ps1(2010)Sol

ps1(2010)Sol - MS&E 246 Game Theory with Engineering...

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Unformatted text preview: MS&E 246: Game Theory with Engineering Applications Problem Set 1 1. (a) (3pts) When s- i = 1, any strategy s i yields a zero payoff. So, there are no strictly dominated strategies. (b) (3pts) s i = 0 is the only weakly dominated strategy. It is weakly dominated by any strategy s i ∈ (0 , 1]. (c) (2+2pts) Any ( s 1 ,s 2 ) with s 1 + s 2 = 1 is a pure strategy NE. The payoff of player i is s i . Given the strategy s- i , player i has no incentive to deviate from the strategy s i = 1- s- i . If he chooses s i < s i his payoff will be only s i . If he chooses s i > s i , his payoff will be zero. (1,1) is also a Nash equilibrium. 2. (a) (6pts) Suppose player i has two weakly dominant strategies, s i and s 00 i . Then u i ( s i ,s- i ) ≥ u i ( s i ,s- i ) for all s- i and for all s i . In particular, u i ( s i ,s- i ) ≥ u i ( s 00 i ,s- i ) for all s- i . Since s 00 i is weakly dominant, there is some s- i such that u i ( s 00 i ,s- i ) > u i ( s i ,s- i ), which is a contradiction. Therefore, a player can not have two weakly dominant strategies. (b) (4pts) According to (a), there can be no such game. 3. (a) (3pts) Playing LL, L or R is quite risky, since we do not know what player 1 will be playing. The risk of obtaining a playoff of-49 is very large compared to the payoff of 1 if player 2 player L, and the risk of obtaining a payoff of -100 is very large compared to the payoff of 2 if player 2 played LL or R. Therefore, it seems ”reasonable” to play M. (b) (4+6pts) The two pure NE of this game are (U,LL) and (D,R). To check for mixed strategy NE, player 1 must mix between U (with probability p) and D (with probability 1-p). Player 2 has 11 possible mixing combinations: { LL,L } , { LL,M } , { LL,R } , { L,M } , { L,R } , { M,R } , { M,L,R } , { LL,M,R } , { LL,L,R } , { LL,L,M } and { LL,L,M,R } . We will show that only the { LL,L } is part of a mixed NE. For player 2 to mix between LL and L (with probabilities q and 1-q respectively), we must have p · (2) + (1- p ) · (- 100) = p · (1) + (1- p ) · (- 49) which yields p = 51 52 . The utility of player 2 from each strategy is then: u 2 ( LL ) = u 2 ( L ) = 1 26 , u 2 ( M ) = 0 and u 2 ( R ) < 0. Then for player 1 to mix between U and D, we 1 must have: q · (100)+(1- q ) · (- 100) = q · (- 100)+(1- q ) · (100) which yields q = 1 2 and u 1 ( U ) = u 1 ( D ) = 0. Therefore, p = 50 51 and q = 1 2 is a mixed NE. We now show that no other mixing combination of player 2 can be part of a mixed NE.combination of player 2 can be part of a mixed NE....
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This note was uploaded on 03/04/2011 for the course MS&E 246 taught by Professor Johari during the Winter '07 term at Stanford.

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ps1(2010)Sol - MS&E 246 Game Theory with Engineering...

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