HW2_Sol - 14.12 Game Theory Prof Muhamet Yildiz Fall 2010...

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14.12 Game Theory Prof. Muhamet Yildiz Fall 2010 Homework 2 Solutions Due on 10/12/2010 (in class) You need to show your work in all questions. 1. Consider the following game: L M N R A (3 ; 2) (0 ; 0) (6 ; 0) (0 ; 0) B (0 ; 0) (2 ; 4) (1 ; 2) (0 ; 0) C (1 ; 4) (1 ; 3) (0 ; 5) ( ° 1 ; 0) D (0 ; 0) (0 ; 0) (0 ; ° 1) (0 ; 0) (a) Compute the set of rationalizable strategies. Solution: C is strictly dominated by 2/5A+3/5C, strike out. Once C has been eliminated, N is strictly dominated by any mixture of L and M. The remaining strategies are rationalizable. (b) Find all Nash equilibria (including those in mixed strategies). Solution: L M N R A (3 ; 2 ) (0 ; 0) (6 ; 0) (0 ; 0) B (0 ; 0) (2 ; 4 ) (1 ; 2) (0 ; 0) C (1 ; 4) (1 ; 3) (0 ; 5 ) ( ° 1 ; 0) D (0 ; 0 ) (0 ; 0 ) (0 ; ° 1) (0 ; 0 ) NE is a strategy pro°le where both players are best responding to each other. The numbers underlined in the table are the best responses of the players for a given strategy of the other player. According to the de°nition of NE, a pure strategy NE corresponds to those strategy pro°les where both numbers are underlined ((A,L), (B,M), and (D,R)). To °nd mixed strategy NE, we have to start considering all possible supports (excluding non rationalizable ones). That is, (A, B, D) for player 1 and (L, M, R) for player 2. Next, note that, in fact, D and R won±t be played in a mixed strategy NE. The argument is as follows. Consider the decision of player 1. For any arbitrary large probability that player 2 plays R (bounded away from 1), player 1 strictly prefers to play any mixture of A and B instead of D. Therefore, in any ²strictly³mixed NE, player 1 does not play D. By the similar argment, in any ²strictly³mixed NE, player 2 does not play R. Let us look for mixed strategies with support A and B, and L and M. Indi/erence condition for player 2: 2 ° ( A ) + 0 ° ( B ) = 0 ° ( A ) + 4 ° ( B ) yields ° ( A ) = 2 = 3 and ° ( B ) = 1 = 3 . Doing the same for player 1 yields the mixed NE (2/3A+1/3B, 2/5L+3/5M). To conclude, the NE are (A,L), (B,M), (D,R), and (2/3A+1/3B, 2/5L+3/5M). 1
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2. Solve the following game using backward induction: Solution: In last nodes player 1 plays a and D respectively. Therefore player 2 plays ± .
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