Maxmin strategy and Nash equilibrium

Question

It's well known fact that maxmin strategy in Nash equilibrium in the two-players zero-sum finite game, but to prove it?

How to show that maxmin strategy is actually Nash equilibrium in the case of zero-sum two-players game.

Answer 1

You are basically asking what we do when we have the following: $w^- < w^+.$ In this case, how is the game played? In a game of poker, if a poker player always bluffs, especially when he/she is holding a weak hand, the "strong-hold" of bluffing disappears. So the player must bluff sometimes and hold a strong hand at others. If we consider model of mixing strategies in a game and allow mixed strategies in a matrix game, it will always have a value and optimal strategies. Now we consider the assumptions of von Neumann theorem to solve for general matrix games.

The sets of points here the min or max is attained by $$B_x = {y^0 \in D : f(x,y^0) = \min f(x,y)} $$ for each fixed $x \in C,$ and $$A_y = {x^0 \in C : f(x^0,y) = \max f(x,y)} $$ for each fixed $y \in D.$ By the assumption on $f, C, D,$ these sets are nonempty, closed, and convex. For instance, here is why $B_x$ is convex. Take $y_1^0, y_2^0 \in B_x,$ and let $\lambda \in (0,1).$ Then $$f(x,\lambda) y_1^0 + (1-\lambda)y_2^0)\leq\lambda f(x,y_1^0) + (1-\lambda)f(x,y_2^0) = \min f(x,y).$$ But $f(x, \lambda y_1^0 + (1-\lambda) y_2^0) \ge \min_{y \in D} f(x,y)$ as well, and so they must equal. This means that $\lambda y_1^0 + (1-\lambda) y_2^0 \in B_x$. Now we define $g(x,y) \equiv A_y$ $x$ $B_x,$ which takes a point $(x,y) \in C$ $x$ $D$ and gives the set $A_y$ $x$ $B_x$. This function satisfies the continuity property for Kakutani's theorem. So the sets $A_y$ $x$ $B_x$ are nonempty, convex, and closed.