Two Players Different Strategies in infinitely repeated game

Question

I am vexed by a Game Theory question regarding infinitely repeated games. Usually, strategies such as Grim Trigger or Tit-for-Tat are discussed in a context where both players have the same strategy.

However, I am wondering how to proceed when two players play different strategies, say Player 1 plays Grim Trigger and Player 2 plays Tit-for-tat.

My question is whether there exists any $\delta\in(0,1)$ for which a profile of {Grim Trigger, Tit-for-tat} constitutes an equilibrium.

The stage game payoffs are as follows:

\begin{array}{|c|c|c|} \hline & C & D \\\hline C & 3,3 & 0,4 \\\hline D & 4,0 & 2,2 \\\hline \end{array}

Answer 1

Just so that we have a common understanding of the strategies you mentioned:

Grim trigger: "Play $C$ unless the outcome of any previous stage is not $(C,C)$; in that case, play $D$."

Tit-for-tat: "Start by playing $C$, and then play the action used by the other player in the previous stage."

Therefore, on equilibrium path, the outcome $(C,C)$ will be observed, and each player gets a discounted payoff of \begin{equation} 3+3\delta+3\delta^2+\cdots=\frac{3}{1-\delta}.\tag{1} \end{equation}

Suppose player 1 who adopts grim trigger deviates in the first stage only. The outcome in stage 1 would be $(D,C)$, and the outcome in all subsequent stages would be $(D,D)$ according to the strategy profile. Thus, player 1's payoff from the deviation would be \begin{equation} 4+2\delta+2\delta^2+\cdots=4+\frac{2\delta}{1-\delta}.\tag{2} \end{equation} $(1)$ and $(2)$ imply that a one-stage deviation is not profitable for player 1 if $\delta\ge\frac12$.

Suppose player 2, who adopts tit-for-tat, deviates in the first stage only. The outcome in stage 1 would be $(C,D)$, and $(D,C)$ in stage 2, $(D,D)$ in all subsequent stages. Player's payoff from the deviation would be \begin{equation} 4+0\delta+2\delta^2+2\delta^3+\cdots=4+\frac{2\delta^2}{1-\delta}.\tag{3} \end{equation} $(1)$ and $(3)$ imply that a one-stage deviation is not profitable for player 2 if $\delta\ge 1-\frac{\sqrt 2}{2}$.

Note that $\frac12>1-\frac{\sqrt2}2$. Hence, as long as $\delta\ge\frac12$, player 1 using grim trigger and player 2 using tit-for-tat can be sustained as a Nash equilibrium of the infinitely repeated prisoner's dilemma. However, that the strategy profile is not a subgame perfect equilibrium (for a similar reason why a symmetric tit-for-tat strategy profile is not subgame perfect).

Answer 2

There is not a ton to go off of with your question, so apologies for the vagueness of my answer:

First, since strategies aren’t symmetric, you will find a $\delta_1$ and $\delta_2$ for respective players.

For each player:

1. Check the payoffs from a one-shot optimal deviation (only deviate this period) and accepting the punishment. Note that the other players strategy tells you what will happen in the future.

For the grim strategy player, the tit for tat player will play punishment next period, then return to original move, then to punishment, etc. (two infinite sums)

For the tit for tat player, the grim strategy will play punishment forever.

This could be confusing but your question doesn’t offer enough to answer.

2. Check that the punishment is incentive compatible. That is, does the punishment constitute a Nash equilibrium. The reason for this is the punishment has to be a “credible threat” or else neither will follow it. (Perhaps think of Mutually Assured Destruction)