# problem-set-2-answers - 1 1 0 = 2’s minmax strategy...

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Economics 703 Advanced Microeconomics Prof. Peter Cramton Problem Set 2: Suggested Answers 1. By backward recursion, the unique subgame-perfect equilibrium is {R, rl}, indicated by in the extensive form. l r L R 1 2 2 4 3 2 l r 1: 3 1 2: 1 4 2 Using the normal form below, it is clear that {R, rl} is also the unique pure-strategy Nash equilibrium: best responses are underlined in the bimatrix below. Only {R, rl} has a pair of mutual best responses. ll lr rl rr L 3, 1 3, 1 1, 4 1, 4 R 2, 3 4, 2 2, 3 4, 2 2. 2 t 1 t 2 t 3 t 4 s 1 2, 2 x, 0 -1, 0 0, 0 s 2 0, x 4, 4 -1, 0 0, 0 s 3 0, 0 0, 0 0, 2 0, 0 1 s 4 0, -1 0,-1 -1,-1 2, 0 Note that (s i ,t i ) for i = 1, 3, and 4 are all Nash equilibria in the stage game. Since the payoffs are symmetric the same calculation works for both players: 4 + 2 x + 0, so x 6.

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2 2. 2 t 1 t 2 t 3 s 1 1, 1 5, 0 0, 0 1 s 2 0, 5 4, 4 0, 1 1 1 s 3 0, 0 1, 0 -1,-1 0=1 ’s minmax strategy
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Unformatted text preview: 1 1 0 = 2’s minmax strategy Supporting (4,4) requires δ and N that satisfy: (i) 4 > 5(1 - δ ) + δ p * , and (ii) p * = (1 - δ N )(-1) + δ N (4) > 0. At δ =.9 and N=1, we have p * = 4(.9) - .1 = 3.5 and .1(5) + .9(3.5) = 3.65 < 4, as required. One period of punishment is needed. As before to support (3/4, 3/4) requires δ and N that satisfy: (i) 3/4 > .5 + .9p * , and (ii) p * = .9 N (3/4) - 1 + .9 N > 0. At N = 1, the punishment payoff p * = .575 > 0, but 3/4 < .5+.9(.575)= 1.02, so (i) is violated. At N = 2, p * = .4175 > 0, but 3/4 < .5+.9(.4175)=.876 so (i) is violated again. However, if we set N = 3, then p * = .27575 > 0 and 3/4 > .5+.9(.27575)=.74818. Hence, a three-period punishment phase is needed to support (3/4, 3/4)....
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## This note was uploaded on 10/21/2009 for the course ECON ECON703 taught by Professor Professorpetercramton during the Fall '09 term at Maryland.

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problem-set-2-answers - 1 1 0 = 2’s minmax strategy...

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