Topics in Mathematical Economics. Atsushi Kajii Kyoto University

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1 Topics in Mathematical Economics Atsushi Kajii Kyoto University 26 June 2018

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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 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 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 and transversality theorem Generic Di erentiability of Demand

4 4 CONTENTS 4.3 Generic Regularity of a competitive equilibrium Structure of sunspot equilibria Order structure 57

5 Chapter 1 Preliminary Mathematics 1.1 Topology We shall review necessary 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. 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. q P Write the norm of a vector x 2 R L by kxk = l (xl ) 2. Obviously kxk = 0, x = 0. Observe that the norm is sublinear; for any positive scalar t, ktxk = t kxk, 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. 5

6 6 CHAPTER 1. PRELIMINARY MATHEMATICS 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. 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. 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 F 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:

7 1.1. TOPOLOGY 7 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. 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. 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. 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.

8 8 CHAPTER 1. PRELIMINARY MATHEMATICS 1.2 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. 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.

9 1.2. LINEAR ALGEBRA 9 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 V x 2 2 determines a unique linear subspace, regardless of the choice of x. The dimension of an a ne space V is de ned by the dimension of this unique linear subspace, de ned by the largest number of linearly independent vectors. 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. 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. Notice that if a convex set C R n has dimension n, then there are some 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. Therefore, C has a non-empty interior. More generally, if C R n has dimension m, C has a nonempty interior with respect to the relative topology of the m dimensional a ne space containing C.

10 10 CHAPTER 1. PRELIMINARY MATHEMATICS 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 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.

11 1.2. LINEAR ALGEBRA 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 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.

12 12 CHAPTER 1. PRELIMINARY MATHEMATICS

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 close and convex, and 0 =2 ^X. It su ces to show that there is p 6= 0 such that p ^X for some > 0. Choose a large number M such that X := fx : kxk Mg \ ^X 6=?. Since ^X is closed, X is compact. So there is p 2 ^X which achieves minfkxk : x 2 Xg. By construction, kpk = minfkxk : x 2 ^Xg. Since 0 62 ^X, p 6= 0 must hold. 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 ; 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. 13

14 14 CHAPTER 2. SEPARATION THEOREMS AND THEIR APPLICATIONS 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 for the previous result, 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 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.

15 2.1. SEPARATION THEOREMS 15 Theorem (supporting hyperplane theorem)let X be a convex set. 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. By choosing some linear translation, X can be identi ed with a subset of R m f0g where 0 2 R n m. Then X R m f0g so x 2 R m f0g. Regard X as a subset of R m and x as a point in R m. Once we nd p 2 R m with the desired property, then p = (p; 0) 2 R n is what we wanted. Thus without loss of generality, we may assume X is a convex set of dimension n, and X o is a non empty subset of R n. We have x =2 X o by de nition. Also X o is convex by Lemma 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 V around 0 so that y +V X. Then for any t 2 (0; 1), (1 t) x+t (y + V ) X since X is convex. Thus (1 t) x + ty 2 X o since tv is an open set around 0 if t > 0. 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. 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.

16 16 CHAPTER 2. SEPARATION THEOREMS AND THEIR APPLICATIONS 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, 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 dual 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. 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.

17 2.2. CONIC HULL AND FARKAS LEMMA 17 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. Observe that Z \ ( Z) = f0g, and any x 2 X can be written as x = w + z where w 2 X \ ( X) and z 2 Z, w z = 0, 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 p x = p (w + z) = p z 0 for any x 2 X. Moreover, if x 2 X n (X \ ( X)), 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. The curious question is thus whether or not if the objective function can go negative 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. Farkas lemma, which characterizes this property, is as follows: 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.

18 18 CHAPTER 2. SEPARATION THEOREMS AND THEIR APPLICATIONS So Farkas lemma says that either y = 0 is not a solution to the optimization problem (i.e., case (2)), or b can be expressed as a convex combination of constraining vectors times some positive scalar (i.e., case (1)). 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 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 n ces to show that if (1) fails, (2) holds. If (1) does not hold, b is not contained in PM o n m=1 x PM o ma m : x 0, which is clearly a convex cone. So if m=1 x ma m : x 0 is closed in addition, then by the closed n cone separation result (Theorem ), PM o there is y 2 R n with y b < 0 and y m=1 x ma m : x 0, and so we see (2) holds by appropriately choosing x 0. Thus the essence of Farkas lemma is 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. 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, we conclude t n = pzn converges to a number t 0. Hence px n y n = z n t n x n must also converge, and its limit must lie in L since L is closed. Hence z 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 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

19 2.2. CONIC HULL AND FARKAS LEMMA 19 for any t 0, x t (1; 0; 1) = (t; 1; t) =2 Y. Now consider t (1; 0; 1) + t; 1 + 1; 1 t = 0; 1 + 1t ; 1 t t 2 X + Y. Since t t t 1 t t! 0, as t! 1, we see that x belongs to the closure of X + Y. t 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 is when co fa 1 ; :::; a M g contains 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. 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. Lemma co (Y ) \ cone (co (X )) =?. Proof. Suppose not, then 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, 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) x k. k x k Choosing b large enough, 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. If X 6=?, by Lemma 2.2.7, cone (co (X )) is a linear subspace, L. If X =?, we set L = f0g. And S := co (Y ) is a compact convex set and S \ L =? by Lemma Then apply Lemma to complete the proof.

20 20 CHAPTER 2. SEPARATION THEOREMS AND THEIR APPLICATIONS 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. 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.

21 2.2. CONIC HULL AND FARKAS LEMMA 21 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 there were vectors x and y 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 since if x T Ay > 0 for some x T A 2 X then t x T Xy could be made arbitrarily large by taking t > 0 while is tx T A 2 X. Next, y cannot have a negative element or else y R n ++ could be arbitrarily small, so y 2 R n +. 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 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 Now 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 P P player 2 chooses a mixed strategy s 2 2 S 2, the expected payo for player 1 is i j si 1s j 2a ij = s T 1 As 2, and that for player 2 is s T 1 As 2. 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. De nition (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.

23 2.3. MINIMAX THEOREM AND ITS APPLICATIONS 23 Thus an equilibrium refers to a pair of strategies (a strategy pro le) where each player does the best against the other. In particular, it does not necessarily take into account the worst case. But we will see that an equilibrium always consists of a maximin strategy and a minimax strategy. Note that min s 2 2S 2 max s 1 2S 1 s T 1 As 2 max s 1 2S 1 min s T 1 As 2 s 2 2S 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. The existence of an equilibrium warrants that the inequality holds with equality, and vice versa. Lemma If there exists an equilibrium then 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) holds. Conversely, if this equation holds, then for any maximin strategy s 1 and minimax strategy s 2, (s 1; s 2) is an equilibrium. Proof. Let (s 1; s 2) 2 S 1 S 2 be any strategy pro le. Then using (2.1), max s T 1 As 2 min max s T 1 As 2 max min s T 1 As 2 s 1 2S 1 s 2 2S 2 s 1 2S 1 s 1 2S 1 s 2 2S 2 min s T 1 As 2. s 2 2S 2 So if (s 1; s 2) 2 S 1 S 2 is an equilibrium, max s1 2S 1 s T 1 As 2 = s T 1 As 2 = min s2 2S 2 s T 1 As 2 holds in addition, thus (2.2) is established. Suppose (2.2) holds. Fix any maximin strategy s 1 and any minimax strategy s 2. Then, using (2.2), min s T 1 As 2 = max s 2 2S 2 s 1 2S 1 min s T 1 As 2 = min max s T 1 As 2 = max s T 1 As 2 s 2 2S 2 s 2 2S 2 s 1 2S 1 s 1 2S 1 By de nition, max s1 2S 1 s T 1 As 2 s T 1 As 2 and s T 1 As 2 min s2 2S 2 s T 1 As 2, and 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 Theorem (minimax theorem) Equation (2.2) always holds. Thus the set of equilibria is exactly the set of all pro les of minimax strategies and maximin strategies.

24 24 CHAPTER 2. SEPARATION THEOREMS AND THEIR APPLICATIONS Proof. To establish (2.2), it su ces to show 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. Let J be the m n matrix whose elements are all one. So s T 1 Js 2 = 1 for any (s 1 ; s 2 ) 2 S 1 S 2. Consider the following subsets of R: X := n 2 R : (s 1 ) T (A o 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 has an upper bound and Y has a lower bound, so let X := sup X and Y := inf Y. It can be readily seen that X is a closed set, and moreover if 2 X then any 0 belongs to X. Thus X = (1; X ]. Similarly, Y = [ Y ; 1). Applying Theorem of Alternatives (Theorem 2.3.1) to each matrix (A J), 2 R, we conclude X [Y = R. Therefore Y X must hold, and hence there is a number 2 X \ Y. Find s 1 2 S 1 and s 2 2 S 2 such that (s 1 ) T (A J) 0 and (A J) s 2 0. Multiply these inequalities with s 2 and (s 1 ) T, and we have (s 1 ) T As 2 > for any s 2 2 S 2 and (s 1 ) T As 2 for any s 1 2 S 1. Therefore, max min s T 1 As 2 min s T 1 As 2 max s T 1 As 2 min max s T 1 As 2 ; s 1 2S 1 s 2 2S 2 s 2 2S 2 s 1 2S 1 s 2 2S 2 s 1 2S 1 as we wanted. 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. a 0 Example For A = where a b, the value of A is 0 b ab if b > 0, and 1+b it is 0 if b 0. 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.

25 2.3. MINIMAX THEOREM AND ITS APPLICATIONS 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 an 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 is NBR. Let b ij := a ij a 1j for each i and j, and let B := (b ij ) be the m n matrix consisting of these elements. 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. 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 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 (small) subset of S 2, a similar result does not hold: one can construct a game where a player has a NBR strategy which is not dominated. 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:

26 26 CHAPTER 2. SEPARATION THEOREMS AND THEIR APPLICATIONS 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 ) and let x be the strategy which chooses strategy m with probability one. Since A is non-negative, X N x T ( A) y = m y m a mn y n n=1 m y m max n = m y m max n = 0: y n y n m N X n=1 a mn! Let 1 be the vector whose elements are all 1, then by construction ( A) 1 = 0, and so x T ( A) 1 = 0. So if x T ( A) 6= 0, at least one element of x T ( A) 6= 0 must be negative, so x T ( A) y < 0 since y minimizes x T ( A) y, a contradiction. Hence x T ( A) = 0, so q = x is the desired vector. Consider a Markov process over N states. From state m, it reaches state n with probability a mn, n = 1; :::; N. So if p = ( ; p m ; ) is the current distribution over states, the probability of reaching state n is P m a mnp m, so the distribution in the next period is ( P m a mnp m ) n. A distribution p is stationary if the distribution of the next period is the same as p. Writing A = (a mn ), a N dimensional square matrix, a distribution p is stationary if p = pa holds. A stationary distribution always exists: Proposition Let A = (a mn ) be an n dimensional square matrix, where each row column is a probability vector. Then there exists an n dimensional probability vector p such that p = pa. Proof. Since P n a mn = 1 by assumption, it follows immediately from Lemma , observing m = 1 for all m Robust Voting Scheme Imagine a community of n individuals and we are interested in a scheme which aggregates the opinions of these members about some agenda for the community. Suppose that each member either supports or objects to the agenda. It is convenient to assign 1 for support, and -1 for objection. So individual i s opinion is represented by x i 2 f1; 1g. Write x 2 X := f1; 1g n, ( ; x i ; ) for a pro le of opinions. Our aggregation scheme is then a function which maps each x 2 X to f1; 1g. An important class of aggregation schemes consists of majority rules: a rule is called a weighted majority rule if there are positive weights w = ( ; w i ; ) such that (x) ( P i w ix i ) = P i w i ( (x) x i ) 0 for any x. That is, it has to choose 1 if w x > 0, i.e., the weighted sum of opinions favors it, and choose 1 if w x < 0, i.e., the weighted sum of opinions is against it, and it may choose anything when w x = 0.

27 2.3. MINIMAX THEOREM AND ITS APPLICATIONS 27 Clearly, it is sensible require that (1; :::; 1) = 1 and ( 1; :::; 1) = 1. Then obviously for any such rule, at any pro le x, there is somebody whose opinion goes through, i.e., (x) = x i. So let s look at the rule from a particular individual i s perspective. He wishes that his opinion goes through but clearly it will not be always the case. For instance, when all but him are against the agenda, it is rather sensible if the rule decides against it. Then will his opinion go through on average? To facilitate this question, imagine that before the content of the agenda is exhibited, the likelihood of opinions is expressed by a probability distribution p on X. Note that from individual i s point of view, his opinion is the same as the community if (x) x i = 1, and it is di erent if (x) x i = 1. Thus, i s opinion goes through on average, i.e., the community opinion tends to agree with individual i s opinion ex ante, if E [ (x) x i ] 0 holds. The plausibility depends on the shape of p, obviously. For instance, if p might assign probability one to the pro le where all but i is against the agenda, then the story is the same as above. If p assigns probability one to some pro le, then obviously there is somebody whose opinion goes through. But what about general p? It seems to sensible to require that there exists an individual i whose opinion goes through on average, whatever p might be. De nition Rule is strongly robust to uncertainty if for any p 2 4 (X), there exists individual i such that E [ (x) x i ] > 0 holds. It is robust if for for any p 2 4 (X), there exists individual i such that E [ (x) x i ] 0 holds. As we argued above if we only look at the deterministic cases (p assigns probability one to some pro le), such an individual always exists for any rule, so it might rst appear that there should be many robust rules. For instance, a weighted majority rule is robust. To see this, rst note that by de nition there are w such that (x) (w x) 0 at any x. Thus for any p, 0 E p [ (x) (w x)] = X i w i E p [x i (x)] and so there must be at least one individual i for whom E p [x i (x)] 0. The same idea shows that there is an individual i such that E p [x i (x)] > 0 for any p which assigns probability zero to x with w x = 0. For instance, if the number of individuals is odd, and w i s are the same (i.e., simple majority) then there cannot be a tie and so any p assigns probability zero to w x = 0. So in this case the rule is strongly robust. It turns out that the weighted majority rules are exactly those rules which are robust. Proposition A rule is robust to uncertainty if and only if it is a weighted majority rule.