Game Theory in Economics Quiz 3 PDF

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Game Theory in Economics ECON 4130 QUIZ 3...

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Quiz 3 (Econ 4130) You have 30 minutes

QUESTION: Determine all of the Nash equilibria (pure-strategy and mixed-strategy equilibria) of the following game: L U 10, 3 C 5, 3 D 7, 2

M 2, 4 6, 6 5, 5

R 6, 5 2, 2 3, 6

SOLUTION: There are two pure strategy Nash equilibria: (U, R) and (C, M ). To find the mixed NE, use IESDS: M strictly dominates L. Clearly U or C are not strictly dominated because they are best response strategies for Player 1. So, the game is reduced to M R U 2, 4 6, 5 C 6, 6 2, 2 D 5, 5 3, 6

Strategy D is not a best response to any pure strategy (belief) of Player 2. However, D is not strictly dominated (verify this). In fact, it is a best response to a mixed strategy of player 2: σ2 = (M 21, R 21). Let’s verify this: Suppose player 2 plays M with probability p. Then, Player 1’s expected payoff of playing U is E1 (U, p) = 2p + 6(1 − p) = 6 − 4p. Player 1’s expected payoff of playing C is E1 (C, p) = 6p + 2(1 − p) = 2 + 4p. Finally, Player 1’s expected payoff of playing D is E1 (D, p) = 5p + 3(1 − p) = 3 + 2p. U is better than D iff E1 (D, p) ≤ E1 (U, p), equivalently p ≤ 1/2. C is better than D iff E1 (D, p) ≤ E1 (C, p), equivalently p ≥ 1/2. Moreover, U is better than C iff E1 (U, p) ≥ E1 (C, p), equivalently p ≤ 1/2. That is, if Player 2 plays a mixed strategy p such that p < 1/2, then Player 1’s best response is U ; if p > 1/2 then Player 1’s best response is C; and finally, if p = 1/2 then Player 1’s best response is all three strategies (i.e., U, C and D). Now, I try to find a NE where Player 1 plays D with a positive probability. Above arguments show that p must be 1/2 in such equilibrium. Because Player 2 will also be mixing, it means she should be indifferent between M and R. Now, suppose Player 1 mixes between U, C and D with probabilities q1 , q2 , and 1 − q1 − q2 , respectively. Therefore, Player 2’s expected payoff of playing M is E2 (M, q1 , q2 ) = 4q1 + 6q2 + 5(1 − q1 − q2 ). Player 2’s expected payoff of playing R is E2 (R, q1 , q2 ) = 5q1 + 2q2 + 6(1 − q1 − q2 ). Therefore, Player 2 is indifferent between M and R iff E2 (M, q1 , q2 ) = E2 (R, q1 , q2 ) iff q2 = 1/5. 1

Hence, there are infinitely many mixed Nash equilibria of this game where D is a part of: 1 (Uq, C , D( 4 − q)), (M 1 , R 1 ), 5 5 2 2 where q ∈ [0, 45 ]. However, there may be other mixed NE where D is not part of (i.e., p 6= 1/2). To find these NE, we basically find the mixed NE of the following 2 by 2 game: U C

M R 2, 4 6, 5 6, 6 2, 2

If you check NE of this game, you will see that a mixed NE exist and it is already a part of NE described above (namely, there is no NE where p 6= 1/2).

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