Levy Process and Infinitely Divisible Law

Transcription

1 Stat205B: Probability Theory (Spring 2003) Lecture: 26 Levy Process an Infinitely Divisible Law Lecturer: James W. Pitman Scribe: Bo Li Levy Processes an Infinitely Divisible Law Well begin the lecture with some classical istribution theory for Levy processes an infinitely ivisible istributions. Definition 1.1 X = {X(t)} t 0 is sai to be a Levy process if 1. X has inepenent increments. 2. X(0) = 0 a.s. 3. X is stochastically continuous (also calle continuous in probability or P-continuous), if, for s 0 X(t + s) X(s) P 0 as t 0 4. X is time homogeneous, i.e., for t 0, L(X(t + s) X(s)) oes not epen on s X is rcll (right continuous with left limits) almost surely. Without 5, X is sai to be a Levy process in law. Its easy to check that both Brownian Motion an Poisson process are Levy processes. Theorem 1.1 The marginal istribution of X(t) is etermine by X(1). Proof: Since X(t) has inepenent increments an is time homogenous, we can write ϕ s+t (u) = Ee iux(s+t) = Ee iu(x(s+t) X(t)) e iux(t) = Ee iu(x(s+t) X(t)) Ee iux(t) = Ee iux(s) Ee iux(t) = ϕ s (u)ϕ t (u) The feature of P continuity implies ϕ t (u) is continuous with respect to t for all u. Hammels theorem implies that ϕ t (u) = exp(tφ(u)), an further ϕ 1 (u) = exp(φ(u)). Hence ϕ t (u) = (ϕ 1 (u)) t. Definition 1.2 Two stochastic processes {Z 1 (t)} t T an {Z 2 (t)} t T are moifications (also calle inistinguishable) of each other, if P {Z 1 (t) = Z 2 (t)} = 1 for all t T 1-1

2 1-2 Levy Process an Infinitely Divisible Law There is a theorem that blurs the ifference between Levy processes in law an Levy processes. Theorem 1.2 Each Levy process in law has a moification that is a Levy process. The general proof is elicate an well skip it. Definition 1.3 A ranom vector Y is infinitely ivisible (i) if, for each n N, there is an i.i.. sequence Y n,1,..., Y n,n so that Y = Y n,1 + + Y n,n The next result provies the basic link between Levy processes an triangular arrays. Theorem 1.3 (Levy processes an infinitely ivisibility) For any ranom vector Y in, these conitions are equivalent: I. Y is infinitely ivisible. II. Y n,1 + + Y n,rn Y for some i.i.. array (Yn,j ) n 1,rn j 1, where r n. III. Y = X 1 for some Levy process X in. Two lemmas are neee for the proof. Lemma 1.4 If Y n,j are such as in II, then Y n,1 P 0. Proof: Let µ an µ n enote the istribution of Y an Y n,1 respectively. Choose r > 0 so small that ϕ µ 0 on [ r, r], an write ϕ µ = e ψ on this interval, where ψ : [ r, r] R is continuous with ψ(0) = 0. For each u [ r, r], since ϕ rn µ n (u) ϕ µ (u), we see for sufficiently large n, ϕ µn (u) 0. Thus, we may write ϕ µn (u) = e ψn(u), where r n ψ n (u) ψ(u). Hence ψ n (u) 0 an therefore ϕ µn (u) 1. Now let ɛ r 1, an note r r 1 ϕ n (u)u = 2r (1 sin rx rx )µ n(x) 2r(1 sin rɛ )µ n ( x ɛ) rɛ As n, the left-han sie tens to 0 by ominate convergence theorem, an we get µ n w δ0. Lemma 1.5 (Kolmogorov consistency) Given istribution functions {{F t : [0, 1]} t T } N, there exists a stochastic process {Z(t)} t T with these istributions as its Fiis, iff, the following two consistency conitions hol F...,ti 1,t j,t i+1,...,t j 1,t i,t j+1,...(..., x i 1, x j, x i+1,..., x j 1, x i, x j+1,... ) = F t (x); lim xk+1 F t,tk+1 (x, x k+1 ) = F t (x). Well skip the proof of this lemma.

3 Levy Process an Infinitely Divisible Law 1-3 Proof: (of Theorem 1.3) I II is trivial. II I. Write m n = [ rn 2 ] an let S 2n = (Y 2n,1 + + Y 2n,mn ) + (Y 2n,n Y 2n,2mn ) = Z n,1 + Z n,2 The ranom vectors Z n,1 an Z n,2 are inepenent an have the same istribution. Lemma 1.4 implies S 2n Y. Then the istribution of Z n,1 is a tight sequence since P (Z n,1 > z) 2 = P (Z n,1 > z)p (Z n,2 > z) P (S 2n > 2z) an similarly P (Z n,1 < z) 2 P (S 2n < 2z). Similarly we see that Z n,2 is also tight. So we can take a subsequence n k so that Z nk,1 Z 1 an Z nk,2 Z 2. Then Y = Z 1 + Z 2. A similar argument shows that Y can be ivie into k 2 pieces. III I. X(1) = X 1 + (X 2 X 1 ) + + (X 1 X n 1 ) yiels the esire result. n n n n I III. We specify the law of F t1,...,t n through that of F t1,t 2 t 1,...,t n t n 1, for 0 < t 1 <... < t n, as that with ch.f. ϕ Y (θ 1 ) t1 ϕ Y (θ n ) t2 t1... ϕ Y (θ n ) tn tn 1. These istributions are consistent. Thus Lemma 1.5 gives us a process X, with these Fiis, that must be a Levy process in law with X(1) = Y. By Theorem 1.1, these exists a Levy process with the same Fiis. The proof is not re- The following result is of the most funamental importance in probability. ally ifficult, but too technical to be worthwhile oing here. Theorem 1.6 (Levy-Khintchine Formula) Let X be a Levy process in. There exists a triplet (A, γ, ν) of A a symmetric non-negative efinite matrix (the Gaussian convariance) γ a constant in ν a measure on with ν0 = 0 an ( y 2 1)ν(y) < (the levy measure) which in that case is uniquely etermine, such that, for all u an t 0, Ee iux(t) = e tψu, where ψ u = 1 2 < u, Au > +i < u, γ > + (e i<u,y> 1 1 ( y 1) i < u, y >)ν(y) If γ 0 = γ yν(y) is well-efine an finite, then we may rewrite the Levy-Khintchine Formula y 1 with a new triplet (A, γ 0, ν) 0 (the rift), as ψ u = 1 2 < u, Au > +i < u, γ 0 > + (e i<u,y> 1)ν(y) If γ 1 = γ + yν(y) is well-efine an finite, then we can rewrite the Levy-Khintchine Formula y >1 with a new triplet (A, γ 1, ν) 1 (the center), as ψ u = 1 2 < u, Au > +i < u, γ 1 > + (e i<u,y> 1 i < u, y >)ν(y) Definition 1.4 A compoun Poisson process is a Levy process with generating triplet (0, 0, λσ) 0, where λ > 0 is a constant an σ a probability measure on with σ{0} = 0. Theorem 1.7 Let {N(t)} t 0 be a Poisson process with rate λ, an {Y k } k=1 i.i.. rv.s, inepenent of N, with L(Y k ) = σ, where σ{0} = 0. Denoting S 0 = 0, S n = Σ n k=1 Y k for n N, X(t) = S N(t) is a compoun Poisson process with generating triplet (0, 0, λσ).

4 1-4 Levy Process an Infinitely Divisible Law Proof: Rcll sample path an X(0) = 0 are immeiate. P continuity follows from P { X(t + s) X(s) > ɛ} P N(s + t) N(s) > 0} = 1 e λ t 0 as t 0 Inepenence an homogeneity of increments come by conitioning on the values of N involve. The generating triplet is the claime one, since Ee i<u,x(1)> = Σ n=0(ee i<u,s1> ) n λn n! e λ = exp{λ (e i<u,y> 1)σ(y)} Definition 1.5 Y is stable if, for each n N, with Y 1,..., Y n i.i.. copies of Y, Y Y n = by +c for some constants b = b(n) > 0 an c = c(n). Y is strictly stable if it is possible to take c(n) = 0 for n N. From the efinition, its straightforwar that stable rvs are i. Stable istributions are among the few most important i istributions. Two reasons are their stability uner aition an the explicitness of their ch.f.(see below). For an i ranom vector Y, Y r enotes an ranom vectores with ch.f. ϕ r Y. Definition 1.6 A stable Y is calle α stable, α (0, 2], whenever Y r = r 1/α Y + c for t > 0 an some constant c = c(r). Y is calle strictly α stable if c(t) = 0 for t > 0. For Y non-trivial (strictly) stable, there exists a unique constant α (0, 2] such that Y is α stable (strictly). Definition 1.7 A Levy process X with X(1) (strictly) α stable is calle a (strictly) α stable Levy motion. Theorem 1.8 Let X be a non-trivial Levy process in R with generating triplet (A, γ, ν). Then X is α stable for some α > 0 iff exactly one of these conitions hols: 1. α = 2 an ν = α (0, 2), A = 0, an ν(x) = (c + 1 (0, ) (x) + c 1 (,0) (x)) x (α+1) x on R for some c +, c 0. Proof: Suppose the generating triplet of Y = X(1) is (A, γ, ν). We know Y is α stable iff Y rα = ry +c for t > 0 an some constant c. Since the characteristics of Y rα an ry +c are r α (A, γ, ν) an (r 2 A, rγ + c, ν Sr 1 ) respectively, where S r : x rx for any r > 0. It follows that X(t) is α stable iff r α A = r 2 A an r α ν = ν Sr 1 for all r > 0. Thus A = 0 when α 2. Writing F (x) = ν[x, ) or ν(, x], then r α ν = ν Sr 1 implies r α F (rx) = F (x) for all r, x > 0, an so F (x) = x α F (1). The conition (x 2 1)ν(x) < implies F (1) < an when α = 2 we have F (1) = 0. This complete the proof.

5 Levy Process an Infinitely Divisible Law 1-5 Reference: [1]: Kallenberg, Founation of Moern Probability [2]: Patrik Albin, From Levy processes to Semimartingales (lecture notes) palbin/levy.html