1 , -1Paper (p12) 1 , -1 0 , 0-1 , 1Scissors (p1

153、3)-1 , 1 1 , -1 0 , 0Dec, 2006, Fudan University 159Game Theory-Lecture 1Example: Rock, paper and scissorsn0 p21+(-1) p22+1 p23= 1 p21+0 p22+(-1) p23 0 p21+(-1) p22+1 p23 = (-1) p21+1 p22+0 p23 p21+ p22+ p23=1 nThe solution is p21= p22= p23=1/3Player 2Rock (p21)Paper (p22)Scissors (p23)Player 1Rock

154、(p11) 0 , 0-1 , 1 1 , -1Paper (p12) 1 , -1 0 , 0-1 , 1Scissors (p13)-1 , 1 1 , -1 0 , 0Dec, 2006, Fudan University 160Game Theory-Lecture 1Example: Rock, paper and scissorsnPlayer 2 is indifferent among her three pure strategies: EU2(Rock, p1)=0 p11+(-1) p12+1 p13EU2(Paper, p1)=1 p11+0 p12+(-1) p13E

155、U2(Scissors, p1)=(-1) p11+1 p12+0 p13 nEU2(Rock, p1)= EU2(Paper, p1)=EU2(Scissors, p1)nTogether with p11+ p12+ p13=1, we have three equations and three unknowns. Player 2Rock (p21)Paper (p22)Scissors (p23)Player 1Rock (p11) 0 , 0-1 , 1 1 , -1Paper (p12) 1 , -1 0 , 0-1 , 1Scissors (p13)-1 , 1 1 , -1

156、0 , 0Dec, 2006, Fudan University 161Game Theory-Lecture 1Example: Rock, paper and scissorsn0 p11+(-1) p12+1 p13=1 p11+0 p12+(-1) p130 p11+(-1) p12+1 p13=(-1) p11+1 p12+0 p13 p11+ p12+ p13=1 nThe solution is p11= p12= p13=1/3Player 2Rock (p21)Paper (p22)Scissors (p23)Player 1Rock (p11) 0 , 0-1 , 1 1

157、, -1Paper (p12) 1 , -1 0 , 0-1 , 1Scissors (p13)-1 , 1 1 , -1 0 , 0Dec, 2006, Fudan University 162Game Theory-Lecture 1Example: Rock, paper and scissorsnPlayer 1: EU1(Rock, p2) = 0 (1/3)+(-1) (1/3)+1 (1/3)=0 EU1(Paper, p2) = 1 (1/3)+0 (1/3)+(-1) (1/3)=0 EU1(Scissors, p2) = (-1) (1/3)+1 (1/3)+0 (1/3)

158、=0nPlayer 2: EU2(Rock, p1)=0 (1/3)+(-1) (1/3)+1 (1/3)=0 EU2(Paper, p1)=1 (1/3)+0 (1/3)+(-1) (1/3)=0 EU2(Scissors, p1)=(-1) (1/3)+1 (1/3)+0 (1/3)=0nTherefore, (p1=(1/3, 1/3, 1/3), p2=(1/3, 1/3, 1/3) is a mixed strategy Nash equilibrium by Theorem 4.Player 2Rock (1/3)Paper (1/3)Scissors (1/3)Player 1R

159、ock (1/3) 0 , 0-1 , 1 1 , -1Paper (1/3) 1 , -1 0 , 0-1 , 1Scissors (1/3)-1 , 1 1 , -1 0 , 0Dec, 2006, Fudan University 163Game Theory-Lecture 1Example: Rock, paper and scissorsnCheck whether there is a mixed strategy Nash equilibrium in which one of p11, p12, p13 is positive, and at least two of p21

160、, p22, p23 are positive.nThe answer is No.Player 2Rock (p21)Paper (p22)Scissors (p23)Player 1Rock (p11) 0 , 0-1 , 1 1 , -1Paper (p12) 1 , -1 0 , 0-1 , 1Scissors (p13)-1 , 1 1 , -1 0 , 0Dec, 2006, Fudan University 164Game Theory-Lecture 1Example: Rock, paper and scissorsnCheck whether there is a mixe

161、d strategy Nash equilibrium in which two of p11, p12, p13 is positive, and at least two of p21, p22, p23 are positive.nThe answer is No.Player 2Rock (p21)Paper (p22)Scissors (p23)Player 1Rock (p11) 0 , 0-1 , 1 1 , -1Paper (p12) 1 , -1 0 , 0-1 , 1Scissors (p13)-1 , 1 1 , -1 0 , 0Dec, 2006, Fudan Univ

162、ersity 165Game Theory-Lecture 1Example: Rock, paper and scissorsnTherefore, (p1=(1/3, 1/3, 1/3), p2=(1/3, 1/3, 1/3) is the unique mixed strategy Nash equilibrium by Theorem 4.Player 2Rock (p21)Paper (p22)Scissors (p23)Player 1Rock (p11) 0 , 0-1 , 1 1 , -1Paper (p12) 1 , -1 0 , 0-1 , 1Scissors (p13)-

163、1 , 1 1 , -1 0 , 0Dec, 2006, Fudan University 166Game Theory-Lecture 1Exercise 138.1 of OsbornenWe first consider pure-strategy Nash equilibria. How many can you find? nIn order to find all Nash equilibria, we need to consider 15 more cases by Theorem 4. nWe first consider complicated cases. Some ca

164、ses are very easy. Player 2L (p21)M (p22)R (p23)Player 1T (p11) 2 , 2 0 , 3 1 , 3B (p12) 3 , 2 1 , 1 0 , 2Dec, 2006, Fudan University 167Game Theory-Lecture 1Exercise 138.1 of OsbornenCase 1: check whether there is a mixed strategy in which p110, p120, p210, p220, p230 By theorem 4, we should have 2

165、 p11+2 p12= 3 p11+1 p12 = 3 p11+2 p12and p11+p12=1.nWe should have2 p21+0 p22+1 p23 =3 p21+1 p22+0 p23 and p21+ p22+ p23 = 1nSolve these equations. If we can get a solution that satisfies p110, p120, p210, p220, p230 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a sol

166、ution or we find a solution that does not satisfies p110, p120, p210, p220, p230, then we do not have such a mixed strategy Nash equilibrium. Player 2L (p21)M (p22)R (p23)Player 1T (p11) 2 , 2 0 , 3 1 , 3B (p12) 3 , 2 1 , 1 0 , 2Dec, 2006, Fudan University 168Game Theory-Lecture 1Exercise 138.1 of O

167、sbornenCase 2: check whether there is a mixed strategy in which p110, p120, p210, p220, p23=0 By theorem 4, we should have 2 p11+2 p12= 3 p11+1 p12 3 p11+2 p12and p11+p12=1.nWe should have2 p21+0 p22+1 p23 =3 p21+1 p22+0 p23 and p21+ p22+ p23 = 1nSolve these. If we can get a solution that satisfies

168、p110, p120, p210, p220, p23=0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p110, p120, p210, p220, p23=0, then we do not have such a mixed strategy Nash equilibrium. Player 2L (p21)M (p22)R (p23)Player 1T (p11)

169、 2 , 2 0 , 3 1 , 3B (p12) 3 , 2 1 , 1 0 , 2Dec, 2006, Fudan University 169Game Theory-Lecture 1Exercise 138.1 of OsbornenCase 3: check whether there is a mixed strategy in which p110, p120, p210, p22=0, p230 By theorem 4, we should have 2 p11+2 p12 = 3 p11+2 p12 3 p11+1 p12 and p11+p12=1.nWe should

170、have2 p21+0 p22+1 p23 =3 p21+1 p22+0 p23 and p21+ p22+ p23 = 1nSolve these. If we can get a solution that satisfies p110, p120, p210, p22=0, p230 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p110, p120, p210, p

171、22=0, p230, then we do not have such a mixed strategy Nash equilibrium. Player 2L (p21)M (p22)R (p23)Player 1T (p11) 2 , 2 0 , 3 1 , 3B (p12) 3 , 2 1 , 1 0 , 2Dec, 2006, Fudan University 170Game Theory-Lecture 1Exercise 138.1 of OsbornenCase 4: check whether there is a mixed strategy in which p110,

172、p120, p21=0, p220, p230 By theorem 4, we should have 2 p11+2 p12 3 p11+1 p12 = 3 p11+2 p12and p11+p12=1.nWe should have2 p21+0 p22+1 p23 =3 p21+1 p22+0 p23 and p21+ p22+ p23 = 1nSolve these. If we can get a solution that satisfies p110, p120, p21=0, p220, p230 then we have a mixed strategy Nash equi

173、librium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p110, p120, p21=0, p220, p230, then we do not have such a mixed strategy Nash equilibrium. Player 2L (p21)M (p22)R (p23)Player 1T (p11) 2 , 2 0 , 3 1 , 3B (p12) 3 , 2 1 , 1 0 , 2Dec, 2006, Fudan Universit

174、y 171Game Theory-Lecture 1Exercise 138.1 of OsbornenCase 5: check whether there is a mixed strategy in which p110, p120, p210, p22=0, p23=0 (Note this implies p21=1) By theorem 4, we should have 2 p11+2 p12 3 p11+1 p12 and 2 p11+2 p12 3 p11+2 p12and p11+p12=1.nWe should have2 p21+0 p22+1 p23 =3 p21+

175、1 p22+0 p23 and p21+ p22+ p23 = 1nSolve these. If we can get a solution that satisfies p110, p120, p210, p22=0, p23=0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p110, p120, p210, p22=0, p23=0, then we do not

176、have such a mixed strategy Nash equilibrium. Player 2L (p21)M (p22)R (p23)Player 1T (p11) 2 , 2 0 , 3 1 , 3B (p12) 3 , 2 1 , 1 0 , 2Dec, 2006, Fudan University 172Game Theory-Lecture 1Exercise 138.1 of OsbornenCase 6: check whether there is a mixed strategy in which p110, p120, p21=0, p220, p23=0 (N

177、ote this implies p22=1) By theorem 4, we should have 3 p11+1 p12 2 p11+2 p12 and 3 p11+1 p12 3 p11+2 p12and p11+p12=1.nWe should have2 p21+0 p22+1 p23 =3 p21+1 p22+0 p23 and p21+ p22+ p23 = 1nSolve these. If we can get a solution that satisfies p110, p120, p21=0, p220, p23=0 then we have a mixed str

178、ategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p110, p120, p21=0, p220, p23=0, then we do not have such a mixed strategy Nash equilibrium. Player 2L (p21)M (p22)R (p23)Player 1T (p11) 2 , 2 0 , 3 1 , 3B (p12) 3 , 2 1 , 1 0 , 2Dec, 2006,

179、 Fudan University 173Game Theory-Lecture 1Exercise 138.1 of OsbornenCase 7: check whether there is a mixed strategy in which p110, p120, p21=0, p22=0, p230 (Note this implies p23=1) By theorem 4, we should have 3 p11+2 p12 2 p11+2 p12 and 3 p11+2 p12 3 p11+1 p12 and p11+p12=1.nWe should have2 p21+0

180、p22+1 p23 =3 p21+1 p22+0 p23 and p21+ p22+ p23 = 1nSolve these. If we can get a solution that satisfies p110, p120, p21=0, p22=0, p230 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p110, p120, p21=0, p22=0, p230

181、, then we do not have such a mixed strategy Nash equilibrium. Player 2L (p21)M (p22)R (p23)Player 1T (p11) 2 , 2 0 , 3 1 , 3B (p12) 3 , 2 1 , 1 0 , 2Dec, 2006, Fudan University 174Game Theory-Lecture 1Exercise 138.1 of OsbornenCase 8: check whether there is a mixed strategy in which p110, p12=0, p21

182、0, p220, p230 (Note this implies p11=1)By theorem 4, we should have 2 p11+2 p12= 3 p11+1 p12 = 3 p11+2 p12and p11+p12=1.nWe should have2 p21+0 p22+1 p23 3 p21+1 p22+0 p23 and p21+ p22+ p23 = 1nSolve these. If we can get a solution that satisfies p110, p12=0, p210, p220, p230 then we have a mixed str

183、ategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p110, p12=0, p210, p220, p230, then we do not have such a mixed strategy Nash equilibrium. Player 2L (p21)M (p22)R (p23)Player 1T (p11) 2 , 2 0 , 3 1 , 3B (p12) 3 , 2 1 , 1 0 , 2Dec, 2006,

184、Fudan University 175Game Theory-Lecture 1Exercise 138.1 of OsbornenCase 9: check whether there is a mixed strategy in which p110, p12=0, p210, p220, p23=0 (Note this implies p11=1)By theorem 4, we should have 2 p11+2 p12= 3 p11+1 p12 3 p11+2 p12and p11+p12=1.nWe should have2 p21+0 p22+1 p23 3 p21+1

185、p22+0 p23 and p21+ p22+ p23 = 1nSolve these. If we can get a solution that satisfies p110, p12=0, p210, p220, p23=0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p110, p12=0, p210, p220, p23=0, then we do not ha

186、ve such a mixed strategy Nash equilibrium. Player 2L (p21)M (p22)R (p23)Player 1T (p11) 2 , 2 0 , 3 1 , 3B (p12) 3 , 2 1 , 1 0 , 2Dec, 2006, Fudan University 176Game Theory-Lecture 1Exercise 138.1 of OsbornenCase 10: check whether there is a mixed strategy in which p110, p12=0, p210, p22=0, p230 (No

187、te this implies p11=1)By theorem 4, we should have 2 p11+2 p12 3 p11+1 p12 3 p11+2 p12and p11+p12=1.nWe should have2 p21+0 p22+1 p23 3 p21+1 p22+0 p23 and p21+ p22+ p23 = 1nSolve these. If we can get a solution that satisfies p110, p12=0, p210, p22=0, p230 then we have a mixed strategy Nash equilibr

188、ium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p110, p12=0, p210, p22=0, p230, then we do not have such a mixed strategy Nash equilibrium. Player 2L (p21)M (p22)R (p23)Player 1T (p11) 2 , 2 0 , 3 1 , 3B (p12) 3 , 2 1 , 1 0 , 2Dec, 2006, Fudan University 1

189、77Game Theory-Lecture 1Exercise 138.1 of OsbornenCase 11: check whether there is a mixed strategy in which p110, p12=0, p21=0, p220, p230 (Note this implies p11=1)By theorem 4, we should have 2 p11+2 p12 3 p11+1 p12 = 3 p11+2 p12and p11+p12=1.nWe should have2 p21+0 p22+1 p23 3 p21+1 p22+0 p23 and p2

190、1+ p22+ p23 = 1nSolve these. If we can get a solution that satisfies p110, p12=0, p21=0, p220, p230 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p110, p12=0, p21=0, p220, p230, then we do not have such a mixed

191、strategy Nash equilibrium. Player 2L (p21)M (p22)R (p23)Player 1T (p11) 2 , 2 0 , 3 1 , 3B (p12) 3 , 2 1 , 1 0 , 2Dec, 2006, Fudan University 178Game Theory-Lecture 1Exercise 138.1 of OsbornenCase 12: check whether there is a mixed strategy in which p11=0, p120, p210, p220, p230 (Note this implies p

192、12=1)By theorem 4, we should have 2 p11+2 p12= 3 p11+1 p12 = 3 p11+2 p12and p11+p12=1.nWe should have2 p21+0 p22+1 p23 3 p21+1 p22+0 p23 and p21+ p22+ p23 = 1nSolve these. If we can get a solution that satisfies p11=0, p120, p210, p220, p230 then we have a mixed strategy Nash equilibrium. Otherwise,

193、 if we can not find a solution or we find a solution that does not satisfies p11=0, p120, p210, p220, p230, then we do not have such a mixed strategy Nash equilibrium. Player 2L (p21)M (p22)R (p23)Player 1T (p11) 2 , 2 0 , 3 1 , 3B (p12) 3 , 2 1 , 1 0 , 2Dec, 2006, Fudan University 179Game Theory-Le

194、cture 1Exercise 138.1 of OsbornenCase 13: check whether there is a mixed strategy in which p11=0, p120, p210, p220, p23=0 (Note this implies p12=1)By theorem 4, we should have 2 p11+2 p12= 3 p11+1 p12 3 p11+2 p12and p11+p12=1.nWe should have2 p21+0 p22+1 p23 3 p21+1 p22+0 p23 and p21+ p22+ p23 = 1nS

195、olve these. If we can get a solution that satisfies p11=0, p120, p210, p220, p23=0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11=0, p120, p210, p220, p23=0, then we do not have such a mixed strategy Nash equ

196、ilibrium. Player 2L (p21)M (p22)R (p23)Player 1T (p11) 2 , 2 0 , 3 1 , 3B (p12) 3 , 2 1 , 1 0 , 2Dec, 2006, Fudan University 180Game Theory-Lecture 1Exercise 138.1 of OsbornenCase 14: check whether there is a mixed strategy in which p11=0, p120, p210, p22=0, p230 (Note this implies p12=1)By theorem

197、4, we should have 2 p11+2 p12 3 p11+1 p12 3 p11+2 p12and p11+p12=1.nWe should have2 p21+0 p22+1 p23 3 p21+1 p22+0 p23 and p21+ p22+ p23 = 1nSolve these. If we can get a solution that satisfies p11=0, p120, p210, p22=0, p230 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not fin

198、d a solution or we find a solution that does not satisfies p11=0, p120, p210, p22=0, p230, then we do not have such a mixed strategy Nash equilibrium. Player 2L (p21)M (p22)R (p23)Player 1T (p11) 2 , 2 0 , 3 1 , 3B (p12) 3 , 2 1 , 1 0 , 2Dec, 2006, Fudan University 181Game Theory-Lecture 1Exercise 1

199、38.1 of OsbornenCase 15: check whether there is a mixed strategy in which p11=0, p120, p21=0, p220, p230 (Note this implies p12=1)By theorem 4, we should have 2 p11+2 p12 3 p11+1 p12 = 3 p11+2 p12and p11+p12=1.nWe should have2 p21+0 p22+1 p23 3 p21+1 p22+0 p23 and p21+ p22+ p23 = 1nSolve these. If w

200、e can get a solution that satisfies p11=0, p120, p21=0, p220, p230 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11=0, p120, p21=0, p220, p230, then we do not have such a mixed strategy Nash equilibrium. Player

201、 2L (p21)M (p22)R (p23)Player 1T (p11) 2 , 2 0 , 3 1 , 3B (p12) 3 , 2 1 , 1 0 , 2Dec, 2006, Fudan University 182Game Theory-Lecture 1Dynamic Games of Complete InformationYongqin Wang,Dec., 2006CCES, FudanOutline of dynamic games of complete informationnDynamic games of complete information nExtensiv

202、e-form representationnDynamic games of complete and perfect informationnGame treenSubgame-perfect Nash equilibriumnBackward inductionnApplicationsnDynamic games of complete and imperfect informationnMore applicationsnRepeated gamesDec, 2006, Fudan University 184Game Theory-Lecture 1Entry gamenAn inc

203、umbent monopolist faces the possibility of entry by a challenger.nThe challenger may choose to enter or stay out.nIf the challenger enters, the incumbent can choose either to accommodate or to fight.nThe payoffs are common knowledge.ChallengerInOutIncumbentAF1, 22, 10, 0The first number is the payof

204、f of the challenger. The second number is the payoff of the incumbent.Dec, 2006, Fudan University 185Game Theory-Lecture 1Sequential-move matching penniesnEach of the two players has a penny. nPlayer 1 first chooses whether to show the Head or the Tail. nAfter observing player 1s choice, player 2 ch

205、ooses to show Head or TailnBoth players know the following rules:If two pennies match (both heads or both tails) then player 2 wins player 1s penny. Otherwise, player 1 wins player 2s penny. Player 1Player 2HT-1, 11, -1HTPlayer 2HT1, -1-1, 1Dec, 2006, Fudan University 186Game Theory-Lecture 1Dynamic

206、 (or sequential-move) games of complete informationnA set of playersnWho moves when and what action choices are available?nWhat do players know when they move?nPlayers payoffs are determined by their choices.nAll these are common knowledge among the players.Dec, 2006, Fudan University 187Game Theory

207、-Lecture 1Definition: extensive-form representationnThe extensive-form representation of a game specifies: the players in the gamewhen each player has the movewhat each player can do at each of his or her opportunities to movewhat each player knows at each of his or her opportunities to movethe payo

208、ff received by each player for each combination of moves that could be chosen by the playersDec, 2006, Fudan University 188Game Theory-Lecture 1Dynamic games of complete and perfect informationnPerfect informationAll previous moves are observed before the next move is chosen.A player knows Who has m

209、oved What before she makes a decisionDec, 2006, Fudan University 189Game Theory-Lecture 1Game treenA game tree has a set of nodes and a set of edges such thateach edge connects two nodes (these two nodes are said to be adjacent)for any pair of nodes, there is a unique path that connects these two no

210、desx0x1x2x3x4x5x6x7x8a nodean edge connecting nodes x1 and x5a path from x0 to x4Dec, 2006, Fudan University 190Game Theory-Lecture 1Game treenA path is a sequence of distinct nodes y1, y2, y3, ., yn-1, yn such that yi and yi+1 are adjacent, for i=1, 2, ., n-1. We say that this path is from y1 to yn

211、. nWe can also use the sequence of edges induced by these nodes to denote the path.nThe length of a path is the number of edges contained in the path.nExample 1: x0, x2, x3, x7 is a path of length 3. nExample 2: x4, x1, x0, x2, x6 is a path of length 4x0x1x2x3x4x5x6x7x8a path from x0 to x4Dec, 2006,

212、 Fudan University 191Game Theory-Lecture 1Game treenThere is a special node x0 called the root of the tree which is the beginning of the gamenThe nodes adjacent to x0 are successors of x0. The successors of x0 are x1, x2nFor any two adjacent nodes, the node that is connected to the root by a longer

213、path is a successor of the other node. nExample 3: x7 is a successor of x3 because they are adjacent and the path from x7 to x0 is longer than the path from x3 to x0x0x1x2x3x4x5x6x7x8Dec, 2006, Fudan University 192Game Theory-Lecture 1Game treenIf a node x is a successor of another node y then y is

214、called a predecessor of x. nIn a game tree, any node other than the root has a unique predecessor.nAny node that has no successor is called a terminal node which is a possible end of the gamenExample 4: x4, x5, x6, x7, x8 are terminal nodesx0x1x2x3x4x5x6x7x8Dec, 2006, Fudan University 193Game Theory

215、-Lecture 1Game treenAny node other than a terminal node represents some player.nFor a node other than a terminal node, the edges that connect it with its successors represent the actions available to the player represented by the nodePlayer 1Player 2HT-1, 11, -1HTPlayer 2HT1, -1-1, 1Dec, 2006, Fudan

216、 University 194Game Theory-Lecture 1Game treenA path from the root to a terminal node represents a complete sequence of moves which determines the payoff at the terminal nodePlayer 1Player 2HT-1, 11, -1HTPlayer 2HT1, -1-1, 1Dec, 2006, Fudan University 195Game Theory-Lecture 1StrategynA strategy for

217、a player is a complete plan of actions.nIt specifies a feasible action for the player in every contingency in which the player might be called on to act.nWhat the players can possibly play, not what they do play.nCf: static gamesDec, 2006, Fudan University 196Game Theory-Lecture 1Entry gamenChalleng

218、ers strategiesInOutnIncumbents strategiesAccommodate (if challenger plays In)Fight (if challenger plays In)nPayoffs nNormal-form representationIncumbentAccommodateFightChallengerIn2 , 10 , 0Out1 , 21 , 2Dec, 2006, Fudan University 197Game Theory-Lecture 1Strategy and payoffnIn a game tree, a strateg

219、y for a player is represented by a set of edges.nA combination of strategies (sets of edges), one for each player, induce one path from the root to a terminal node, which determines the payoffs of all players Dec, 2006, Fudan University 198Game Theory-Lecture 1Sequential-move matching penniesnPlayer

220、 1s strategiesHeadTailnPlayer 2s strategiesH if player 1 plays H, H if player 1 plays T H if player 1 plays H, T if player 1 plays T T if player 1 plays H, H if player 1 plays T T if player 1 plays H, T if player 1 plays T Player 2s strategies are denoted by HH, HT, TH and TT, respectively.(n x m)De

221、c, 2006, Fudan University 199Game Theory-Lecture 1Sequential-move matching penniesnTheir payoffsnNormal-form representationPlayer 2HHHTTHTTPlayer 1H-1 , 1-1 , 1 1 , -1 1 , -1T 1 , -1-1 , 1 1 , -1-1 , 1Dec, 2006, Fudan University 200Game Theory-Lecture 1Nash equilibriumnThe set of Nash equilibria in

222、a dynamic game of complete information is the set of Nash equilibria of its normal-form.Dec, 2006, Fudan University 201Game Theory-Lecture 1Find Nash equilibriumnHow to find the Nash equilibria in a dynamic game of complete informationConstruct the normal-form of the dynamic game of complete informa

223、tionFind the Nash equilibria in the normal-formDec, 2006, Fudan University 202Game Theory-Lecture 1Nash equilibria in entry gamenTwo Nash equilibria ( In, Accommodate ) ( Out, Fight )nDoes the second Nash equilibrium make sense? nNon-creditable threatsIncumbentAccommodateFightChallengerIn2 , 10 , 0O

224、ut1 , 21 , 2Dec, 2006, Fudan University 203Game Theory-Lecture 1Remove nonreasonable Nash equilibriumnSubgame perfect Nash equilibrium is a refinement of Nash equilibriumnIt can rule out nonreasonable Nash equilibria or non-creditable threatsDec, 2006, Fudan University 204Game Theory-Lecture 1Entry

225、gamenAn incumbent monopolist faces the possibility of entry by a challenger.nThe challenger may choose to enter or stay out.nIf the challenger enters, the incumbent can choose either to accommodate or to fight.nThe payoffs are common knowledge.ChallengerInOutIncumbentAF1, 22, 10, 0The first number i

226、s the payoff of the challenger. The second number is the payoff of the incumbent.Dec, 2006, Fudan University 205Game Theory-Lecture 1Strategy and payoffnA strategy for a player is a complete plan of actions.nIt specifies a feasible action for the player in every contingency in which the player might

227、 be called on to act.nIt specifies what the player does at each of her nodesPlayer 1Player 2HT-1, 11, -1HTPlayer 2HT1, -1-1, 1a strategy for player 1: Ha strategy for player 2: H if player 1 plays H, T if player 1 plays T (written as HT)Player 1s payoff is -1 and player 2s payoff is 1 if player 1 pl

228、ays H and player 2 plays HT Dec, 2006, Fudan University 206Game Theory-Lecture 1Nash equilibrium in a dynamic gamenWe can also use normal-form to represent a dynamic gamenThe set of Nash equilibria in a dynamic game of complete information is the set of Nash equilibria of its normal-formnHow to find

229、 the Nash equilibria in a dynamic game of complete informationConstruct the normal-form of the dynamic game of complete informationFind the Nash equilibria in the normal-formDec, 2006, Fudan University 207Game Theory-Lecture 1Entry gamenChallengers strategiesInOutnIncumbents strategiesAccommodateFig

230、htnPayoffs nNormal-form representationIncumbentAccommodateFightChallengerIn2 , 10 , 0Out1 , 21 , 2ChallengerInOutIncumbentAF1, 22, 10, 0Dec, 2006, Fudan University 208Game Theory-Lecture 1Nash equilibria in entry gamenTwo Nash equilibria ( In, Accommodate ) ( Out, Fight )nDoes the second Nash equili

231、brium make sense? nNon-creditable threatsnLimitations of this approach and the need for better oneIncumbentAccommodateFightChallengerIn2 , 10 , 0Out1 , 21 , 2ChallengerInOutIncumbentAF1, 22, 10, 0Dec, 2006, Fudan University 209Game Theory-Lecture 1Remove nonreasonable Nash equilibriumnSubgame perfec

232、t Nash equilibrium is a refinement of Nash equilibriumnIt can rule out nonreasonable Nash equilibria or non-creditable threatsnWe first need to define subgameDec, 2006, Fudan University 210Game Theory-Lecture 1SubgamenA subgame of a game tree begins at a nonterminal node and includes all the nodes a

233、nd edges following the nonterminal nodenA subgame beginning at a nonterminal node x can be obtained as follows: remove the edge connecting x and its predecessor the connected part containing x is the subgame-1, 1Player 1Player 2HT1, -1HTPlayer 2HT1, -1-1, 1a subgameDec, 2006, Fudan University 211Gam

234、e Theory-Lecture 1Subgame: examplePlayer 2EFPlayer 1GH3, 11, 20, 0Player 1CD2, 0Player 2EFPlayer 1GH3, 11, 20, 0Player 1GH1, 20, 0Dec, 2006, Fudan University 212Game Theory-Lecture 1Subgame-perfect Nash equilibriumnA Nash equilibrium of a dynamic game is subgame-perfect if the strategies of the Nash

235、 equilibrium constitute a Nash equilibrium in every subgame of the game.nSubgame-perfect Nash equilibrium is a Nash equilibrium. Dec, 2006, Fudan University 213Game Theory-Lecture 1Entry gamenTwo Nash equilibria ( In, Accommodate ) is subgame-perfect.( Out, Fight ) is not subgame-perfect because it

236、does not induce a Nash equilibrium in the subgame beginning at Incumbent.ChallengerInOutIncumbentAF1, 22, 10, 0IncumbentAF2, 10, 0Accommodate is the Nash equilibrium in this subgame.Dec, 2006, Fudan University 214Game Theory-Lecture 1Find subgame perfect Nash equilibria: backward inductionnStarting

237、with those smallest subgamesnThen move backward until the root is reachedChallengerInOutIncumbentAF1, 22, 10, 0The first number is the payoff of the challenger. The second number is the payoff of the incumbent.Dec, 2006, Fudan University 215Game Theory-Lecture 1Find subgame perfect Nash equilibria:

238、backward inductionnSubgame perfect Nash equilibrium (DG, E)Player 1 plays D, and plays G if player 2 plays EPlayer 2 plays E if player 1 plays CPlayer 2EFPlayer 1GH3, 11, 20, 0Player 1CD2, 0Dec, 2006, Fudan University 216Game Theory-Lecture 1Existence of subgame-perfect Nash equilibriumnEvery finite

239、 dynamic game of complete and perfect information has a subgame-perfect Nash equilibrium that can be found by backward induction.Dec, 2006, Fudan University 217Game Theory-Lecture 1Sequential bargaining (2.1.D of Gibbons)nPlayer 1 and 2 are bargaining over one dollar. The timing is as follows:nAt th

240、e beginning of the first period, player 1 proposes to take a share s1 of the dollar, leaving 1-s1 to player 2.nPlayer 2 either accepts the offer or rejects the offer (in which case play continues to the second period)nAt the beginning of the second period, player 2 proposes that player 1 take a shar

241、e s2 of the dollar, leaving 1-s2 to player 2.nPlayer 1 either accepts the offer or rejects the offer (in which case play continues to the third period)nAt the beginning of third period, player 1 receives a share s of the dollar, leaving 1-s for player 2, where 0s 1.nThe players are impatient. They d

242、iscount the payoff by a fact , where 0 1Dec, 2006, Fudan University 218Game Theory-Lecture 1Sequential bargaining (2.1.D of Gibbons)Player 2acceptrejectpropose an offer ( s2 , 1-s2 )Period 1Player 1acceptpropose an offer ( s1 , 1-s1 ) s1 , 1-s1Player 1s2 , 1-s2s , 1-sPeriod 2Period 3rejectPlayer 2De

243、c, 2006, Fudan University 219Game Theory-Lecture 1Solve sequential bargaining by backward inductionnPeriod 2:Player 1 accepts s2 if and only if s2 s. (We assume that each player will accept an offer if indifferent between accepting and rejecting) Player 2 faces the following two options:(1) offers s

244、2 = s to player 1, leaving 1-s2 = 1-s for herself at this period, or(2) offers s2 s to player 1 (player 1 will reject it), and receives 1-s next period. Its discounted value is (1-s) Since (1-s)1-s, player 2 should propose an offer (s2* , 1-s2* ), where s2* = s. Player 1 will accept it.Dec, 2006, Fu

245、dan University 220Game Theory-Lecture 1Sequential bargaining (2.1.D of Gibbons)Player 2acceptrejectpropose an offer ( s2 , 1-s2 )Period 1Player 1acceptpropose an offer ( s1 , 1-s1 ) s1 , 1-s1Player 1s2 , 1-s2s , 1-sPeriod 2Period 3rejectPlayer 2 s , 1- sDec, 2006, Fudan University 221Game Theory-Lec

246、ture 1Solve sequential bargaining by backward inductionnPeriod 1:Player 2 accepts 1-s1 if and only if 1-s1 (1-s2*)=(1- s) or s1 1-(1-s2*), where s2* = s. Player 1 faces the following two options:(1) offers 1-s1 = (1-s2*)=(1- s) to player 2, leaving s1 = 1-(1-s2*)=1-+s for herself at this period, or(

247、2) offers 1-s1 (1-s2*) to player 2 (player 2 will reject it), and receives s2* = s next period. Its discounted value is s Since s a cNote: Osborne used b2(q1) instead of R2(q1)Dec, 2006, Fudan University 233Game Theory-Lecture 1Stackelberg model of duopolynSolve firm 1s problem. Note firm 1 can also

248、 solve firm 2s problem. That is, firm 1 knows firm 2s best response to any q1. Hence, firm 1s problem isMax u1(q1, R2(q1)=q1(a(q1+R2(q1)c)subject to 0 q1 +That is,Max u1(q1, R2(q1)=q1(aq1c)/2subject to 0 q1 +FOC: (a 2q1 c)/2 = 0 q1 = (a c)/2Dec, 2006, Fudan University 234Game Theory-Lecture 1Stackel

249、berg model of duopolynSubgame-perfect Nash equilibrium( (a c)/2, R2(q1) ), where R2(q1) = (a q1 c)/2 if q1 a c = 0 if q1 a c That is, firm 1 chooses a quantity (a c)/2, firm 2 chooses a quantity R2(q1) if firm 1 chooses a quantity q1.The backward induction outcome is ( (a c)/2, (a c)/4 ). Firm 1 cho

250、oses a quantity (a c)/2, firm 2 chooses a quantity (a c)/4.Dec, 2006, Fudan University 235Game Theory-Lecture 1Stackelberg model of duopolynFirm 1 produces q1=(a c)/2 and its profit q1(a(q1+ q2)c)=(ac)2/8nFirm 2 produces q2=(a c)/4 and its profit q2(a(q1+ q2)c)=(ac)2/16nThe aggregate quantity is 3(a

251、 c)/4.Dec, 2006, Fudan University 236Game Theory-Lecture 1Cournot model of duopolynFirm 1 produces q1=(a c)/3 and its profit q1(a(q1+ q2)c)=(ac)2/9nFirm 2 produces q2=(a c)/3 and its profit q2(a(q1+ q2)c)=(ac)2/9nThe aggregate quantity is 2(a c)/3.Dec, 2006, Fudan University 237Game Theory-Lecture 1

252、MonopolynSuppose that only one firm, a monopoly, produces the product. The monopoly solves the following problem to determine the quantity qm.nMax qm (aqmc)subject to 0 qm +FOC: a 2qm c = 0 qm = (a c)/2nMonopoly produces qm=(a c)/2 and its profit qm(aqmc)=(ac)2/4Dec, 2006, Fudan University 238Game T

253、heory-Lecture 1DiscusssionnThe first-mover advantagenStrategic substitutes and commitment (threat)nStackelberg modelnStrategic complements and commitmentnBertrand model (your task)nThe curse of knowledgenIt is a good thing for rational one-agent decision problemnNot for a person with self-controlnTh

254、e last leaf (O.Henry)Dec, 2006, Fudan University 239Game Theory-Lecture 1Sequential-move Bertrand model of duopoly (differentiated products)nTwo firms: firm 1 and firm 2. nEach firm chooses the price for its product. The prices are denoted by p1 and p2, respectively.nThe timing of this game as follo

255、ws.Firm 1 chooses a price p1 0.Firm 2 observes p1 and then chooses a price p2 0.nThe quantity that consumers demand from firm 1: q1(p1, p2) = a p1 + bp2.nThe quantity that consumers demand from firm 2: q2(p1, p2) = a p2 + bp1.nThe cost to firm i of producing quantity qi is Ci(qi)=cqi.Dec, 2006, Fuda

256、n University 240Game Theory-Lecture 1Sequential-move Bertrand model of duopoly (differentiated products)nSolve firm 2s problem for any p1 0 to get firm 2s best response to p1. Max u2(p1, p2)=(a p2 + bp1 )(p2 c)subject to 0 p2 + FOC: a + c 2p2 + bp1 = 0 p2 = (a + c + bp1)/2Firm 2s best response, R2(p

257、1) = (a + c + bp1)/2Dec, 2006, Fudan University 241Game Theory-Lecture 1Sequential-move Bertrand model of duopoly (differentiated products)nSolve firm 1s problem. Note firm 1 can also solve firm 2s problem. Firm 1 knows firm 2s best response to p1. Hence, firm 1s problem isMax u1(p1, R2(p1)=(a p1 +

258、b R2(p1) )(p1 c)subject to 0 p1 + That is,Max u1(p1, R2(p1)=(a p1 + b (a + c + bp1)/2 )(p1 c)subject to 0 p1 + FOC: a p1 + b (a + c + bp1)/2+(1+b2/2) (p1 c) = 0 p1 = (a+c+(ab+bcb2c)/2)/(2b2)Dec, 2006, Fudan University 242Game Theory-Lecture 1Sequential-move Bertrand model of duopoly (differentiated

259、products)nSubgame-perfect Nash equilibrium(a+c+(ab+bcb2c)/2)/(2b2), R2(p1) ), where R2(p1) = (a + c + bp1)/2Firm 1 chooses a price (a+c+(ab+bcb2c)/2)/(2b2), firm 2 chooses a price R2(p1) if firm 1 chooses a price p1.Dec, 2006, Fudan University 243Game Theory-Lecture 1Perfect information: illustratio

260、n (sequential matching pennies)nEach of the two players has a penny. nPlayer 1 first chooses whether to show the Head or the Tail. nAfter observing player 1s choice, player 2 chooses to show Head or TailnBoth players know the following rules:If two pennies match (both heads or both tails) then playe

261、r 2 wins player 1s penny. Otherwise, player 1 wins player 2s penny. Player 1Player 2HT-1, 11, -1HTPlayer 2HT1, -1-1, 1Dec, 2006, Fudan University 244Game Theory-Lecture 1Dynamic games of complete and imperfect informationnImperfect informationA player may not know exactly Who has made What choices w

262、hen she has an opportunity to make a choice.Example: player 2 makes her choice after player 1 does. Player 2 needs to make her decision without knowing what player 1 has made.Dec, 2006, Fudan University 245Game Theory-Lecture 1Imperfect information: illustrationnEach of the two players has a penny.

263、nPlayer 1 first chooses whether to show the Head or the Tail. nThen player 2 chooses to show Head or Tail without knowing player 1s choice, nBoth players know the following rules:If two pennies match (both heads or both tails) then player 2 wins player 1s penny. Otherwise, player 1 wins player 2s pe

264、nny. Player 2Player 1Player 2HT-1, 11, -1HTHT1, -1-1, 1Dec, 2006, Fudan University 246Game Theory-Lecture 1Information setnGibbons definition: An information set for a player is a collection of nodes satisfying:the player has the move at every node in the information set, andwhen the play of the gam

265、e reaches a node in the information set, the player with the move does not know which node in the information set has (or has not) been reached.nAll the nodes in an information set belong to the same playernThe player must have the same set of feasible actions at each node in the information set.Dec

266、, 2006, Fudan University 247Game Theory-Lecture 1Information set: illustrationPlayer 1LRPlayer 2LR2, 2, 3Player 2LR3L”R”3L”R”3L”R”3L”R”1, 2, 03, 1, 22, 2, 12, 2, 10, 1, 11, 1, 21, 1, 1an information set for player 3 containing three nodesan information set for player 3 containing a single nodetwo in

267、formation sets for player 2 each containing a single nodeDec, 2006, Fudan University 248Game Theory-Lecture 1Information set: illustrationnAll the nodes in an information set belong to the same playerPlayer 1CDPlayer 2EF3, 0, 22, 1, 3Player 3GH1, 3, 10, 2, 2This is not a correct information setDec,

268、2006, Fudan University 249Game Theory-Lecture 1Information set: illustrationnThe player must have the same set of feasible actions at each node in the information set.Player 1CDPlayer 2EF3, 02, 1Player 2GH1, 30, 21, 1An information set cannot contains these two nodes KDec, 2006, Fudan University 250

269、Game Theory-Lecture 1Represent a static game as a game tree: illustrationnPrisoners dilemma (another representation of the game in Figure 2.4.3 of Gibbons. The first number is the payoff for player 1, and the second number is the payoff for player 2)nStatic game is a game of imperfect informationPri

270、soner 1Prisoner 2Prisoner 1MumFink4, 45, 0MumFinkMumFink0, 51, 1Dec, 2006, Fudan University 251Game Theory-Lecture 1Example: mutually assured destructionnTwo superpowers, 1 and 2, have engaged in a provocative incident. The timing is as follows.nThe game starts with superpower 1s choice: either igno

271、re the incident ( I ), resulting in the payoffs (0, 0), or to escalate the situation ( E ).nFollowing escalation by superpower 1, superpower 2 can back down ( B ), causing it to lose face and result in the payoffs (1, -1), or it can choose to proceed to an atomic confrontation situation ( A ). Upon

272、this choice, the two superpowers play the following simultaneous move game.nThey can either retreat ( R ) or choose to doomsday ( D ) in which the world is destroyed. If both choose to retreat then they suffer a small loss and payoffs are (-0.5, -0.5). If either chooses doomsday then the world is de

273、stroyed and payoffs are (-K, -K), where K is very large number.Dec, 2006, Fudan University 252Game Theory-Lecture 1Example: mutually assured destruction(think of Cuba crisis)1IE0, 02BA1, -112RD-0.5, -0.5-K, -KRDRD2-K, -K-K, -KDec, 2006, Fudan University 253Game Theory-Lecture 1Perfect information an

274、d imperfect informationnA dynamic game in which every information set contains exactly one node is called a game of perfect information.nA dynamic game in which some information sets contain more than one node is called a game of imperfect information.Dec, 2006, Fudan University 254Game Theory-Lectu

275、re 1Strategy and payoffnA strategy for a player is a complete plan of actions.nIt specifies a feasible action for the player in every contingency in which the player might be called on to act.nIt specifies what the player does at each of her information setsPlayer 1Player 2HT-1, 11, -1HTPlayer 2HT1,

276、 -1-1, 1a strategy for player 1: Ha strategy for player 2: TPlayer 1s payoff is 1 and player 2s payoff is -1 if player 1 plays H and player 2 plays T Dec, 2006, Fudan University 255Game Theory-Lecture 1Strategy and payoff: illustration1IE0, 02BA1, -112RD-0.5, -0.5-K, -KRDRD2-K, -K-K, -Ka strategy fo

277、r player 1: E, and R if player 2 plays A, written as ER a strategy for player 2: A, R, if player 1 plays E, written as AR Dec, 2006, Fudan University 256Game Theory-Lecture 1Nash equilibrium in a dynamic gamenWe can also use normal-form to represent a dynamic gamenThe set of Nash equilibria in a dyn

278、amic game of complete information is the set of Nash equilibria of its normal-formnHow to find the Nash equilibria in a dynamic game of complete informationConstruct the normal-form of the dynamic game of complete informationFind the Nash equilibria in the normal-formDec, 2006, Fudan University 257G

279、ame Theory-Lecture 1Remove nonreasonable Nash equilibriumnSubgame perfect Nash equilibrium is a refinement of Nash equilibriumnIt can rule out nonreasonable Nash equilibria or non-creditable threatsnWe first need to define subgameDec, 2006, Fudan University 258Game Theory-Lecture 1SubgamenA subgame

280、of a dynamic game tree begins at a singleton information set (an information set contains a single node), and includes all the nodes and edges following the singleton information set, and does not cut any information set; that is, if a node of an information set belongs to this subgame then all the

281、nodes of the information set also belong to the subgame.Dec, 2006, Fudan University 259Game Theory-Lecture 1Subgame: illustration1IE0, 02BA1, -112RD-0.5, -0.5-K, -KRDRD2-K, -K-K, -Ka subgamea subgameNot a subgameDec, 2006, Fudan University 260Game Theory-Lecture 1Subgame-perfect Nash equilibriumnA N

282、ash equilibrium of a dynamic game is subgame-perfect if the strategies of the Nash equilibrium constitute or induce a Nash equilibrium in every subgame of the game.nSubgame-perfect Nash equilibrium is a Nash equilibrium. Dec, 2006, Fudan University 261Game Theory-Lecture 1Find subgame perfect Nash e

283、quilibria: backward induction1IE0, 02BA1, -112RD-0.5, -0.5-K, -KRDRD2-K, -K-K, -Ka subgamea subgameStarting with those smallest subgamesThen move backward until the root is reachedOne subgame-perfect Nash equilibrium( IR, AR )Dec, 2006, Fudan University 262Game Theory-Lecture 1Find subgame perfect N

284、ash equilibria: backward induction1IE0, 02BA1, -112RD-0.5, -0.5-K, -KRDRD2-K, -K-K, -Ka subgamea subgameStarting with those smallest subgamesThen move backward until the root is reachedAnother subgame-perfect Nash equilibrium( ED, BD )Dec, 2006, Fudan University 263Game Theory-Lecture 1Dynamic games

285、 of complete informationnPerfect informationA player knows Who has made What choices when she has an opportunity to make a choicenImperfect informationA player may not know exactly Who has made What choices when she has an opportunity to make a choice.Dec, 2006, Fudan University 264Game Theory-Lectu

286、re 1Information set: illustrationPlayer 1LRPlayer 2LR2, 2, 3Player 2LR3L”R”3L”R”3L”R”3L”R”1, 2, 03, 1, 22, 2, 12, 2, 10, 1, 11, 1, 21, 1, 1an information set for player 3 containing three nodesan information set for player 3 containing a single nodetwo information sets for player 2 each containing a

287、 single nodeDec, 2006, Fudan University 265Game Theory-Lecture 1Subgame-perfect Nash equilibriumnA Nash equilibrium of a dynamic game is subgame-perfect if the strategies of the Nash equilibrium constitute or induce a Nash equilibrium in every subgame of the game.nA subgame of a game tree begins at

288、a singleton information set (an information set containing a single node), and includes all the nodes and edges following the singleton information set, and does not cut any information set; that is, if a node of an information set belongs to this subgame then all the nodes of the information set al

289、so belong to the subgame.Dec, 2006, Fudan University 266Game Theory-Lecture 1Find subgame perfect Nash equilibria: backward induction1IE0, 02BA1, -112RD-0.5, -0.5-K, -KRDRD2-K, -K-K, -Ka subgamea subgameStarting with those smallest subgamesThen move backward until the root is reachedOne subgame-perf

290、ect Nash equilibrium( IR, AR ) (treat last subgame as a static game)Dec, 2006, Fudan University 267Game Theory-Lecture 1Find subgame perfect Nash equilibria: backward induction1IE0, 02BA1, -112RD-0.5, -0.5-K, -KRDRD2-K, -K-K, -Ka subgamea subgameStarting with those smallest subgamesThen move backwar

291、d until the root is reachedAnother subgame-perfect Nash equilibrium( ED, BD ) Dec, 2006, Fudan University 268Game Theory-Lecture 1Find subgame perfect Nash equilibria: backward inductionnWhat is the subgame perfect Nash equilibrium?Player 1LRPlayer 2LR2, 2, 0Player 2LR3L”R”3L”R”3L”R”3L”R”1, 2, 33, 1

292、, 22, 2, 12, 2, 10, 0, 11, 1, 21, 1, 1Dec, 2006, Fudan University 269Game Theory-Lecture 1Bank runs (2.2.B of Gibbons)nTwo investors, 1 and 2, have each deposited D with a bank.nThe bank has invested these deposits in a long-term project. If the bank liquidates its investment before the project matu

293、res, a total of 2r can be recovered, where D r D/2.nIf banks investment matures, the project will pay out a total of 2R, where RD.nTwo dates at which the investors can make withdrawals from the bank.Dec, 2006, Fudan University 270Game Theory-Lecture 1Bank runs: timing of the gamenThe timing of this

294、game is as followsnDate 1 (before the banks investment matures)Two investors play a simultaneous move gameIf both make withdrawals then each receives r and the game endsIf only one makes a withdrawal then she receives D, the other receives 2r-D, and the game endsIf neither makes a withdrawal then th

295、e project matures and the game continues to Date 2.nDate 2 (after the banks investment matures)Two investors play a simultaneous move gameIf both make withdrawals then each receives R and the game endsIf only one makes a withdrawal then she receives 2R-D, the other receives D, and the game endsIf ne

296、ither makes a withdrawal then the bank returns R to each investor and the game ends.Dec, 2006, Fudan University 271Game Theory-Lecture 1Bank runs: game tree1WNW2WNW12WNWWNWWNW2a subgameOne subgame-perfect Nash equilibrium( NW W, NW W )Wr, rNWDate 1Date 2W: withdrawNW: not withdraw2D, 2rD2rD, DR, RD,

297、 2RD2RD, DR, RDec, 2006, Fudan University 272Game Theory-Lecture 1Bank runs: game tree1WNW2WNW12WNWWNWWNW2One subgame-perfect Nash equilibrium( W W, W W )Wr, rNWDate 1Date 2W: withdrawNW: not withdraw2D, 2rD2rD, DR, RD, 2RD2RD, DR, Ra subgameDec, 2006, Fudan University 273Game Theory-Lecture 1Tariff

298、s and imperfect international competition (2.2.C of Gibbons)nTwo identical countries, 1 and 2, simultaneously choose their tariff rates, denoted t1, t2, respectively.nFirm 1 from country 1 and firm 2 from country 2 produce a homogeneous product for both home consumption and export. nAfter observing

299、the tariff rates chosen by the two countries, firm 1 and 2 simultaneously chooses quantities for home consumption and for export, denoted by (h1, e1) and (h2, e2), respectively.nMarket price in two countries Pi(Qi)=aQi, for i=1, 2. nQ1=h1+e2, Q2=h2+e1.nBoth firms have a constant marginal cost c.nEac

300、h firm pays tariff on export to the other country.Dec, 2006, Fudan University 274Game Theory-Lecture 1Tariffs and imperfect international competitionDec, 2006, Fudan University 275Game Theory-Lecture 1Tariffs and imperfect international competitionDec, 2006, Fudan University 276Game Theory-Lecture 1

301、Backward induction: subgame between the two firmsDec, 2006, Fudan University 277Game Theory-Lecture 1Backward induction: subgame between the two firmsDec, 2006, Fudan University 278Game Theory-Lecture 1Backward induction: whole gameDec, 2006, Fudan University 279Game Theory-Lecture 1Tariffs and impe

302、rfect international competitionDec, 2006, Fudan University 280Game Theory-Lecture 1Repeated gamenA repeated game is a dynamic game of complete information in which a (simultaneous-move) game is played at least twice, and the previous plays are observed before the next play. nWe will find out the beh

303、avior of the players in a repeated game.Dec, 2006, Fudan University 281Game Theory-Lecture 1Two-stage repeated gamenTwo-stage prisoners dilemmaTwo players play the following simultaneous move game twiceThe outcome of the first play is observed before the second play beginsThe payoff for the entire g

304、ame is simply the sum of the payoffs from the two stages. That is, the discount factor is 1.Player 2L2R2Player 1L11 , 15 , 0R10 , 54 , 4Dec, 2006, Fudan University 282Game Theory-Lecture 1Game tree of the two-stage prisoners dilemma1L1R12L2R22L2R2L1R12L2R22L2R2L1R12L2R22L2R2L1R12L2R22L2R2L1R12L2R22L

305、2R21+11+11+51+01+01+51+41+411115+10+15+50+05+00+55+40+40+15+10+55+00+05+50+45+44+14+14+54+04+04+54+44+4Dec, 2006, Fudan University 283Game Theory-Lecture 1Informal game tree of the two-stage prisoners dilemma1L1R12L2R22L2R2L1R12L2R22L2R2L1R12L2R22L2R2L1R12L2R22L2R2L1R12L2R22L2R2115005441111(1, 1)(5,

306、 0)(0, 5)(4, 4)115005441150054411500544Dec, 2006, Fudan University 284Game Theory-Lecture 1Informal game tree of the two-stage prisoners dilemma1L1R12L2R22L2R2L1R12L2R22L2R2L1R12L2R22L2R2L1R12L2R22L2R2L1R12L2R22L2R2115005441111(2, 2)(6, 1)(1, 6)(5, 5)115005441150054411500544Dec, 2006, Fudan Universi

307、ty 285Game Theory-Lecture 1two-stage prisoners dilemmanThe subgame-perfect Nash equilibrium(L1 L1L1L1L1, L2 L2L2L2L2) Player 1 plays L1 at stage 1, and plays L1 at stage 2 for any outcome of stage 1.Player 2 plays L2 at stage 1, and plays L2 at stage 2 for any outcome of stage 1.Player 2L2R2Player 1

308、L11 , 15 , 0R10 , 54 , 4Dec, 2006, Fudan University 286Game Theory-Lecture 1Finitely repeated gamenA finitely repeated game is a dynamic game of complete information in which a (simultaneous-move) game is played a finite number of times, and the previous plays are observed before the next play. nThe

309、 finitely repeated game has a unique subgame perfect Nash equilibrium if the stage game (the simultaneous-move game) has a unique Nash equilibrium. The Nash equilibrium of the stage game is played in every stage.Dec, 2006, Fudan University 287Game Theory-Lecture 1What happens if the stage game has m

310、ore than one Nash equilibrium?Two players play the following simultaneous move game twiceThe outcome of the first play is observed before the second play beginsThe payoff for the entire game is simply the sum of the payoffs from the two stages. That is, the discount factor is 1.Question: can we find

311、 a subgame perfect Nash equilibrium in which M1, M2 are played? Or, can the two players cooperate in a subgame perfect Nash equilibrium?Player 2L2M2R2Player 1L11 , 15 , 00 , 0M10 , 54 , 40 , 0R10 , 00 , 03 , 3Dec, 2006, Fudan University 288Game Theory-Lecture 1Informal game tree1L1R122L2R2M2L2R2M2L2

312、R2M22L1R122L2R2M2L2R2M2L2R2M22M1(1, 1)(5, 0)(0, 5)(4, 4)(0, 0)M1(0, 0)(0, 0)(0, 0)(3, 3)1(1, 1)(5, 0)(0, 5)(0, 0)(0, 0)(0, 0)(0, 0)(3, 3)(4, 4)Dec, 2006, Fudan University 289Game Theory-Lecture 1Informal game tree and backward induction1L1R122L2R2M2L2R2M2L2R2M22L1R122L2R2M2L2R2M2L2R2M22M1(1, 1)(5, 0

313、)(0, 5)(4, 4)(0, 0)M1(0, 0)(0, 0)(0, 0)(3, 3)1(1, 1)(5, 0)(0, 5)(0, 0)(0, 0)(0, 0)(0, 0)(3, 3)(4, 4)(1, 1)(1, 1)(1, 1)(3, 3)(1, 1)(1, 1)(1, 1)(1, 1)(1, 1)+Dec, 2006, Fudan University 290Game Theory-Lecture 1Two-stage repeated gamePlayer 2L2M2R2Player 1L11 , 15 , 00 , 0M10 , 54 , 40 , 0R10 , 00 , 03

314、, 3nSubgame perfect Nash equilibrium:player 1 plays M1 at stage 1, and at stage 2, plays R1 if the first stage outcome is ( L1, L2 ), or plays L1 if the first stage outcome is not ( L1, L2 ) player 2 plays M2 at stage 1, and at stage 2, plays R2 if the first stage outcome is ( L1, L2 ), or plays L2

315、if the first stage outcome is not ( L1, L2 ) Dec, 2006, Fudan University 291Game Theory-Lecture 1Two-stage repeated gamePlayer 2L2M2R2Player 1L12 , 26 , 11 , 1M11 , 67 , 71 , 1R11 , 11 , 14 , 4nSubgame perfect Nash equilibrium:At stage 1, player 1 plays M1, and player 2 plays M2.At stage 2, player 1

316、 plays R1 if the first stage outcome is ( M1, M2 ), or plays L1 if the first stage outcome is not ( M1, M2 ) player 2 plays R2 if the first stage outcome is ( M1, M2 ), or plays L2 if the first stage outcome is not ( M1, M2 ) The payoffs of the 2nd stage has been added to the first stage game.Dec, 2

317、006, Fudan University 292Game Theory-Lecture 1An abstract game: generalization of the tariff game (general form)nFour players: 1, 2, 3, 4. The timing of the game is as follows. nStage1: Player 1 and 2 simultaneously choose actions a1 and a2 from feasible action sets A1 and A2, respectively.nStage 2:

318、After observing the outcome (a1, a2) of the first stage, Player 3 and 4 simultaneously choose actions a3 and a4 from feasible action sets A3 and A4, respectively.nThe game ends. nThe payoffs are ui(a1, a2, a3, a4), for i=1, 2, 3, 4Dec, 2006, Fudan University 293Game Theory-Lecture 1An abstract game:

319、 informal game treeplayer 1Player 1 action set A1Stage 1Stage 2player 2player 3Player 2 action set A2Player 3 action set A3Player 4 action set A4player 4a1a2a3a4A smallest subgame following (a1, a2)Dec, 2006, Fudan University 294Game Theory-Lecture 1Backward induction: solve the smallest subgameDec,

320、 2006, Fudan University 295Game Theory-Lecture 1Backward induction: back to the rootplayer 1Player 1 action set A1Stage 1Stage 2player 2Player 2 action set A2a1a2Dec, 2006, Fudan University 296Game Theory-Lecture 1Backward induction: back to the rootDec, 2006, Fudan University 297Game Theory-Lecture

321、 1Tariffs and imperfect international competition (2.2.C of Gibbons)nTwo identical countries, 1 and 2, simultaneously choose their tariff rates, denoted t1, t2, respectively.nFirm 1 from country 1 and firm 2 from country 2 produce a homogeneous product for both home consumption and export. nAfter ob

322、serving the tariff rates chosen by the two countries, firm 1 and 2 simultaneously chooses quantities for home consumption and for export, denoted by (h1, e1) and (h2, e2), respectively.nMarket price in two countries Pi(Qi)=aQi, for i=1, 2. nQ1=h1+e2, Q2=h2+e1.nBoth firms have a constant marginal cos

323、t c.nEach firm pays tariff on export to the other country.Dec, 2006, Fudan University 298Game Theory-Lecture 1Tariffs and imperfect international competitionnThis model fits the abstract model. nCountry 1 and 2 are player 1 and 2, respectivelynFirm 1 and 2 are player 3 and 4, respectivelyCountry 1Fi

324、rm 1Q1= h1 + e2 P1(Q1)=aQ1Country 2Firm 2Q2= h2 + e1 P2(Q2)=aQ2t2t1Stage 1Stage 2Dec, 2006, Fudan University 299Game Theory-Lecture 1Backward induction: subgame between the two firmsDec, 2006, Fudan University 300Game Theory-Lecture 1Backward induction: subgame between the two firmsDec, 2006, Fudan

325、University 301Game Theory-Lecture 1Backward induction: back to the rootDec, 2006, Fudan University 302Game Theory-Lecture 1Tariffs and imperfect international competitionDec, 2006, Fudan University 303Game Theory-Lecture 1Infinitely repeated gamenA infinitely repeated game is a dynamic game of compl

326、ete information in which a (simultaneous-move) game called the stage game is played infinitely, and the outcomes of all previous plays are observed before the next play. nPrecisely, the simultaneous-move game is played at stage 1, 2, 3, ., t-1, t, t+1, . The outcomes of all previous t-1 stages are o

327、bserved before the play at the tth stage.nEach player discounts her payoff by a factor , where 0 0.5, Chris best response is operaA Bayesian Nash equilibrium: (opera, (opera if happy and prize fight if unhappy)Dec, 2006, Fudan University 356Game Theory-Lecture 1Battle of the sexes with incomplete in

328、formation (version one) contdnBest responseIf Chris chooses prize fight then Pats best response: prize fight if he is happy, and opera if he is unhappySuppose that Pat chooses prize fight if he is happy, and opera if he is unhappy. What is Chris best response?If Chris chooses opera then she get a pa

329、yoff 0 if Pat is happy, or 2 if Pat is unhappy. Her expected payoff is 00.5+ 20.5=1If Chris chooses prize fight then she get a payoff 1 if Pat is happy, or 0 if Pat is unhappy. Her expected payoff is 10.5+ 00.5=0.5Since 10.5, Chris best response is opera(prize fight, (prize fight if happy and opera

330、if unhappy) is not a Bayesian Nash equilibrium.Dec, 2006, Fudan University 357Game Theory-Lecture 1Cournot duopoly model of incomplete information (version three) contdnFirm 2s cost depends on some factor (e.g. technology) that only firm 2 knows. Its cost can be HIGH: cost function: C2(q2)=cHq2.LOW:

331、 cost function: C2(q2)=cLq2.nFirm 1s cost also depends on some other (independent or dependent) factor that only firm 1 knows. Its cost can be HIGH: cost function: C1(q1)=cHq1.LOW: cost function: C1(q1)=cLq1.Dec, 2006, Fudan University 358Game Theory-Lecture 1Cournot duopoly model of incomplete info

332、rmation (version three) contdDec, 2006, Fudan University 359Game Theory-Lecture 1Cournot duopoly model of incomplete information (version three) contdDec, 2006, Fudan University 360Game Theory-Lecture 1Cournot duopoly model of incomplete information (version three) contdDec, 2006, Fudan University 3

333、61Game Theory-Lecture 1Cournot duopoly model of incomplete information (version three) contdu1(q1, q2(cH); cH)u1(q1, q2(cL); cH)Dec, 2006, Fudan University 362Game Theory-Lecture 1Cournot duopoly model of incomplete information (version three) contdu1(q1, q2(cH); cL)u1(q1, q2(cL); cL)Dec, 2006, Fuda

334、n University 363Game Theory-Lecture 1Cournot duopoly model of incomplete information (version three) contdu2(q1(cH), q2; cH)u2(q1(cL), q2; cH)Dec, 2006, Fudan University 364Game Theory-Lecture 1Cournot duopoly model of incomplete information (version three) contdu2(q1(cH), q2; cL)u2(q1(cL), q2; cL)D

335、ec, 2006, Fudan University 365Game Theory-Lecture 1Cournot duopoly model of incomplete information (version three) contdDec, 2006, Fudan University 366Game Theory-Lecture 1Cournot duopoly model of incomplete information (version three) contdDec, 2006, Fudan University 367Game Theory-Lecture 1Normal-

336、form representation of static Bayesian gamesDec, 2006, Fudan University 368Game Theory-Lecture 1Normal-form representation of static Bayesian games: payoffsDec, 2006, Fudan University 369Game Theory-Lecture 1Normal-form representation of static Bayesian games: beliefs (probabilities)Dec, 2006, Fudan

337、 University 370Game Theory-Lecture 1StrategyDec, 2006, Fudan University 371Game Theory-Lecture 1Bayesian Nash equilibrium: 2-playerDec, 2006, Fudan University 372Game Theory-Lecture 1Bayesian Nash equilibrium: 2-playerplayer 1s best response if her type is t1iplayer 2s best response if her type is t

338、2jIn the sense of expectation based on her beliefIn the sense of expectation based on her beliefDec, 2006, Fudan University 373Game Theory-Lecture 1Static Bayesian gamesDec, 2006, Fudan University 374Game Theory-Lecture 1Bayesian Nash equilibrium: 2-playerDec, 2006, Fudan University 375Game Theory-L

339、ecture 1Bayesian Nash equilibrium: 2-playerplayer 1s best response if her type is t1iplayer 2s best response if her type is t2jIn the sense of expectation based on her beliefIn the sense of expectation based on her beliefDec, 2006, Fudan University 376Game Theory-Lecture 1Battle of the sexesnAt the

340、separate workplaces, Chris and Pat must choose to attend either an opera or a prize fight in the evening. nBoth Chris and Pat know the following:Both would like to spend the evening together. But Chris prefers the opera.Pat prefers the prize fight.PatOperaPrize FightChrisOpera2 , 10 , 0Prize Fight0

341、, 01 , 2Dec, 2006, Fudan University 377Game Theory-Lecture 1Battle of the sexes with incomplete information (version two)nPats preference depends on whether he is happy. If he is happy then his preference is the same.nIf he is unhappy then he prefers to spend the evening by himself.nChris cannot fig

342、ure out whether Pat is happy or not. But Chris believes that Pat is happy with probability 0.5 and unhappy with probability 0.5nChris preference also depends on whether she is happy. If she is happy then her preference is the same.nIf she is unhappy then she prefers to spend the evening by herself.n

343、Pat cannot figure out whether Chris is happy or not. But Pat believes that Chris is happy with probability 2/3 and unhappy with probability 1/3.Dec, 2006, Fudan University 378Game Theory-Lecture 1Battle of the sexes with incomplete information (version two) contdnCheck whether (Opera if happy, Opera

344、 if unhappy), (Opera if happy, Fight is unhappy) is a Bayesian NEChris is happy Pat is happyPatOperaFightChrisOpera2 , 10 , 0Fight0 , 01 , 2Chris is happy Pat is unhappyPatOperaFightChrisOpera2 , 00 , 2Fight0 , 11 , 0Chris is unhappy Pat is happyPatOperaFightChrisOpera0 , 12 , 0Fight1 , 00 , 2Chris

345、is unhappy Pat is unhappyPatOperaFightChrisOpera0 , 02 , 2Fight1 , 10 , 0Dec, 2006, Fudan University 379Game Theory-Lecture 1Battle of the sexes with incomplete information (version two) contdDec, 2006, Fudan University 380Game Theory-Lecture 1Battle of the sexes with incomplete information (version

346、 two) contdDec, 2006, Fudan University 381Game Theory-Lecture 1Battle of the sexes with incomplete information (version two) contdDec, 2006, Fudan University 382Game Theory-Lecture 1Battle of the sexes with incomplete information (version two) contdDec, 2006, Fudan University 383Game Theory-Lecture

347、1Battle of the sexes with incomplete information (version two) contdChris is unhappyPat (0.5, 0.5)(O,O)(O,F)(F,O)(F,F)ChrisO0112F11/21/20Chris is happyPat (0.5, 0.5)(O,O)(O,F)(F,O)(F,F)ChrisO2110F01/21/21Chris believes that Pat is happy with probability 0.5, unhappy 0.5Chris expected payoff of playi

348、ng Fight if Chris is happy and Pat plays (Opera if happy, Fight if unhappy)Dec, 2006, Fudan University 384Game Theory-Lecture 1Battle of the sexes with incomplete information (version two) contdPat is happyPatOFChris(2/3, 1/3)(O,O)10(O,F)2/32/3(F,O)1/34/3(F,F)02Pat is unhappyPatOFChris(2/3, 1/3)(O,O

349、)02(O,F)1/34/3(F,O)2/32/3(F,F)10Pat believes that Chris is happy with probability 2/3, unhappy 1/3Pats expected payoff of playing Opera if Pat is unhappy and Chris plays (Fight if happy, Fight if unhappy)Dec, 2006, Fudan University 385Game Theory-Lecture 1Battle of the sexes with incomplete informat

350、ion (version two) contdnCheck whether (Fight if happy, Opera if unhappy), (Fight if happy, Fight is unhappy) is a Bayesian Nash equilibrium. Dec, 2006, Fudan University 386Game Theory-Lecture 1First-price sealed-bid auction (3.2.B of Gibbons)Dec, 2006, Fudan University 387Game Theory-Lecture 1First-

351、price sealed-bid auction (3.2.B of Gibbons) contdDec, 2006, Fudan University 388Game Theory-Lecture 1First-price sealed-bid auction (3.2.B of Gibbons) contdDec, 2006, Fudan University 389Game Theory-Lecture 1First-price sealed-bid auction (3.2.B of Gibbons) contdDec, 2006, Fudan University 390Game T

352、heory-Lecture 1First-price sealed-bid auction (3.2.B of Gibbons) contdDec, 2006, Fudan University 391Game Theory-Lecture 1First-price sealed-bid auction (3.2.B of Gibbons) contdDec, 2006, Fudan University 392Game Theory-Lecture 1For further studiesDec, 2006, Fudan University 393Game Theory-Lecture 1