Topics in Mathematical Economics. Atsushi Kajii Kyoto University

Transcription

1 Topics in Mathematical Economics Atsushi Kajii Kyoto University 25 November 2018

2 2

3 Contents 1 Preliminary Mathematics Topology Linear Algebra convex sets A ne spaces and the dimension of a convex set convex functions cones Separation Theorems and Their Applications Separation Theorems Separation of convex sets Topological properties of convex sets Separation of convex cones Conic hull and Farkas lemma Minimax Theorem and its applications The minimax theorem Characterization of dominated strategies Existence of stationary distribution Robust Voting Scheme Impossibility of rejecting a random forecast Blackwell s Information Theorem Bayesian decision problems Comparison of experiments Blackwell s theorem Cooperative games and non-emptiness of Core Choquet Integral and Multiple priors Multiple priors Choquet Integral Fundamental Theorems of Welfare Economics and No speculation theorem Model of Pure Exchange Fundamental Theorems of Welfare Economics Characterization of Speculative Trade

4 4 CONTENTS 3 Fixed Point Theorem Fixed point theorems and partition of unity Existence of a Nash equilibrium in discontinuous games Existence of a competitive equilibrium Genericity Argument Di erential manifold Implicit Function Theorem and Di erential Manifold Application: Di erentiability of demand Application: Structure of Pareto e cient allocations Transversality theorem Application: Almost everywhere Di erentiability of Demand Genericity Generic Regularity of a competitive equilibrium Set up and basic properties Regular economies Genericity of Regular Economies Structure of sunspot equilibria set up sunspot equilibria Order structure 83

5 Chapter 1 Preliminary Mathematics 1.1 Topology We shall review some topological concepts in R L. Let X R be any non empty set. A number b is said to be an upper bound of X if x 2 X implies x b. We use the following as an axiom: if the set of upper bounds is non empty, then there exists the smallest upper bound, which is denoted by sup (X). That is, sup (X) itself is an upper bound, and if b is an upper bound for X, then sup (X) b. A lower bound is de ned similarly, and denote by inf (X) the largest lower bound. When sup (X) 2 X, then by construction sup (X) is the largest number, the maximum, of X. So we write max (X) for sup (X), in such a case. By convention we say max (X) does not exist, when sup (X) =2 X or X has no upper bound. Also when X is a nite set X = fx 1 ; x 2 ; :::; x n g, then we may write max (x 1 ; x 2 ; :::; x n ) instead of max (fx 1 ; x 2 ; :::; x n g) by convention. The minimum of set X, min (X), is de ned similarly. Example (1) sup ((0; 1)) = 1; since any x 2 (0; 1) satis es x 1, 1 is an upper bound. If 1 were not the smallest, then there is b < 1 such that x 2 (0; 1) implies x b, which is impossible since (0; 1) contains a number arbitrarily close to 1. (2) for any two non empty sets X and Y in R, sup (X [ Y ) = max (sup (X) ; sup (Y )); if b is an upper bound for X [Y, then it is clearly an upper bound for both X and Y. Since sup (X [ Y ) is an upper bound for X [Y, we have sup (X [ Y ) sup (X) and sup (X [ Y ) sup (Y ), hence sup (X [ Y ) max (sup (X) ; sup (Y )). Choose any z 2 X [ Y. If z 2 X, z sup (X) since sup (X) is an upper bound of X. Similarly, if z 2 Y, z sup (Y ). Hence for any z 2 X [ Y, z max (sup (X) ; sup (Y )), so max (sup (X) ; sup (Y )) is an upper bound of X [ Y. Therefore, sup (X [ Y ) max (sup (X) ; sup (Y )) since sup (X [ Y ) is the smallest upper bound. Exercise (1) Let X = f0g. Show that sup (X) = inf (X) = 0. (2) Let X R be any non empty set. Show that if sup (X) = inf (X), then there is x such that X = fxg. Write the norm of a vector x 2 R L by kxk = q P l (xl ) 2. Obviously kxk = 0, x = 0. Observe that the norm is sublinear; for any positive scalar t, ktxk = t kxk, 5

6 6 CHAPTER 1. PRELIMINARY MATHEMATICS and kx + yk kxk + kyk. Then in particular d (x; y) := kx yk is positive and d (x; y) = 0, x = y, and moreover it satis es the triangular inequality d (x; z) d (x + y) + d (y; z). So d is called the (Euclidean) distance in R L. A sequence in R L is a function from the set of natural numbers to R L, which we shall denote as x (n), n 2 N, (x n ) 1 n=1, x n, n = 1;, etc. Let f : N!N be an increasing function. Then the composite function x f is a sequence whose range is contained in that of x (n), n 2 N. So x f (n), n 2 N, is called a subsequence 1 of x (n), n 2 N. Writing n q := f (q), q = 1; 2; :::, we may as well write x, or nq q=1 x nq, q = 1;. A sequence (x n ) 1 n=1 in RL is said to converge to x 2 R L if for any " > 0, there is n 2 N such that kx n xk < " for any n n. In this case, point x is called the limit of sequence (x n ) 1 n=1. We write lim n!1 x n = x to mean (x n ) 1 n=1 converges to x. Example (1) x n = 1 converges to 0; for any " > 0, choose n such that n n" > 1, or equivalently, " > 1. Then for any n n, x n n = 1 < 1 < ". n n (2) Let X R L be a set such that for any integer n, there is x 2 X with kxk < 1. Then there is a sequence fx n ng in X such that lim n!1 x n = 0; for each n, by assumption, there exists x 2 X with kxk < 1. So let x n n be such a point x (there might be many of them, but just pick one of them and call it x n ). Then for each " > 0, choose n such that " > 1=n, we have kx n 0k = kx n k < 1 < 1 < " so n n lim n!1 x n = 0. Exercise (1) Show that the limit of sequence (x n ) 1 n=1 is unique: that is, if sequence (x n ) 1 n=1 converges to x and x0, then x = x 0 must hold. (2) Show that if sequence (x n ) 1 n=1 converges to x, and subsequence of sequence (x n ) 1 n=1 converges to x. Exercise (1) Let fx n : n = 1; 2; :::g R be a non-decreasing sequence with an upper bound b: i.e., x n x n+1 b, n = 1; 2; :::. Show that it converges to sup fx n : n = 1; 2; :::g. What happens if there is no upper bound? (2) Let X R be any non empty set. Show that there is a non-decreasing sequence fx n : n = 1; 2; :::g in X, such that lim n!1 x n = sup X. De nition A point x is said to be adherent to set X R L if there is a sequence in X which converges to x. The set of all adherent points to X is said to be the closure of X, denoted by X. A set is said to be closed if X = X, and a set is said to be open if its complement X c is closed. The interior of set X, denoted by X o, is de ned as X o := XnX c. The boundary of set, denoted is de ned := XnX o : Note that set X is open if and only if X = X o. Exercise Prove that (1) for any " > 0, fx : kxk < "g is an open set; (2) A set X is open if and only if for any x 2 X, there is " > 0 such that fx 0 : kx 0 xk < "g X.

7 1.1. TOPOLOGY 7 Exercise Prove that (1) for a family of closed sets, ff : 2 g where is a set of indices (with arbitrary cardinality), the intersection \ 2 F is closed; (2) for a family of open sets, fo : 2 g where is a set of indices (with arbitrary cardinality), the union [ 2 O is open. Exercise Prove that (1) x 2 X if and only if for any open set O such that x 2 O, O \ X 6=?; (2) x 2 X o if and only if there exists an open set O such that x 2 O X. The next concerns the relative topology: De nition Let X be a subset of R L, and Y X. The closure of set Y relative to X is the set of all of its adherent points contained in X; that is, Y \ X. Y is said to be closed in X if Y = Y \ X, and Y is said to be open in X if its complement in X; Y n X; is closed in X. The relative interior of Y in X is Y nx n Y. The relative boundary of Y in X is Y \ V n Y nx n Y : Example Consider set X = f(x 1 ; 0) : x 1 2 (0; 1)g. X is neither closed nor open in R 2, and X = f(x 1 ; 0) : x 1 2 [0; 1]g. Now consider V = f(x 1 ; 0) : x 1 2 Rg. Then V n X is closed in V, because V n X \ V = V n X holds. Thus X is open in V, and so the relative interior of X is X itself. The relative boundary is f(0; 0) ; (1; 0)g. De nition Let X R L. A function f : X! R M is continuous provided for any sequence (x n ) 1 n=1 in X, if x := lim n x n 2 X, then lim n f (x n ) = f (x). Proposition Let X R L and the following statements about a function f : X! R M are equivalent: (1) f is continuous; (2) for any x 2 X and any " > 0, there is > 0 such that kx 0 xk < and x 0 2 X imply kf (x 0 ) f (x)k < "; (3) for any open set O R M, f 1 (O) is open in X; (4) for any closed set F R M, f 1 (F ) is closed in X. Exercise Prove the result above. Exercise (1) Show that f (x) = kxk is continuous. (2) De ne f (x) = x sin x 1 if x > 0, and f (0) = 0. Show that f is a continuous function on [0; 1) (you may take it for granted that sin is continuous, and 1 sin (y) 1 for any y). De nition A set X is said to be bounded if there is a number m such that kxk m for any x 2 X. De nition A collection of sets fo : 2 g is an open cover of a set X if each O is open and X [ O. A set X is said to be compact if any open cover fo : 2 g has a nite subcover; there are nitely many 1,..., K 2 such that fo k : k = 1; :::; Kg is an open cover of X.

8 8 CHAPTER 1. PRELIMINARY MATHEMATICS An equivalent condition for compactness can be stated using closed sets. A collection of subsets of X, ff : 2 g, is said to have a nite intersection property if for any choice of nitely many 1,..., K 2, \ K k=1 F k 6=?. Then a set X is compact if and only if for any collection of closed subsets of X with nite intersection property, \ 2 F 6=?. The notion of compactness above is de ned for a general topological space. Because of the special structure of R L, compact sets in R L can be characterized as follows: Theorem The following conditions about a set X in R L are equivalent: (1) X is compact; (2) X is closed and bounded; (3) any sequence (x n ) 1 n=1 in X has a subsequence converging to a point in X. Exercise Let x n 2 R L, n = 1; 2; ::::; be a sequence with non-increasing norm: kx n+1 k kx n k, n = 1; 2; :::. Show that a subsequence of fx n g must converge to a point x, and kxk = inf fkx n k : n = 1; 2; :::g. An important implication of compactness and continuity: Theorem If X is compact and f : X! R M is continuous, f (X) is compact. Thus if X is a compact set and f is a continuous function on X, f attains a maximum and a minimum on X. Exercise Prove Theorem Linear Algebra convex sets First we note the convention of linear operation on sets: De nition If X is a subset of R L and t is a real, then tx := ftx : x 2 Xg. If X 1 and X 2 are subsets of R L, X 1 + X 2 := fx 1 + x 2 : x 1 2 X 1 ; x 2 2 X 2 g. When x 2 R L, X R L, we shall simply write x + X instead of fxg + X. It can be readily shown that if X 1 and X 2 are open sets, then so is X 1 + X 2. Also, if X 1 is compact and X 2 is closed, then X 1 + X 2 is closed. However, the sum of two closed sets is not necessarily closed. For instance, let L = 2, and consider X 1 = f(x 1 ; x 2 ) : x 1 x 2 1, x 1 0g and X 2 = f(z; 0) : z 0g. De nition A set C R L is said to be convex if for any pair of points x; y 2 C; tx + (1 t)y 2 C holds for any number t 2 [0; 1]. By de nition, a one point set is convex, so is the empty set. It can be readily veri ed that (1) if C R L is convex, so is tc; and (2) if C 1 and C 2 are convex, so is C 1 + C 2. The intersection of an arbitrary collection of convex sets is convex. The convex hull of a set X. denoted by co(x), is the smallest convex set containing X. Then by de nition, co(x) = fc R L : C is convex and X Cg.

9 1.2. LINEAR ALGEBRA 9 Exercise Let X = fx 1 ; :::; x K g be a nite set. co (X), is a compact set. Show that its convex hull, A convex combination of ( nitely many) points, x 1,..., x K is P K k=1 t kx k where t k 0 for each k = 1; :::; K and P K k=1 t k = 1. To simplify notation, we write t 2 4 to mean that t k 0 for each k = 1; :::; K and P K k=1 t k = 1. The convex hull can also be characterized as the set of convex combinations of nitely many points; that is: Lemma co (X) = f P K k=1 t kx k : x k 2 X and t 2 4g Exercise Prove Lemma Lemma implies that if x 2 co (X), x is expressed as a convex combination of nitely many points in X. But this expression is not unique in general. For example, let L = 1 and consider X = f 1; 0; 1g A ne spaces and the dimension of a convex set An a ne space is a convex set of special interest: De nition A set V is an a ne space if for any pair of points x; y 2 V; sx + ty 2 V holds for any numbers s; t 2 R with s + t = 1. Recall that a linear subspace ^V requires that for any pair of points x; y 2 ^V; sx + ty 2 ^V holds for any numbers s; t 2 R. A linear subspace of R L is clearly an a ne space. An a ne (sub)space V of R L is a linear subspace if 0 2 V. Indeed, let V be an a ne space and pick any x; y 2 V, x any numbers s; t 2 R. Since 0 2 V, 2sx = 2sx + (1 2s) 0 2 V and 2ty 2 V. So sx + ty = 12sx + 12ty 2 V. 2 2 If V is an a ne space, so is V x := fv x : v 2 V g for any xed vector x. Thus if x 2 V, V x is a linear subspace because 0 2 V x. Moreover, V x = V y whenever x and y are both contained in V. Indeed, if z 2 V x, there is v 2 V such that z = v x = (v x + y) y. Since 2v x 2 V and 2y x 2 V, v x + y = 1 (2v x) + 1 (2y x) 2 V, and so z 2 V y. This observation implies that 2 2 V x determines a unique linear subspace, regardless of the choice of x 2 V. The dimension of an a ne space V is de ned by the dimension of this unique linear subspace, i.e., the largest number of linearly independent vectors in V x. A set of vectors, x 1 ; :::; x K are a ne independent if whenever P K k=1 kx k = 0 for some scalers ; k ; such that P k k = 0, then k = 0 for every k. So x 1 ; :::; x K are a ne independent if and only if K 1 vectors x k x K, k = 1; ::; K 1, are linearly independent. Thus the dimension of an a ne space V is n if and only if the largest number of a ne independent vectors in V is n + 1. Alternatively, notice that x 1 ; :::; x K are a ne independent if and only if L + 1 simultaneous equations P K k=1 kx k = 0 and P K k=1 k = 0 have a unique solution = 0. Thus x 1 ; :::; x K are a ne independent if and only if L + 1 dimensional vectors (x k ; 1), k = 1; :::; K, are linearly independent.

10 10 CHAPTER 1. PRELIMINARY MATHEMATICS Exercise Show that for any nite set X of vectors, there are a ne independent vectors x 1 ; :::; x K 2 X such that X is contained in the smallest a ne space containing x 1 ; :::; x K. If V, 2, is a collection of a ne spaces, their intersection \V is an a ne space. The dimension of a convex set C is by de nition the dimension of the smallest a ne space containing C, which is \ fv : V is a ne and C V g. Thus a convex set of dimension n can be identi ed with a subset of R n containing 0, with n linearly independent elements. Example Consider set X = f(x 1 ; 0) : x 1 2 (0; 1)g. X is a convex set in R 2, but neither closed nor open. The smallest a ne subspace containing X is V = f(x 1 ; 0) : x 1 2 Rg. Thus the dimension of X is one. Note that X can be naturally identi ed with (embedded into) an open interval (thus a convex subset) of the real line. Lemma If a convex set C R n has dimension n, C has a non-empty interior. Proof. By assumption there are n+1 a ne independent vectors x 1 ; :::; x n ; x n+1 in C. The set co (fx 1 ; :::; x n ; x n+1 g) has the same structure as the n-dimensional simplex in R n (i.e., they are linearly isomorphic) and its interior is P n+1 k=1 t kx k : t 2 4; 8k; t k > 0 which is non empty. Exercise Prove that if C R n has dimension m, C has a non-empty interior with respect to the relative topology of the m dimensional a ne space containing C. We have seen that the convex hull can be identi ed with convex combinations of nitely many points. As a matter of fact, the number of points is related to the dimension of the convex hull. Theorem (Carathéodory) If the dimension of co (X) is n, then any point in co (X) can be expressed by at most n + 1 a ne independent points in X. Proof. Since the dimension is n, after some linear transformations, we might as well regard X R n. Let x 2 co (X), thus for some x k 2 X; k = 1; :::; K, we can write x = P K k=1 kx k with 2 4. We shall show that if x 1 ; :::; x K are not a ne independent, then we can re-express x = P K k=1 0 k x k with where one of k 0 is zero. This establishes the result since we can reduce K one by one till x is a convex combination of a ne independent vectors, and we know there can be at most n + 1 a ne independent vectors. If any of k s is zero, we already have the desired expression. So assume that >> 0 and x 1 ; :::; x K are not a ne independent. Then there is a non zero vector

11 1.2. LINEAR ALGEBRA 11 2 R K, such that KX k x k = 0; k=1 KX k = 0: Note that k > 0 for some k because P K k=1 k = 0 and 6= 0. Thus k=1 s := max fs : k i s k 0 for all kg = min ; i: i >0 i is well de ned. De ne for each k 0 k = k s k. So by construction, k 0 0 for every k, and 0 k above). And, = 0 for some k (i.e., at the minimizer KX kx 0 k = k=1 = = KX ( k s k ) x k k=1 KX k x k k=1 KX k x k ; k=1 KX s k x k k=1 as we wanted convex functions De nition Let X R L be convex. A real valued function f on X is convex if f (tx + (1 t) y) tf (x) + (1 t) f (y) for any x; y 2 X and t 2 [0; 1]. f is said to be concave if f is convex. f is convex if and only if its epigraph f(x; ) : f (x) g is convex f is concave if and only if its hypograph f(x; ) : f (x) g is convex for any collection ff : 2 g of convex functions on X, x 7! sup 2 f (x) is a convex function cones Cones: A set X R L is said to be a cone with vertex x if y 2 X implies x + t (y x) 2 X for any t 0. If X is a cone with vertex x, X x is a cone with

12 12 CHAPTER 1. PRELIMINARY MATHEMATICS vertex 0, and vice versa. So unless otherwise stated, a cone with vertex 0 will be called a cone, for simplicity. For instance, an a ne space A is a cone with vertex x for any x 2 A, and a linear subspace of R L is a cone. These are convex sets, but a cone is not necessarily convex: consider for instance X = f(x 1 ; 0) : x 1 0g [ f(0; x 2 ) : x 2 0g.

13 Chapter 2 Separation Theorems and Their Applications For set X; Y R L and a vector p 2 R L, we write p X to mean that p x holds for all x 2 X, and p X p Y to mean p x p y holds for all x 2 X and y 2 Y. Similarly, we write p X > to mean that p x > holds for all x 2 X, and p X > p Y to mean p x > p y holds for all x 2 X and y 2 Y. Notice that by de nition p X p Y is equivalent to p (X Y ) 0 and p X > p Y is equivalent to p (X Y ) > 0. When p X p x holds for some x 2 X, we say the vector p supports X at x. The hyperplane fz : p z = p xg is referred to as the supporting hyperplane. 2.1 Separation Theorems Separation of convex sets We show a series of separation theorems. Theorem (strict separation) Let X R L be a non-empty, closed and convex set, and x 2 R L such that x =2 X. Then there is a non-zero vector p 2 R L and a number such that p x < < p X. Proof. Set ^X = X + f xg. Then ^X is closed and convex, and 0 =2 ^X. It su ces to show that there is p 6= 0 such that p ^X for some > 0. Consider the set D := fkxk : x 2 ^Xg R. Clearly, 0 is a lower bound of D, so inf D 0. One can then nd a sequence fx n g in ^X such that kx n k! inf D. In particular, the sequence is bounded and so it has a convergent subsequence. Since ^X is closed, the limit point of this subsequence, which denote by p, belongs to ^X. That is, there exists p 2 ^X such that kpk = minfkxk : x 2 ^Xg. Since 0 62 ^X, p 6= 0 must hold. Then it su ces to show that p ^X p p: Choose any x 2 ^X. For any t 2 [0; 1], (1 t)p + tx = p + t (x p) 2 ^X since ^X is convex. Then by the construction of p, kp + t (x p) k 2 = kpk 2 + 2t (p (x p)) + t 2 kx pk 2 ; 13

14 14 CHAPTER 2. SEPARATION THEOREMS AND THEIR APPLICATIONS must be minimized at t = 0. That is, viewing above as a function of t, its derivative, 2 (p (x p))+2t kx pk 2, must be non-negative at t = 0. So we conclude px pp. Corollary If X and Y are non empty convex sets with X \ Y =? and if X Y is closed, then there is a non-zero vector p 2 R L such that p Y < p X. Corollary If X and Y are non empty closed convex sets with X \ Y =?, and if one of them is compact, then there is a non-zero vector p 2 R L such that p Y < p X. Exercise Prove the corollaries above. The separation properties so far utilized the topological property of convex sets. But the separation theorem holds without topological assumptions. Theorem (separation) Let X R L be a non-empty and convex set, and x 2 R L such that x =2 X. Then there is a non-zero vector p 2 R L such that p x p X. Proof. Since we can translate the set as in the proof of Theorem 2.1.1, we might as well assume x = 0 to start with. For each x 2 X, let P (x) := p 2 R L : kpk = 1; p x 0. The theorem is proved if \ x2x P (x) 6=?. Since each P (x) is a closed subset of a compact set fq 2 R L : kqk = 1g, it su ces to show that fp (x) : x 2 Xg has the nite intersection property. So choose x 1 ; :::; x K 2 X arbitrarily, and set F := co fx 1 ; :::; x K g. F is closed and convex. Also F X since X is convex, so 0 =2 F. Applying Theorem 2.1.1, there exists a vector p with kpk = 1 such that p F > 0. Then p x k > 0, for k = 1; :::; K; i.e., p 2 \ K k=1 P (x k), so \ K k=1 P (x k) 6=? is shown. Corollary If X and Y are non empty convex sets with X \ Y =?, then there is a non-zero vector p 2 R L such that p Y p X. Exercise Let X be a convex cone such that X \ ( there is p 6= 0 such that p X 0. X) = f0g. Show that Topological properties of convex sets The separation theorems tell us that a convex set has some topological property automatically. Let X R L and write X for the closure of X and X o for the interior of X. The boundary of X is X n X o by de nition. Lemma If X is convex, both X and X o are convex sets. Exercise Prove Lemma above

15 2.1. SEPARATION THEOREMS 15 Exercise Assume that X is convex and let x 2 X o. Show that for any x 2 X; (1 t) x + tx 2 X o. Supply an example that this property does not hold if X is not convex. Theorem (supporting hyperplane theorem)let X be a convex set in R L. If x belongs to the boundary of X, then there exists p 6= 0 such that p X p x. Proof. Let the dimension of X be m, and X is contained in an a ne space V of dimension m. Since V is closed, X V. If m < L, then there is a non zero vector p (in the orthogonal subspace of V x) such that p (V x) 0, so p X p x. So assume m = L, and then X o is a non empty, convex subset of R n by Lemma and Lemma By assumption x =2 X o, so applying Theorem 2.1.5, nd p 6= 0 such that p X o p x. Now we show p X p x. Choose any x 2 X. Fix y 2 X o, i.e., there is an open set A around 0 so that y + A X. Then for any t 2 (0; 1), (1 t) x + t (y + A) = ((1 t) x + ty) + ta X since X is convex. Since ta is an open set around 0 if t > 0, this means that (1 t) x + ty 2 X o. So p ((1 t) x + ty) p x for any t 2 (0; 1), and letting t! 0 we conclude p x p x. Corollary If X is a convex set, its boundary coincides with the boundary of X. Proof. Applying Theorem it can be readily seen that a point in the boundary of X cannot be an interior point of X. Exercise Provide the detail of the proof above. Supply an example of a non convex set where the boundary of X is di erent from that of X. Separation Theorem implies that a closed convex set is characterized by half spaces containing it. Let X R L be a non empty convex set, and X 6= R L. For each p 2 R L, p 6= 0, and a number 2 R L, let H (p; ) := x 2 R L : p x, i.e., a half space. Clearly, H (p; ) is closed and convex. Let := f(p; ) : X H (p; )g and let Y = \ (p;)2 H (p; ). Set Y is closed and convex since it is the intersection of such sets. In fact Y is the smallest closed and convex set which contains X; that is, Y itself is a closed and convex set containing X, and if Y 0 is another closed and convex set containing X, then Y Y 0. Exercise Show that Y = \ (p;)2 H (p; ) is the smallest closed and convex set which contains X Separation of convex cones For closed convex cones, we get some sharper separation results, in particular, the duality result. First we provide versions of the separation theorems for convex cones.

16 16 CHAPTER 2. SEPARATION THEOREMS AND THEIR APPLICATIONS Theorem (cone separation) Let X and Y be convex cones, and X\Y = f0g. If X n f0g or Y n f0g is non empty and convex, then there is p 6= 0 such that p X 0 p Y. Proof. Say X n f0g is non empty and convex. Then X n f0g and Y are disjoint convex sets, so by the separation theorem (Theorem 2.1.5), there is p 6= 0 such that p (Xn f0g) p Y. Since 0 2 Y, p (Xn f0g) 0 and so p X 0. If there is y 2 Y with p y > 0, then ty 2 Y for any t > 0 so p (ty) could get arbitrarily large, contradicting to p (Xn f0g) p Y. So 0 p Y. Remark The condition that X n f0g or Y n f0g is non empty and convex is indispensable above. For instance consider X = f(x; 0) : x 2 Rg and Y = f(0; y) : y 2 Rg. For a closed convex cone, a strict separation is possible: Theorem (strict cone separation) Let X be a convex and closed cone, and z =2 X. Then there is p 6= 0 such that p X 0 > p z. Proof. By the strict separation (Theorem 2.1.1), there is p 6= 0 such that px > pz. Since 0 2 X; 0 > p z must hold. If x 2 X, p x < 0 is impossible since tx 2 X for any t > 0 and p (tx) < p z if t is large enough. So p X 0. The closedness is essential above. Consider X = R 2 ++ [ f0g and z = (1; 0). De nition For X R L ; X := p 2 R N : p X 0 is called the dual cone of X. Note that X = \ x2x p 2 R N : p x 0, i.e., X can be written as the intersection of some closed half spaces. Thus in particular, X is a closed cone. Here is the duality result. Theorem If X is a non empty, closed and convex cone, then X = (X ). Proof. Let x 2 X. Then for any p 2 X, p x 0 holds. This means x 2 (X ), and so X (X ) is shown. Let z 2 (X ), and suppose that z =2 X. Applying Theorem , there is p 2 R N such that p X 0 > p z. But if p X 0, p 2 X follows and so p z 0 must hold since z X 0. A contradiction. Suppose that X and Y are closed convex cones, and X \ Y = f0g. If X n f0g or Y n f0g are convex, we conclude p X 0 p Y for some non zero vector p by Theorem But since the intersection of the two sets is f0g, this inequality seems loose, not taking advantage of closedness. It seems possible to choose p so that px > 0 > py for non zero elements of X and Y. But consider X = f(x; 0) : x 2 Rg and Y = f0g. Then p X 0 p Y implies p X = 0. That is, if the cone in question contains a line, the strict inequality cannot hold. But when it does not, we can establish such a strict separation result.

17 2.2. CONIC HULL AND FARKAS LEMMA 17 Lemma Let X be a closed convex cone, and X \ ( X) = f0g. Then there exists p 2 R n ; p 6= 0, such that p x > 0 for any x 2 X n f0g. Proof. Claim: the dual cone X has a non empty interior. If not, since X is convex, it must be contained in a linear subspace of dimension less than n. Note that any non zero vector q in the orthogonal complement of this subspace, i.e., q X = 0, belongs to (X ), hence (X ) \ ((X ) ) contains such a vector. By Theorem , (X ) = X, so X \ ( X) 6= f0g, violating the assumption. So let p belong to the interior of X. By de nition of the dual cone, p X 0. If there is x 2 X, x 6= 0 with p x = 0, we would be able to nd p 0 2 X close enough to p such that p 0 x < 0, a contradiction. Consequently, p x > 0 for all x 2 X n f0g. Theorem Let X be a closed convex cone. Then there is p 2 R n ; p 6= 0, such that p X 0 and p x > 0 for any x 2 X n (X \ ( X)). Proof. Note that X \ ( X) is a linear subspace contained in X. Let L be the orthogonal complement of this linear space and let Z := X \ L. If both z and z belong to X \L, then z 2 X \( X) in particular, and so z L = 0. And from z 2 L; we conclude z = 0. Consequently, Z \ ( Z) = f0g, and any x 2 X can be written as x = w + z where w 2 X \ ( X) and z 2 Z, and moreover if x =2 X \ ( X) then z 6= 0. Applying Lemma above to Z as a subset of L, nd p 2 L such that p z > 0 for all z 2 Z n f0g. Then for any x 2 X, p x = p (w + z) = p z 0. Moreover, if x 2 X n (X \ ( X)), x = w + z with z 6= 0 and so p x = p z > 0 as we wanted. Exercise Extend Theorem to the case where X and Y are both closed convex cones and X \ Y = f0g. 2.2 Conic hull and Farkas lemma Let a 1 ; :::; a M 2 R n and b 2 R n and consider a linear constrained optimization problem: min b y y2r n subject to y a m 0 for m = 1; :::; M Clearly y = 0 satis es the constraints and so it is clear that the minimum is not positive if it exists. The curious question is thus whether or not if the objective function can take a negative value under the constraint, i.e., if there exists a vector y 2 R n with y b < 0 such that y a m 0 for m = 1; :::; M. Since if y satis es the constraints, ty also satis es the constraints for any t > 0, thus in this case, the objective function can be made arbitrarily small, thus the minimum is 1. Farkas lemma, which characterizes this property, is stated in the following:

18 18 CHAPTER 2. SEPARATION THEOREMS AND THEIR APPLICATIONS Theorem (Farkas Lemma) Let a 1 ; :::; a M 2 R n and b 2 R n. Then exactly one of the following properties holds: (1) there exists a non negative vector x 2 R M + ; x = (x 1 ; :::; x M ), such that P M m=1 x ma m = b (2) there exists a vector y 2 R N such that y b < 0 and y a m 0 for m = 1; :::; M. So Farkas lemma says that either y = 0 is not a solution to the optimization problem and the objective function can be made arbitrarily small (i.e., case (2)), or b can be expressed as a convex combination of constraining vectors times some positive scalar (i.e., case (1)). With matrix notation, the conditions (1) and (2) in Farkas lemma can be succinctly stated: if we write A for the M n matrix whose rows are given by a T 1,, a T M, P M m=1 x ma m = b can be written as x T A = b T and y a m 0 for m = 1; :::; M can be written as Ay 0. To see why Farkas lemma (Theorem 2.2.1) holds, rst note that if (1) holds, then y P M m=1 x ma m and y b has the same sign, so (2) cannot hold since x 2 R M +. Thus it su ces to nshow that if (1) fails, (2) holds. If (1) does not hold, b is not PM o contained in C := m=1 x ma m : x 0, which is clearly a convex cone. So if C is closed in addition, then by the strict separation theorem (Theorem 2.1.1), there is y 2 R n and 2 R with y b < y C. Since 0 2 C, 0 must be the case. Also since C is a cone, y C 0 must hold. 1 So we see (2) holds since a m 2 C for each m. Thus the essence of Farkas lemma lies, in addition to the separation argument above, in the property that the cone generated by nitely many vectors is closed. Although this fact can be shown more directly, it is instructive to learn a bigger picture: so in the rest of this section, we take a detour to study the closedness of cones generated by some set of vectors. For X R L, let cone (X) be the cone generated by X: cone (X) := ftx : t > 0; x 2 Xg. cone (X) is referred to as the conic hull of X (or the conical hull of X). It is readily con rmed that cone (X) is convex if X is convex. Exercise Show that cone (X) is convex if X is convex. We are interested in the general question of when cone (X) is closed. Lemma Suppose S is compact and convex, and let L be a linear subspace of R n such that S \ L =?. Then cone (S) + L is closed. Proof. Since S \ L =? and S is compact and convex, and L is closed and convex, by the strict separation theorem (Theorem 2.1.1), there is a non zero vector p 2 R n such that p S > p L. Since L is a linear subspace, p L = 0 must hold. Let z n = t n x n + y n for n = 1; 2; :::, where t n 0, x n 2 S and y n 2 L, and suppose z n! z as n! 1. We want to show z 2 cone (S) + L. We might as well assume x n! x 2 S since S is compact. Then from p z n = p (t n x n + y n ) = p (t n x n ) and p x n > 0 for all n, and also p x > 0, we conclude t n = pzn converges px n 1 This can be seen as a simple instance of the closed cone separation result (Theorem ).

19 2.2. CONIC HULL AND FARKAS LEMMA 19 to t := pz px 0. Hence y n = z n t n x n must also converge to a point y, and y 2 L since L is closed. Hence z = t x + y 2 cone (S) + L follows. Looking at the particular case where L = f0g, we have: Corollary If S is compact and convex and 0 =2 S, then cone (S) is closed. Exercise Give an example of a compact convex set X where cone (X) is not closed. Remark The sum of two closed cones, even if they are also convex, may not be closed: in R 3, let X = ft (1; 0; 1)g and Y = f(x; y; z) : z > 0; x 2 + y 2 z 2 g. They are closed convex cones. Then x = (0; 1; 0) is not in X +Y because for any t 0, x t (1; 0; 1) = (t; 1; t) =2 Y. Now consider t (1; 0; 1) + t; 1 + t 1; t t 0; 1 + 1t ; 1 t t 2 X + Y. Since t t! 0, as t t t! 1, we see that x belongs to the closure of X + Y. So in Lemma 2.2.3, it is important that L is a linear space, not just a closed convex cone = Coming back to Farkas lemma, if co fa 1 ; :::; a M g does not contain 0, Lemma above gives us the closedness property we wanted, by taking L = f0g. So the tricky case we shall handle from now on is when co fa 1 ; :::; a M g contains 0. Notice in the proof of 2.2.3, it is crucial that p x > 0, which is not warranted if S contained 0. Say a nite set X = fx 1 ; :::; x K g is positively dependent if there are positive weights k > 0; k = 1; :::; K, such that 0 = P k x k. Lemma If a nite set X is positively dependent, cone (co (X)) is a linear subspace. Proof. It su ces to show that if z 2 cone (co (X)), then z 2 cone (co (X)). If z = P t k x k, then z = P t k x k = P t k x k + b P k x k = P (t k + b k ) x k for any b > 0 since P k x k = 0. So choosing a large enough b so that t k + b k > 0 for every k, we see z 2 cone (co (X)). If both X 0 and X 00 are positively dependent then so is X 0 [ X 00. Thus for any nite set X, there is the largest positively dependent subset X X. Re-labeling indices without loss of generality, X = fx r+1 ; ; x K g and 0 = P K k=r+1 kx k with k > 0 for every k. Write Y := XnX = fx 1 ; :::; x r g. If X =?, then write co (X ) = f0g. Lemma co (Y ) \ cone (co (X )) =?. Proof. If Y =?, the statement holds trivially, so we may as well assume that Y 6=?. Suppose that co (Y ) \ cone (co (X )) 6=?.

20 20 CHAPTER 2. SEPARATION THEOREMS AND THEIR APPLICATIONS If X =?, then 0 2 co (Y ), i.e., we can write P r k=1 t kx k = 0 where t k 0 for all k, and P r k=1 t k = 1. Since some t k is positive, some of Y s elements are positively independent, a contradiction to X =?. If X 6=?, there is z 2 co (Y ) \ cone (co (X )). Then z = P r k=1 t kx k = P K k=r+1 t kx k where t k 0 for all k, and P r k=1 t k = 1. Since 0 = P K k=r+1 kx k, we obtain for any number b, 0 = = rx t k x k k=1 rx t k x k + k=1 K X k=r+1 KX k=r+1 t k x k + b KX k=r+1 ( t k + b k ) x k. k x k Choose b large enough so that t k + b k > 0 for every k > r. Since some of t k, k = 1; ::; r, must be positive, we would nd a set of positively dependent vectors larger than X, a contradiction. Now we are ready to establish the main result, which is essentially Farkas lemma as we have argued. Theorem Let X be any nite set in R n. Then cone (co (X)) is closed. Proof. Since cone (co (X)) = cone (co (Xn f0g)), we might as well assume that 0 =2 X, and also we may decompose X into Y and X as above. Then cone (co (X)) = cone (co (Y ))+cone (co (X )) by construction (where cone (co (Y )) = f0g if Y =?). By Lemma 2.2.7, cone (co (X )) is a linear subspace, L. And S := co (Y ) is a compact convex set and S \ L =? by Lemma Then apply Lemma to complete the proof. Exercise Show that if S is compact and 0 =2 S, cone (S) is closed. (Thus in the case L = f0g in Lemma 2.2.3, the convexity of S is not essential) Remark The condition 0 =2 S is however indispensable in the exercise above: consider S = (x 1 ; x 2 ) : (x 1 1) 2 + x in R 2. Finally we discuss a variant of Farkas lemma. Note that in (1) if P M m=1 x ma m b, then we can nd a non negative vector such that P M m=1 x ma m + = b, and P M m=1 x ma m = b can be thought as a special case of = 0. Also in (2), there is no sign condition for y but if fa m g contains all the unit vectors, y a m 0 for all m e ectively impose non negativity of y. So one can write an equivalent form of Farkas lemma, which treat the two conditions more symmetrically, as follows: Theorem (Farkas lemma in inequality form) Let a 1 ; :::; a M 2 R n and b 2 R n. Then exactly one of the following properties holds: (1) there exists a non negative vector x 2 R M + ; x = (x 1 ; :::; x M ), such that P M m=1 x ma m b (2) there exists a non negative vector y 2 R n + such that y b < 0 and y a m 0 for m = 1; :::; M.

21 2.2. CONIC HULL AND FARKAS LEMMA 21 Let s rst see how Farkas lemma (Theorem 2.2.1) gives Theorem Like in Theorem 2.2.1, it is clear (1) and (2) are mutually exclusive. Let a M+j 2 R n be the unit vector whose jth element is 1 and the other elements are all zero, for j = 1; :::; n. If (2) in Theorem does not hold, then there is no vector y 2 R n such that y b < 0 and y a m 0 for m = 1; :::; M; M + 1; :::; M + n, i.e., (2) of Theorem fails for this enlarged set of vectors. So (1) of Theorem must hold, so there exists x 2 R M+n + such that P M m=1 x ma m + P n P j=1 x M+ja M+j = b. Since n j=1 x M+ja M+j 0 (2 R n ), (1) of Theorem holds. Next, let s see how Theorem can be derived directly from Theorem Suppose (2) in Theorem fails. Then for any pair of non negative vectors z and w, A (z w) 0 implies b T (z w) 0. Write A for the M n matrix whose rows are given by a T 1,:::, a T M and set A ~ = (A; A), which is a M 2n matrix, and ~ b T = b T ; b z T 2 R 2n. Then, for any non negative vector ~y = 2 R w +, A~y ~ = Az Aw = A (z w) 0 implies ~ b T ~y = b T z b T w = b T (z w) 0. So (2) of Theorem fails for A. ~ Then there exists x 2 R M + such that x T A ~ ~ b, which implies both x T A b T and x T A b T. Thus x T A = b must hold, i.e., (1) of Theorem holds. References: R. Tyrrell Rockafellar 1979 Convex Analysis Princeton University Press - this is THE classic of this topic. Leonard D. Berkovitz 2001 Convexity and Optimization in R n. - There are many variant proofs for the separation theorems. The one I provided above is based on a proof for Theorem 3.2 on page 51 of this book. Chandler Davis, Theory of positive linear dependence American Journal of Mathematics, Vol. 76, No. 4 (Oct., 1954), pp the discussion of Farkas Lemma is inspired by this paper.

22 22 CHAPTER 2. SEPARATION THEOREMS AND THEIR APPLICATIONS 2.3 Minimax Theorem and its applications The minimax theorem First we provide a simple application of the separation theorem (Theorem 2.1.5): Theorem (Theorem of alternatives) Let A be an m n matrix. Then exactly one the following conditions holds: (1) 9x 2 R m +, x 6= 0, such that x T A >> 0; (2) 9y 2 R n +, y 6= 0, such that Ay 0: Proof. First, we show that (1) and (2) cannot hold simultaneously. Indeed, if x and y are vectors for which both (1) and (2) hold, from x T A >> 0, x T Ay > 0 because y has at least one positive element. On the other hand, from Ay 0, x T Ay 0 because x is a non negative vector, leading to a contradiction. So it su ces to show that (2) holds if (1) does not hold. Consider X := x T A : x 2 R m +n f0g, which is a non empty convex set. If (1) does not hold, then X \ R n ++ =?. Then by the separation theorem, 9y 2 R n, y 6= 0, such that y X y R n ++. This inequality in particular implies y X 0 and y 2 R n + since y R n ++ contains an arbitrary small positive number and y R n ++ could be arbitrarily small if y has a negative element. Finally, if vector Ay 2 R m has a positive element, then by choosing x 2 R m + to be the unit vector for this element, we would have x T Ay > 0. But by construction x T A 2 X, a contradiction to y X 0. Hence Ay 0, and we have shown that (2) holds. Remark The vectors x and y above can be chosen to be probability vectors: i.e., x can be chosen from S 1 := x 2 R m + : P m n i=1 xi = 1 and y can be chosen from S 2 = y 2 R n + : P o n j=1 xj = 1 Let A be an m n matrix. We view A as the payo matrix for a zero sum game: its ij element a ij is the payo for player 1, and a ij is the payo for player 2, when player 1 chooses strategy i and player 2 chooses strategy j. So when player 1 chooses a mixed strategy s 1 2 S 1 and player 2 chooses a mixed strategy s 2 2 S 2, the expected payo for player 1 is P m P n i=1 j=1 si 1s j 2a ij = s T 1 As 2, and that for player 2 is s T 1 As 2. We shall consider maximization and minimization of payo s, but rst we shall argue the maximum and the minimum are well de ned in this setup. Note that since the payo is continuous on S 1 S 2 and S 2 is compact, for any s 1 2 S 1, min s2 2S 2 s T 1 As 2 exists. The compactness of S 1 then implies that max s1 2S 1 min s2 2S 2 s T 1 As 2 exists. Indeed, since min s2 2S 2 s T 1 As 2 : s 1 2 S 1 is bounded above, there is a sequence fs n 1 : n = 1; 2; :::g such that min s2 2S 2 (s n 1) T As 2 is non decreasing in n and lim n!1 min s2 2S 2 (s n 1) T As 2 = sup min s2 2S 2 s T 1 As 2 : s 1 2 S 1. The compactness of S 1 assures that fs n 1 : n = 1; 2; :::g has a limit point, say s 1, in S 1. Then by construction, min s2 (s 1 ) T As 2 min s2 (s 1 ) T As 2 must hold for any s 1 2 S 1. Similarly, min s2 2S 2 max s1 2S 1 s T 1 As 2 exists.

23 2.3. MINIMAX THEOREM AND ITS APPLICATIONS 23 De nition A strategy s 1 2 arg max s1 2S 1 min s2 2S 2 s T 1 As 2 is called a maximin strategy. A strategy s 2 2 arg min s2 2S 2 max s1 2S 1 s T 1 As 2 is called a minimax strategy. In interpretation, a maximin strategy and a minimax strategy are very conservative strategies which maximize the payo s in the worst case scenario. Equivalently, they are the best strategies for a player with the understanding that the other player chooses a strategy after his mixed strategy is chosen. Note that min s 2 2S 2 max s 1 2S 1 s T 1 As 2 max s 1 2S 1 min s 2 2S 2 s T 1 As 2 (2.1) holds immediately by construction. Indeed, for any s 1 and s 2, max s1 2S 1 s T 1 As 2 s T 1 As 2, so min s2 2S 2 max s1 2S 1 s T 1 As 2 mins2 2S 2 s T 1 As 2 for any s1, and hence min s2 2S 2 max s1 2S 1 s T 1 As 2 maxs1 2S 1 min s2 2S 2 s T 1 As 2. This inequality is intuitively plausible: lhs is the highest payo when player 1 choose a strategy after player 2 s choice, while rhs is that for player 1 moves rst. The inequality asserts that in the zero sum environment, moving rst is never a good idea, and there is some built in advantage for the follower. Theorem (minimax theorem) The following equation holds: min max s T 1 As 2 = max min s T 1 As 2 s 2 2S 2 s 1 2S 1 s 1 2S 1 s 2 2S 2 (2.2) Proof. Since (2.1) always holds, it su ces to show that min s2 2S 2 max s1 2S 1 s T 1 As 2 max s1 2S 1 min s2 2S 2 s T 1 As 2. Let J be the m n matrix whose elements are all one. Note that s T 1 Js 2 = 1 for any (s 1 ; s 2 ) 2 S 1 S 2. Consider the following subsets of R: o X := n 2 R : (s 1 ) T (A J) >> 0 for some s 1 2 S 1 Y := f 2 R : (A J) s 2 0 for some s 2 2 S 2 g Clearly, X and Y are non empty. Fix any 2 R, and apply Theorem of Alternatives (Theorem 2.3.1) to matrix (A J). Then either (1) 9s 1 2 S 1, such that s T 1 (A J) >> 0 (and so 2 X) or (2) 9s 2 2 S 2 such that (A J) s 2 0 (and so 2 Y ) must hold, but not both. So X [ Y = R and X \ Y =?. and if 2 X then any 0 belongs to X. Similarly, if 2 Y, any 0 belongs to Y. Since S 2 is compact, Y = [; 1) for some and hence X = ( 1; ). Since 2 Y, we can nd s 2 2 S 2 such that (A J) s 2 0. Then for any s 1 2 S 1, we have s T 1 (A J) s 2 0, and so s T 1 As 2. Thus, max s1 2S 1 s T 1 As 2 min s2 2S 2 max s1 2S 1 s T 1 As 2. So we are done if max s1 2S 1 min s2 2S 2 s T 1 As 2. Suppose on the contrary that ^ := max s1 2S 1 min s2 2S 2 s T 1 As 2 <. Then ^ 2 X, and so there is a strategy ^s 1 2 S 1 with (^s 1 ) T (A ^J) >> 0, and so in particular min s2 ^s T 1 As 2 > ^ = max s1 2S 1 min s2 2S 2 s T 1 As 2, which is impossible.

24 24 CHAPTER 2. SEPARATION THEOREMS AND THEIR APPLICATIONS Remark Since (2.2) holds, appeared in the proof above must be uniquely determined i.e., Y = X must hold. One can observe this directly: if we set X 0 := f 2 R : (s 1 ) T (A J) >> 0 for some s 1 2 S 1 g alternatively, then X 0 = ( 1; sup X 0 ) and Theorem of Alternatives says X 0 and Y form a partition of R. So sup X 0 = inf Y, and this is the common value we want. De nition The common value attained in (2.2) is referred to as the value of game A. Example For A = it is 0 if b 0. a 0 0 b where a b, the value of A is ab 1+b if b > 0, and Since a zero sum game is a special case of a non-cooperative game, one can analyze it from the perspective of Nash equilibrium: De nition a pair of strategies (s 1; s 2) 2 S 1 S 2 is an equilibrium if s 1 2 arg max s T 1 As 2 and s 2 2 arg min s T 1 As 2 s 1 2S 1 s 2 2S 2 hold. That is, max s1 2S 1 s T 1 As 2 s T 1 As 2 min s2 2S 2 s T 1 As 2 hold. Thus an equilibrium refers to a pair of strategies (a strategy pro le) where each player does the best against the other, when strategies are chosen simultaneously. In particular, it does not necessarily take into account the worst case. Interestingly enough, an equilibrium always consists of a maximin strategy and a minimax strategy. The existence of an equilibrium warrants that the inequality holds with equality, and vice versa. Corollary A strategy pro le (s 1; s 2) is an equilibrium if and only if s 1 is a maximin strategy and s 2 is a minimax strategy. Proof. Fix any maximin strategy s 1 and any minimax strategy s 2. Using (2.2) which holds by the minimax theorem min s T 1 As 2 = max min s T 1 As 2 = min max s T 1 As 2 = max s T 1 As s 2 2S 2 s 1 2S 1 s 2 2S 2 s 2 2S 2 s 1 2S 1 s 1 2S 1 2, so we have hence (s 1; s 2) is an equilibrium. s T 1 As 2 min s T 1 As 2 = max s T 1 As s 2 2S 2 s 1 2S 1 2; s T 1 As 2 max s T 1 As 2 = min s T 1 As 2 ; s 1 2S 1 s 2 2S 2

25 2.3. MINIMAX THEOREM AND ITS APPLICATIONS 25 Conversely, suppose (s 1; s 2) is an equilibrium. Then by de nition and (2.2), s T 1 As 2 = max s 1 2S 1 s T 1 As 2 min s 2 2S 2 max s 1 2S 1 s T 1 As 2 min s T 1 As 2 = s T 1 As s 2 2S 2 2, = max min s T 1 As 2 s 1 2S 1 s 2 2S 2 so the terms above are all equal. In particular, max s1 2S 1 min s2 2S 2 s T 1 As 2 = mins2 2S 2 s T 1 As 2 so s 1 is a maximin strategy, and min s2 2S 2 max s1 2S 1 s T 1 As 2 = maxs1 2S 1 s T 1 As 2 so s 2 is a minimax strategy. Exercise Provide an example where a pro le (s 1 ; s 2 ) attains the value of game A, but it is not necessarily an equilibrium of game A. References Von Neumann, J. Zur Theorie der Gesellschaftsspiele, Math. Annalen. 100 (1928) rst paper to state and prove the minimax theorem. Newman, D. J., Another proof of the minimax theorem, Proc. Amer. Math. Soc, proof given in this note is based on this paper Characterization of dominated strategies Let A be an m n matrix. Interpret its ij element a ij is the payo when a player chooses strategy i and a designated opponent chooses strategy j. So when the player chooses a mixed strategy s 1 2 S 1 and the opponent chooses a mixed strategy s 2 2 S 2, the expected payo for the player is P P i j si 1s j 2a ij = s T 1 As 2. De nition Strategy i is dominated if there is s 1 2 S 1 such that P j a ijs j 2 < s T 1 As 2 for any s 2 2 S 2. De nition Strategy i is never best response (NBR) if for any s 2 2 S 2, there is s 1 2 S 1 such that P j a ijs j 2 < s T 1 As 2. Clearly, if strategy i is dominated, then it is NBR: just consider s 1 which dominates strategy i in the de nition of NBR. But the converse is also true. Proposition Strategy i is dominated if and only if it is NBR. Proof. It su ces to show that if strategy i is NBR, it is dominated. So WLOG assume strategy 1 (the rst row of A) is NBR. Let A 1 be the m m matrix whose rows are all equal to the rst row of A, and let B = A A1. So for s 1 2 S 1 and s 2 2 S 2, s T 1 Bs 2 = s T 1 As 2 Pj a 1js j 2 by construction.

26 26 CHAPTER 2. SEPARATION THEOREMS AND THEIR APPLICATIONS By the minimax theorem, there is an equilibrium (s 1 ; s 2 ) of the zero sum game whose payo s are given by B. Notice that since strategy 1 is NBR, for any s 2 2 S 2, there is s 1 2 S 1 such that s T 1 Bs 2 > 0, so the value of this game must be positive, i.e., s 1 Bs 2 > 0. From the equilibrium condition, s 1 Bs 2 s 1 Bs 2 = s T 1 As 2 Pj a 1js j 2 holds for any s 2 2 S 2. Therefore, s T 1 As 2 Pj a 1js j 2 > 0 holds for any s 2 2 S 2 which shows that strategy 1 is dominated by strategy s 1. Remark Randomization over S 1 is essential. Consider 2 A = Playing the third row is never a best response, since if the rst column is chosen with probability 1, then the rst row gives a higher payo and if the second column 2 is chosen with probability 1 then the second row gives a higher payo. But the 2 third row is dominated by neither the rst row nor the second row; that is, there is no pure strategy i which dominates it. The third row is however dominated, for instance, by the strategy which chooses the rst and the second with equal probability. 3 5 Remark For 2 person games, the set S 2 corresponds to the set of mixed strategies of the other player. But for a game with 3 players or more, an element of set S 2 allows coordinated randomization of strategies by the other players. If one requires that the randomization must be independent across the players, which in e ect restricts attention to a smaller subset of S 2 and hence NBR gets weaker, a similar result does not hold: one can construct a game where a player has a NBR strategy which is not dominated. Exercise De ne the domination and NBR property for a mixed strategy ^s 2 S 1. Examine if a result analogous to Proposition holds. Pearce, D. (1984) Rationalizable Strategic Behavior and the Problem of Perfection, Econometrica 52: Existence of stationary distribution Lemma Let A = (a mn ) be an n dimensional, non-negative square matrix. Let m := P n a mn and be the diagonal matrix whose mth element is m. Then there exists an n dimensional probability vector q such that q T = q T A: Proof. Consider the zero sum game with payo matrix ( A), and let (x; y) be an equilibrium. Observe that the value of this game must be non-negative. Indeed, let m 2 arg max n (y n ) (i.e., y m is the largest probability in y) and let x be the strategy