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)