# Is there always at most one full-support extreme symmetric equilibrium?

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 $s_2$.

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 $s_2$.
• No doubt that $s$ is an equilibrium itself. What's not obvious to me is why is $s$ necessarily symmetric? Dec 8, 2014 at 22:48
• 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. Dec 8, 2014 at 23:17