Determining strict competitiveness: -

Are either of the two games in Figure 1 strictly competitive (a) if we restrict attention to pure strategies and (b) if we allow mixed strategies?

We saw above that in any game a player's Nash equilibrium payoff is at least her max minimized payoff. I now show that for a strictly competitive game that possesses a Nash equilibrium, the two payoffs are the same: a pair of actions is a Nash equilibrium if and only if the action of each player is a max minimizer.

Denote player i's vNM payoff function by Ui and assume, without loss of generality, that U2 = -U1. Though the proof may look complicated, the ideas it entails are very simple; the arguments involve no more than the manipulation of inequalities.

The following fact is used in the argument. The maximum of any function f is equal to the negative of the minimum of - f : max x f(x) = - minx(- f(x)). It follows that