To determine the critical such that Apologize-and-Restitute is an equilibrium, we consider the possible positions that the Row player could be in and compare the present value of sticking with the SSP versus choosing the most attractive alternative. By standard arguments Row’s present value is calculated under the assumption that Column uses the SSP, that Row deviates at the first stage and returns to the SSP immediately after the deviation.
Suppose players are on the initial path. The sequence of play will be TT, TT, TT, . . . and Row’s payoff stream will be 6, 6, 6, . . . for a present value of 6/(1 - ). If Row deviates to W on the first move the play will be WT, SW, TW, TT, TT, . . . and Row’s stream will be 12, -1, -6, 6, 6 . . . for a present value of 12 - 1 - 6 2 + 6 3/(1 - ).As long as > .473, the former value will be greater and Row will not be tempted to deviate. There are four other positions that Row may face during the game – at the first or second move or Row’s own punishment path, or of Column’s path, and each of these yields a condition on for compliance with the equilibrium. All of these conditions must hold for the strategies to be a subgame perfect equilibrium, and the strictest is found to be > .564. The robustness of Apologize-and-Restitute
There is a finite number of pure-strategy symmetrical equilibria that produce cooperation and restore it after a violation in two stages. Row has three possible moves for each of the first two stages of Row’s punishment path, and so does Column, yielding 81 possibilities. By considering cases it can be shown that for these particular payoffs, Apologize-and-Restitute has a lower threshold for maintaining cooperation than any other strategy in the set.
Equilibria of the partial transparency game Player 1’s payoff is increasing in b, and b has no other effect within the model. Therefore if in equilibrium Player 1 promises for a given b, he will promise for all higher values of b as well. It follows that generically, 1’s strategy can be represented by a function of c, f(c), which states the cutoff for b above which 1 will promise. Also if the probability of promising is positive, then f(0) = 0 since a player with c = 0 can only gain by promising. We will first consider the case of a function such that 0 < f(c) < 0 for 0 < c < 1.