Gaussian Random Fields

Transcription

1 Gaussian Random Fields Mini-Course by Prof. Voijkan Jaksic Vincent Larochelle, Alexandre Tomberg May 9, 009 Review Defnition.. Let, F, P ) be a probability space. Random variables {X,..., X n } are called jointly Gaussian with variance matrix D D ij ] > 0 if dµ X...X n π) n/ det D] / e /<x,d x> dx where x x,..., x n ) R n, dx dx dx n Remark... {X,..., X n } are Gaussian with variance D D ij ] if and only if C X...X n Ee it X +...+it nx n ) e /<t,dt> n < t, Dt > t i t j D ij, t t,..., t n ) i,j. D ij EX i X j ), EX i ) 0 3. {X,..., X n } are Gaussian if and only if α i R, α X α n X n is Gaussian. 4. Let {X,..., X n } be Gaussian with variance I n n. { EX m X m 0 if l is odd Xn mn ) if l l where l m m n l )! l )l )!

2 Gaussian Random Fields Defnition.. Let G be a countable set. The family of random variables {X n } is called a Gaussian Random Field GRF), if for any finite subset {n,..., n k } G, the random variables are jointly Gaussian. {X n,..., X nk } Remark... G could be finite, or G could be a singleton, in which case we have a single random variable. However, we only care about the case when G is infinite.. Same definition applies for an arbitrary G e.g. G 0, ) for Brownian motion, say). The only real difference is that some measure theoretic aspects are more delicate.. Uniqueness Given a GRF {X n }, we have D nm EX n X m ), and D : G G R, Dn, m) D nm Theorem.3. Let, F, P ) and, F, P ) be two probability spaces, and {X n } and {X n} be two GRF on these spaces with variances D nm and D nm. Suppose D nm D nm, and assume also, that F, F are minimal σ-fields with respect to which {X n } and {X n} are measurable. Then,, F, P ) and, F, P ) are isomorphic, and under this isomorphism, X j corresponds to X j. Remark.4. Equivalent formulation: such that W, a unitary map, W : L, dp ) L, dp ). W, W map bounded functions to bounded functions.. W XY ) W X) W Y ) for bounded X, Y. 3. W X n ) X n

3 . Existence and Basic Model Let D : G G R be a map, Dn, m) D nm, such that n,..., n k G, D ni n j ] i,j k is strictly positive definite. Set R G R, ω, {ωn)}, ωn) R and let F be the Borel σ-algebra generated by cylinders, i.e. sets of the form {ω : ωn ) B,..., ωn k ) B k }, where B i are Borel in R If a n > 0, a n < and dω, ω ) a n ωn) ω n) + ωn) ω n) then d is a metric on, and, d) is a complete and separable metric space. Moreover, the Borel σ-field generated by the sets open with respect to d is precisely F. Given a cylinder C {ω : ωn ) B,..., ωn k ) B k }, set P C) µ n,...,n k C) π) k/ det D c ) / e x, Dc x dx k i B i where D c D ni n j ] i,j k. This a good definition does not depend on the way we write C) and if we take a field consisting of finite disjoint unions of cylinders, P is a countably additive set function. Hence, by Caratheodory Extension Theorem or by Kolmogorov Theorem), P extends uniquely to a probability measure on, F ). Thus,, F, P ) is a probability space such that, by our construction, X n ω) ωn), n G is a random variable, and the joint distribution of {X n,..., X nk } is µ n,...,n k, and so, they are Gaussian with variance D D ij ]. From now on, we will work only with that model: R G F Borel σ-field P Gaussian measure induced by the marginals µ n,...,n k.3 Support properties of P Proposition.5. Let A n > 0, n G be a sequence such that A n D nn < 3

4 and let {ω : A n ω n) < } Then the following holds:. is measurable.. P ) Proof. Property ) is trivial. To prove ), let F ω) A n ω n), F : 0, ], F 0 Then F ω)dp ω) A n ω n)dp ω) A n ω n)dp ω) A n D nn < F ω) < P -a.e. ω.4 Hilbert Spaces Consider the Hilbert space l G) {ω : R ωn) < } equipped with the usual inner product ω, ω ωn)ω n) Remark.6. Usually, l G) denotes the complex version of this Hilbert space. However, in this document, we are mostly considering lr, and we decided to simplify our notation by dropping the subscript R. Hence, to avoid confusion, we will always write lc when referring to the complex space. The matrix D D nm ] defines a linear map, which naturally extends to a possibly unbounded) self-adjoint linear operator on l G). We will assume that both D and D are bounded. This amounts to say that, if Dω)n) m G D nm ωm) 4

5 then the sum on the right hand side is convergent, and also assuming that for some M 0 Dω)n) M ωn), ω l G)G) This assures that D is bounded and self-adjoint, or more explicitly, that { ω, Dω M ω ω, Dω Dω, ω To insure the existence and boundedness of D, we further need to assume that m > 0, ω, Dω m ω D m In terms of {D nm }, D is bounded if sup { D nm } < m G In particular, n, m D nn M, so that A n D nn < A n <.5 Explicit Computations with GRF Let α l G), and suppose that only finitely many coordinates are non-zero. The vector space of such α s is denoted f l R G). For α f l G), set f l G) {α l G) : αn) 0 for finitely many n} Φ α ω) αn)ωn) αn)x n ω) } {{ } finite sum! Hence, Φ α ω) is a Gaussian random variable and since Φ α ω) dp ω) αn)αm) ωn)ωm)dp n,m G Φ α ω) has variance α, Dα. n,m G αn)αm)d nm α, Dα 5

6 Remark.7. If δ n m) is the Kronecker delta on l G), i.e. { if n m δ n m) 0 if n m Then obviously, δ n f l G), and Φ δn X n The map f l G) α Φ α L, dp ) is linear because Φ α is a finite linear combination), and Φ α L,dP ) Φ α ω) dp ω) α, Dα D α lr G) This implies that the map α Φ α is uniformly continuous. Then, by Extension by Uniform continuity Theorem MATH-354, Analysis 3), it extends uniquely to a bounded linear map l G) L, dp ) and if α n f l G) is such that α n α, then Φ αn Φ α in L, dp ). Claim.8. For all α l G), Φ α is a Gaussian random variable with variance α, Dα. Proof. We know that α n f l G), e itφαn dp e αn,dαn Furthermore, Φ αn Φ α in L, dp ) implies that there exists a subsequence α nk α, such that Φ αnk Φ α P -a.e. ω. Then, as Φ αnk is real, e itφαn k, and thus, by Dominated Convergence Theorem, lim e itφαn k dp e itφα dp k And obviously, lim k e α nk,dα nk e α,dα So that we have e itφα dp lim k e itφαn k dp lim k e α nk,dα nk e α,dα Note also that, by the same argument, Φ αdp 0 6

7 Exercise. Show that Φ α δ n, α Φ δn where the sum on the right is converging in L -sense. Solution. If α f l G), then the sum is finite, and the result holds trivially. Now let α l G) be fixed and let {g, g,...} be a numbering of elements of G. Define { αgi ) if i k α k g i ) 0 else Then clearly, α k f l G) and α k α. Hence, Φ αk Φ α in L, dp ) and thus, δ n, α k Φ δn Φ α in L, dp ) Finally, notice that, the k th partial sum of δ n, α Φ δn is δ n, α Φ δn δ n, α k Φ δn g,...,g k Hence, the sequence of partial sums converges in L -sense as required. Proposition.9. If {α,..., α n } are linearly independent in l G), then {Φ α,..., Φ αn } are jointly Gaussian with variance matrix D ij ] α i, Dα j ] i,j n Proof. The matrix α i, Dα j ] i,j n is strictly positive definite since n n n γ i γ j α i, Dα j γ i α i, Dγ j α j i,j i j i j n n γ i α i, D γ j α j n D γ i α i > 0 unless i γ iα i 0 because D m > 0 by assumption). But, by the linear independence of α i s, this is possible only if γ i 0, i. Hence e i n tnφαn dp ω) e iφ n tnαn dp ω) e n tnαn,d n tnαn e i i,j t it j α n,dα j Fact.0. For any z C, e zφα L, dp ), and e zφα dp e zx e x α,dα dx e z α,dα R 7

8 3 Trace Class Background reading Barry and Simon, Functional Analysis Volume I, Chapter VI last section). 3. Trace Class Operators Defnition 3.. Let H be a real Hilbert space, and let A : H H be a bounded self-adjoint operator as H is real, self-adjoint simply means symmetric). We say that A is a trace class operator if A λ n f n, f n where λ n R, λ n <, and {f n } form an orthonormal basis of H Note that from this definition it follows that f n is the eigenvector associated to eigenvalue λ n, as Af n λ n f n Defnition 3.. Also define the trace of a trace class operator A: and the trace norm TrA) A λ n λ n Note that we will continue to use the standard operator norm, and we will denote it as usual, while will now denote the trace norm. Also, we say that A is Hilbert-Schmidt if λ n <. For a more detailed introduction to traces and trace norms, see Trace ideals and its applications by Barry Simon. If A is a trace class, the standard operator norm A sup λ n n Note that the supremum is actually achieved, as λ n < λ n 0 8

9 Moreover, A inf n λ n, meaning that ψ H ) ψ, Aψ inf λ n ψ, ψ n Now, let A be trace class such that A >, i.e. inf n λ n > or equivalently, n, λ n >. Then, deti + A) lim n n + λ k ) Claim 3.3. The limit on the right hand side exists. Proof. Write Now, as n + λ k ) e n ln+λ k) ln + x) lim by L Hospital Rule) x 0 x For small enough x, we have that x ln + x) x Now, as λ n 0, we have that for large enough n, and thus, λ n ln + λ n ) λ n λ n < ln + λ k ) < Hence, we have convergence, and we can write deti + A) + λ k ) 3. Variance Induced Inner Products As before, consider l G), and let D be self-adjoint and positive definite. Define, D, D ) D ), D ) Then,, D is an inner product on l G), and l G),, D ) is a Hilbert space. We will denote it by l D G). 9

10 Note that, if D is bounded the resulting norm is equivalent to the Euclidean norm. Also, l IG) l G) Now, let A be a trace class operator on ld G), and write A λ n f n, D f n where {f n } is an orthonormal basis of l D G), and λ n <. Consider D AD λ n f n, D ) D fn λ n f n, D ) D D fn λ n D fn, Now, the set {D f n } is an orthonormal basis of l G), as D fn, D fm f n, f m D δ nm D fn And thus, D AD is trace class on l G) with eigenvalues λ n and eigenvectors D f n. Conversely, if A is trace class on l G), D AD is trace class on ld G) with eigenvectors D f n. 3.3 Example Consider ld G) and let A be a self-adjoint trace class operator such that A >. Write A λ n f n, D f n where {f n } are orthonormal w.r.t., D. Now, set F ω) λ n Φ f n ω) Claim 3.4. F L, dp ) Proof. By Monotone convergence Theorem, we have F ω) dp ω) λ n Φ f n ω)dp ω) 0 λ n Φ f n ω)dp ω)

11 λ n f n, Df n λ n < Example 3.5. Show that e F ω) dp ω) deti + A) ] / Solution. We wish to compute e F ω) dp ω). To this end, let us first compute e n λ kφ f ω) k dp ω) By Proposition.9, {Φ fk } n are jointly Gaussian with variance Hence, e n λ kφ f k ω) dp ω) π) n/ R n e f i, Df j ] i,j n I n n π) n/ e R n n n λ k+)x k dx Taking n in the above, we get that n λ kx k e n x k dx + λ k ) / lim e n λ kφ f ω) k dp ω) + λ k ) / deti + A)] / n Now, since λ n 0, there exists n 0 such that n n 0, λ n <. So, set F ω) n 0 λ k Φ f k ω) kn 0 λ k Φ f k ω) As before, F L, dp ), and since F ω) F ω), we have that e F ω) dp ω) e F ω) dp ω) As the second sum in F ω) is monotone, we can use the Monotone Convergence Theorem and we have ) e F dp exp n 0 λ k Φ f k ω) exp λ k Φ f k ω) dp ω) kn 0

12 ) MCT lim exp n 0 n λ k Φ n f k ω) exp λ k Φ f k ω) dp ω) kn 0 n0 ] / / n lim + λ k ) λ k ) n kn 0 Since, λ k <, k n 0, the second product is non zero, and as λ k < and λ n < /, the limit exists by Claim 3.3. And thus, e F ω) dp ω) < e F ω) dp ω) < making e F ω) integrable. Now, as F ω) lim N N λ nφ f n ω), with the sum converging in L -sense, there exists a subsequence N k such that But F ω) lim k e N k k λ n Φ f n ω) P -a.e. ω Nk λnφ fn ω) e F L So, by Dominated Convergence Theorem, e F ω) dp ω) lim e Nk λnφ fn ω) dp ω) k deti + A) ] / Case D I: As seen before, in this case, A is trace class on l G), and so, A λ n f n, f n Then, for ω l G), we have that ω, Aω λ n f n, ω This leads us to define: ω, Aω def λ n Φ f n ω) λ n Φ f n ω) F ω) Hence, we can rewrite the result of Example 3.5 as e ω,aω dp ω) deti + A) ] /

13 3.4 Variance Induced Measures Let A be trace class on ld G) and suppose A >. As usual, write, A λ n f n, D f n λ n < where {f n } is an orthonormal basis of l D G). We were prompted to define And we showed that ω, Aω D def λ n Φ f n ω). ω, Aω D L, dp Claim 3.4). exp ω, Aω ) D dp ω) deti + A)] / Example 3.5) We can now introduce the probability measure and discuss its effect on the GRF. dq deti + A)] / e ω,aω D dp Claim 3.6. {Φ fn,..., Φ fnk } are jointly Gaussian w.r.t. dq, with variance matrix δij + λ ni ) ] i,j K Proof. Consider E dq e i t Φ fn t K Φ fnk )] deti + A)] / exp and m N, ) K exp i t j Φ fnk exp i ) K t k Φ fnk exp e i t Φ fn t K Φ fnk ) dq ) λ k Φ f k ω) dp m λ k Φ f k ω)) exp m λ k Φ f k ω)) Recall that in Exercise 3.5, we have shown that there exists a subsequence m j such that m j λ k Φ f k ω) lim λ k Φ f k ω) j P -a.e. ω 3

14 and j, Then obviously, exp lim j exp m j ) λ k Φ f k ω) i exp ) K t k Φ fnk exp i exp ) F ω) L ) λ k Φ f k ω) ) K t k Φ fnk exp m j )] λ k Φ f k ω) Thus, by Dominated Convergence Theorem, we get that )) exp i t Φ fn t K Φ fnk dq K lim deti + A)] / exp i t k Φ fnk m j ) ] λ k Φ f j k ω) dp mj lim + λ k) / K j π) m exp i t k x nk m j ) ] + λ k )x j/ R m k dx j Now, we integrate out all the coordinates except {x n,..., x nk }, and for all m j max{n k : k K} we get mj + λ k) / π) j/ exp i R m j K t k x nk m j ) + λ k )x k dx K + λ n k ) / K π) K/ exp it k x nk + λ ] ) n k x n R K k dx n dx nk And thus, by uniqueness of characteristic function, {Φ fn,..., Φ fnk } are jointly Gaussian w.r.t. dq, with variance matrix δij + λ ni ) ] i,j k Now consider the random variables Φ δn ω) over dq. Proposition 3.7. {Φ δn } DI + A). form a GRF over, F, dq), with variance 4

15 Proof. First of all, notice that {Φ δn } form a GRF over, F, dp ) with variance D, essentially by construction. It is obvious, when we apply Proposition.9 to any finite subcollection of δ n s. Now, over dq, consider t Φ δn t K Φ δnk Expanding in f n, we have that δ nk j f j, δ nk D f j and thus, K t k δ nk K f j, t k δ nk j where δ t K t kδ nk. And so, D f j f j, δ t D f j j K t k Φ δnk Φ K t kδ nk Φ δt f j, δ t D Φ fj j with the sum on the right hand side converging in L, dq). Therefore, there exists a subsequence J l such that Now, consider J l Φ δt lim f j, δ t D Φ fj Q-a.e. l j E dq e i t Φ δn +...+t K Φ δnk )] ) e i t Φ δn +...+t K Φ δnk dq e iφ δ t dq By Dominated convergence Theorem, bounded by ), we get e iφ δ t dq lim e i J l j f j,δ t D Φ fj dq l But by previous claim, Φ fj s have variance δ ij + λ ni ) ] w.r.t. dq, i,j k 5

16 so lim e i J l j f j,δ t D Φ fj dq l lim exp J l ] fj, δ t l D + λj ) j lim exp J l ] I + A) / f j, δ t l D j lim exp J l ] f j, I + A) / δ t l D j exp ] f j, I + A) / δ t D j exp ) I + A) / δ t D exp ) I + A) / δ t, I + A) / δ t D exp ) δt, I + A) δ t D exp δt, DI + A) ) δ t exp t i t j δni, DI + A) δ nj i,j And thus, {Φ δn } form a GRF w.r.t. dq, with variance DI + A). 3.5 Mutual Absolute Continuity Let D, D be two variances under usual assumptions, and let P, P be the corresponding Gaussian measures on, F ). That is {Φ δn } are jointly Gaussian w.r.t. P i with variance D i, i,. Question: When are P and P mutually absolutely continuous? Theorem 3.8 Shale Theorem). P and P are mutually absolutely continuous if and only if the operator D D I) is trace class on l D. Remark 3.9. This is a fundamental result Remark 3.0. ω, D D I)ω D ω, D D D I)ω ω, D D )ω sym D D )ω, ω D D D I)ω, ω D D I)ω, ω D That is D D I) is self-adjoint on ld, and the statement of the theorem makes sense. 6

17 Remark 3.. Trace class assumption means: D D I λ n f n, D f n D / D D / ] D / D D / D / λ n f n, D D / f n λ n D / f n, D / f n As {D / f n } form an orthonormal basis of l G),D / D D / I) is trace class on l G). This implies that D / D D )D / is trace class. By properties of trace class operators, this yields that D D ) is trace class. So, the natural formulation of Shale Theorem is P and P are mutually absolutely continuous if and only if D D ) is trace class on l G). Exercise. Express the D D ) is trace class condition in terms of matrix elements. In particular, show that is necessary. n,m G D ] nm D ] nm < Proof of Shale Theorem. We will prove only one direction, namely If D D I) is trace class on l D, then P and P are mutually absolutely continuous. So, write D D I) λ n f n, D f n where {f n } form an orthonormal basis of ld, and λ n <. Consider now the random variables {Φ fn } on L, dp ). Φ fn Φ fm dp f n, D f m f n, f m D δ nm Then, as {Φ fn } are jointly Gaussian w.r.t. dp, we have that E dp e i ] N tnφ fn e i N tnφ fn dp 7

18 e i N π) N/ tnxn e N ] x n dx dx N R N Now, D D I) is trace class, so λ n δ nm f n, D D I)f m D f n, D D D I)f m f n, D f m f n, D f m f n, D f m δ nm f n, D f m + λ n )δ nm So, if we consider {Φ fn } w.r.t. dp, they have variance D D nm ], D nm + λ n )δ nm. Also, notice that f n, Df n > 0 λ n >, n. We can now compute E dp e i ] N tnφ fn ] / π) N/ N + λ n) Consider K + λ n k ) ] } {{ } P K P K π) K/ P K π) K/ Thus, lim N R N e i N tnφ fn dp e i ] N tnxn e N x n +λn dx dx N e i K t kφ fnk e K Φ +λn K k fn k e Φ fn k dp e i K t kx nk K e R K e i K t kx nk K e R K E dp e i K ] tn k Φ fn k N ] / + λ n) x +λn n k k e K x n k e K x n k }{{} dx x +λn n k k dx e i K tn k Φ fn k dp e i K tn k Φ fn k dp e i K t kφ fnk e N λn +λn Φ fn dp since for N > max{n k : k K} we simply integrate out all the extra coordinates. Now, by Claim 3.3, { lim N N / exists since n)] λ n < + λ λ n > 8

19 Hence, consider the function F ω) e λn +λn Φ fn We will show that F L, dp ), so that the limit and the integral can be interchanged by Dominated Convergence Theorem. So, as λ n Φ f + λ n n dp λ n Φ λ n f + λ n n dp + λ }{{} n < λn +λn Φ fn F is well defined. Hence, set F N e N As λ n <, there is N 0 > 0 such that n N 0, λ n < 3. So for N > N 0, consider F N exp N 0 λ n Φ f + λ n + N λ n Φ f n + λ n n nn 0 + Then, F N is monotonically increasing, and F N dp π) N/ e N0 R N π) N/ exp R N And lim N N nn 0 + lim N N0 N 0 + λ n ) λn +λn x n e N nn λ n x n ] / N nn 0 + λ ) n / lim + λ n N nn 0 + N λn +λn x n e N x ndx N nn 0 + λ ) n + λ n λ ) n / + λ n N nn 0 + x n dx λ ) n / /3 3 λ ) n / < by Claim 3.3) as 3 λ n < and λ n < /3. Thus, we have a uniform bound F N dp N0 + λ n ) ] / 9 N nn 0 + λ ) n / < + λ n

20 Hence, as F N is monotonically increasing, by Monotone Convergence Theorem, we have that: ] F N dp F N dp F dp < lim N lim N Now, as F N F N F L, dp ) for all N > N 0, we apply Dominated Convergence Theorem to get F N dp F dp F dp < F L, dp ) So that lim N e i K t kφ fnk dp We will now show that + λ n ) ] F ω) dp + λ n)] / dp First, consider δ nk j f j, δ nk D f j, and Φ δnk f j, δ nk D Φ fj j e i K t kφ fnk F ω)dp is converging in L -sense in both L, dp ) and L, dp ), because the expansion is in the same space, namely ld. Then, by taking two successive subsequences, we get J m such that K J m K t k Φ δnk lim f j, t k δ nk Φ fj m j both P -a.e. ω and P -a.e. ω, and thus, exp i K t kφ δnk ) dp lim m lim m exp i + λ n ) / J m j f j, exp i K t k δ nk J m j f j, D Φ fj D K t k δ nk dp D Φ fj ] / ) K + λ n ) exp i t k Φ δnk F dp 0 F dp

21 That is the characteristic functions of the marginals of the measures ) dp ) and + λ n ) / F dp are the same, which implies that the marginals are the same. Thus, these measures coincide on the cylinders, and are therefore the same. Exercise 3. Show that F L p, dp ) for some p >, and find the optimal p in terms of λ n s. 4 Weak Convergence of Gaussian Measures 4. General Definition Let M, d) be a complete and separable metric space, F - the Borel σ-algebra on M, and {P t } t>0 or {P n } n N ) a family of Borel probability measures on M, F ). Defnition 4.. We say that P t converges weakly to P as t if for any bounded continuous function F : M R, F dp t F dp We then write P t P. lim t M For a more detailed discussion of weak convergence, see the first 8 pages of Convergence of Probability Measures by Patrick Billingsley. 4. The case of GRF Suppose now, for our setting: R G F Borel σ-field P t Gaussian measures with variances D t) Φ δn GRF over P t with variance D t) Φ δn X n ) For α f l G), the map M P hi α ω) αn)ωn) is continuous, and so, e iφαω) is continuous as well.

22 So, if P t P, we must have e iφαω) dp t e iφαω) dp lim e / α,dt) α e t iφαω) dp We don t want the integral on the right to be neither 0, nor, so we assume that {D t) } are uniformly bounded, i.e. that there exist 0 < m M such that m D t) M m α α, D t) α M α Then, By polarization, lim α, D t) α t lim α, D t) β t exists exists for α, β f l G). By continuity and boundedness, the limit exists α, β f l G). If α δ n and β δ m, D nm lim t D nm t) exists. Let D be an operator with matrix D nm ]. Then, D is self-adjoint and bounded, namely Moreover, α, β f l G)m Hence, D is a good variance, and m D M lim α, D t) β α, Dβ t lim e / α,dt) α e / α,dα t so that, P is Gaussian with variance D. Theorem 4.. Let P t be a family of Gaussian measures with variances D t) such that for some M, m > 0, m D t) M Then P t P if and only if D t) D, and in that case, P variance D. Proof. We have just shown the direction. Conversely, if D t) D, e iφαω) dp t e / α,dt) α e / α,dα From Billingsley, this implies that for every cylinder C And this implies that P t P. lim P tc) P C) t e iφαω) dp

23 Corollary 4.3. Let {P t } t>0 be a family of Gaussian measures with variances D t) satisfying, 0 < m D t) M Then there exists a subsequence t n such that P t P for a Gaussian P. Proof. By diagonal argument, extract a subsequence t n such that and by theorem, we are done. D tn) D 3