Consider a 2-player symmetric game given by a payoff matrix $A\in [0,1]^{n,n}$ for the row player (i.e. the column player matrix is $A^t$).

Define the social welfare as the sum of payoffs for both players, i.e. $$SW(i,j)=A(i,j)+A(j,i)$$

Define the social-welfare of a (possibly mixed) equilibrium in a straight forward manner: $$SW(s_1,s_2) = \sum _{i\in [n]}\sum_{j\in [n]}SW(i,j)\Pr_{s_1}(i)\Pr_{s_2}(j)$$

Define $Sup(s)=Sup(s_1)\cup Sup(s_2)$, where $Sup(s_i)$ is the set of strategies in $s_i$ that has a strictly positive probability.

• Let $s=<s_1,s_2>$ be an equilibrium for the game, and let $<s',s'>$ be a symmetric equilibrium. Is it true that $Sup(s)=Sup(s')$ imply $SW(s')\leq SW(s)$?

• If the statement is true, does it extend to symmetric games with any number of players?

The intuition is that one may always resort to playing the symmetric equilibrium, hence if the other player would make more in asymmetric equilibrium it has to benefit them both.

For example, consider the following simple game, for some $x>0$:

$A= \left( \begin{array}{ccc} 0 & 1 \\ x & 0 \\ \end{array} \right) $

And the column player profit, given by $A^t$ is:

$A^t= \left( \begin{array}{ccc} 0 & x \\ 1 & 0 \\ \end{array} \right) $

There exists a asymmetric equilibrium $s$ where some player plays strategy $a$ and the other plays $b$. This gives a social welfare of $1+x$.

The symmetric equilibrium $s'$ is reached when both play strategy $a$ with probability $p_a=\frac{1}{1+x}$ and $b$ otherwise, giving a social welfare of: $$SW(s')=1\cdot(2p_ap_b)+x(2p_ap_b)=2(1+x)p_ap_b=\frac{2x}{1+x}=1+x-\frac{1+x^2}{1+x}<SW(s)$$