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Herr K.
Herr K.
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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 $(2)$$(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).

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 $(2)$ 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).

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).

Source Link
Herr K.
Herr K.
  • 16.1k
  • 5
  • 29
  • 53

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 $(2)$ 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).