<p>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 $a$.</p>

<p>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.</p>

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

CC BY-SA 3.0
2015-01-01T07:56:14.083Z
Love the non-singular counter example, will be using it for some application soon !
R B
CC BY-SA 3.0
2014-12-31T14:08:31.973Z
2014-12-31T15:50:50.907Z
2014-12-31T15:50:50.907Z
Giskard
<p>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).</p>

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

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

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

<hr>

<p>Formulating the claim in linear algebra:</p>

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

<p>Denote by $\Delta $ the set of probability distributions over $[n]$.</p>

<p>Let $a,b\in \Delta$ be two distribution vectors.</p>

<p>Does</p>

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

<p>Imply
$$a^tAa\leq b^tBb?$$</p>

<p>If not, does the answer change if $A,B$ are symmetric?</p>

<hr>

<p>Notice that without the support condition, the claim is false, for example-</p>

<p>$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) $</p>

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

<game-theory><nash-equilibrium>
Are symmetric equilibria monotone?
R B
# Are symmetric equilibria monotone?

(Asked by: username_0 on 2014-12-30)

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) $

**Tags:** game-theory, nash-equilibrium

---

## 1 answers

### Answer by username_1 on 2014-12-31. Score: 2 | Accepted answer

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

> **Comments:**
> **username_0**: Love the non-singular counter example, will be using it for some application soon !  

---

