Question

1. Consider the following normal form game 112 L CR T|10 1012 1210 13 M 12 25 5 0 (0 B113 0100 (a) (Level A) First suppose this game is played only once. What are the pure strategy Nash equilibria? (b) (Level B) Now suppose this game is played twice. Players observe the actions chosen in the first period prior to the second period. Each player's total payoff is the sum of his/her payoff in the two periods. Consider the following strategy profile: Player 1: plays ·T in period 1 . M in period 2 if the action profile played in period 1 is (T, L); and B otherwise Player 2: plays . L in period 1 . C in period 2 if the action profile played in period 1 is (T, L); and R otherwise Argue that the above strategy profile constitutes a subgame perfect equilib- rium of the two-period game (c) (Level C) Recall from Lecture 8 that, in a finitely repeated Prisoner's Dilemma, the only subgame perfect equilibrium involves playing the stage game Nash equilibrium at every history. In the above subgame perfect however, the play Date: 24th Jan 2017 in period 1 is not a Nash equilibrium of the stage game. What is the difference between the above game and Prisoner's Dilemma?

Answer 1

Using the credible threat approach to find SPNE in the finitely repeated games ,where multiple pure strategy NE exist in the one stage of game

Pure St NE of Stage Game rightarrow < M, C > & < B, R > Ans Twice Repeated Game rightarrow Now If both cooperate, each gets 10, Equation is < T, L > So using Pareto Inferior NE = < B, R > , as a credible threat (credible since < B, R > is NE of Stage Game > . In case of deviation from playing < M, C > in the 2nd period, thus we can sustain the Mentioned strategy as SPNE, (*) BWZ total payoff from cooperation (play T in first pd, & Min 2nd) by P_1 = 10 + 5 = 15 If P_1 deviates to play B in 1st pd, then play NE < B, R > in 2nd period, total payoff = 13 + 1 = 14 Thus higher payoff from playing St. Hence proved. (Similarly for P_2 rightarrow by symmetry) difference b/w prisoner's dillemma & given Game is that only one NE exist in pD, PD so no other NE exists that can be used as a credible threat. Thus can't sustain any other o\c (which is not NE) as SPNE in repeated