ECON60361st semester 05-06

• Format of final exam• Same as the mid term • Material not covered in final exam• fixed point theorem—both proof and application• purification of mixed strategy• rationalizability, dominance solvability• repeated game – overtaking criteria, limit of

means criteria, Maskin-Tirole theorem• Cho-Kreps Intuitive criteria

• Material covered in final exam• strategic game• --Nash equilibrium• Bayesian game• --Bayesian equilibrium • extensive game with perfect information• --subgame perfect equilibrium• bargaining game• repeated game—subgame perfect equilibrium, trigger strategy,

minmax value• extensive game with imperfect information• --perfect Bayesian equilibrium• --sequential equilibrium

Exercise 211.1 (Timing claims on an investment)

• An amount of money accumulates; in period t (= 1, 2, ..., T) its size is $2t.

• In each period t two people simultaneously decide whether to claim the money. If only one person does so, she gets all the money; if both people do so, they split the money equally; either case, the game ends.

• If neither person does so, both people have the opportunity to do so in the next period; if neither person claims the money in period T, each person obtains $T.

• Each person cares only about the amount of money she obtains.

• Formulate this situation as an extensive game with perfect information and simultaneous moves, and find its SPE.

Equilibrium: immediate claiming

• Claim: in the SPE, each player always claims money whenever she is asked to move

• Proof: When t=T, it is the strictly dominant action for each to claim (by claiming, she gets T rather than 0 if the other also claims; she gets 2T rather than T if the other doesn’t) => each always claims money at t=T

• Assume each always claims money at t=k+1,…,T• Then at t=k, each claiming is also best response (by

claiming, she gets k rather than 0 if the other also claims; she gets 2k rather than (k+1) if the other doesn’t).

• (There exists another SPE in which in the first period neither claims money but in any subsequent period both claim money.)

EXERCISE 227.3 (Sequential duel)

• In a sequential duel, two people alternately have the opportunity to shoot each other; each has an infinite supply of bullets.

• On each of her turns, a person may shoot or refrain from doing so. Each of person i’s shots hits (and kills) its intended target with probability p, (independently of whether any other shots hit their targets).

• Each person cares only about her probability of survival (not about the other person's survival).

• Model this situation as an extensive game with perfect information and chance moves.

• Show that the strategy pairs in which neither person ever shoots and in which each person always shoots are both subgame perfect equilibria.

No shooting SPE

• Claim: each never shoots (whether or not somebody has ever shot)

• Proof: According to the prescripts, each’s survival probability is already one, and cannot be further increased. Hence, no beneficial unilateral deviation.

Shooting SPE

• Claim: each always shoot• Proof:

– We argue that deviating one is not beneficial. Suppose now it is player 1’s turn to to move in period t.

– Let Q<1 be 1’s payoff (survival probability) conditional on both players exist in period t+1 and they act according to the prescripts thereafter.

– If 1 shoots in period t and both act according to the prescripts thereafter, 1’s payoff is p1+(1-p1)Q=p1(1-Q)+Q. If he does not shoot in period t and both act according to the prescripts, his payoff is Q. Clearly, the deviation is NOT beneficial.

Example 473.1 (One-sided offers)

• Consider the variant of the bargaining game of alternating offers in which only player 1 makes proposals.

• In every period, player 1 makes a proposal, which player 2 either accepts, ending the game, or rejects, leading to the next period, in which player 1 makes another proposal.

• Consider the strategy pair in which player 1 always proposes (x1,1-x1) and player 2 always accepts a proposal (y1,y2) if and only if y2 ≥ 1-x1.

• Find the value(s) of x1 for which this strategy pair is a subgame perfect equilibrium.

Equilibrium

• SPE: 1 always proposes (x1,1-x1); 2 always accepts a proposal giving her at least 1-x1 and rejects any inferior proposal.

• Claim: 1-x1=0.• Proof: (use one stage deviation)

– Suppose not (so that 1-x1>0). Consider the history in which a proposal (z1,1-z1) is proposed so that δ(1-x1)<1-z1<(1-x1).

– By accepting this offer, 2 gets 1-z1 now. By rejecting this offer, 2 will get δ(1-x1)< 1-z1.

– Hence, 2 should accept the proposal which is strictly inferior than (x1,1-x1). But according to her prescript, she should not accept such an inferior proposal. A contradiction.

EXERCISE 445.1 (Tit-for-tat as a subgame perfect equilibrium)

• Consider the infinitely repeated Prisoner's Dilemma in which the payoffs of the component game are those given in the Figure.

• Show that (tit-for-tat, tit-for-tat) is a subgame perfect equilibrium of this infinitely repeated game with discount factor δ if and only if y-x=1 and δ= 1/x.

C D

C x,x 0,y

D y,0 1,1

Note: 1 < x < y

Tit for Tat

• Tit-for-tat: do whatever the other did to you in the previous period

• Four types of histories to check: those ending with (C,C), (C,D), (D,C), (D,D).

• Need to show player 1 does not gain by one deviation

(C,C)

2

2 4

Following equilibrium strategy, period outcomes will be (C,C), (C,C),...

and 1 will get

(1 ...)1

If 1 deviates (to D), period outcomes alternate ./. (D,C) and (C,D) and

1 will get

(1 ...)

xx

y

2

2

1

1 does not deviate if and only if 1 1

or (1 ) .

y

x y

x y

C D

C x,x 0,y

D y,0 1,1

C,D

2 42

2

By following equil strategy, the outcomes will alternate ./. (D,C) and (C,D)

and 1 will get

(1 ...)1

By deviating once (to C), the outcomes will become (C,C),(C,C),...

1 will get

(1 ...)

yy

x

2

1

Hence 1 does not deviate if and only if 1 1

or (1 )

x

y x

y x

C D

C x,x 0,y

D y,0 1,1

(D,C)

32

2

By following equil strategy, the outcomes will alternate ./. (C,D) and (D,C)

and 1 will get

( ...)1

By deviating once (to D), the outcomes will become (D,D),(D,D),...

1 will get

1(1 ...)

1

yy

2

1Hence 1 does not deviate if and only if

1 1or (1 )

y

y

C D

C x,x 0,y

D y,0 1,1

(D,D)

2

3

By following equil strategy, the outcomes will become (D,D),(D,D),...

1 will get

1(1 ...)

1By deviating once (to C), the outcomes will alternate ./. (C,D) and (D,C)

and 1 will get

( ...)1

yy

2

2

1Hence 1 does not deviate if and only if

1 1or 1+

y

y

C D

C x,x 0,y

D y,0 1,1

To summarize

• Conditions for no deviations:

• CC: y(1+δ)x• CD: y≥(1+δ)x• DC: δy≥1+δ• DD: δy1+δ

• Hence, y=(1+ δ)x and δy=1+δ.

• Finally, δ=1/x and y-x=1.

• Both using tit-for-tat is SPE if and only if these two conditions hold.

• Very stringent conditions indeed!

EXERCISE 331.1 (Selten's horse)

• Find the perfect Bayesian equilibria of the game in Figure 331.2 in which each player's strategy is pure.

• Hint: Find the pure strategy Nash equilibria, then determine which is part of a weak sequential equilibrium

• Two pure strat Nash equil: (D,c,L) and (C,c,R)

• The 1st one is NOT part of a PBE. Foreseeing 3 will choose L, 2 should choose d to earn 4 rather than c to earn 1. Not sequential rationality. Hence not PBE.

• The 2nd one is part of a PBE.

c d

C 1,1,1 4,4,0

D 3,3,2 3,3,2

c d

C 1,1,1 0,0,1

D 0,0,0 0,0,0

L R