# Proving the existence of mixed-strategy NE for 2-player zero-sum symmetric game

I am trying to prove the existence of mixed-strategy NE for 2-player zero-sum symmetric game, under the condition that given they have $$I$$ pure strategies and for the pay-off matrix $$A$$, $$\exists x\in R_+^I$$ such that $$xA\ge0$$. I know that I can use Nash's theorem directly, but I want to prove it explicitly.

So far what I have done is that I proved the payoff must be 0 since $$A=-A^T$$. And if $$xA$$ always has negative element(s), the best response of each players must be some pure strategy (by decomposing $$xAy$$). But it seems that all these are irrelevant to a direct proof the existence of NE. Please share with me if you have any idea!

• You stipulate "$\exists x\in R_+^I$ such that $xA\ge0$" but then go on to write "if $xA$ always has negative element(s)". Don't these two contradict? Oct 12, 2023 at 13:20
• Sorry for the confusion. I just mean that I happened to prove the "If NOT $xA\ge 0$ then NO mixed strategy NE" case, but just don't know how to deal with prove the original statement. Oct 13, 2023 at 0:49

For a mixed strategy $$x$$ (a column vector in the $$I$$-simplex) define a pure strategy $$i$$'s excess payoff against $$x$$ as $$k_i(x)=\max\{(Ax)_i,\,0\}$$. Let $$\bar k(x)=\sum_{j\in I} k_j(x)$$ be the sum of excess payoffs.
Now consider the system of differential equations known as the Brown-von Neumann-Nash dynamics (BNN dynamics), $$\dot x_i=k_i(x)-x_i\bar k(x)$$.
Then it is fairly straightforward to show: The function $$V(x)=\sum_{j\in I} k_j(x)^2$$ is a global Ljapunov function for the BNN dynamics and therefore all solutions converge to the set where $$V(x)=0$$. This set is the set of Nash equilibria, which is therefore nonempty.