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Some applications of Choquet’s integral representation theorem to probability Padayachee, Krishna 1981

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SOME APPLICATIONS OF CHOQUET'S INTEGRAL REPRESENTATION THEOREM TO PROBABILITY by • KRISHNA PADAYACHEE B.Sc.(Hons.) Brock University, 1978 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE in THE FACULTY OF GRADUATE STUDIES (Department of Mathematics) We accept this thesis as conforming to the required standard THE UNIVERSITY OF BRITISH COLUMBIA November 1981 © Krishna Padayachee, 1981 I n p r e s e n t i n g t h i s t h e s i s i n p a r t i a l f u l f i l m e n t o f t h e r e q u i r e m e n t s f o r an advanced degree a t t h e U n i v e r s i t y o f B r i t i s h C o l u m b i a , I a g r e e t h a t t h e L i b r a r y s h a l l make i t f r e e l y a v a i l a b l e f o r r e f e r e n c e and s t u d y . I f u r t h e r agree t h a t p e r m i s s i o n f o r e x t e n s i v e c o p y i n g o f t h i s t h e s i s f o r s c h o l a r l y p u r p o s e s may be g r a n t e d by t h e head o f my department o r by h i s o r h e r r e p r e s e n t a t i v e s . I t i s u n d e r s t o o d t h a t c o p y i n g o r p u b l i c a t i o n o f t h i s t h e s i s f o r f i n a n c i a l g a i n s h a l l n o t be a l l o w e d w i t h o u t my w r i t t e n p e r m i s s i o n . Department of ^ &AV> g-J=-i c s The U n i v e r s i t y o f B r i t i s h C o l u m b i a 2075 Wesbrook P l a c e V ancouver, Canada V6T 1W5 Date - 3 / 0^ W^ ^ $ 1 DE-6 (2/79) - i i -ABSTRACT In this thesis we look at the applications of Choquet's integral representation to probability theory. Applications of Choquet's theorem are given to obtain a representation of superharmonic functions on the Martin Boundary, a representation theorem for invariant measures with respect to a family of transformations T and fi n a l l y to symmetric measures on a product space. In order to obtain the desired representation theorem in the above mentioned/applications we need to consider an appropriate topology on the spaces. In the case of the Martin boundary our underlying space is R°° equipped with the product topology. The set of a l l superharmonic functions is shown to be a compact convex metrizable subset of R°°. Furthermore the extreme points are isolated and they turn out to be the minimal harmonic functions. With regards to the other two applications we consider the space of measures on an appropriate topological space. The"probability measures invariant with respect to a family of transformations T form a Compact convex set in the weak-star topology and the extreme points are the ergodic measures. In the case of the symmetric measures on the product space the symmetric probability measures form a compact convex set in the weak-star topology and the extreme points are the product probability measures. - i i i -TABLE OF CONTENTS Page ABSTRACT i i TABLE OF CONTENTS H i ACKNOWLEGEMENT. I ' i v INTRODUCTION 1 CHAPTER 1 6 The Martin Boundary and the representation of superharmonic functions on the Martin Boundary CHAPTER 2. 25 Application of Choquet's theorem to invariant and ergodic measures with respect to a family of trans-formations T' CHAPTER 3 34 Part I Symmetric Measures on a Product Space 34 Part I I Exchangeable processes need not be mixtures 48 of independent and i d e n t i c a l l y distributed random variables. BIBLIOGRAPHY 58 - i v -ACKNOWLEGEMENT I would l i k e to thank P ro fe s so r Ghoussoub f o r suggest ing t h i s t o p i c as a t h e s i s sub jec t and f o r p r o v i d i n g me w i t h adv ice i n the p r epa r a t i on of t h i s t h e s i s . I would a l so l i k e to extend my thanks to P ro fe s so r Pe rk in s who prov ided a great dea l of e n l i g h t e n i n g c r i t i c i s m which helped to c l a r i f y the contents of some of the p roo f s . INTRODUCTION The aim of this thesis is to apply Choquet's integral representa-tion theorem to some areas of probability- We consider the following applications: i . The Martin Boundary provides for the represent-ation of superharmonic functions as integrals of "Martin Kernels". Here the Martin-Doob-Hunt representation is obtained via Choquet's theorem. i i . The set of a l l probability measures invariant with respect to a family of measurable trans-formations T. i i i . The set of a l l symmetric measures on a product space. In the sequel we compare the theorems of Choquet and Krein-Milman and make clear how the Choquet theorem generalizes the Krein-Milman theorem. Fir s t we need some definitions: ( 0 . 1 ) Definition: Suppose X is a locally convex space (l.c.s.) and K c X a compact convex subset and that u is a probability measure on K (i.e. a non-negative regular Borel measure with u(K) = 1 ) . A point xeX is called the barycentre of u (or is represented by u) i f f(x) = / f d u V continuous linear function f on K. ( 0 . 2 ) Definition: If p is a non-negative regular Borel measure on the compact Hausdorff space K and S is a Borel subset of K we say u is supported by S i f y(K\S) = 0 . We may now consider the following questions: If K is a compact convex subset of a l.c.s. X, and xeK, does there exist a probablity measure u on K supported by the extreme points of K which has x as i t s barycentre? If y exists is i t unique? - 2 -Under the hypothesis that X is metrizable Choquet has shown that the answer to the f i r s t question is yes. A positive answer to the second question depends on a geometrical property of K. The following proposition gives us a characterization of the closed convex hull of a compact set in terms of measures and their bary centres. The proposition also allows us to reformulate the Krein-Milman theorem as an integral representation theorem. (0.3) Proposition: Suppose that Y is a compact subset of a l.c.s. X. A point xeX is in the closed convex hull Z of Y i f f . 3 a probability measure u on Y which is the barycentre of x. Proof: If u is a probability measure on Y which represents x, then for each f in X* (dual of X), f(x) = u(f) < sup f(Y) < sup f(Z). Since Z is closed and convex, i t follows that x £ Z (by the Hahn-Banach separation theorem). Conversely i f xeZ, there exists a net in the convex hull of Y which converges to x. Equivalently 3 points y a of the form n a . a a , , 0 : . - v - » a i a  y a = i=l X i x i ( X i > °» E X i = L' x i £ Y and a in some directed set.) which converge to x. We may represent each y a by the probability OL measure u a = £ \^ e a ( e a Dirac measure) i i By the Riesz theorem and the Banach-Alaoglu theorem the set of prob-abi l i t y measures on Y may be identified with a w*-compact convex subset of C(Y)*, and hence there exists a subset (ug) of (u a) converging - 3 -(in the weak* topology of C(Y)*) to a probability measure u on Y. In particular, each f in X* is (when restricted to Y) in C(Y), so 11m f ( y g ) - lim /fdy g = / f d u . Since y a converges to x, so does the subnet y^, and hence f(x) = / yfdy forVfeX*, which completes the proof. £ The above proposition makes i t easy to reformulate the Krein-Milman theorem. Recall the statement: If K Is a compact convex subset of a l.c.s., then K is the closed convex hull of its extreme points. (0.4) The reformulation is the following: Every point of a compact convex subset K of a l.c.s. is the barycentre of a probability measure on K which is supported by the closure of the extreme points of K (0.5) An easy use of proposition (0.3) shows the equivalence of these two assertions. Now i t is clear that any representation by means of measures supported by the extreme points (rather than their closures) is a sharpening of the Krein-Milman theorem. We now discuss some preliminaries which w i l l lead up to the ver-sion of Choquet's theorem that w i l l be used. Suppose Ks X (l.c.s.) and K is compact convex, the question of the uniqueness of the representing measure is most naturally studied when K is the base of a convex cone C, with vertex at the origin. This entails assuming that K is contained in a closed hyperplane missing the origin. We embed X as the hyperplane X x {l} i n X x R (with the product - 4 -topology. Thus K is mapped to K x {l} which is affinely homeomorphic to K, (recall K is convex). When K is contained in a hyperplane which misses the origin we may always define a convex cone (with vertex at the origin) for which K is a base. Take C = K where K = {ax|a > 0 , x e K} is the cone generated by K. A cone C is proper i f CH(-C) = { o } . Certainly Kfl (-K) = { o } . Since K - K is a vector space and K is a proper pointed cone we have that there exists a unique partial order on K - K making i t into an ordered vector space for which K is the positive cone, viz., x ^ y i f f . x-y £ K (the proof is a straightforward check -reference Choquet Vol. 1 Ch. 1 0 , p. 171 . } ( 0 . 6 ) Definition: If a compact convex set K is a base of a cone K we c a l l K a simplex i f f the space K - K is a lattice in the ordering induced by K. We note K - K is a vector lattice i f f . K is a lattice. Proof - (Phelps Sec. 9 , p. 6 0 ) . We now state Choquet's theorem, the proof of which may be found in Phelps or Choquet, Lectures on Analysis, Volume II. ( 0 . 7 ) Theorem - Suppose K is a compact convex subset of a locally convex space X. Furtheremore i f K is metrizable or the extreme points of K (ext(K)) i s closed in X thanVx Q £ K 3 a regular Borel measure u representing x Q. If K is a simplex the representing measure u is unique. ( 0 . 8 ) We w i l l also have occasion to use the following: - 5 -let X be a t.v.s. then X* with the weak-star topology is a l.c.s. (Rudin,Theorem 3.10) (0.9) Milman's "converse" to the Krein-Milman theorem. Suppose that K is a compact convex subset of a locally convex space and Z £ K and further K = Co(Z). Then ext(K) c cl(Z). - 6 -CHAPTER 1 The Martin Boundary and the representation of superharmonic  functions on the Martin Boundary i n the Markov Chain case Before we begin with the representation of superharmonic func-tions on the Martin Boundary we give some preliminaries. Consider the discrete parameter stochastic process (", F, Fn,Xn,Pr). Here (ft.F.Pr) i s probability t r i p l e and (F n) i s an increasing sequence of o-algebras contained i n the a-algebra •V n»Xn:fi + S i s F n - measurable. Here S consists of a count-able number of elements with each element being measurable. We say (^,F,Fn,Xn,Pr) i s a Markov Chain i f P r [ X n + l = Jn+l| x 0 = J 0 , - - - , x n = jJ = p r [ x n + l = jn+l|xn = j n ] I f further Pr[xn+1 = jn+i|xn = j n ] = Pr[xi=jn+i|x0 = j n ] process i s called a time homogeneous Markov Chain. We w i l l only be concerned with time homogeneous Markov Chains. Henceforth we w i l l denote the Markov Chain by ( X n ) . Let P be the t r a n s i t i o n probability matrix: i.e. P = (P^j) where p = P r[Xi. = j | x 0 = i ] , V i , j e s . We assume that P i s substochastic i.e. PI <_ 1. Here we are supposing the existence of a c o f f i n state A appended to S such that P J A = 1 - £ P. . (V ieS). i A jeS ! J A Markov Chain i s said to be transient i f 7r[T± = +°° |x0 = i ] > 0 V i where T^  i s the f i r s t time the chain h i t s i . In the ensuing - 7 -discussion we w i l l be dealing with transient Markov Chains. +°° n Let G = £ P , G is called the Green's kernel. n=l The probabilistic interpretation of G is as folows: G^j is the expected number of times the Markov Chain starting from i is in j . Since (X n) is transient G = [ g ^ ] < + °°. Let TT = [ir(i)j be the i n i t i a l distribution i.e. ir(i) = P r [ x 0 = l ] . We choose IT so that TTG > 0. This assumption w i l l be used in obtaining the desired representa-tion theorem for superharmonic functions defined on the state space S of the Markov Chain (X n). To summarize ( f i ,F,F n,X n,Pr) is a transient Markov Chain with substochastic transition matrix P. G is the Green's kernel such that G < + 0 0 and IT an i n i t i a l distribution such that TTG > 0 . Martin Boundary Theory for Markov Chains 1. Introduction To motivate the introduction of the Martin Boundary for Markov Chains with only transient states, we consider an open unit disc of 2-dimensional Euclidean space. 2 1 In R the boundary of the disc i.e. the circle S has the property that there is a 1-1 correspondence between the non-negative harmonic functions h(re^®) on the disc and the non-negative Borel measures y on the cir c l e . The correspondence is h(re i e) = / i P ( r e i 6 , t ) y h ( t ) (1.1) where P(re i 0,t) is the Poisson kernel, 1 - r 2 l-2rcos(6-t) + r 2 The purpose of the transient Markov Chain boundary theory is to seek an analogous representation of non-negative harmonic functions defined on the state space of the Markov Chain. Now in the case of the disc in R , a calculation using Green's identities shows that any kernel P(re l 9,t) giving rise to the J . 2u correspondence and satisfying / P(re l 9,t)dt = 1, must be the 0 normal derivative at t of the Green's function for the disc relative to the point re* 9. i6 9 10 That is P(re ,t) = [g^G(«,re ) ] t (The Green's function G is defined as follows i8 16 1 G(z,re ) = H(z,re ) + log -j z - r e 1 9 where the function H satisfies and H(z,rei 6) = l o g | z - r e i e | on S 1) An application of l'Hopital's rule yields lim G( Z,re 1 6) _ r 3 . * 8w3 r,. P N 1 z+t G(z,p) " fe G ( ' r e > f e G ( * ' P ) t ! = P(re i 6,t)/P(p,t). - 9 -Here p is the fixed reference point in the unit disc. Hence except for a positive factor P(p,t) which depends on t but not on r e l ^ , the Poisson kernel is equal to lim G(z,re 1 6) z+t G(z,p) (1.2) Therefore the above function may be used in place of the Poisson kernel, the distinction between the kernels is just the normalizing factor (depending on t) which may be absorbed by changing the measures. Now the above considerations apply equally well to any domain i n n-dimensional space with sufficiently smooth boundary. Although the explicit form of the kernel w i l l vary from region to region i t w i l l always be connected to the Green's function in the same way we have described above. R.S. Martin (1941) made use of these observations to describe an ideal boundary for an arbitrary domain in Euclidean space. If the Green's function for the region is denoted G(z,y) he noted that points t on the boundary of the region did not necessarily have the property that l i - G i i j r i e x i s t s . z+t G(z,p) He suggested that distinct ideal boundary points u should be associated to subsequences {zn} which yield distinct values for the limits lim G(z n,y) z n+t G(z n, P) K C y ' u ) He went on to show that the desired representation theorem i s indeed obtained in terms of this boundary and the kernel K(y,u). - 10 -Doob (1959) taking advantage of the fact that the G matrix for a transient Markov Chain i s the analog of the Green's function showed that Martin's approach could be used to obtain a boundary for Markov Chains. This enables us to obtain a representation of non-negative superharmonic functions defined on the state space of a Markov Chain. As the analog of Martin's kernel he used l i m i t s on j of expressions of the form G^j/G 0j, assuming G Qj > 0 V j . In this respect we s h a l l not follow him, we simply use l i m i t s of G i i / ""i G±i when n i s a probability vector such that IT G > 0, J ies Now the introduction of TT i n place of 0 i t s e l f leads to a problem. The representation w i l l have to be re s t r i c t e d to tr-integrable superharmonic functions. We now give a brief sketch of the Martin-Doob-Hunt theory to which Choquet's theory w i l l be applied. F i r s t we require the following:-D e f i n i t i o n A function f or S (the state space of (X n)) i s P-superharmonic i f Pf <_ f i.e. Vf± < f± V ieS P-harmonic i f Pf = f We say f i s a (pure) potential i f f i s superharmonic and +00 n P f = lim P f = 0 n A l l measures p on S and a l l functions on S are f i n i t e and non-negative. The value of the function f at the state ieS i s denoted by or f ( i ) P f i = E P i j f j a n d u i P = 1 P j i > jes jes - 11 -I f f and g are column vectors than f _< g i f f . fj. <. gi V ieS Sim i l a r l y for matrices over the same index set we have A ;< B i f f . A ± j < B j j V i , jeS The Riesz decomposition of a non-negative superharmonic function f i s given by f = Gc + P+°°f where c i s the charge of f and P+°°f i s the harmonic function. The above decomposition i s unique i.e. Gc and P + 0 0f are unique (See Kemeny, Snell and Knapp). Furthermore since we are dealing with a transient Markov Chain we have G < +00 and thus i t i s easy to see that the charge c associated to f i s unique. Define for arbitrary jeS, K(»,j) = g../g X J TT j where g = Z ir.g. .(> 0) ( TTG > 0) 11 j ies 1 Note K(*,j) = Gc"' where c? = SJ ./g i i j 5 T T j . r _ r l i f i=j where o± A - [ N , ^ X>J l0 elsewhere Thus v J £S K.(»,j) i s a potential with a point charge . The Martin-Doob-Hunt Theory The details of the following results may be found i n Kemeny, Snell and Knapp. a. We may define a f i n i t e metric d2 on S such that a sequence of states { j n } i s d2-Cauchy i f f . VieS the sequence {K(i,J n)} i s a Cauchy sequence of real numbers. The metric d2 i s defined as follows: - 1 2 -where I w.g < +°° (w^ positive r e a l s ) , jes J * j J Let S* be the completion of S under the metric dz- Here we note that the characterization of Cauchy sequences i n S given above shows that the nature of the space S* does not depend on the choice of the weights (w^). S* turns out to be a compact metric space and S i s a dense subset of S*. The Martin Boundary i s the set B = S*\S. Note that the d e f i n i t i o n of S* does depend on the starting d i s t r i b u t i o n IT. For different starting distributions we may obtain a different Martin Boundary. (c). V d^-Cauchy sequence { j n l S, such that j -*• x eB, l e t K(»,x) be the function defined by K(i,x) = lim K ( i , j ) V ieS. n ti K(«,x) exists by the d e f i n i t i o n of d^• The above d e f i n i t i o n of the Martin boundary apart from enabling us to obtain a representation theorem for superharmonic functions on S also gives information about the long range behaviour of the Markov Chain. We state the following theorem, the proof i s found i n Kemeney, Snell and Knapp, (p.339). (1.4) Theorem: Let (X n) be a transient Markov Chain with i n i t i a l d i s t r i b u t i o n such that TT G > 0. For each u> e ft l e t v(o)) be the supremum of the n such that Xn((o) e S. Then a.e. either v(o>) < +°° and X(a>) e S or v(co) = +°° and Xn(u>) v(u)) converges to a point X(w) e S* as n •*- +00. v(a>) - 13 -(1.5) Theorem To each superharmonic function f on S with irf < +°° a unique Borel measure u f on S = S U B e c s * where B e = ext(S*\S) such that f ( i ) = J K(i,x) dy f(x) (ieS) with S yf(S) = irf, and this representation corresponds to the Riesz decomposition f ( i ) = u ( i ) + r ( i ) on S where the integral over S yields the potential. u ( i ) = S g c ( j ) (ieS) with charge c(j) = y f ( j ) / g w (ieS) and the integral over B e yields the harmonic function: r ± = J K(i,x) dyf(x) (ieS) I f irf = l , y f i s a Borel probability measure. Our aim i s to obtain the above representation using Choquet's integral representation theorem. We now develop the machinery which w i l l enable us to obtain the Doob-Martin-Hunt representation via the Chocjuet integral representation theorem. Recall we are dealing with a transient Markov Chain (X n) taking values i n the countable state space S, with probability t r a n s i t i o n matrix P, Green's kernel G and i n i t i a l d i s t r i b u t i o n TT such that TT G > 0. We consider as our underlying space the space R which i n the product topology i s a l o c a l l y convex metrizable space. Let K = {f e R+ |pf < f and irf < l } . K C R and we show that K i s a compact convex metrizable subset of RS. - 14 -Furthermore consider K = {f eR+ Pf < f and Trf < -H»} K i s the cone generated by the base K. We show that ^ i s a l a t t i c e i n the cone order and therefore ^ i s a simplex. Lastly we isolate the extreme points of K. S Since the space R i s equipped with the product topology (topology of pointwise convergence) we have that the projection maps r^(f) = f ( i ) are continuous linear functionals. We have a sequence {f n} i n S S R converges to a point f i n R i f f . F i ( f n ) •»• T i(f) (V ieS). (1.6) Proposition: The set K i s a compact convex metrizable subset of RS. Proof: K. = {feR S Pf _< f and iff <_ l } . Clearly K i s convex. Since S S ~*~ K c R and R i s metrizable, K i s metrizable. Thus we need only show that K i s compact. Suppose { f n } c K and f n •»• f e R^ . ( i . e . r ± ( f n ) + r ± ( f ) V ieS). Then by Fatou's lemma and the fact that each f n i n superharmonic we have Pf = P(lim inf f n ) < lim inf P f n < lim inf f n = f. Also irf <^  lim inf ivf^- _< 1 (Fatou's lemma). So f e K. Therefore K i s closed. I f we show that K i s contained i n a s compact subset of R then the proof w i l l be complete. I t i s here that we use the assumption that irG > 0. So i f we show that V keS 3 M k > 0 such that irf <^  1 implies f(k)< M (VkeS) then K c II [0,Mk] k keS which is a compact subset of RJ (Tychonov's theorem). Since nG > 0 we have Tf k G = Z T ^ g ^ > 0 V keS. i e S Thus V keS ak'eS such that Tr k», g^, ^ > 0 and since [g, - - G = E rtP,3m>0 such that P . , . > 0. L i j J n=0 y k ,k m Now i f f is superharmonic, then P f < f so P k'ik < jig P k \ j | f ( J > < f(k') d.7) and i f irf <_ 1 then in particular T r k t f ( k ' ) _< 1 (1.8) Using (1.7) and (1.8) above we need only choose M = [ i r k , P^jJ 1« ThusVfeK we have f(k) e [0,Mk] Vk. Therefore, since K i s closed, K is compact. • Now we examine the lattice structure of K, the cone generated by K. We prove that K is a lattice in the cone order. The cone order is defined as follows: V f,ge K f « g i f f g-f e K. The key to proving that K is a lattice is provided by the Riesz decompo-sition of a superharmonic function. We recall that i f f is a super-harmonic function then f = Gc(f) + P+°°f where c(f) is the unique charge of f and P + 0 Of is the harmonic part of f. (1.9) Lemma: Let f,g e k. Then f < g i f f . c(f) < c(g) and P+°°f < P+^g i.e. c ( f ± ) < c ( g i ) and P+^fi < P+^gi V i£S. Proof: Suppose c(f) _< c(g) and P+°°f _< P+°°g - 16 -Note: P(P+°°g - P+^g) = P P+^g - P P+°°f = P+°°g - P+°°f C".* P+°°g and P + c o f are harmonic). So P+0°g - F^^f is harmonic and is in K.. P(g-f) = Pg - Pf = P(Gc(g) + P+°°g) - P(Gc(f) + P+°°f) = Gc(g) - Gc(f) + P+°°g - P + 0 0f - l(c(g) - c(f)) < Gc(g) + P+°°g - (Gc(f) + P+°°f) = g-f /. g-f £ K. Conversely: Suppose g-f e K. Since the Riesz decomposition is unique we have 0 < P+^g-f) = P+°°g - P+^f and G(c(g-f)) = Gc(g) - Gc(f). By the uniqueness of the charge we have c(g-f) = c(g) - c ( f ) . (1.10) Proposition: ? f , g e K we have fAg = G(c(f) A c(g)) + P+»(P+»f A P+°°g) (1.11) Here c(f) A c(g)(i) = c ( f ± ) A c(g±) and (P + 0 0f A P + 0 0g) i P+^fiA P*"^-Proof; The r.h.s. of (1.11) makes sense since i t is clear that c(f) A c(g) is a charge and (P+°°f A P+°°g) is a superharmonic function (since both P ^ f and P + 0 0g are harmonic). Let * = G(c(f) A c(g)) + P + 0 o(P + 0 0f AP + 0 0g) Note: <)) e K, c(f) A c(g) < c(g) and P+°°f A P+°°g < P+^g. Hence by Lemma (1.9) i t follows that <(> < g. Similarly <p <SC f so (f is a lower bound in K for f and g. Let i|> be any other K-lower bound for f and g. - 17 -It follows easily that c(i|0 < c(f) A c(g) and P + 0> < P+"f A P + 0 0g .*. * « +. Thus (j» = f A g. • The above proposition implies that K is a lattice in the cone order so we have that ^ is a simplex. We isolate now the extreme points of the set K. Recall K = {feR^/Pf < f and irf < 1} . Definition: A non-negative superharmonic function f is said to be minimal i f for any non-negative superharmonic function g such that f-g is non-negative superharmonic we have that g = af for 0 _< a <_ 1. Recall that K(« j) = Gc^ where c^ = g 1 6_, . . ' J i 6 T f j i j Thus V jeS, K(*,j) is a potential with a point charge denote the extreme points of K by ext(K). ( 1 .12 ) Proposition: A function f e ext(K)\{u} i f f . f is minimal and irf = 1. Proof: Suppose fe ext(K)\{u}. Now i f fe ext (K)\{o} we must have irf = 1 since i f irf < 1 , TTf(|jO + (1 " ' f ) (0) = f . * Suppose that we have a non-negative superharmonic function g such that f-g is superharmonic. If f is not minimal we have that - 18 -But (1.13) contradicts the fact that f is extreme so f must be minimal. Conversely suppose f is minimal and irf = 1. Suppose f = y g + y h g,h e K. (1.14) Since f was assumed to be minimal | g - o f , y h = 3f 0 < a; 3 < 1. By (1.6) f = j g + j h = (a+3)f so ct+3 = 1. \ 2 \ ^8 = a i r f = a a n d y > y ^h = 3 (g,h e K) But a+8 = 1, .". a = 8 - y h = g = f. So f is extreme. H Proposition; If f is minimal then f is either a potential or is harmonic. Proof: By the Riesz decomposition we have f = Gc + h and since Gc, h are superharmonic, f-Gc and f-h are superharmonic and we have by the definition of minimality Gc = ctf, h = 8f. Now i f a 3 * 0 we have Gc = h/B. But this cannot be by the uniqueness of the Riesz a decomposition, so we have that either a or 3 = 0 . Thus f is either harmonic or a potential. Proposition (1.15): Furthermore a non-zero potential is an extreme point of K i f f i t is of the form K(»,j) for some j . Proof: Suppose that Gc(c > 0) is an extreme point of K (Gc \ 0) - 19 -Gc = [ Z g. .c.]. _ jeS J J Note C l 8 F F I K(.,l) = ^ g . ^ (K(.,j) = and Gc - c]_g K(*,l) i s superharmonic because p[Gc-c l 8 i r iK(.,D] = P[Gc-c l 8 i r i Gc J] = ( G - I ) c - c l g i i i ( G - I ) c ] = [ G c - c i g n G c J ] - l [ c + c l g n c J ] < Gc-c l 8 i r iK(.,l) Since Gc i s minimal and TTGC = 1 (v Gc i s assumed to be extreme) C l g K(.,l) = oGc 0 < a < 1 implies Gc = ^  8 K(.,l) ( u * ) c l c l 1 = TTGC = — 8 irK(-,l) = — 8 . Therefore g^ = a/^. So i n (1.16), Gc = K(«,l). Conversely consider K(»,j) V j = 1,2,... Suppose there exists a superharmonic f such that K(«,j)-f i s superharmonic. Let f = Gc + h; c > 0 (charge); h i s harmonic. Since K(*,j) i s a f i n i t e potential and h i s bounded above by K(«,j), we claim h = 0. [Proof: Let Gg be a f i n i t e potential and h <^  Gg h > 0 and harmonic with respect to P. n n +°° k h = P h < P G g = ( Z P )g ->- 0 — k=n ( S t r i c t l y decreasing sequence of f i n i t e functions bounded below by 0), - 20 -So f = Gc and K(«,j) - Gc > 0 = GCc^-c) > 0 So c -c > 0 i.e. c.-c(i) > 0 i . e . — - c ( i ) > O V i V 3 implies c i s a const, multiple of c"^ . Thus from the above i t follows that ext(K)\{o} = set of a l l minimal potentials K(*,j); jeS and minimal harmonic functions h such that irh = 1. Let P = JK(» j)|jes} and l e t H be the set of a l l minimal m I m harmonic functions such that trh = 1 The Martin Boundary v i a Choquet's Theorem Recall that the non-zero members of ex( K) are the minimal potentials K(»,j); jeS which are called P and the minimal harmonic m functions H . m Since P C K, the set P* = c l . ( P ) (closure of P i n K) i s a compact set. The mapping S > P* defined by j •*• K(«,j) V jeS, i d e n t i f i e s S m X with the dense subset P^ of P^. (The mapping i s 1-1 by the uniqueness of charge). We define the Martin Boundary to be the set B = P* \ P and m m and late r show that this d e f i n i t i o n of the Martin boundary coincides within that given by the Martin-Doob-Hunt theory. With this d e f i n i t i o n we have to show tf^C P*. This i s a corollary to the next proposition. - 21 -1.17 Proposition: K = co(P {J {o}) m Proof: Let sets J denote subsets of S and l e t A^-^ denote the j t h . column (jeS) of the matrix A over S. The proof i s i n two parts: we f i r s t show ueK, u a potential, implies ue co (P U {0}, and then we show that every superharmonic c function m i s the l i m i t of an increasing net of potentials. I f ueK i s a potential with charge c, then u = Gc = Z G ( j ) C i = Z K(«, j ) g 7 r C i , jes jes j and 1 > TTU = Z irK( • i)e c.= Z g c J£s J jes J Set a-! = g^ c^ (jeS) then Z a* < 1 V j . J j J jeJ J ~ So U = Z a.K(» . j) + (1 - S a.)«0 i s i n co(P U {o} ( V J C S; J J jeJ 3 jeJ 2 f i n i t e ) Hence u = lim Uj e Co(P U {o}) J m Now, l e t f be a superharmonic function. Note f A u i s a potential, for any potential u (f A u < u, .*. P + 0 0(fAu) < P + 0 0u = 0). Let | j | be the car d i n a l i t y of the f i n i t e set J. Since g. . > 1 . V jeS (G = z"^ P n) J J n=0 we have K ( j , j ) > 0 V jeS. The function " J Z K(«,j) i s a potential V J c S (J f i n i t e ) jeS .'. {f A. <I>T : J c s} i s an increasing sequence of potentials such that - 22 -f A * ~^ f (weakly) since • j ( j ) becomes unbounded for every jes. F i n a l l y i t follows from Milman's "converse" to the Krein-Milman theorem that e x ( K ) c c l (P<J{o}) But the P U H = ex(K)\{o} C c l ( P ) = P*. m m m m (1.18) Theorem: To each superharmonic function f on S with irf < +°° there corresponds a unique Borel measure yf on P* with support m P U tf such that m m f ( i ) = / gidu f(g) (ieS) (1.19) Pm U H m with yf(P U H ) = irf and this representation corresponds to m m the Riesz decomposition f ( i ) = y ( i ) + r ( i ) on S where the integral over P m yields the potential u(i) = £ g l 1 C 1 (i£S) (1.20) jeS J J c-j = yf(j)/g„ (ieS) (1.21) j where y f ( j ) = y f ( K ( . , j ) ) , and the integral over ^ yields the harmonic function r ( i ) = / gidpf(g) (ieS) (1.22) H m Proof By Choquet's theorem V superharmonic f i n i t e feK 3 a unique Borel measure yf on K with support i n ext(K) such that L ( f ) = ' tVs L(8)dy f(g) = / , ; ,L(g)dy f(g) ex(K) ex(K)\tO} S for every continuous linear functional L on R . In par t i c u l a r , g since the projections on R are continuous linear functions and since a ex(K)\{o} = P U H , m m - 23 -f ( i ) = r (f) = J p u r ( g ) d y f ( g ) = / P M H g,du f(g) 1 r i u "in 1 rm nm 1 To show that the representation holds for any superharmonic f with Trf<+°° note that <(> = f/irfeK (may assume irf >^  0 i f irf = 0 then f = 0 result t r i v i a l . ) Thus •(!) = Jp u H g d y*(g) = ^ mm- 1-Setting yf = T T f y ' P defines a representing measure for f which i s unique since i f v also represents f, the v / i r f represents <|> so v = TTf. = yf Let {J n} be a s t r i c t l y increasing sequence of f i n i t e subsets of S Then i r f = lim Z rr.r.(f) = lim Z n./p 0 „ g i d y f ( g ) n j£>Jn ^  1 n J e J n ^ m m = /P mU H m d ^ f ( g ) = /p m*J n V ^ C g ) (Monotone Convergence Theorem). = y f ( P U ' tf ) m m Thus (1.19) i s proved. Let f ( i ) = u ( i ) + r ( i ) = Z gi-ic(j) + r ( i ) be the Riesz decompo-ses s i t i o n of f on S. The function v ( i ) = J p gdy f(g) = Z K ( i , j ) y f ( j ) = Z g, jfy^-rm jeS jeS X J gffj i s a potential, and the function h ( i ) = / g(i)dyf(g) i s harmonic since Ph. = lim Z j P i j g ( j ) d y f ( g ) = / Pg.dy f(g) 1 n J £ J n Hm «m 1 = L g±dyf(g) = ^  V i e s Hm - 24 -By the uniqueness of the Riesz decomposition u(i) = v ( i ) and r ( i ) = h ( i ) V ieS. By the uniqueness of charge we have c(j) = y f(j)|g_ . This proves (1.20), (1.21) and (1.22). • 1 11 j We conclude this section by showing that the Martin boundary as described by the Martin-Doob-Hunt theory i s equivalent to the d e f i n i t i o n that was given using the Choquet theory. Recall that S* was defined to be the completion of S under the metric (1.23) Proposition The mapping x •*• K(.,x) i s a uniform isomorphism of S* onto P*. m Proof: Since K(.,j) = Gel, we have by the uniqueness of charge that the mapping <|>:S •*• P m; <Kj) = K(.,j) i s one-to-one. From (a) of the sketch of the Martin-Doob-Hunt theory, the mapping <|> induces a b i j e c t i o n between Cauchy sequences i n S and Cauchy sequences i n Pm. Now we r e c a l l a theorem which states that i f ty:A •*• Y i s a mapping from a dense subset of a metric space X into a complete metric space Y which carries Cauchy sequences to Cauchy sequences (and hence i s continous) then ty extends to a continuous function on X. [Royden, Ch. 7, Sec. 6]. Since <\> and <|>-1 have this property <\> and <t>-1 extend to continuous function S* and P* respectively, m The extension i s given as follows xeS*\S {jn} C S such that j n x then <|>(x) = <(>(lim j n ) = K(»,lim j n ) n and <t>~1(K(,,lim j )) = lim jn. n n Since S* and P are compact; <p, <f>~ are uniformly continuous. - 25 -CHAPTER 2 Application of Choquet's theorem to invariant  and ergodic measures Our aim i n this section i s to obtain a representation theorem for the set of a l l invariant probability measures with respect to a family of transformations defined on (S,4) when S i s an appropriate topological space and -6 a a-algebra of subsets of S. When we say invariant probability measures we w i l l always mean invariant probability measures with respect to a family of measurable transformations f. I f X i s the set of invariant probability measures we show that under appropriate conditions V ueX g a unique Borel probability measure m supported on the extreme points of X such that U(f) = / f dm,V feC(S) (2.0) ext(X) Here the ext(X) turn out to be the ergodic measures. As to the topology considered on the space S we consider S to be a compact Hausdorff space and -6 the Borel a-algebra of subsets of S. We show that the set X i s a w*-compact convex set. Let P be the cone generated by the set of invariant probability measures X, i.e. P = {cty|uex,a > o}, i n proposition (2.7) we show that P i s a l a t t i c e i n the cone order. This implies that X i s a simplex. Proposition (2.14) shows that the extreme points of X are ergodic measures. - 26 -Theorem (2.16) gives us the desired representation (2.0) Let S be a set, i> a a-ring of subsets of S and T a family of measurable functions from S into S. Then V TeT we have T:S •*• S and T - 1(A)e4 whenever Ae-4. De f i n i t i o n (1) A non-negative f i n i t e measure y on 4 i s said to be invariant (T invariant) i f y(T - 1A) = y(A) V A£4 and T e T . Def i n i t i o n (2) Suppose y i s a measure on '6. An element A of -6 i s said to be invariant (modu) i f u(AAT - 1A) = OVTeT. (AAB = A\B U B\A). Denote the family of a l l such sets by 4 p ( T ) or 6 . A l i t t l e computation shows that y i s a sub-o-ring of -6. Lemma 2.1 Let y and v be measures on -6. Suppose u i s invariant and v i s absolutely continuous with respect to u (with dv/dy = f a.e.)' Then v i s invariant i f f . f = foT [y] a.e. for a l l T i n T. Proof: I f f = foT [y] a.e. y for a l l T i n T, and i f A&6, then V TeT v(T _ 1A) = / fdy = / foT dy = / f dyoT - 1 T - 1A T~ A A = / fdy = v(A) A (V y i s invariant) To prove the converse Suppose V o T - 1 = v for some T i n T (2.2) V r real l e t A = {x: f(s) < r}, l e t B = T~1A \ A and l e t C = A \ T - 1A. Then on B, f-r > 0. So v(B) - ry(B) = / ( f - r ) dy > 0 (2.3) B with equality i f f . yB = 0. - 27 -Now v(C) = / fdu < ruC (2.4) C ~ Also v(B) = v(T _ 1A) - v ( T _ 1 A n A) (N.B. B = TT1A\ A). = v(A) - v ( T - 1 A n A) by assumption (2.2) = V(C). S i m i l a r l y y(B) = y(C) Combining (2.3), (2.4), we have v(B) > ry(B) = ry(C) >_ v(C) = v(B). So equality holds throughout. I t follows that y(B) = 0 and u(C) = 0 (by 2.3). Thus, for any r, {x:f(x) _< r} and T - 1{x:f(x) _< r} d i f f e r by a set of y measure zero. (2.5) Suppose now that g and h are real-valued functions then we have {x:g(x) > h(x)} = U {x:g(x) > r > h(x)} reQ = u [{x:r > h ( x ) } ] \ [{x:r > g(x)}] reQ (Q i s the set of rationals i n R). Let g = f and h = foT i n the above identity and using (1.4) we see that f < foT [u] a.e. Interchanging f and foT i.e. l e t g = foT and h = f we have f > foT, [y] a.e. So f = foT, [y] a.e. Corollary 2.6 If y and v are invariant measures and y = v on -6 , then y = v on -4. y+v Proof: Let f = dy/d(y+v), g = dv/d(y+v) (Since y € y+v etc.) - 28 -Here f, g e L^y+v). We w i l l have y(A) = v(A) for a l l A i n -i i f uA = / f d(y+v) = / g d(y+v) = vA i.e. f = g [y+v] a.e. A A Now f and g are 4 measurable functions on S and i n fact they are /i>v+v measurable. To see t h i s , choose arbitrary TeT, then since y,v and y+v are invariant, lemma (2.1) implies that foT = f and goT = g a.e. [y+v]. This implies immediately that f and g are -^ + v measurable. Since / f d(y+v) = / g d(y+v) V Ae 4 A ) A y (by assumption y(A) = v(A),V Ae S^+v ) We have f = g, [y+v] a.e. • Let P = {ayjet > 0, yex} where X i s the set of invariant probability measures. P i s a cone with base X. Proposition 2.7 The cone P of a l l f i n i t e non-negative measures i s a l a t t i c e ( i n i t s own ordering). Proof: In order to show P i s a l a t t i c e i t suffices to produce a greatest lower bound i n P for any two non-negative invariant measures y and v. Note: y < y+v and v « y+v So dy = f d(y+v), dv = g d(y+v); f, gs L X(y+v) (2.8) Let h = f A g e L X(y+v). Note: Y.(A) > / h d(y+v) where Yi = y; Y? = v. 1 - A Define d(yAv) = h d(y+v); y^v i s a measure and yAV € y+v. Since V TeT, (fAg)oT(x) = inf{foT(x), goT(x)} = foT A goT(x) = fAg(x) = h(x) a.e. [y+v] [y,v are invariant]. By lemma (2.1) yAV is invariant. We now show that yAV defined above is indeed the infimum. Suppose a >_ yAV i.e. a (A ) >_ yAv(A) V AE4 and a < y; a < v (2.9) By the Radon-Nikodym theorem a f^, f y such that da = f^dy = f ydv (2.10) and by (2.9) a ( A ) = / f dy < y ( A ) Ae4. A U = > 0 1 f y 1 1 [w] a.e.; similarly 0 < f y < 1 [v] a.e. Also 3 f Q such that dyAV = f^da (since yAV < a) . (2.11) and 0 <^  f < 1 [a] a.e. (by 2.9). But dyAV = h d(y+v) (by definition). By (2.11) f adc = h d(u+v). So h d(y+v) = f Q f f d(y+v) = f ^ f ^ g d(y+v) by (2.8), (2.10) (2.11) h = f a f y f = Vv g tp+v^ a - e - (2-12) implies f f = f v g > h (0 < f q < 1). But f f = f v g <^  {f and since h = fAg f f = f v g = h so in (2.12) we conclude that f^ = 1 [y+v] a.e. and hence [a] a.e. (v y < y+v). Now we have dyAV = f do and since f =1 [al a.e. a a L J yAV = a on 4. Now we verify that yAV is indeed the infimum in the cone order. Let P - 30 -be the cone of non-negative measures generated by the invariant probability measures. Define the order u < v i f f . p-v e P. If we suppose that 3 aeP such that a > UAV and a < u; a < v. Then a-pAv, p - o , v-a e P. i.e. a(A) > pAv(A); a(A) < p(A); a(A) < v(A))VAe4. By the discussion above one has o = pA\V. I Thus the above implies that P - P is a vector lattice and so X is a simplex. Definition (2.13) We call an invariant measure p ergodic if p(A) equals 0 or 1 VAeip. Recall -6^  consists of all Ae-4 such that p(AAT_1A) = 0 V TeT. Proposition (2.14) Suppose that p is a member of the set X of a l l invariant probability measures on -6. Then p is an extreme point of X if and only if p is ergodic. Proof: Suppose that p is an invariant probability measure and that 0 < p(A) < 1 for some A in ^ Define p^B) = p(B R A)/p(A) and p 2(B) = p(B \ A)/[l-p(A)]; then p x H p, p = p(A) ]i\ + (l-p(A)) p 2 each p ± is a probability measure, and moreover, each p ^ is invariant. [This uses the facts that p is invariant and that AAT-1(A) has P measure zero, together with the identity Cin(C 2AC 3) = (CiO C 2) A (Ci(l C 3)]. - 31 -To prove the converse suppose u(A) = 0 or p(A) = 1 for each A'e-4 , and suppose 2p = p^ + p2 where Pi and P2 are invariant probability measures• I t follows easily that p =? P-s on & . i = 1,2. P+Pi Thus by corollary (2.6) P = Pi on -6^  i = 1,2. So p i s extreme. To use the above results to obtain a representation theorem we must define a l o c a l l y convex topology on P - P (the subspace generated by the cone P) under which the convex set X of invariant probability measures i s compact. Let S be a compact Hansdorff space. & the a-algebra of Borel subsets of S. Let T be any family of continuous maps T:S •*• S. Thus T i s measurable with respect to Via the Riesz Representation Theorem the space of a l l regular Borel measures on -4 can be indentified with the dual space C(S)* of C(S). We consider the w* topology on C(S)*. Now V TeT the map p -*• poT - 1 i s a continuous linear transformation which carries the w* compact convext set K of probability measures into i t s e l f . The mapping p ->• poT - 1 i s linear. To show the map i s continuous l e t (pg) be any net converging i n the w* topology to p. (3 i n some directed set). Then V fe C(S) foT e C(S). So foT(pg) foT(p). i.e. / foT dp * / foT dp V fe C(S). S 0 S / i foT dp. = / f dp QoT - 1 •»• / fdPoT - 1, V feC(S) T _ 1(S) . B S G S i.e. p RoT - 1 * poT - 1 i n the w* topology. - 32 -So the map i s w* continuous for each TeT. i t i s easy to see that the map induced by each TeT maps K into i t s e l f . The set X of invariant probability measures i s precisely the set of common fixed points for the family of transformations of K into i t s e l f induced by T. To see t h i s , note y an invariant probability measure, y o T _ 1 ( f ) = / f dyoT"1 = / foTdy = / fdy S T _ 1(S) T - 1(S) = / f dy = y ( f ) . S Since f = foT [y] (a.e.) TeT by lemma (2.1) (yoT - 1 € y). Since the induced maps y yoT - 1 are w* continuous for each TeT we have that X i s closed i n the w* topology and hence i s a w* compact set since X c K. If we suppose that X i s non-empty then X has extreme points. (Krein Milman theorem). Further on assuming that X i s metrizable we may apply Choquet's theorem to obtain the following result: (2.16) Theorem; If S i s a compact Hansdorff space, T a family of continuous functions from S into S; then to each element y of the set X of T-invariant probability Borel measures. There exists a unique probability measure m supported on the ergodic probability measures (extreme points)of X such that y(f) = / f dm V fe C(S). ext(X) Remark: If the set X i s empty the above theorem holds vacuously. However, to ensure that X i s non-empty we impose additional constraints on the family T. - 33 -I f T i s a commuting family of continuous transformations we have by the Markov-Kakutani fixed point theorem that X w i l l be non-empty. We state the Markov-Kakutani theorem. Theorem: (Markov-Kakutani) Let Y be a l o c a l l y convex space, K C Y a compact convex subset and T = {T|T:K + K; T affine continuous} We assume T i s a commuting family ( i . e . TiT 2= T 2 T i T L T 2 e f ) Then 3 kn eK such that 1 k„ = k„ V TeT. - 34 CHAPTER 3 PART I Symmetric Measures on a Product Space  The Problem: Let (S,F) be a measure space -f-oo -j-oo -j-oo (S , F ) = (S,F) the usual product space. Let S* denote the class of a l l p r o b a b i l i t i e s 6 on (S,F). Consider the following a-algebra on S* i.e. the a-algebra generated by a l l sets of the form {0eS* 6(F) < t} where F £ F and 0 < t < 1 We c a l l t his the "weak-star" a-algebra F*. For each 9eS* l e t 9 + c o be the product probability on (S + c o, F 4 " 0 0 ) . The correspondence 9 + 9 + c o i s cl e a r l y 1-1. A permutation TT on the positive integers N i s f i n i t e i f Tr(n) = n for a l l but a f i n i t e number of the n i. e . n i s a 1-1 map from N •*• N having a l l but a f i n i t e number of the n, unchanged. Let TT be the induced transformation defined as follows: Tr:S+ 0 0 -»• S+00 S( x i,x 2,...) = ( x m ) , x^. ,...). I t i s clear that if i s a measurable transformation with respect to the a-algebra f^ 0 0. A probability Pe(S + c 0)* i s exchangeable i f P i s invariant under a l l IT, i.e. P ( T T _ 1 ( A ) ) = P ( A ) V A £ F+0° and a l l IT. Suppose y i s a probability on F * and define P as follows P ( A ) = / 9 + 0 0 ( A ) du(9), V A e r^ ™ (3.1) P u S* - 35 -is a probability. Since each 6+0° i s exchangeable we have that P^ i s exchangeable. Using the terminology of Hewitt and Savage we say that P^ i s presentable. Formula (3.1) indicates that a presentable probability i s i n a certain sense a mixture of elements of S*. The question may be posed: i f P i s exchangeable on S+0° what sort of topological structure i s necessary on (S,F) so that 3 yeF* with P = P sa t i s f y i n g (3.1)? Hewitt and Savage have shown that i t i s enough to assume that S is compact Hausdorff and F is the Baire a - f i e l d . Our aim here i s to obtain the representation (3.1) together with the uniqueness of the representing measure u via Choquet's integral representation theorem. The topology on S w i l l be discussed l a t e r . (3.2) Theorem The set of a l l product p r o b a b i l i t i e s on (S + c o, p+°°) forms the extreme points of M, the space of exchangeable probability measures on (S+00, F+~). To prove the theorem we need two lemmas. (3.3) Lemma Let n be a positive integer, E i , . . , E n elements of F, and l e t aeM. Then {a(E 1 x E 2 x... x En x S x S x . . . ) } 2 <_ o(Ei x E 2... x En x E^ x E 2 x ... x En x S x S x ...) If we l e t the cylinder set E^ x E 2 x... x E n x S xS... = C(E ]_,..,En) and Ei x E 2 x . . . x E n x E j x... x E n x S xS... = C(Ei,.., En, Ej,..,En) - 36 -Then the above result reduces to o[c(E!,.., En, E i , . . , En)] >{(j[c(Ej.,.., En)]} 2 (3 Proof: Let C(E 1 }.., En, E i , . . , En) = A and C(Ei,.., En) = B. Let Xr (f = 1,2,...) be the characteristic function of the cylinder {a|ai+( r-l)n e E i , i = 1,.., n} Then / ^  X r(a) da(a) = o(B) by exchangeability. s r  m S o V m J -H» r ? i X r(a) do(a) = m0(B) s . m 2 Furthermore J , ( T. x ) °(a) , m m = /s+00 A X r(a) • ^  X s(a) da(a) m m , = A sh ! s + m X r(a) x 8(a) do(a) / g +oo X1(a) do(a) + m(m-l) J X l ( a ) x 2 ( a ) da(a) = m = ma(B) + m(m-l) a(A). Using the Cauchy-Schwartz inequality with f = vh X r a n d 8 = 1 5 ( ^ f g ) 2 < (/f2)(/§2)-i-e- J i X r da(a)) 2 < ^ ( j ^ x / a ) ) 2 do(a) i.e. m 2{o(B)} 2 < ma(B) + m(m-l) a(A) < ma(B) + m2a(A) o(«) > [a( B ) ] 2 - v .-. a(A) > [a(B)] 2. m m - 37 -(3.5) Lemma: Let a be an element of M such that equality holds i n (3.4) for any positive integer n and arbitrary. E^,.., En e F . Then a i s an extreme point of M. Proof: I f oeM and i s not an extreme point there exists a', a" e M and a, 0 < a < 1 such that a 1 * a" and a = a a'+ (1-a)a. Since a l l measures on (S+c°, F + c°) are determined by their measures on cylinders, 3 a cylinder B = C(El,.., En) such that a*B * a"B. Let A = C(Ei,.., En, E i , . . , En). Then aA = aa*(A) + (1-a) a"(A) > a(o'(B)) 2 + ( l - a ) ( ( a " B ) ) 2 . Applying the Cauchy Schwartz inequality [ao'(B) + (1-a) a"(B)] 2 < a(a'(B)) 2 + ( l - a ) ( a " ( B ) ) 2 o'B fO,a) 1 1 (Let X = { (/ X d t ) 2 < / X 2 dt) o"B (a,l) 0 0 We have s t r i c t inequality above since X t const.. We obtain aA > [aa'(B) + (1-a) a"BJ 2 = (aB) 2. Thus s t r i c t inequality holds i n (3.4). • Proof of Theorem 3.2: Let 6+00 be a product probability on ( S + 0 0 ) F + C 0 ) . We clea r l y have equality i n (3.4) so 6+" i s extreme by lemma (3.5). To show that the product p r o b a b i l i t i e s are the only extreme points we consider aeM, a i s exchangeable and not a product probability. So 3 sets E i , F i , . . , F R e F such that o[C(E 1,F 1,..,Fj * aC(Ei) aC(F!,..,Fj (3.6) - 38 -Consider TT:N -*• N; ir(n) = n+1 Vn. The induced transformation TT' i s a measurable transformation from (S+0°, F4"0") to (S + c o, F+°°). N.B. V AeF1"00 T T - 1 ( A ) = {a|(a 2, a3,...) £A, a£ S + 0 0}. Also we claim a?- 1 (A) = a(A),V A e . F+00 (This i s proved l a t e r ) . Condition (3.6) may-be rephrased i n terms of TT as follows: 3 B = C(F.1,..,F ) such that 1 n o[c(Ei)n * _ 1 ( B ) ] * oC(E 1) o(B) (3.7) In view of (3.7) i t i s impossible that either a(c(Ei)) or o ( c ( S \ E i ' ) ) vanish. Define the conditional p r o b a b i l i t i e s a' = a(»|c(Ei)) and a" = o(«|c(S\ E i ' ) ) a = a[c(E 1)] a' + [ l - a C ( S \ E 1 ' ) ] a " It i s clear from (3.7) above that a,a',a" are a l l d i s t i n c t and since a i s exchangeable o' and a" £ M. Proof of the claim i n the above proof: (3.8) Theorem: Consider the transformation ir:N •*• N, ir i s 1-1. Here TT i s any 1-1 transformation not necessarily a f i n i t e permutation. Let TT be the induced transformation defined on S+°° by ir(a) = ( a T r ( i ) , 3 ^ ( 2 ) , . . . ) . Then V aeM aif- 1(A) = a(A) where A £ - p + c o. Proof: Note that IT i s (S+0°, p4"00) measurable, since i f C i s a cylinder i n F*00, - 39 -+00 i.e. C = TT, E where E = S for a l l but a f i n i t e number of n, then n=l n n T T - 1 ( C ) i s also a cylinder /. T r - ^ r ^ ) c F+00 Consider the probability aeM confined to the semi-algebra of cylinders C, by exchangeability we have o = a f f - 1 on C. The set function defined on F+0° by a i f " 1 (A) VAeF+0° i s an extension of a and a n - 1 on F+00. Since F+00 i s the smallest a-algebra containing C we have by the uniqueness of the Caratheodory extension that a (A) = a-if-^A) V AeF + c o. We now consider the topology on S. Let S be a compact Hausdorff space and F the Borel a-algebra on S. (Later we extend the result to a l o c a l l y compact Hausdorff space.) Then S + c o i s a compact Hausdorff space i n the product topology (Tychonov's theorem and the direct product of Hausdorff spaces i s Hausdorff). F4"00 i s the Borel a-algebra on S + c o. Consider Y = C(S+0°) the space of a l l continuous real-valued functions on S+°°. Let Y* = C*(S+°°) be the dual space endowed with the weak* topology. Y* i s a l . c . s . i n the w* topology (see 0.7). Via the Riesz representation theorem we have a 1-1 correspondence between Y* and the set of a l l non-negative regular Borel measures on S+00. - 40 -Thus the set of exchangeable pr o b a b i l i t i e s M i s a subset of {y*eY* lly*H < l} which i s w* compact. (A consequence of the Banach-Alaoglu Theorem). Clearly M i s convex, we need to show that M i s w* closed. Let (eg) be a net i n M such that ag a i n the w* topology where a i s a probability measure. We need to show a i s exchangeable. Since the 0g are a l l exchangeable, ag(A) = ag(TT _ 1(A)) V A e p1"00, where TT:N •*• N i s a f i n i t e permutation and Tr:S+°° S+°° the induced transformation. Note also TT i s continuous with respect to the product topology on S+°° (3.9) (3.9) implies ag = a g ^ - 1 ) (3.10) Vfe C(S+°°) ag(f) - a(f) (3.11) Thus | + o o f da g * - l ( x ) - /~_i ( s + 0 0 ) f *<*> d^g = !+„ d o g * l + c o ^ = ^ + 0 0 f dOTf"1 So ag i r - 1 •»• OTT-1 i n the w* topology. (3.12) Using (3.10), (3.11), and (3.12) we have ai r - 1 = a .*. a i s exchangeable. So M i s w* closed and hence w* compact convex. • We now show that the extreme points of M form a w* closed set provided we r e s t r i c t a l l the measures i n M to the Baire sets i n F+°° (the Borel a - f i e l d ) . +00 +oo w * Let (6g ) be a net of product p r o b a b i l i t i e s such that 9g •»• a, aeM. - 41 -Consider arbitrary f eC(S) and define f*:S+0° •+ R by f * ( x i , X 2 , . . - ) = f ( x i ) V xeS + C 0, x = ( x i , x2,...) I t i s easily seen that f* i s well defined and continuous with respect to the product topology of S + c o. Also 6^ °° (f*) = / + c o f * ( x ) d9*°° (x) = | f ( X l ) d 6 g ( X l ) = 6 g ( f ) ; V g. +00 Now 6 o ( f * ) o ( f * ) . p We define a map 9:C(S) + R as follows: 9(f) = a(f*)V feC(S); 9 i s a bounded linear functional on C(S) and 9(1) =1. So 9 corresponds to a unique probability measure defined on the Borel sets of S. Sim i l a r l y V f e C(S 2) define f * ( x l 5 x2,...) = f ( x 1 ? x 2) +00 +00 +00 So f* e C(S ), then 8 (f*) = 0^(f) and 9 D ( f * ) •> a(f*) p p p Then we have o(f*) = lim 6 2 ( f ) V feC(S 2). B B Fubini's theorem gives us a(f*) = 9 2 ( f ) v feC(S 2). By induction we obtain o(f*) = 9 n ( f ) V feC(S n). +00 We claim that for every set of the form A x ^ S, A a Baire set +oo n a(A x ^ S) = 9 (A) V n (see below) Therefore we have that I . -^-CO a = 9"*" on a l l Baire sets i n F (See Halmos Sec. 38, Theorem B) so i t follows that ext(M) are weak-star closed. We now prove the claim referred to above, v i z . , Suppose Q\ and 9 2 are two measures on the measure space (S,F), - 42 -[For our purposes S i s compact Hausdorff, F a Borel a-algebra] such that 01(f) = 62(f) V feC(S). Then 6^  = 9 2 on the Baire sets in F. Proof of Claim: V BeF, B a compact G^  a a sequence of continuous functions (f ) i n C(S) such that n fn + 1 B (Royden p. 304) 9 X(B) = lim / f d&i = lim / fnd6 2 = 0 2(B) n n (by Lebesgue's convergence theorem). Then by the monotone class theorem we have 6j_ = 9 2 on a l l Baire sets i n F. • B Since the r e s t r i c t i o n of a l l the measures i n M to the Baire sets i n F" .gives us that the ext (M) are weak-* closed, we have by Choquet's theorem V aeM 3 a regular Borel probability measure u on M, supported on the extreme points of M such that u represents a. .*. f«0 = / e x t ( M ) f ( e + " ) duO) V fe C(S). Also we have a(A) = / e x t ( M ) 9+0°(A) dy(9) V A a Baire set i n (Just use the same argument i n the proof of the claim above.) So i f we r e s t r i c t our Borel measure to the Baire sets we have that a i s presentable. To see whether our, representing measure i s unique we have to show that M i s a simplex. To this end l e t C be the positive cone generated by the exchangeable (symmetric) measures i.e . C = {cta|ct > 0,aeM} - 43 -We need to show C-C i s a vector l a t t i c e i n the cone order or equivalently that C i s a l a t t i c e i n the cone order. Since the set M of symmetric probability measures is invariant with respect to the transformations {TT | IT:N •> N, TT i s 1-1} we have by proposition (2.7) that C i s a l a t t i c e i n the cone order. Therefore M i s a simplex and the representing measure i s unique. In the above we have proved the presentability of every symmetric (exchangeable) probability on (S+c°, F1"00) where S i s a compact Hausdorff space and F i s the Baire a-algebra. We consider now the case where S i s a l o c a l l y compact Hausdorff space. F i r s t we have a d e f i n i t i o n . (3.9) D e f i n i t i o n : Consider the space (S, F). We say the a-algebra f i s presentable i f a l l the exchangeable p r o b a b i l i t i e s on (S+c°, F+°°) are presentable. (3.10) Lemma: Let H be a presentable a-algebra of the set G. Let S be any non-empty set i n H and define F = {HnS |Hetf} i.e. F i s a sub a-algebra of H. Then F i s a presentable a-algebra-Proof: Let M denote the set of a l l exchangeable p r o b a b i l i t i e s on (S+0°, F+00) -4-co +°°\ V aeM extend a to a probability on (G1" » H ) as follows: define a(A) = 0 V A e h*00, A£ G+~ \ S+00. I t i s easy to check that a i s indeed a probability on (G+00,H+°°). Furthermore, a i s an exchangeable probability on (G+co,H+GO). Suppose TT i s any f i n i t e permutation on N and A e ff" 0 0. - 44 -We need only show that a(ir(A 0 G"t\°S+0°)) = a(A f] ( G + \ ° S + 0 ° ) ) = 0 (3.11) Since A D S+°° e F1"00 and a i s exchangeable on F1"00. (Note: Use of the monotone class theorem gives F*° = { B O S + 0 0|(BC + 0 0 } . ) . But i r " 1 (A H G+00\ S+0°) C G+0° \ S+" by d e f i n i t i o n of the extension of 0,(3.11) holds. Since o" i s exchangeable on (G + c o, K*"00) we have that a i s presentable. Therefore F i s presentable. • (3.12) Theorem: Let S be a l o c a l l y compact Hausdorff space, F the a-algebra of Baire. subsets of S then F i s presentable-Proof: Let G be the one point compactification of S. Let q be the "point at i n f i n i t y " of G. Now the open sets of G consist of open sets i n S and complements of compact sets i n S. G i s a compact Hausdorff space. Let H be the Baire a-algebra on G. We distinguish two cases: ( i ) Suppose S i s a-compact (e.g. R). +00 Then S = U. Sn where Sn i s compact i n S and hence closed i n G. n=l Since G i s compact Hausdorff 3 continuous functions (f ) on G such that n f (S ) = 0 and f f{q}l = 1. n n n 1 ' +00 So S = U.{f < 1} which i s Baire. n=l n Since S G i t follows that F i s presentable. ( i i ) I f S i s not a-compact then {q} i s not a set by de f i n i t i o n of the topology on G. Hence i s not a Baire set (see Halmos p. 221 Thm. D). Hence S i s not Baire. - 45 -However there does exist an intimate connection between the Baire sets of S and those of G. Let F be any compact G^  of G. Then 3 fe C(G) such that f = O o n F a n d O < f < l o n G \ F . (See Royden proposition 9.20) So F = {yeGJf(y) = 0} Let A = {x£S|f(x) * f(q)} Now {x£S|f(x) > f(q)} = U{f(x) > f(q) + l) n n {xeSlf(x) < f(q)} = U{f(x) < f(q) - l } I n IT and for each n, {f(x) > f(q) + 1_} and {f(x) < f(q) - 1_} n n are compact Gg's. A i s a union of compact G^ 's and hence i s a Baire set of S.(a) I f f(q) * 0 then F c A and F i s a Baire set of S. (b) I f f(q) = 0 then G \ F = A and S \ A = S \ F which i s Baire since A i s Baire. The above shows that for any compact Gj., F of G either FeF or F D S e F We claim that for any Baire set B of G, i.e. Be , B S i s Baire i n S, i.e. B D S e F. (3 .12) To prove the above claim we use the monotone class theorem. Let M be the c o l l e c t i o n of a l l BeK that have the property (3.12). I t i s t r i v i a l that M i s a monotone class. Since a l l compact Gg's i n G have property (3.12) i t follows by the monotone class theorem that H G M, proving the claim. - 46 -We thus have a map <\>:H •*• F given by <|>(B) = B 0 S,V Be H. We claim that <)> i s 1-1, onto and preserves the operations of countable unions and intersections and 4>(Bi\B2) = <KB l A K ^ ) • Since {q} i s not Baire i t follows easily that <|> i s 1-1. V DeF either D or Du{q} (but not both) i s a Baire set i n H. Thus <|> i s onto. The rest of the claim i s easy. Let G* and S* denote a l l the Baire measures on G and S respectively. Thus <|> induces a 1-1, onto map $ from G* onto S*, defined by $VF) = y 6 * ( F ) ; PG £ G* & F e F ' Since H i s a presentable a-algebra we have that F i s presentable. Thus i f S i s a l o c a l l y compact Hansdorff space the Baire o-algebra F i s presentable. M As an i l l u s t r a t i o n we have that on R the Borel a-algebra i s presentable. (In R the Borel a-algebra = Baire a-algebra.) This i s de F i n e t t i ' s result. Let Xn:(ft ,F ) (R, B ) and (Xn) a sequence of exchangeable random variables i.e. V TT a f i n i t e permutation on N di s t (X ,.. . ,Xn) = d i s t ( X 1 T i t , . . . . X ^ J J ) Then V n, p[x± E i < n] = / , | B(Hi) dF(6); H e B where 6 = d i s t ( Y i ) and (Yi) i s a sequence of i . i . d . random variables. F i s the unique Borel measure supported on a l l such 6. Example: Suppose (Xn) i s an i n f i n i t e sequence of exchangeable random variables taking only the values 0 and 1. - 47 -Then P[XJ. = 1 X k = 1, X k + i =0,.., Xn = 0] ,1 = J 0 r k ( l - r ) n - k dF( r) where r = 6[Yi = l ] , (Yi) are i . i . d . and F i s the unique Borel measure supported on [ 0 , l ] . This shows that the d i s t r i b u t i o n of exchangeable random variables taking values 0 and 1 i s obtained as a mixture of the i . i . d B e r n o u i l l i random variables. (Reference: F e l l e r Volume I I , Chapter v i i , Section 4). - 48 -CHAPTER 3 PART I I Example to show that exchangeable processes need not be  mixtures of i . i . d . random variables. Hewitt and Savage raised the question whether i n the absence of topology on the space S the exchangeable probability on (S°°, F°°) i s presentable. Dubins and Friedman i n .1979 gave a counterexample answering this question i n the negative. There exists a separable metric space equipped with a Borel a - f i e l d which i s not presentable. We give the construction of such a space i n d e t a i l . The Construction Let I = [ 0 , l ] , equip I with the usual Borel a - f i e l d . For t e l , l e t t j be the j-th d i g i t i n the binary expansion of t, 0 0 j - t = E t ./2 , t . = 0 or 1 J=l J J For 0 _< p _< 1 l e t 6 p be the probability on (1,8) which makes the t j ' s independent with common d i s t r i b u t i o n : 6 {t.=l} = p and 6 {t.=0} = 1 - p (3.13) P J P J Let Q = /J 6^  X(dp); (3.14) A represents Lebesgue measure on (1,8). Thus Q i s cle a r l y an exchangeable probability on (I00,8°°). - 49 -lim 1 n Let Z(t) = n > + 0 0 — ^ t j on the subset L of I where the l i m i t exists N.B. Let Z (t) - - l i , t . n n j=l j , since lim inf Z N and lim sup Z N are Borel measurable on I, {t|lim i n f Z f l(t) - lim sup Z n ( t ) = u} = L i s Borel measurable. Thus Z defined on L i s a Borel measurable function. Furthermore by the strong law of large numbers we have 9 P(Z=p) = 1 7V p e l (3.16) Let x = (x^, x^,...) be a t y p i c a l point i n I°°. (3.17) Lemma: For CeB, Q{X|Z ( X ; l ) e c} = X(C). Proof: Note {x|z ( X l ) e c} e 800 since Z i s a Borel measurable function. 1 °° Q { x | Z ( X l ) e c} = / 0 6 p lx| z ( X l ) e c} X ( d p ) = ^0 6 p i x l | z ( x l ) e c} X ( d P ) = / c X(dp) by (3.16) • (3.18) Lemma: Let Tel and Card(T) < c. (c being the car d i n a l i t y of the r e a l s ) . 00 Let T = U\ = 1 T j , where Tj i s the set of a l l x i n I 0 0 with XjeL and Z(XJ) e T, then T has Q-measure 0. Proof: Since card (T) < c we have T i s countable. Thus X(T) = 0. Since L and T are Borel sets we have Tj e g00. Therefore -M -j-oo oo T = U . , T . e 8 • 1=1 1 - 50 -Now Q(T-s) = Q { X | Z ( X , ) G T } = X(T) = 0 by lemma (3.17) * -|-00 Therefore Q(T ) < E . . _ ^ Q ( T ) = 0. Since Q i s a nonnegative measure we have Q ( f ) =0. • Henceforth the symbol Q * w i l l denote the outer measure of Q . (3.19) Proposition Define Q and Z as i n (3.14) and (3.15). Then there i s a subset S of the unit i n t e r v a l I with the following two properties: Q*(S°°) = 1 (3.20) S R {z=p} i s countable for each p e l (3.21) Proof Let K be the set of ordinals of card i n a l i t y s t r i c t l y less than c. Let K be the c o l l e c t i o n of a l l A e 800 of positive Q-measure. Now card (K) = c (This follows from 3.17(b)). Both K and K are well ordered sets and are isomorphic. Hence there i s a one-to-one map o + A a of K onto K. For each a e K , choose a point yet £ Aa as follows: f i x 3 e K, and suppose by induction that the yet have been chosen for a l l a < 3 . Consider for a < yet e I°° and l e t yet be i t s j-th j coordinate. Define Tg as follows: Tg = { t e l j t = Z(y a..) for some ct < 3, j = 1,2... with y a^ e L } Claim: Card (Tg) < c Since 3 < c and Card (N) =)^ o» Card (Tg) < c - 00 Define T^  = U\_^(Tg)j where as i n lemma 3.18 (Tg)j = {x e I 0 0 , X j e L, Z ( X j ) e (Tg)} - 51 -By that result Tg has Q-measure zero. So A3 - Tg i s non-empty. Now choose y^ e A^-Tg Having chosen the ya for a l l a e K l e t 00 S "aHk U j = l ^ } Then y a e S T ) A a so S°° intersects each A e B°° of positive Q-measure. Therefore Q*(S°°) = 1. Also p e [0,l] Z(y a^) = p for at most one a because by construction we chose y a e A a-T a So S f) {z = p} i s countable.. * The next two lemmas w i l l be useful i n obtaining the contradiction. (3.22) Lemma Let (X,E) be an abstract measurable space- YC X, Y not necessarily i n £. Let £y= Y 0 £ be the o - f i e l d of subsets of Y of the form Y D B, with B e E (a). Let <|> be a probability on (Y,E Y). Then •)> induces a probability ri<p on (X,E) by the rule n (j) (B) = <t>(Y O B) for B e E. And (Tl<f>)*(Y), = 1. ('b) Let 9 be a probability on (x,E) with 9*(Y) = 1. Then 9 has a trace probability p9(Y Cl B) = 9(B) for B e E (c) The map n, defined in (a) i s one-to-one, i t s range i s the set of pro b a b i l i t i e s assigning outer measure 1 to Y and i t s inverse i s p as defined in (b). (d) Consider n as acting only on the set Y* of p r o b a b i l i t i e s on (Y,£y) and p as acting only on Y = {9: 9 e X* and 9*(Y) =1} - 52 -where 9 e X* i s a probability on (X,E). * * * * Then n i s (Ev,£ ) - measurable, and p i s (Y O £ ,E^) - measurable. Proof:-(a) n<p as defined i n (a) i s clea r l y a probability on (X,E). V B e 8, Y c B ^ T V K B ) = <p(Y O B) = 1 So n<)>*(Y) = 1 (b) F i r s t p9 i s well defined. I f B 0 ) Bi e £ and Y 0 Bo=YO B I then B Q A B I E E and ( B Q A B I ) f] Y So since 9*(Y) = 1 9 (B 0ABi) = 0, this implies 9(B 0) = 9(B]_). To see whether p9 i s countably additive CO Consider (Y f) B ). ; d i s j o i n t -n 1 Then YD Bi and Y 0 B 2 are di s j o i n t - So since B i D B 2 e E and i s di s j o i n t from Y, 6(BiD B 2 ) = 0 (V9*(Y) = 1). Therefore p9(Y D ( B i U B 2 ) ) = p9(YO Bi)+p9(YO B 2)-p9(Yfl B i f ) B 2 ) =e(B 1 ) -e(B 2 ) - e(B,n B 2 ) . I t follows by induction that p9(Yfl B ) = \ = 1 E ( \ ) for any n. -f-oo -J-co Therefore 6( U B, ) = E 9(B, ) k=l k k=l k The rest of the axioms to check whether p9 i s a probability follow easily. (c) Suppose f i s a probability on (Y,Ey). Then n.<|> i s a probability on (X,E) by (a) and p(n.<|>) i s a probability on (Y,E Y). - 53 -We claim p(n9) = <f> because Y B e £, pn<|>(Y 0 B) = n<f>(B) = <|>(Y 0 B) which implies that p = n - 1 . To check whether n i s one-to-one i s easy. The rest follows from (a), (d) To show n i s measurable, f i x B e E and 0 < t < 1 Then n— 1{0: 6 e X* and 6(B) < t} = {<j>: <(> e Y* and K Y f l B) < t} e E^ (Using (a), (b) and ( c ) ) . Also P~l{i>: • e Y* and <p(Y H B) < t} = {6: 0 e Y, 0(B) < t} e Y O E* (3.23) Lemma (a) S~D B°° = (S fl B)°° (b) n°y° = (nc)))00 for <t> e S* (c) If A e 8°°, then <|> •»• (n00.))00) (A) i s F* measurable Let v be a probability on (S*,F*) and l e t P = /«. *"v(d*) (3.24) S* .be an exchangeable probability on (S°°, F°°). Then P induces an exchangeable probability n°°P on (I°°,W) and n°°P = / (n<p)°° v(d*) S* where n<f> i s the probability induced by <j> on (1,8). Proof (a) Clearly (Sf) B ) " c S°°n 830. B e B°°, S°°n B i s generated by sets of the form 00 -{-co S n n. A, where a l l but a f i n i t e number of the n=l k A k = I, for those A k * I; A k e B-- 54 -OO 00 Now these sets S D \ = 1 \ belong to (SO 8)°°. Therefore S°°n B°° = (S D 8)°°. (b) F i x n, and B ,. . .,Bn e 8. Let A = {x: x e I 0 0 and x^ e B^, 1=1,...,n} Then n " * " ( A ) = •"(S-f) A) - ( n * ) - ( A ) Note that n°V°(B) = n s " f ) B^V B e 8". Also n 0^ 0 0 i s a probability measure on ( I ^ . B 0 0 ) - Thus since n 0 0^ 0 0 and (n<t>)°° agree on the algebra of sets generating 8°° we have n.00^00 = (r)d>)°°, by the Monotone Class theorem. (c) We f i r s t show that ((J™* n^dJ^CA) i s (F°°)* measurable (3.26) Consider {<t>°°| T W ( A ) ' < t} = { r | r < s " n A> < t} e(S°°n 8 5 0)* = {(SO B ) 0 0 } * = ( F 0 0 ) * by (a). Now consider the mapping <j> •*• d)00 which i s F* measurable To see this consider arbitrary A e F and S x S x . . . x S x A x S .. =F (A i n the n-th position.) (* *) {<t>|<t>°°(F) < t} = { * | * ( A ) < t} e F* Since a l l sets of the form (**) generate F 0 0 <|> -*• <t>°° i s ( F * , (F°°)*) measurable (3.27) Combining (3.26) and (3.27) we have that <}> ->- rfd/"^) i s F* measurable. F i r s t we ve r i f y that n°°P i s exchangeable-(n°°P) i s a probability on (I°° B°°), P an exchangeable probability on ( S " F ~ ) . - 55 -r)0oP(-if-1A) = P C T T - ^ O S°°) = PCfr-^A D S°°)) = P(Afl S°°) = n°°P(A) To establish (3.25) f i x A e B°° Then n°°P(A) = P(S°°n A) = / ^ ( S ^ n A) v (d<(.) by (3.24) S* = / (nd>)°°(A)v(d<|>) by (b) 0 S* The next theorem shows that there i s a separable metric space S whose Borel a - f i e l d i s not presentable Indeed l e t S be the subset of I = [0,l] constructed i n proposition (3.19). S i s separable i n the re l a t i v e metric since every subspace of a separable metric space i s separable. F = SD B is the Borel a - f i e l d of S. Define the exchangeable probability Q on (I^.B 0 0) by (3.13) and (3.14). Let P be the trace of Q on ( S ^ F " ) : this i s possible since Q*(S°°) = 1 by (3.20) and 3.22(b). (3.28) Theorem: The probability P on (S00,F°°) i s exchangeable but cannot be presented i n the form (3.1). Proof Suppose P were presentable P = / <t)°°v(d<t)) (3.29) S* By (3.23) Q = n P = / ( T i d > ) ° ° v ( d d ) ) (3.30) S* - 56 Let R be the range of the mapping 3P I •> I * defined by p * 9 Claim (1) R e 5* To see this we endow I * with the weak star topology. Note I*C C(I)* and I* i s a closed subset of {u|liull <1 } P e C(I)*}, I * i s a weak-star compact set- Since C(I) i s a separable topological vector space and I * i s weak-star compact we have that I* i s metrizable i n the weak-star topology. So I* i s compact metric and B* i s the Borel a - f i e l d i n I*. I f we show the map p •»• 9 p i s continuous (0 <^  p _< 1) then the range R i s a compact subset of I*. To v e r i f y that that p -»• 9 p i s continuous we consider p r + p as r •*• +°° (r an integer) and consider open intervals of the form (a n,a m) = ( E m t k / 2 k E n t k / 2 k ) , n > m; k=l k=l (a m,a n) e I (3.31) Calculation using (3.13) shows that 9p r( am> an) * 6p(am. an)-Now i t i s easy to show for any open i n t e r v a l A C I that 8 p r(A) •»• 9 p, therefore by the montone class theorem 6p r(B) * e p( B) V B £ 8. Hence the map i s continuous. Since we are assuming that P i s presentable we have that u i s unique. (3.32) Comparing (3.14) and (3.30) .1 co i.e. Q = J 9 X(dp) 0 p - 57 -and Q = n°°p = J ^  (Ti«j) cov(dd>) s We note that the v d i s t r i b u t i o n of <f> •»• (ni))) coincides with the X-d i s t r i b u t i o n of p •*• 9 p by (3.32). In particular v(n - 1R) = 1. Therefore there exists at least one <|> e S* and p e (0,1) such that n<J> = 6p. This i s a contradiction since (n.<|>)*(S) = 1 by (3.22a) and 9 p(s) = 0 by (3.21) and (3.16). • - 58 -BIBLIOGRAPHY 1. G. Choquet - Lectures i n Analysis, Volume (1) and (2). W.A. Benjamin Inc. (Amsterdam 1969). 2. Dubins and Friedman - Exchangeable Processes need not be mixtures of independent and i d e n t i c a l l y distributed random variables. Z e i t s c h r i f t Fur Wahrscheinlich-keitstheorie. Vol. 48, pp. 115 - 132 (1979). 3. W. F e l l e r - Introduction to Probability Theory and i t s Applications, Vol. I I , Wiley (Second E d i t i o n ) . 4. P. Halmos - Measure Theory; D. van Nostrand Co., Ltd. 5. Hewitt and Savage - Symmetric Measures on Cartesian Products; A.M.S. Transactions; Vol. (80), 1955, pp. (470-501) 6. Kemeny, Snell and Knapp - Denumerable Markov Chains; D. van Nostrand Co., Inc. 7. R. Phelps - Lectures on Choquet's Theorem, D. van Nostrand Co., Ltd. 8. H.L. Royden - Real Analysis; McMillan Publishing Co., Inc. 9. W. Rudin - Functional Analysis, McGraw-Hill Series i n Higher Mathe-matics . 

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