PracticeProblems-1_solutions

PracticeProblems-1_solutions - Economics 101 practice...

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Unformatted text preview: Economics 101: practice problems 1 solutions 1. (Nicholson-Snyder, 8.1) Consider the following game: D E F A 7 , 6 5 , 8 , B 5 , 8 7 , 6 1 , 1 C , 1 , 1 4 , 4 a. Find the pure-strategy Nash equilibria (if any). Solution: we compute pure-strategy Nash equilibria in the usual way, underlining conditional best responses. D E F A 7 , 6 5 , 8 , B 5 , 8 7 , 6 1 , 1 C , 1 , 1 4 , 4 The only sell with co-occurring underlines is ( C,F ); this is the unique pure-strategy Nash equi- librium. b. Find the mixed-strategy Nash equilibrium in which each player randomizes over just the first two actions. Solution: let σ j i be the probability that player i plays strategy j . We use indifference conditions to find E [ u 1 ( A,σ 2 )] = E [ u 1 ( B,σ 2 )] σ D 2 u 1 ( A,D ) + σ E 2 u 1 ( A,E ) = σ D 2 u 1 ( B,D ) + σ E 2 u 1 ( B,E ) 7 σ D 2 + 5 σ E 2 = 5 σ D 2 + 7 σ E 2 2 σ D 2 = 2 σ E 2 1- σ E 2 = σ E 2 = ⇒ σ E 2 = 1 2 E [ u 2 ( D,σ 1 )] = E [ u 2 ( E,σ 1 )] σ A 1 u 2 ( D,A ) + σ B 1 u 2 ( D,B ) = σ A 1 u 2 ( E,A ) + σ B 1 u 2 ( E,B ) 6 σ A 1 + 8 σ B 1 = 8 σ A 1 + 6 σ B 1 2 σ B 1 = 2 σ A 1 1- σ A 1 = σ A 1 = ⇒ σ A 1 = 1 2 Then the unique mixed-strategy Nash equilibrium is [( 1 2 , 1 2 ) , ( 1 2 , 1 2 )]. c. Compute players’ expected payoffs in the equilibria found in parts (a) and (b). Solution: the expected payoffs from the pure-strategy Nash equilibrium ( C,F ) are (4 , 4), as presented in the payoff matrix. We compute the expected payoffs from the mixed-strategy Nash January 14, 2011 1 Economics 101: practice problems 1 solutions equilibrium as E [ u 1 ( σ * 1 ,σ * 2 )] = σ A 1 σ D 2 u 1 ( A,D ) + σ A 1 σ E 2 u 1 ( A,E ) + σ B 1 σ D 2 u 1 ( B,D ) + σ B 1 σ E 2 u 1 ( B,E ) = 1 4 (7) + 1 4 (5) + 1 4 (5) + 1 4 (7) = 6 E [ u 2 ( σ * 2 ,σ * 1 )] = σ D 2 σ A 1 u 2 ( D,A ) + σ D 2 σ B 1 u 2 ( D,B ) + σ E 2 σ A 1 u 2 ( E,A ) + σ E 2 σ B 1 u 2 ( E,B ) = 1 4 (6) + 1 4 (8) + 1 4 (8) + 1 4 (6) = 7 Then player 1’s expected payoff is 6 and player 2’s expected payoff is 7. Notably, they are both better off in this mixed-strategy Nash equilibrium than in the pure-strategy Nash equilibrium. 2. (Nicholson-Snyder, 8.2) The mixed-strategy Nash equilibrium in the Battle of the Sexes may depend on the numerical values for the payoffs. To generalize this solution, assume that the payoff matrix for the game is given by B allet B oxing B allet K, 1 , B oxing , 1 ,K where K ≥ 1. Show how the mixed-strategy Nash equilibrium depends on the value of K . Solution: we apply indifference conditions to solve for mixed-strategy Nash equilibrium. To simplify notation, let L represent Ba L let, and X represent Bo X ing. E [ u 1 ( L,σ 2 )] = E [ u 1 ( X,σ 2 )] σ L 2 u 1 ( L,L ) + σ X 2 u 1 ( L,X ) = σ L 2 u 1 ( X,L ) + σ X 2 u 1 ( X,X ) σ L 2 ( K ) = σ X 2 Kσ L 2 = 1- σ L 2 = ⇒ σ L 2 = 1 K + 1 Appealing to the nearly symmetric structure of player 2’s payoffs, we see that player 1’s mixture would have σ X 1 = 1 K +1 . Then the mixed-strategy Nash equilibrium of this game, as a function of....
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PracticeProblems-1_solutions - Economics 101 practice...

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