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corr_minimax_p4 - Minimax Theorem Notice first that...

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Unformatted text preview: Minimax Theorem Notice first that both (Pg) and (PM) are feasible problems. We can choose any p E P and sufficiently small 0: to get a feasible solution to (P1), and any q E Q and sufficiently large {3 to get a feasible solution to (PM). Moreover, they are dual problems. Their coefficient matrices are transposes of one another and their right-hand sides and objective vectors switch. The unconstrained (.1 corresponds to the equality constraint Z (,3 = 1, etc. Since both are feasible, strong duality tells us that both have optimal solutions, and their optimal values are equal. Hence player I has a maximin mixed strategy 13‘, II has a minimax mixed strategy 9‘. and r=a*='p“Aq* =l3*=fi- This is John von Neumann’s Minimax Theorem (1928): Every 11'}. x n two-person zero-sum game with payoff matrix A has a value in mixed strategies, 1‘; := min max wk; 2 u := max min pAq. qEQ pEP pep QEQ Further results We can use other results we know from LP to help our understanding of zero-sum game theory. For instance, the complementary slackness theorem tells us: Suppose p* is a max irnin strategy for I, and p‘AJ- > n“ for somej (n‘ = 1.1 = i7 is the value of the game). Then there is strict inequality in the jth constraint of (Fir) at optimality. Hence the jth dual variable is zero in every optimal solution of (PH), so or; = [1 for every minimax strategy q‘ for II; 11 doesn’t use her jth pure strategy. Note in the example, where p‘ = (U,U.4,U.6) and U" = 0.2, that $3013 = 0.4 > n“ = 0.2. Hence we could have known immediately that in the minjmax strategy (1* we would have q; = 0. Similarly, #144 = 0.5 > 1." = 0.2; hence q; = 0. 3 ...
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