# When can tit-for-tat be sustained?

Consider the following infinitely repeated hawk-dove game:

$\hspace{1cm}\hspace{1cm}\hspace{1cm}$

For what discount rate $\delta$ can tit for tat sustain $D,D$? My professor says the answer is $\delta=1$ (discounts are applied by multiplication e.g. payoff of 1 in the next period yields $\delta$).

It seems to me like two cases need to be considered:

1) Suppose we start with $D,D$, then tit-for-tat sustains this $\iff$ $\frac{4}{1-\delta}>8+\frac{\delta}{1-\delta} \iff \delta>4/7$.

2) Suppose we start at $H,H$, under tit-for-tat, player 1 would restore $D,D$ if $\frac{1}{1-\delta}<0+\frac{4\delta}{1-\delta} \iff \delta>1/4$.

What am I doing wrong?

History $(D,D)$ (Yes, this is with some abuse of notation.):

On Path: Each player gets an average payoff of $4$.

Deviating: Say player $1$ deviates and instead chooses $H$. Then the resulting sequence of actions is the alternating sequence, $(H,D)$, $(D,H)$, $(H,D)$,...

Accordingly, player $1$’s sequence of payoffs is $(8,0)$, $(0,8)$, $8,0$... Player $1$'s average payoff is $(1-\delta)\big(8 + 0 + \delta^{2}8 + 0 + \delta^{4}8 + \cdots\big)$, which is $$(1-\delta)\frac{8}{1-\delta^{2}}=(1-\delta)\frac{8}{(1-\delta)(1+\delta)}=\frac{8}{1+\delta}$$ Hence, there is no profitable deviation here iff

$$\begin{split} 4 &\geq \frac{8}{1+\delta}\\ 4\delta &\geq 4\\ \delta &\geq 1 \end{split}$$

Thus, $\delta \geq 1$ is a necessary condition (not necessarily sufficient because we still need to check that following tit-for-tat is still optimal for the other histories. I'll leave that to you!)

• @denesp Good suggestion! Edited as suggested. – user11305 Mar 29 '18 at 17:50
• Why did you not discount the on path payoff $4: 4+\delta 4+ \delta^24=\frac{4}{1-\delta}$ – user526463 Mar 29 '18 at 18:35
• @user526463 It’s convention to multiply payoffs by $(1-\delta)$ in order to work with average payoffs. – user11305 Mar 29 '18 at 18:53