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### Assignment4_MATH4321_12S

Course: MATH 4321, Spring 2012
School: HKUST
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310 Game MATH Theory Spring 2012 Assignment 4 1. Problems II.1.5.1, II.1.5.2, II.1.5.3 2. Problems II.2.6.2, II.2.6.4, II.2.6.5, II.2.6.6, II.2.6.7, II.2.6.8, II.2.6.10 3. Prove that for an mxn matrix game, any two saddle points have the same value. Mission : Proof of the Minimax Theorem Let A be the payoff matrix of a 0-sum 2-person game. Let V ( A) denote the upper value and V ( A) denote the lower value of...

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310 Game MATH Theory Spring 2012 Assignment 4 1. Problems II.1.5.1, II.1.5.2, II.1.5.3 2. Problems II.2.6.2, II.2.6.4, II.2.6.5, II.2.6.6, II.2.6.7, II.2.6.8, II.2.6.10 3. Prove that for an mxn matrix game, any two saddle points have the same value. Mission : Proof of the Minimax Theorem Let A be the payoff matrix of a 0-sum 2-person game. Let V ( A) denote the upper value and V ( A) denote the lower value of game with payoff matrix A . We have proved that V ( A) V ( A ) . Minimax Theorem (1928, John von Neumann): V ( A) = V ( A) Follow the following steps to prove the Minimax Theorem. Definition: gap( A )= V ( A) - V ( A ) . It is clear that gap( A )is nonnegative. Theorem: (Minimax Theorem) gap( A )=0. Lemma: If A has more than one element, then there exists a row or a column such that gap( A ) gap( A ), where A is obtained from A by deleting the given row or column. We break down the proof of the Lemma into steps. Let X *, Y * be the set of mixed strategies for Player I and Player II respectively. Let p* X *, q* Y * be the safety strategy for Player I and Player II respectively. Suppose is A an m n matrix. Let e1 , , em be unit vectors in m representing the pure strategies of Player I. Let f1 , , f n be unit vectors in n representing pure strategies of Player II. Exercise 1: For i=1,m, we have eiT Aq* V ( A) . (*) For j=1,,n, we have p* T Af j V ( A) . (**) Exercise 2: Show that if equalities hold for all inequalities in (*) and (**) then gap( A )=0. Moreover, gap( A )=0 for any A obtained from A by deleting one row or one column from the matrix A . Therefore, Lemma is valid in this case. Suppose one of the inequalities is strict. In particular, let us assume that th p* T Afn > V ( A ) . Let A be obtained from A by deleting the n column from A . Exercise 3: Show that V ( A) V ( A) (***) Exercise 4: Show that (****). Then, we obtain the result of Lemma 1 in this case. This completes the proof of Lemma 1. Then, we apply the Lemma repeatedly to delete rows and columns of A until we get to a 1x1 matrix A . Clearly, gap( A )=0 in this case. Thus, we have 0 gap( A ) 0 . This completes the proof of the Minimax Theorem. V ( A) V ( A)
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