# 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

1. $\forall x\in \Delta: x^tAa\leq a^tAa \ \$ (i.e. $a$ is an equilibrium for $A$)
2. $\forall x\in \Delta: x^tBb\leq b^tBb \ \ \$ (i.e. $b$ is an equilibrium for $B$)
3. $\forall i\in [n]: a_i>0 \iff b_i>0 \ \$ ($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-

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

$a= \left( \begin{array}{ccc} 1 \\ 0 \\ \end{array} \right)$ $b= \left( \begin{array}{ccc} 0 \\ 1 \\ \end{array} \right)$

For a simple counterexample let $$A = B = \left( \begin{array}{cc} 1 & 0 \\ 1 & 0 \end{array} \right).$$ 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: $$A = \left( \begin{array}{cc} 4 & 0 \\ 6 & -4 \end{array} \right), \ B = \left( \begin{array}{cc} 4 & 1 \\ 6 & 0 \end{array} \right)$$ $$a^t = \left(\frac{2}{3}, \ \frac{1}{3} \right), b^t = \left(\frac{1}{3}, \ \frac{2}{3} \right).$$ $$A \cdot a = \left( \begin{array}{c} \frac{8}{3} \\ \frac{8}{3} \end{array} \right) > \left( \begin{array}{c} 2 \\ 2 \end{array} \right) = B \cdot b$$