2
$\begingroup$

Given a ($n$-player) symmetric game and two equilibriums $s_1,s_2$, is it true that if the support (the set of strategies with positive probabilities) of $s_1$ is identical to the support of $s_2$ then $\frac{s_1+s_2}{2}$ is also a symmetric equilibrium?


By $\frac{s_1+s_2}{2}$ I mean that a strategy $i$ will be selected by a player with a probability which equals the average of the probability of $i$ under $s_1$ and the probability of $i$ under $s_2$.

$\endgroup$
2
$\begingroup$

This seems obvious. If $\sigma_1,\sigma_2$ are in the joint support of $s_1,s_2$ and $\tau$ is not, then (writing $P$ for the first player's payoff) you need $$P(\tau,s)\le P(\sigma_1,s)=P(\sigma_2,s)$$

where $s$ is a convex combination of the $s_i$. But if you replace $s$ with $s_1$ or $s_2$, this holds, so it still holds after you average over $s_1$ and $s_2$.

What am I missing?

$\endgroup$
  • $\begingroup$ No doubt that $s$ is an equilibrium itself. What's not obvious to me is why is $s$ necessarily symmetric? $\endgroup$ – Herr K. Dec 8 '14 at 22:48
  • $\begingroup$ Kevin C: The question asks if $s$ is "also" symmetric, which led me to assume that the OP intended to assume the $s_i$ are symmetric. Without that assumption, of course, the answer is certainly no. $\endgroup$ – Steven Landsburg Dec 8 '14 at 23:17

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.