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

<p>First, since strategies aren’t symmetric, you will find a <span class="math-container">$\delta_1$</span> and <span class="math-container">$\delta_2$</span> for respective players. </p>

<p>For each player:</p>

<ol>
<li>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. </li>
</ol>

<p>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) </p>

<p>For the tit for tat player, the grim strategy will play punishment forever.</p>

<p>This could be confusing but your question doesn’t offer enough to answer.</p>

<ol start="2">
<li>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)</li>
</ol>

CC BY-SA 4.0
2019-07-15T16:04:00.203Z
2019-07-15T16:04:00.203Z
Kyle Butts
<p>Just so that we have a common understanding of the strategies you mentioned: </p>

<blockquote>
  <p><strong>Grim trigger</strong>: "Play <span class="math-container">$C$</span> unless the outcome of any previous stage is not <span class="math-container">$(C,C)$</span>; in that case, play <span class="math-container">$D$</span>." </p>
  
  <p><strong>Tit-for-tat</strong>: "Start by playing <span class="math-container">$C$</span>, and then play the action used by the other player in the previous stage." </p>
</blockquote>

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

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

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

<p>Note that <span class="math-container">$\frac12&gt;1-\frac{\sqrt2}2$</span>. Hence, as long as <span class="math-container">$\delta\ge\frac12$</span>, player 1 using grim trigger and player 2 using tit-for-tat can be sustained as a <em>Nash equilibrium</em> 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). </p>

CC BY-SA 4.0
2019-07-17T05:29:14.943Z
Thank you very much! This has made the problem much clearer to me. As I understand it, the reason why the above strategy profile is not subgame-perfect is because for a profile to be subgame-perfect it has to to induce a Nash-Equilibrium in every subgame, but here only for values $\delta > 1/2$. Is my reasoning correct here?
JAspen
CC BY-SA 4.0
2019-07-17T17:11:30.713Z
@JAspen: The profile is not subgame perfect because it violates the [one-stage deviation principle](https://en.wikipedia.org/wiki/One-shot_deviation_principle). For instance, player 1 can play D, then C forever, and this will yield a payoff higher than deviating only in stage 1.
Herr K.
CC BY-SA 4.0
2019-07-18T20:20:12.103Z
Could you elaborate on that point? You pointed out that by deviating in stage 1 player 1 would get a payoff of $4 + 2*\delta/(1-\delta)$. How would he get a higher payoff by playing D first, then C forever after?
JAspen
CC BY-SA 4.0
2019-07-18T21:25:31.627Z
@JAspen: Holding player 2's tit-for-tat strategy unchanged, player 1 playing $DCCC\dots$ would induce the outcome sequence $(D,C),(C,D),(C,C),(C,C),\dots$, and hence the discounted payoff $4+0\delta+3\delta^3+\cdots=4+\frac{3\delta^3}{1-\delta}$, which is greater than $4+\frac{2\delta}{1-\delta}$ if $\delta$ is sufficiently close to $1$.
Herr K.
CC BY-SA 4.0
2019-07-15T22:32:18.667Z
2019-07-18T21:26:34.997Z
2019-07-18T21:26:34.997Z
Herr K.
<p>I am vexed by a Game Theory question regarding infinitely repeated games. Usually, strategies such as <em>Grim Trigger</em> or <em>Tit-for-Tat</em> are discussed in a context where both players have the same strategy. </p>

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

<p>My question is whether there exists any <span class="math-container">$\delta\in(0,1)$</span> for which a profile of {<em>Grim Trigger</em>, <em>Tit-for-tat</em>} constitutes an equilibrium.</p>

<p>The stage game payoffs are as follows: </p>

<p><span class="math-container">\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}</span></p>

<game-theory>
Two Players Different Strategies in infinitely repeated game
JAspen
# Two Players Different Strategies in infinitely repeated game

(Asked by: username_0 on 2019-07-15)

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}

**Tags:** game-theory

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## 2 answers

### Answer by username_2 on 2019-07-15. Score: 4 | Accepted answer

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).

> **Comments:**
> **username_0**: Thank you very much! This has made the problem much clearer to me. As I understand it, the reason why the above strategy profile is not subgame-perfect is because for a profile to be subgame-perfect it has to to induce a Nash-Equilibrium in every subgame, but here only for values $\delta > 1/2$. Is my reasoning correct here?  
> **username_2**: @JAspen: The profile is not subgame perfect because it violates the [one-stage deviation principle](https://en.wikipedia.org/wiki/One-shot_deviation_principle). For instance, player 1 can play D, then C forever, and this will yield a payoff higher than deviating only in stage 1.  
> **username_0**: Could you elaborate on that point? You pointed out that by deviating in stage 1 player 1 would get a payoff of $4 + 2*\delta/(1-\delta)$. How would he get a higher payoff by playing D first, then C forever after?  
> **username_2**: @JAspen: Holding player 2's tit-for-tat strategy unchanged, player 1 playing $DCCC\dots$ would induce the outcome sequence $(D,C),(C,D),(C,C),(C,C),\dots$, and hence the discounted payoff $4+0\delta+3\delta^3+\cdots=4+\frac{3\delta^3}{1-\delta}$, which is greater than $4+\frac{2\delta}{1-\delta}$ if $\delta$ is sufficiently close to $1$.  

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### Answer by username_1 on 2019-07-15. Score: 0

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)


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