Two Players Different Strategies in infinitely repeated game 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
&amp; C &amp; D \\\hline 
C &amp; 3,3 &amp; 0,4 \\\hline
D &amp; 4,0 &amp; 2,2 \\\hline
\end{array}
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 $\frac12&gt;1-\frac{\sqrt2}2$. 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).