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I am wondering for a two person zero sum symmetric game, is it true that, for any fixed mixed strategy $$p = [p_1, ... , p_n]^T$$ of player 1, is it true that $$max_q\space \pi_2(p,q)\geq0$$ where$$q=[q_1, ... , q_n]^T$$ denotes a mixed strategy for player 2, and$$max_q\space \pi_2(p,q)$$denotes the payoff function of player 2.

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Yes. Note that if a zero-sum game $(A,B)=(A,-A)$ is symmetric, then $A$ is skew-symmetric:

$$A^\top = -A,$$

so $a_{ij} = -a_{ji}$ for all $i$ and $j$. Now fix $p$ as a mixed strategy of the row player as in your question and consider what the column player gets by playing $p$ too:

$$p^\top A p = \sum \sum p_i a_{ij} p_j = \sum \sum p_i (-a_{ji}) p_j = - - \sum \sum p_i a_{ji} p_j = -p^\top A p,$$

so $p^\top A p = 0$. Thus indeed we have that $\max_q \pi_2(p,q) \ge 0$.

BTW, actually a matrix $A$ is skew-symmetric if and only if $x^\top Ax = 0$ for all $x$.

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  • $\begingroup$ Thanks! it helps! but why we calculate the payoff when the column player plays p too? And how do we get that $$max_q\space\pi_2(p,q)\geq0$$ by getting $$p^TAp=0$$? $\endgroup$ – Leo Li Oct 2 '17 at 23:04
  • $\begingroup$ You want to compute a maximum over all choices of $q$ and I showed you that if you choose $q$ as $p$ then the column player gets $0$. There the maximum much be at least $0$, which is exactly what you wanted to prove! $\endgroup$ – Rahul Savani Oct 3 '17 at 5:09

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