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>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) $