1 Solutions to Homework 4

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1 1 Solutions to Homework Q1 Let A be the event that the contestant chooses the door holding the car, and B be the event that the host opens a door holding a goat. A is the event that the contestant does not choose the door holding the car. If P (A B) < 1 2, then the contestant should switch; if P (A B) = 1 2, the contestant is indifferent, and if P (A B) > 1 2, the contestant should not switch. According to Bayes Rule, P (A B) = In this problem, P (A) = 1 3, P ( A) = 2 3. P (B A)P (A) P (B A)P (A) + P (B A)P ( A) Suppose the host always reveals a goat, regardless of whether the contestant chose the door with the car or not. Then P (B A) = 1, P (B A) = 1, and P (A B) = 1 3, so the contestant should switch. Suppose the host opens a door randomly. Then P (B A) = 1, P (B A) = 1 2, and P (A B) = 1 2, so the contestant is indifferent. Suppose the host only reveals a goat when the contestant picks a goat. Then P (B A) = 0, P (B A) = 1, and P (A B) = 0, so the contestant should switch (it is certain that the car is behind the other door). Suppose the host only reveals a goat when the contestant picks a car. Then P (B A) = 1, P (B A) = 0, and P (A B) = 1, so the contestant should not switch (it is certain that the car is behind the contestant s door). 1.2 Q2 The players pure strategies and expected payoffs are shown in this matrix: MM MB BM BB UU 10,2.5 10,2.5 6,3.5 6,3.5 UD 6.25, , , ,3.5 DU 9.5, , , ,3.5 DD 5.75, , , ,3.5 Player 1 s best responses are: B 1 (MM) = UU, B 1 (MB) = UU, B 1 (BM) = DU, B 1 (BB) = UU. Player 2 s best responses are: B 2 (UU) = {BM, BB}, B 2 (UD) = MB, B 2 (DU) = BM, B 2 (DD) = {MB, BB}. Therefore, the pure NE are (UU, BB) and (DU, BM). Let p U, 1 p U be Player 2 s belief that Player 1 s type is H, L, conditional on U being played; likewise, let p D, 1 p D be Player 2 s belief that Player 1 s type is H, L, conditional on D being played. Suppose Player 1 s behavioral strategy is: on H, play (q H, 1 q H ) where q H is the probability of playing U 1

2 on L, play (q L, 1 q L ) By Bayes Rule, we have: p U = Let s look at each NE: q H p q H p + q L (1 p), p D = (1 q H )p (1 q H )p + (1 q L )(1 p) (DU, BM): Since Player 1 plays both D and U, both of Player 2 s information sets will be reached with positive probability, so beliefs must follow Bayes Rule. Using q H = 0, q L = 1 we get p U = 0, p D = 1. We must also check that Player 2 s actions are optimal given his beliefs; this is easily seen in this case since these beliefs are consistent with a pure strategy and Player 2 is playing his best response to pure strategies. (UU, BB): Player 1 does not play D, so at that information set, any beliefs are consistent. At the information set following U, Player 2 s belief matches Nature s distribution: p U = p = 1 4. We want to check that Player 2 s optimal actions, given beliefs, match the NE. This is simple for the information set following U; Player 2 is using his best response. For the information set following D, we compute the expected payoff to M and B: E 2 (M D) = 10p D + 0(1 p D ) = 10p D, E 2 (B D) = 5p D + 3(1 p D ) = 3 + 2p D We want to find beliefs that would justify choosing pure B. This is optimal if E 2 (B D) > E 2 (M D), or if p D 3 8. Therefore, the set of weak sequential equilibria are: Strategies: (DU, BM), Player 2 s beliefs: p U = 0, p D = 1 Strategies: (UU, BB), Player 2 s beliefs: p U = 1 4, 0 p D 3 8 In the weak sequential equilibrium with (DU, BM), Player 1 chooses a different action based on his type, so this is a separating equilibrium. In the equilibrium with (UU, BB), Player 1 chooses the same action regardless of his type, so this is a pooling equilibrium. 2

3 (Strategies in variants of card game and entry game) 3

4 (Subgame perfect equilibrium of finitely repeated Prisoner s Dilemma) (Strategies in an infinitely repeated Prisoner s Dilemma) (a): (b): (c): (Nash equilibria of an infinitely repeated Prisoner s Dilemma) (a) Suppose both players do not deviate. Then the outcome path will be (C, C) in every period, which gives a discounted average of 2 for each player. Now suppose Player 1 4

5 deviates by playing D in the first period. Player 2 will play C, then D, D, D,... The best response to this is to play D, D, D,... So if Player 1 deviates and then plays his best response, the outcome path will be: (D, C), (D, C), (D, D), (D, D),... which gives payoffs of (3, 0), (3, 0), (1, 1), (1, 1),... Player 1 s discounted average is: ) (1 δ)(3 + 3δ + δ 2 + δ = (1 δ) (3(1 + δ) + δ2 = 3 2δ 2 1 δ It is not profitable to deviate if 2 3 2δ 2, or if δ 1 2. (b) If both players do not deviate, the outcome is (C, C) in every period, giving a discounted average payoff of 2. Suppose Player 1 deviates by playing D once, then playing C every period. The outcome path will be (D, C), (C, C), (C, C),... giving payoffs (3, 2, 2,...) to Player 1. This is always better for Player 1 than (2, 2, 2,...), so there is no value of δ for which this is a NE. (c) Suppose both players do not deviate. Then the outcome path will be (C, C) in every period, which gives a discounted average of 2 for each player. If Player 1 deviates, his best response is to play D in every period, which generates the outcome path (D, C), (D, D), (D, C), (D, D),... Here, the outcome alternates between (D, C) and (D, D). Player 1 s payoffs are 3, 1, 3, 1,... with a discounted average of ( 3 (1 δ)(3 + δ + 3δ 2 + δ ) = (1 δ) 1 δ 2 + δ ) 3 + δ 1 δ 2 = (1 δ) (1 δ)(1 + δ) = 3 + δ 1 + δ which is greater than 2 for any value of δ. So it cannot be a NE when both players play this strategy using the standard payoffs we have assumed so far. Are there any other payoffs for the Prisoner s Dilemma that could support this strategy profile as a NE? Suppose the payoffs are: C D C x,x 0,y D y,0 1,1 with 1 < x < y. Then the condition for the strategy profile to be a NE is which is satisfied if δ y x x 1. y + δ 1 + δ x 5

6 (Deviations from grim trigger strategy) (Tit-for-tat as a subgame perfect equilibrium) 6

7 (Selten s Horse) The pure strategy NE are: (C, c, R) and (D, c, L). Let s find the set of beliefs that are consistent with each strategy profile: (C, c, R): Player 3 s information set is not reached, so any beliefs are consistent. However, Player 3 s action of R must also be optimal given beliefs. Suppose Player 3 s belief over the histories D, Cd is (p, 1 p). Calculate the expected payoffs to each strategy: R will be optimal if 1 p 2p, or if p 1 3. E 3 (L) = 2p, E 3 (R) = 1 p (D, c, L): The only beliefs that are consistent is p = 1. Player 3 s action is optimal given this belief, since it matches the pure strategies of the NE. However, Player 2 s action does not satisfy the sequential rationality condition, since if p = 1, Player 2 s optimal choice is d. Therefore, this is not part of a weak sequential equilibrium (Sir Philip Sydney game) Suppose there is a separating equilibrium in which Player 2 (the parent) plays the pure strategies Squawk Give, Quiet Keep. Player 1 (the child) s expected payoffs, conditional on his type, are: Hungry-type: Not Hungry-type: E 1 (S H) = 1 t + rs, E 1 (Q H) = r E 1 (S N) = 1 t + rs, E 1 (Q N) = V + r 7

8 In order for the equilibrium to be separating, it must be optimal for the Hungry types to choose Squawk, and for the Not Hungry types to choose Quiet. The second condition requires that E 1 (Q N) E 1 (S N), or V + r > 1 t + rs. Rearranging, we get t + r(1 S) 1 V. Suppose we are given that r < (1 V )/(1 S). Then r(1 S) < (1 V ), which can only be consistent with the second condition if t > 0 (note that not all values of t > 0 will make the conditions true). Now, let s look at Player 2 (the parent) s expected payoffs. After Squawk: E 2 (G S) = p(s + r(1 t)) + (1 p)(s + r(1 t)) = S + r(1 t) E 2 (K S) = p + (1 p)(1 + rv (1 t)) After Quiet: E 2 (G Q) = p(s + r) + (1 p)(s + r) = S + r E 2 (K Q) = p + (1 p)(1 + rv ) In order for the equilibrium to be pooling, it must be optimal for the parent to choose Keep even if the child chooses Squawk. This requires that E 2 (K S) E 2 (G S), or p + (1 p)(1 + rv (1 t)) S + r(1 t). Rearranging, we get r(1 t) 1 S 1 (1 p)v Suppose we are given that r < (1 S)/1 (1 p)v. By assumption, 0 t 1, so the optimality condition is satisfied (Minmax payoffs) In the dividing money game, suppose Player 1 chooses x. Player 2 s best response is: If x 5, any y 10 x is a best response, which gives Player 2 a payoff of 10 x. The worst case gives a payoff of 5. If 5 < x <= 10, Player 2 s best response is x 1, which gives a payoff of x 1. Worst case is 5. If 10 < x, Player 2 s best response is 10. Worst case: 10. Therefore, the minmax payoff for Player 2 is 5. By symmetry, this is also the minmax payoff for Player 1. In Cournot duopoly, we are given that P (Q) = 0 for a large enough Q = q 1 + q 2. It is possible for Player 1 to choose q 1 large enough such that P = 0, in which case the best response.of Player 2 would be to choose q 2 = 0, with zero profits. Player 2 cannot be forced to accept a negative profit; it is always possible to choose q 2 = 0. Therefore, 0 is the minmax payoff. 8

9 Hotelling s electoral competition model with 2 candidates: Let x 1, x 2 be the candidates chosen positions. We know the best response of Player 2 is: if x 1 = m, the median, then x 2 = m, giving a probability of winning = 1 2. Otherwise, if x 1 m, then Player 2 can choose x 2 closer to m than x 1, giving a probability of winning = 1. The minimax payoff is 1 2. Hotelling s electoral competition model with 3 candidates: Suppose Players 1 and 2 choose x 1 = 1 4, x 2 = 3 4. It is not possible for Player 3 to win the election in this case. If Player 3 chooses a position between x 1 and x 2, the most he can get is 1 4 of the vote, while at least one of the other candidates will get > 1 4. If Player chooses a position x 1 or x 2, the most he can get is also 1 4, while one of the other candidates gets > 1 4. Payoff is the probability of winning, so the minmax payoff is 0. 9