Extension of Minimax to Infinite Matrices

Transcription

1 Extension of Minimax to Infinite Matrices Chris Calabro June 21, 2004 Abstract Von Neumann s minimax theorem is tyically alied to a finite ayoff matrix A R m n. Here we show that (i) if m, n are both inite, then the conclusion of the minimax theorem may fail, and (ii) if at least one of m, n is finite and the entries of A are bounded, then the conclusion of the minimax theorem holds. 1 Definitions Consider a 2-layer zero-sum game G between layers row R and column C where R chooses an integer in m and C chooses an integer in n. The layers choose simultaneously, or euivalently, seuentially but secretly. Then C ays A i,j to R, where A R m n is the ayoff matrix. R wants to maximize the ayoff while C wants to minimize the ayoff. For a countable set α, let D α = { R α i α i 0 X i α i = 1} denote the set of distributions on α. A ure strategy i m (j n) for R (C) is to choose a row i (column j) with robability 1. A mixed strategy D m ( D n) for R (C) is to choose each row i m (column j n) with robability i ( j). If R, C use mixed strategies,, then the exected ayoff is T A = P i,j ijai,j. 2 Basic results When develoing a strategy, R may consider what would haen if he were to give C the following advantage: R chooses first, reveals his choice to C, and then lets C choose. If i is a best ure strategy in this game G and R uses i in the original game G, then R can do no worse in G than he does in G, and he could do exactly as bad. In other words, using i maximizes the minimum ossible ayoff to R, and is therefore a very conservative strategy. C could use the same strategy, minimizing the maximum ossible ayoff in the game G where C must move first. This leads us naturally to comare su i m j n A i,j with j n su i m A i,j. 1

2 Lemma 1. Let X, Y be nonemty sets and f : X Y R. Then su f(x, y) x X y Y su y Y x X even if one or both exressions are inite. f(x, y), Proof. Let ɛ > 0 and choose x X, y Y such that Then su x X y Y y Y f(x, y) su f(x, y) ɛ y Y x X su f(x, y ) x X su y Y x X f(x, y) ɛ y Y f(x, y) f(x, y ) f(x, y) + ɛ. su f(x, y ) x X Since ɛ was arbitrary, the ineuality follows. su y Y x X f(x, y) + ɛ. This makes sense since in G, R has a disadvantage comared to his osition in G. Corollary 2. For A R m n where m, n ω, su Ai,j i m j n su A i,j. j n i m «1 0 The ineuality above could be strict. For examle, if A =, 0 1 then the lhs is 0 while the rhs is 1. We may abstract a bit and consider a choice for R to be that he fixes a distribution D m. We again can consider what haens to the exected ayoff if one layer moves before the other, revealing his choice. 2.1 Defining the exected ayoff But there is a roblem. We need to define the exected ayoff. As long as P m, n are finite, there is no roblem, but if either is inite, then T A = i,j ijai,j could deend on the order in which the terms are added. One way to resolve this is to demand that A be nonnegative. Then the summation is either absolutely convergent or roerly divergent to. Either way, the sum does not deend on the order. Another other way to resolve the roblem is to demand that A be bounded. The first solution is asymmetric since the exected ayoff is then between 0 and, which seems to favor R. Indeed, suose A is nonnegative and unbounded. If column j is unbounded, say k ω A ik,j k with i 0 < i 1 < i 2 <, and C chooses j with ositive robability, then R can use the strategy ik = 6 1 to get an inite exected ayoff. In fact, π 2 k 2 R can do this obliviously: R does not need to know which unbounded column C chooses with ositive robability, only that some such column 2

3 exists. So it would be a bad idea for C to choose any unbounded column with ositive robability. If every column is unbounded, then the roblem is uninteresting since R can get an inite exected ayoff. So now suose that C restricts himself to bounded columns. The same argument as before shows that he must restrict his robabilities so that the roduct of j and the suremum of column j is bounded, as a function of j. To avoid all of these difficulties and to restore symmetry between layers R, C, we will simly reuire that A be bounded. We will soon see that the actual bounds do not matter so that we can choose bounds of 0, 1 wlog. 2.2 Translation and scaling Let A R m n where m, n ω and let c R. Let A + c = (A i,j + c). Then D m, D n T (A + c) = T A + c, and so the game is essentially the same, but the exected ayoff is translated by c. If c > 0, then T (ca) = c( T A). If we scale by 1, then maximization and minimization are interchanged and we get su D m T ( A) = D n D m su T A T. (1) D n By using translation and scaling, it is easy to translate results about matrices with entries coming from [0, 1] into results for arbitrary bounded matrices. (In fact, we do not need scaling by 1 to conclude this, but it will come in handy later since it lets us transose the matrix.) 2.3 Von Neumann s theorem Corollary 3. For A R m n where m, n ω and A is bounded, su D m T A D n D n su T A. D m Proof. Define f : D m D n R by f(, ) = T A, which is well-defined since A is bounded, and aly lemma 1. Von Neumann s minimax theorem, which we do not rove here, reverses the above ineuality in the case that m, n are finite, giving us the somewhat unexected result that R has the same advantage using a mixed strategy in both games G and G. Theorem 4 (von Neumann). For A R m n where m, n are finite, su D m T A = D n D n su T A. (2) D m 3

4 2.4 Counterexamle for inite matrices We now show that (2) may fail when both m, n are inite. Define ( 1 if i j A i,j =. 0 else I.e. the lower triangle is all 1 s while the uer triangle is all 0 s. Claim 5. su D m D n T A = 0 (3) D n su T A = 1. (4) D m Proof. We first show (3). Let ɛ > 0 and let D m. k ω P i < ɛ. Set = e k, the standard basis vector with a 1 in osition k and 0 s elsewhere. Then T A = X i,j i ja i,j = X i ia i,k = = X ia i,k X i < ɛ, which shows (3). The roof of (4) is analogous. 3 Main result In light of the revious section, it is surrising that (2) does hold when one of m, n is finite and the other is inite. The roof of this will use several techniues. At the heart of the argument, we will reduce the roblem to the finite case, but getting there will be tricky. Many times, we will use arguments of the form, real numbers x, y satisfy ɛ > 0 x y + ɛ, and therefore x y. Obviously the conclusion holds if we change the hyothesis to ɛ > 0 x y + cɛ where c is an absolute constant not deending on ɛ. Now let us comlicate this. Fix x, y R. Suose we can show that c 2,..., c k R ɛ > 0 a 1,..., a k R with x = a 1, y = a k such that a 1 a 2 + c 2ɛ a 2 a 3 + c 3ɛ. a k 1 a k + c k ɛ. 4

5 P Then we could conclude that x y. To see this, note that x y + k i=2 ciɛ. The advantage of this last form of the argument is that the intermediate values a i may deend on ɛ as long as the c i do not. With this in mind, once we have an ɛ in our context, we define the notions of almost ineuality and almost euality. Say that x is almost less than or eual to y (x y) iff there is an absolute constant c, not deending on ɛ (even though x, y may deend on esilon), such that x y +cɛ. Also x is almost eual to y (x y) iff x y x y. Notice that, are both reflexive and transitive and that is symmetric. Also if x, y do not deend on ɛ and x y, then x y. This notational trick will simlify life a great deal in the following roof. Theorem 6. For A R m n where m, n ω, at least 1 of m, n is finite, and A is bounded, su T A = D n D ω D n su T A. D ω Proof. If both m, n are finite, this is just theorem 4. If m is finite and n is inite, we can reduce to the case where m is inite and n is finite by using (1) twice: su T A = su T ( A T ) = su T ( A T ) = su T A. So wlog, assume m = ω. By using translation and scaling, wlog assume that each A i,j [0, 1]. Half of the euality was shown in corollary 3. It remains to show su T A su T A. (5) Our next big ste is to aroximate each distribution in D n by one of a finite number of distributions. Later this will let us exloit the continuity in, of the function T A. In articular, fix ɛ > 0 and choose Q = { 0,..., N 1 } D n such that Q = N < D n l N l 1 < ɛ where l 1 = P i n i l i. Any such Q will do, but if one needs a concrete examle, take the integer lattice in R n 1 ɛ, scale it by, then rotate it and translate it in R n to lie n in the affine sace defined by 3/2 P x i = 1. Finally let Q be the intersection of this with D n. 5

6 It is easy to see that for each D n, if we choose l N such that l 1 < ɛ, then D ω T A T A l X j j l j X i ia i,j X j j l j = l 1 < ɛ, and so T A T A l. (6) For each strategy in Q, we choose a counterstrategy. Secifically, l N l D ω and so lt A l su T A l ɛ, lt A l su T A l. (7) This finite number of counterstrategies makes use of only a finite number of rows w.h.. and the remaining rows are used with only negligible robability. Secifically, l N k ω P l i < ɛ, and so, taking the maximum k over all l, we have k ω l N X l i < ɛ. (8) Let B R k n be the first k rows of A. We now define 2 secific column strategies: Then D n D n su su T A T B su T A (9) su T B. (10) l N l 1 < ɛ (11) l N l 1 < ɛ. (12) Given l D ω, we can define a similar distribution l D k by l = l P k. i k l i 6

7 Then we have that D n, lt A lt X B i k,j X i k,j l i = 1 X i k l i ja i,j l X i ja i,j + Pi k l i 1 1 j + ɛ Pi k l i l i + ɛ < 2ɛ, i k,j l i ja i,j and so lt A lt B. (13) Finally we can combine all of this ormation. We have su su T A su T A by (9) T A su T A l by (12) and (6) lt A l by (7) lt B l by (13) su su T B l su T B by (12) and (6) su T B by (10) T B su T A ; and since we have come full circle, we conclude that all of the intermediate exressions are almost eual. In articular, the 9 th and 10 th exressions are almost eual: su T B su T B. (14) We finish the roof with another almost ineuality chain, the middle of which uses theorem 4. su T A su T B = su T B by theorem 4 su T B by (14) su T B l l B l l T A l by (13) su T A l by (7) su T A by (6) su T A by (9). Since the first and last exressions do not deend on ɛ, we conclude (5). 7