# Why is the symmetric grim trigger not a Nash?

Consider the stage game: Let $\delta\in(0,1)$ be the discount factor. Let $G$ be the symmetric grim trigger strategy profile. The payoffs are then

$$E_{A}(G) = E_{B}(G) = \sum_{i=0}^{\infty}3\delta^{i} = \frac{3}{1-\delta}.$$

If a player were to defect from $G$ at time $n$, they would be playing $N$ at time $n$ and every round after they would play $Y$ because that would maximize their stage game payoff given the other player is going to play $N$ forever. Let $G_{n}^{A}$ and $G_{n}^{B}$ be the strategy profiles were $A$ and $B$, respectively, defect from $G$ at time $n$ in this way. By symmetry, we have \begin{align*} E_{A}(G_{n}^{A}) = E_{B}(G_{n}^{B}) = \sum_{i=0}^{n-1}3\delta^{i} + 7\delta^{n} + \sum_{i=n+1}^{\infty}\delta^{i} &= 3\frac{1-\delta^{n}}{1-\delta} + 7\delta^{n} + \frac{1}{1-\delta} - \frac{1-\delta^{n+1}}{1-\delta} \\ &= \frac{3+\delta^{n}(4-6\delta)}{1 - \delta}. \end{align*} If $\delta \geq \frac{2}{3}$, then $E_{A}(G_{n}^{A}) \leq E_{A}(G)$ (and $E_{B}(G_{n}^{B}) \leq E_{B}(G)$) for all $n$ so it's not clear to me why the symmetric grim trigger strategy profile is not a Nash in this case.

My professor claims that $G$ is not a Nash for any $\delta$. I do not believe it.

• Nothing restricts the defecting player to play N forever. They are defecting, they are free to play whatever they like. They should play $Y$ starting at $n+1$ because getting 1 is better than getting $0$, even if it means the other player then gets $7$. This is an infinitely repeated game, I am reading it from the homework now, but notice that it is not a PD because $NN$ is not a Nash, the Nashes are $NY$ and $YN$. – user16268 Feb 14 '18 at 3:29
• Yes, you're right. I wasn't reading the game too carefully. Sorry. The word "defecting" somehow got me into thinking that this was a PD. – Herr K. Feb 14 '18 at 4:26
• Does your professor have the same understanding of $G$ as you? In a typical prisoner's dilemma, $G$ would be "play $N$ forever after the first $N$ is observed", whereas you are (reasonably) taking $G$ as meaning "if $i$ (alone) is the first to play $N$ then we will play the Nash of $i=Y,j=N$ thereafter". – Ubiquitous Feb 14 '18 at 8:24
• Also, a trivial point that you might know already: one can very quickly compute the net gain from deviation under grim trigger as $G-L\frac{\delta}{1-\delta}$, where $G$ is the one-shot gain from deviation and $L$ is the loss in future periods. In this example, $G=7-3=4$ and $L=3-1=2$. It at least saves a little algebra – Ubiquitous Feb 14 '18 at 8:28
• I think I am understanding the grim trigger, if the other guy ever plays $N$, then you play $N$forever. However in this case, the player defecting to $N$ has no reason keep playing N after the first time, they would rather play $Y$. Since they are the defecting player, they can play whatever they like. They can play $Y$ or $N$. I choose them to play $Y$ to show that even the best defection can still be worse than grim trigger if $\delta$ is big enough. – user16268 Feb 14 '18 at 12:08