Are symmetric equilibria monotone? Assume a two player symmetric game is given by $n\times n$ payoff matrix $A$ for the row player (and $A^t$ for the column player).

Let $B$ be a matrix such that $\forall i,j\in [n]:B_{i, j}\geq A_{i,j}$.

Assume $a$ is a symmetric equilibrium in $A$, $b$ is a symmetric equilibrium in $B$ and that $a$ and $b$ has the same support.

Does it mean player's payoff under $a$ (playing matrix $A$) can not be larger than the payoff of playing $b$ in game $B$?



Formulating the claim in linear algebra:

Let $A,B\in [0,1]^{n\times n}$ such that $\forall i,j\in [n]:B_{i, j}\geq A_{i,j}$.

Denote by $\Delta $ the set of probability distributions over $[n]$.

Let $a,b\in \Delta$ be two distribution vectors.

Does


$\forall x\in \Delta: x^tAa\leq a^tAa \ \ $ (i.e. $a$ is an equilibrium for $A$)
$\forall x\in \Delta: x^tBb\leq b^tBb \ \ \ $ (i.e. $b$ is an equilibrium for $B$)
$\forall i\in [n]: a_i&gt;0 \iff b_i&gt;0  \ \ $ ($a,b$ have the same support)


Imply
$$a^tAa\leq b^tBb?$$

If not, does the answer change if $A,B$ are symmetric?



Notice that without the support condition, the claim is false, for example-

$A=
 \left( \begin{array}{ccc}
1 &amp; 0 \\
0 &amp; 0.1 \\
\end{array} \right) $
$B=
 \left( \begin{array}{ccc}
1 &amp; 0 \\
0 &amp; 0.5 \\
\end{array} \right) $

$a=
 \left( \begin{array}{ccc}
1 \\
0  \\
\end{array} \right) $
$b=
 \left( \begin{array}{ccc}
0 \\
1  \\
\end{array} \right) $
For a simple counterexample let
$$
A = B = \left( \begin{array}{cc}
1 &amp; 0 \\
1 &amp; 0 \end{array} \right).
$$
In this game any strategy pair will constitute a Nash-equilibrium. Let
$$
a^t = \left(\frac{2}{3}, \ \frac{1}{3} \right), b^t = \left(\frac{1}{3}, \ \frac{2}{3} \right).   
$$ 
Then the expected payoffs are larger in the symmetric equilibrium associated with $a$.

If you wish for $B$ to be strictly larger than $A$ you can easily modify this example, simply multiply all of $B$'s elements with a number larger than 1 but smaller than 2.

A counterexample where neither $A$ nor $B$ are singular:
$$
A = \left( \begin{array}{cc}
4 &amp; 0 \\
6 &amp; -4 \end{array} \right), \ 
B = \left( \begin{array}{cc}
4 &amp; 1 \\
6 &amp; 0 \end{array} \right)
$$
$$
a^t = \left(\frac{2}{3}, \ \frac{1}{3} \right), b^t = \left(\frac{1}{3}, \ \frac{2}{3} \right).   
$$
$$
A \cdot a = \left( \begin{array}{c}
\frac{8}{3}  \\
\frac{8}{3}  \end{array} \right) &gt;
\left( \begin{array}{c}
2  \\
2  \end{array} \right) = B \cdot b 
$$