<p>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
$$P(\tau,s)\le P(\sigma_1,s)=P(\sigma_2,s)$$ </p>

<p>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 $s_2$.</p>

<p>What am I missing?</p>

CC BY-SA 3.0
2014-12-08T22:48:30.213Z
No doubt that $s$ is an equilibrium itself. What's not obvious to me is why is $s$ necessarily **symmetric**?
Herr K.
CC BY-SA 3.0
2014-12-08T23:17:23.513Z
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.
Steven Landsburg
CC BY-SA 3.0
2014-12-07T14:27:49.690Z
2014-12-07T14:27:49.690Z
Steven Landsburg
<blockquote>
  <p><strong>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?</strong></p>
</blockquote>

<hr>

<p>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 $s_2$.</p>

<mathematical-economics><game-theory><general-equilibrium>
Is there always at most one full-support extreme symmetric equilibrium?
R B
# Is there always at most one full-support extreme symmetric equilibrium?

(Asked by: username_0 on 2014-12-07)

> **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?**

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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 $s\_2$.

**Tags:** mathematical-economics, game-theory, general-equilibrium

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## 1 answers

### Answer by username_1 on 2014-12-07. Score: 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 $s\_2$.

What am I missing?

> **Comments:**
> **username_2**: No doubt that $s$ is an equilibrium itself. What's not obvious to me is why is $s$ necessarily **symmetric**?  
> **username_1**: 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.  

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