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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 $A^t$).

Let $s=<s_1,s_2>$ be a (possibly mixed-strategies, not-necessarily symmetric) equilibrium for the game.

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.

Is there always a symmetric equilibrium $<s',s'>$ such that $Sup(s')\subseteq Sup(s)$?

(note that there always exist a symmetric equilibrium for the game, but I'm interested in knowing whether any equilibrium imply a symmetric equilibrium with subset support).

For example (yes, kinda pointless if $n=2$, but yet), consider the following simple game:

$A= \left( \begin{array}{ccc} 1/3 & 2/3 \\ 1/3 & 1/6 \\ \end{array} \right) $

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

$A^t= \left( \begin{array}{ccc} 1/3 & 1/3 \\ 2/3 & 1/6 \\ \end{array} \right) $

There exists exactly two equilibriums here, both of which are in pure-strategies, in the first both players play strategy 1, and in the second some player plays 1 and the other plays 0.

The support of the symmetric equilibrium is $\{1\}$ which is a subset of the support of the asymmetric equilibrium which as $\{1,2\}$ as it's support.

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 $A^t$).

Let $s=<s_1,s_2>$ be a (possibly mixed-strategies, not-necessarily symmetric) equilibrium for the game.

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.

Is there always a symmetric equilibrium $<s',s'>$ such that $Sup(s')\subseteq Sup(s)$?

(note that there always exist a symmetric equilibrium for the game, but I'm interested in knowing whether any equilibrium imply a symmetric equilibrium with subset support).

For example (yes, kinda pointless if $n=2$, but yet), consider the following simple game:

$A= \left( \begin{array}{ccc} 1/3 & 2/3 \\ 1/3 & 1/6 \\ \end{array} \right) $

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

$A^t= \left( \begin{array}{ccc} 1/3 & 1/3 \\ 2/3 & 1/6 \\ \end{array} \right) $

There exists exactly two equilibriums here (up to symmetry), both of which are in pure-strategies, in the first both players play strategy 1, and in the second some player plays 1 and the other plays 0.

The support of the symmetric equilibrium is $\{1\}$ which is a subset of the support of the asymmetric equilibrium which as $\{1,2\}$ as it's support.

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 $A^t$).

Let $s=<s_1,s_2>$ be a (possibly mixed-strategies, not-necessarily symmetric) equilibrium for the game.

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.

Is there always a symmetric equilibrium $<s',s'>$ such that $Sup(s')\subseteq Sup(s)$?

If this is true, does it also apply for infinite symmetric games?

(note that there always exist a symmetric equilibrium for the game, but I'm interested in knowing whether any equilibrium imply a symmetric equilibrium with subset support).

For example (yes, kinda pointless if $n=2$, but yet), consider the following simple game:

$A= \left( \begin{array}{ccc} 1/3 & 2/3 \\ 1/3 & 1/6 \\ \end{array} \right) $

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

$A^t= \left( \begin{array}{ccc} 1/3 & 1/3 \\ 2/3 & 1/6 \\ \end{array} \right) $

There exists exactly two equilibriums here, both of which are in pure-strategies, in the first both players play strategy 1, and in the second some player plays 1 and the other plays 0.

The support of the symmetric equilibrium is $\{1\}$ which is a subset of the support of the asymmetric equilibrium which as $\{1,2\}$ as it's support.

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 $A^t$).

Let $s=<s_1,s_2>$ be a (possibly mixed-strategies, not-necessarily symmetric) equilibrium for the game.

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.

Is there always a symmetric equilibrium $<s',s'>$ such that $Sup(s')\subseteq Sup(s)$?

(note that there always exist a symmetric equilibrium for the game, but I'm interested in knowing whether any equilibrium imply a symmetric equilibrium with subset support).

For example (yes, kinda pointless if $n=2$, but yet), consider the following simple game:

$A= \left( \begin{array}{ccc} 1/3 & 2/3 \\ 1/3 & 1/6 \\ \end{array} \right) $

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

$A^t= \left( \begin{array}{ccc} 1/3 & 1/3 \\ 2/3 & 1/6 \\ \end{array} \right) $

There exists exactly two equilibriums here, both of which are in pure-strategies, in the first both players play strategy 1, and in the second some player plays 1 and the other plays 0.

The support of the symmetric equilibrium is $\{1\}$ which is a subset of the support of the asymmetric equilibrium which as $\{1,2\}$ as it's support.

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 $A^t$).

Let $s=<s_1,s_2>$ be a (possibly mixed-strategies, not-necessarily symmetric) equilibrium for the game.

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.

Is there always a symmetric equilibrium $<s',s'>$ such that $Sup(s')\subseteq Sup(s)$?

If this is true, does it also applies for infinite symmetric games?

(note that there always exist a symmetric equilibrium for the game, but I'm interested in knowing whether any equilibrium imply a symmetric equilibrium with subset support).

For example (yes, kinda pointless if $n=2$, but yet), consider the following simple game:

$A= \left( \begin{array}{ccc} 1/3 & 2/3 \\ 1/3 & 1/6 \\ \end{array} \right) $

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

$A^t= \left( \begin{array}{ccc} 1/3 & 1/3 \\ 2/3 & 1/6 \\ \end{array} \right) $

There exists exactly two equilibriums here, both of which are in pure-strategies, in the first both players play strategy 1, and in the second some player plays 1 and the other plays 0.

The support of the symmetric equilibrium is $\{1\}$ which is a subset of the support of the asymmetric equilibrium which as $\{1,2\}$ as it's support.

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 $A^t$).

Let $s=<s_1,s_2>$ be a (possibly mixed-strategies, not-necessarily symmetric) equilibrium for the game.

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.

Is there always a symmetric equilibrium $<s',s'>$ such that $Sup(s')\subseteq Sup(s)$?

If this is true, does it also apply for infinite symmetric games?

(note that there always exist a symmetric equilibrium for the game, but I'm interested in knowing whether any equilibrium imply a symmetric equilibrium with subset support).

For example (yes, kinda pointless if $n=2$, but yet), consider the following simple game:

$A= \left( \begin{array}{ccc} 1/3 & 2/3 \\ 1/3 & 1/6 \\ \end{array} \right) $

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

$A^t= \left( \begin{array}{ccc} 1/3 & 1/3 \\ 2/3 & 1/6 \\ \end{array} \right) $

There exists exactly two equilibriums here, both of which are in pure-strategies, in the first both players play strategy 1, and in the second some player plays 1 and the other plays 0.

The support of the symmetric equilibrium is $\{1\}$ which is a subset of the support of the asymmetric equilibrium which as $\{1,2\}$ as it's support.