PSet 2 Solutions v3.pdf

# PSet 2 Solutions v3.pdf - Problem Set 2 Solutions Econ...

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Problem Set 2 Solutions Econ 2020b / HBS 4011 / API-112 1. Pollution Game (a) There are four pure strategy Nash equilibria: Three in which two of the firms clean and one does not, and one in which none of the firms clean. (b) To confirm a MSNE, we need to ensure that each player is best responding to the other. First, fix player I as playing C and player II as playing C with probability p=2/3 . (The game is symmetric so which player we assign the strategies to is irrelevant). The question then is whether, for player III, playing C with probability p=2/3 is now a best response. This mixed strategy will only be a best response where player III is indifferent between playing C and N: If player III plays C, they get a payoff of 2 for sure. If player III plays N, they get a payoff of 2/3*(3) + 1/3*(0) = 2 Hence, player III is indifferent and is willing to play a mixed strategy best response (including p=2/3). The reasoning is exactly the same for player II’s best response, who is in the same position as player III. What about player I; are they best responding to the other two players’ mixed strategy of p=2/3? If player I plays C, there are four possibilities, depending on how the randomization of players II and III turns out. 4/9ths of the time (2/3 * 2/3), they both play cooperate and player I gets 2; another 4/9ths of the time one of them plays cooperate (2/3 * 1/3 + 2/3*1/3) and player I also gets 2. 1/9 th of the time they both play N and player I gets -1 (minus one). Hence, the expected payoff to player 1 is: (2/3*2/3)*2 + ((2/3*1/3) + (2/3 + 1/3))*2 + (1/3*1/3)*(-1) = 15/9 If player I plays N, they get zero unless both other players happen to choose C, which happens 4/9ths of the time, in which case they get 3: (2/3*2/3)*3 + ((2/3*1/3) + (2/3 + 1/3))*0 + (1/3*1/3)*(0) = 12/9 Since 15/9 > 12/9, player I’s best response is to play C. This confirms the Nash equilibrium originally proposed. (c) In a symmetric MSNE, each player will use the same mixing probability, p . Since the problem is the same for all players we only need to make one calculation, which is to derive the p at which a player is indifferent between playing C and N. Following the model of assessing the four probabilistic cases described in the previous question, the expected payoffs to, say, player I are: C: p 2 * 2 + p(1-p)*2 + p(1-p)*2 + (1-p) 2 *(-1) = 2p 2 + 4p(1-p) + (1-p) 2 N: p 2 * 3 + p(1-p)*0 + p(1-p)*0 + (1-p) 2 *(0) = 3p 2 At what p will these expected payoffs be indifferent? 2p 2 + 4p(1-p) + (1-p) 2 = 3p 2

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6p 2 6p + 1 = 0 Solving this quadratic expression provides two values of p that induce indifference for every player: p=0.211, p=0.789 These correspond to two symmetric mixed strategy Nash equilibria. (d) To verify all the possible NE, consider all the possible types of equilibria: Equilibrium Type Equilibria 1 Pure Strategies Part (a) 4 equilibria 2 Symmetric Mixed Strategies Part (c) 2 equilibria Asymmetric Strategies: 3 One player plays a pure strategy of C and the others mix Part (b) 3 equilibria ( Any p≠2/3 doesn’t induce indifference in the other player playing a mixed strategy, so these must be the only equilibria) 4 One player plays a pure
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