A_Course_in_Game_Theory_-_Martin_J._Osborne 38

A_Course_in_Game_Theory_-_Martin_J._Osborne 38 - Page 23...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Page 23 for all and hence for all , so that . Thus u 1 ( x * , y * ) = max x min y u 1 ( x, y ) and x * is a maxminimizer for player 1. An analogous argument for player 2 establishes that y * is a maxminimzer for player 2 and u 2 ( x * , y * ) = max y min x u 2 ( x, y ), so that u 1 ( x * , y * ) = min y max x u 1 ( x, y ). To prove part (c) let v * = max x min y u 1 ( x, y ) = min y max x u 1 ( x, y ). By Lemma 22.1 we have max y min x u 2 ( x, y ) = - v * . Since x * is a maxminimizer for player 1 we have for all ; since y * is a maxminimizer for player 2 we have for all . Letting y = y * and x = x * in these two inequalities we obtain u l ( x * , y * ) = v * and, using the fact that u 2 = - u 1 , we conclude that ( x * , y * ) is a Nash equilibrium of G . Note that by part (c) a Nash equilibrium can be found by solving the problem max x min y u 1 ( x, y ). This fact is sometimes useful when calculating the Nash equilibria of a game, especially when the players randomize (see for example Exercise 36.1).
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/10/2011 for the course DEFR 090234589 taught by Professor Vinh during the Spring '10 term at Aarhus Universitet, Aarhus.

Ask a homework question - tutors are online