3
$\begingroup$

Consider the following infinitely repeated hawk-dove game:

$\hspace{1cm}\hspace{1cm}\hspace{1cm}$enter image description here

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?

$\endgroup$
4
$\begingroup$

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!)

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.