# Thread:

1. ## Proof problem

Hi everyone, I have trouble proving this problem:
the question is:

Show that if a zero-sum game has a (pure) Nash equilibrium (i∗ , j∗ ), then the van
Neumann value of the game is $a_i^*j^*$

Hint. Show that $max_P min E_j (P) >= a_i^*j^*$ and that $min_Q max F_i (Q) <= a_i^*j^*$
, and then use Minimax Theorem.

Now i understand that a Pure Nash Equilibrium means:

If the the first player plays strategy i and the 2nd player plays the strategy j then (i, j) is a Nash Equilibrium if every other entry in the same column as i is less than the strategy i and every entry on the same row as the strategy j for the second player is bigger than j. meaning none of the player can improve their strategy to gain better outcome.

the minmax theorem determines the von neumann value, which is basically the value of the game. and the theorem says: for the payoff matrix A of a zero sum game, there exists optimal mixed strategies P and Q for the row and column player respectedly, such that

$max_p min E_j (P) = min_Q max F_i(Q)$ for 0<=j<=n and 0<=i<=m and P and Q are mixed Nash Equilibrium.

$E_j (p)$ = Sum $(p_i) (a_ij)$ (sum = sigma for 0 <= i <= m )
$F_i (Q)$= Sum $(q_i) (a_ij)$ (sum = sigma for 0 <= j <= n )

I don't know how to show that $max_P min E_j (P) >= a_i^*j^*$. i know that i can pick any u that is u <= $E_j (p)$ but I can't see how that would help. Should i take another approach?
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