. Then V l on K l, and so. e e 1.

Transcription

1 Sanov s Theorem Let E be a Polish space, and define L n : E n M E to be the empirical measure given by L n x = n n m= δ x m for x = x,..., x n E n. Given a µ M E, denote by µ n the distribution of L n under µ n. Lemma. For each M 0, there is a compact set K M M E such that n n log µ ne \ K M M. Proof: Choose a non-decreasing sequence {K j : j } of compact subsets of E so that µe \ K j e 2j, and set E = j= K j and V = j= E\K j. Then V l on K l, and so E µ[ e V ] = e V dµ = e V dµ = e V dµ E l K l K l+ \K l l= e e. At the same time, V l off K l, and so V, ν R = νe \ K l R l. In addition, because V is l.s.c., K R = {ν : V, ν R} is closed. Hence, for each R > 0, K R is compact in M E. Finally, µ n KR n e nr e n V,ν µ n dν = e nr e dµ V = e nr+a, where A = log e V dµ. Lemma 2. Set Λ µ ϕ = log E µ[ e ϕ,ν ] for ϕ C b E; R and Then Λ µ is a rate function and Λ µν = sup { ϕ, ν Λ µ ϕ : ϕ C b E; R }. n n log µ na inf Λ A for all A B M E. Proof: Let Bν, r denote the Lévy ball of radius r around ν M E. Because of Lemma, it suffices to show that r 0 n n log µ n Bν, r Λ ν for each ν M E. But, for each ϕ, there exist ɛ ϕ r 0 such that µ n Bν, r e n ϕ,ν +nɛ ϕ r E µ [ n e n ϕ,ν ] = exp n ϕ, ν + Λ µ ϕ + ɛ ϕ r, and so n n log µ n Bν, r ϕ, ν + Λϕ + ɛϕ r. Now take the it as r 0 and then the supremum over ϕ C b E; R. Lemma 3. For ν M Σ, define Hν µ = { dν f log f dµ if ν µ and f = Σ dµ. otherwise. Then Λ µν = Hν µ. In particular, ν Hν µ is l.s.c. and convex. In fact, if Hν µ Hν 2 µ < and θ 0,, then H θν + θν 2 < θhν µ + θhν 2 µ.

2 Proof: The final assertion follows immediately from the strict convexity of x [0, x log x R. If ν µ and ν θ θµ + θν for θ [0, ], then Hν µ = θ 0 Hν θ µ. To see this, set f = dν dµ and f θ = θ + θf. Since x [0, x log x is convex, Jensen s inequality says that Hν θ µ = f θ log f θ dµ θ f log f dµ = θhν µ. At the same time, since x [0, log x is non-decreasing and concave, log f θ dominates both log θ and θ log f; and therefore Hν θ µ = θ log f θ dµ + θ f log f θ dµ θ log θ + θ 2 Hν µ. After combining these two, one clearly gets the asserted convergence. I next show that if ν µ, then Λ µν Hν µ. In view of the preceding and the obvious fact that ν M Σ Λ µν is lower semi-continuous, I may and will assume that f = dν dµ θ for some θ 0,. In particular, by Jensen s inequality, [ ] [ ] ϕ exp[ϕ] exp ϕ dν Hν µ = exp log f dν dν = exp[ϕ] dµ; f from which it is clear that Λ µν Hν µ. As a consequence of the preceding, all that remains is to show that if Λ µν <, then dν = f dµ and * Λ µ ν f log f dµ. Given ν with Λ µν <, one has 4 ϕ dν log exp[ϕ] dµ Λ µν < for every bounded continuous ϕ. Since the class of ϕ s for which 4 holds is closed under bounded point-wise convergence, 4 continues to be true for every bounded B E -measurable ϕ. In particular, one can now show that ν µ. Indeed, suppose that Γ B E with µγ = 0. Then, by 4 with ϕ = r Γ, rνγ Λ µν, r > 0; and therefore νγ = 0. Knowing that ν µ, set f = dν dµ. If f is uniformly positive and uniformly bounded, then * is an immediate consequence of 4 with ϕ = log f. If f is uniformly positive but not necessarily uniformly bounded, set f n = f n and use 4 together with Fatou s Lemma to justify f log f dµ = log f dν n log f n dν Λ µν + log n f n dµ = Λ µν. Finally, to treat the general case, define ν θ and f θ = θ + θf for θ [0, ] as in the first paragraph of this proof. By the preceding, f θ log f θ dµ Λ µν θ as long as θ 0,. Moreover, since θ [0, ] Λ µν θ is bounded, lower semi-continuous, and convex on [0, ], it is continuous there. In conjunction with the result obtained in the first paragraph, this now completes the proof. As a consequence of Lemma 3 and 4, one knows that 5 ϕ, ν Hν µ + log E µ[ e ϕ] for any B E -measurable ϕ : E R which is bounded below. 2

3 Theorem 6 Sanov. The map ν M E Hν µ [0, ] is a good rate function and for all A B M E. inf Hν µ ν Ao n n log µ na n n log µ na inf Hν µ ν A Proof: In view of Lemmas, 2, and 3, it suffices to prove that if G is open in M E and ν G with Hν µ < then n n log µ ng Hν µ. To this end, suppose that ν G with Hν µ <, and let f = dν dµ. For n, set F nx = n m= fx m for x E n and A n = {x E n : L n x G and F n x > 0}. Then, because t log t e, Jensen s inequality implies log µ n G log A n F n σ νn dσ log ν n A n ν n log F n σ ν n dσ A n A n log ν n A n eν n A n ν n log F n σ ν n dσ A n Σ n = log ν n A n eν n A n n Hν µ ν n A n as long as ν n A n > 0. Finally, by the Strong Law of Large Numbers, ν n A n as n. Cramér vs. Sanov Let E be a separable, real Banach space, and assume that µ M E satisfies E µ[ e α x E ] < for all α 0. Next, set Λ µ x = log E µ[ e x,x ] for x E, and define Λ µx = sup { x, x Λ µ x : x E } for x E. Finally, let µ n M E denote the distribution of x E n S n n n x m E under µ n. m= The goal here is to show that Λ µ is good, that {µ n : n } satisfies the full large deviations principle with respect to Λ µ, and that 7 Λ µx = J µ x inf { Hν µ : x dν < and } y νdy = x. Let I be the set ν M E such that E ν [ x ] <. Then I F σ M E and µ n I = for all n. Lemma 8. There is a l.s.c. V : E [0, such that V x x E, x E V x x E =, and E µ[ e ] V <. In particular, if 0 R l is chosen so that V x l x E for x E R l, then for each M > 0 there is an C M 0, such that n n log µ n M E \ F M M where { F M ν : x E dν C M x E <R l l 3 for all l Z + }.

4 Proof: For each l Z +, choose R l 0, so that x E R l e l x E µdx 2 l. Without loss in generality, assume that R l. Set V x = x E + Rl, x E. l= Then V is l.s.c., V x x E for all x E, V x l x E when x E R l, and V x l x E when x E R l. Hence, E µ[ e V ] e + e l+ x E µdx 2e, R l < x E R l+ and so l= µ n V, ν C e nc E µ n [ e n V,ν ] = exp nc + n log E µ[ e V ] e nc log 2e. Therefore, if C M = M +log 2e, then n n log µ n V, ν M. Since V, ν CM = ν F M, this completes the proof. Continuing with the notation in Lemma 8, note that each of the sets F M is closed in M E and contained in I. Next, define Ψ : I E so that Ψν = x νdx. Then Ψ F M is bounded and continuous for each M. Indeed, the boundedness is obvious. To prove continuity, choose η C R; [0, ] so that ηt = if t 0 and ηt = 0 if t. Then, for each l Z +, the function Ψ l : M E E given by Ψ l ν = η x E R l x νdx is bounded and continuous. Furthermore, as l, Ψ l Ψ uniformly on F M. Lemma 9. For each M 0, there is a K M E such that n n log µ ne \ K M M. Proof: Choose K M M E as in Lemma, and set K M = Ψ K M F M. Then, because K M F M M E and Ψ F M is continuous, K M E. In addition, by Lemmas and 8, n n log µ ne \ K M = n n log µ n M E \ K M F M M. Lemma 0. The function J µ is a good rate function which is convex. Moreover, if x E and J µ x <, there exists a unique ν I such that Ψν = x and Hν µ = J µ x. Proof: Suppose that J µ x <. Then I can find {ν k : k } I such that Ψν k = x and Hν k µ J µ x + k. In particular, {ν k : k } is relatively compact and, by 5 with ϕ equal to the function V in Lemma 8, sup k V, ν k J µ x++log E µ[ e ] V, which means that {ν k : k } F M for some M <. Thus, without loss in generality, I may assume that ν k = ν for some ν F M. Since Ψ F M is continuous and H µ is l.s.c., this implies that Ψν = x and that Hν µ Λ µx, which means that Hν µ = Λ µx. Further, if ν, ν 2 were distinct elements of I satisfying Ψν = x = Ψν 2 and Hν µ = J µ x = Hν 2 µ, then one would have that ν +ν 2 2 I, Ψ ν +ν 2 2 = x, and H ν +ν 2 2 < Jµ x, which is impossible. To prove that {J µ L} E, suppose {x k : k } E with J µ x k L. For each k, choose ν k I so that Ψν k = x k and Hν k µ = J µ x k. Then, because H µ is good, {ν k : k } is relatively compact. In addition, just as above, {ν k : k } F M for some M <. Finally, choose a subsequence {ν km : m } so that ν km = ν. Then ν F M and, because Ψ F M is continuous, x km = Ψν km x = Ψν. Because Hν µ m Hν km µ L, J µ x L. To prove that J µ is convex, suppose that x, x 2 E with J µ x J µ x 2 <, and choose ν, ν 2 I so that Ψν i = x i and J µ x i = Hν i µ for i {, 2}. Then Ψ θν + θν 2 = θx + θx 2 and J µ θx + θx 2 H θν + θν 2 θjµ x + θj µ x 2. 4

5 Lemma. For any closed F E, n n log µ nf inf F J µ. Proof: Refer to the notation in Lemma 8 and the paragraph following the lemma. For any M > 0, µ n F = µ n {ν I : Ψν F } µn {ν FM : Ψν F } + µ n M E \ F M, and so, by Sanov s Theorem and Lemma 8, [ n n log µ nf inf { Hν µ : ν F M & Ψν F } ] [ ] M inf J µ M. F Lemma 2. For each open G E, n n log µ ng inf G J µ. Proof: Again refer to Lemma 8. Let ν 0 I with Hν 0 µ < and x 0 = Ψν 0 G be given, and choose r > 0 so that B E x 0, 2r G. By 5, C V, ν 0 Hν 0 µ + log E µ[ e ] V <. Hence ν 0 F M for any M with C M C. Choose M > Hν 0 µ + 2 so that C M C and l Z + so that C M l < r. Then Ψν B E x 0, 2r if ν F M and Ψ l ν B E x 0, r, and so µ n G µ n BE x 0, 2r µ n {ν FM : Ψ l ν Bx 0, r} µ n Ψl ν B E x 0, r µ n M E \ F M. Finally, since Ψ l ν 0 B E x 0, r, Sanov s Theorem and Lemma 8 say that, for each 0 < δ <, µ n Ψl ν B E x 0, r e nhν 0 µ+δ and µ M E \ F M e nm δ for all sufficiently large n s. Hence, n n log µ ng Hν 0 µ δ. By combining Lemmas and 2, one sees that {µ n : n } satisfies the full large deviations principle with the good rate function J µ. As a consequence of this and Varadhan s Theorem, Λ µ x = sup x, x J µ x. x E Since J µ is l.s.c. and convex, this means that 7 holds. Theorem 3 Cramér. The rate function Λ µ is good, and {µ n : n } satisfies the full large deviations principle with respect to it. In addition, 7 holds. Gibbs Measures Given x E, define the Gibbs measure γ x M E by 4 γ x dy = Lemma 5. For each x E, γ x I and M µ x e y,x µdy where M µ x = 6 Hγ x µ = Ψγ x, x Λ µ x = Λ µ Ψγx. In particular, Hν µ > Hγ x µ for any ν I \ {γ x } with Ψν = Ψγ x. 5 e y,x µdy,

6 Proof: The first equality in 6 is obvious. To prove the second, suppose that ν I with Ψν = x Ψγ x and Hν µ <. Set f = dν dγ and g = dγ x x dµ, and note that Hν µ = log f dν+ log g dν = Hν γ x + y, x νdy Λ µ x x, x Λ ν x = Hγ x µ. Hence, by 7, Hγ x µ = Λ µx, and equality in the preceding hold only if ν = γ x. Theorem 7. Given x E, there exists a ν I such that Ψν = x and Hν µ < if and only if Λ µx <. Moreover, if Λ µx <, then there exists a unique ν I such that Ψν = x and Hν µ = Λ µx. Finally, x E satisfies Ψγ x = x if and only if x, x Λ µ x = Λ µx, in which case Hγ x µ = Λ µx. Proof: The only assertion yet to be proved is that x, x Λ µ x = Λ µx = Ψγ x = x. But x, x Λ µ x = Λ µx implies that y E F y = x, y log e y,y dµdy achieves a maximum at x, and therefore, by the first derivative test, x Ψγ x = x y γ x dy = DF x = 0. Corollary 8. If H, E, W is an abstract Wiener space and ν M E, then Hν W < implies that x 2 E νdx < and that Ψν H. Furthermore, for any x E and x E, x = h x if and only if x = Ψγ x. Thus, if x E, then y E e y x,y Wdy 0, achieves a minimum if and only if x = h x for some x E. Proof: By Fernique s Theorem, A = E W[ ] e α x 2 E < is for some α > 0. Thus, if Hν W <, then, by 5, α x 2 E dν Hν µ + log A <. Furthermore, by 7, W Λ Ψν <, and therefore Ψν H. The second assertion is simply that observation that γ x = T hx W and therefore that Ψγ x = h x. Given the earlier ones, the final assertion is an easy application of the last part of Theorem 7. 6