Homework #6 : final examination Due on March 22nd : individual work
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1 Université de ennes Année Master 2ème Mathématiques Modèles stochastiques continus ou à sauts Homework #6 : final examination Due on March 22nd : individual work Exercise Warm-up : behaviour of characteristic exponent Let ηu be the characteristic exponent of a real-valued Lévy process : ηu = ibu [ 2 Γu2 + e iuy ] iuy { y <} νdy. ηu [. Show that lim u u 2 = Γ. One can show that lim e iuy ] u u 2 iuy { y <} νdy =. 2. Show that X has bounded variations on every time interval a.s. if and only if Γ = and y νdy <. Use Lévy-Itô decomposition or the exponential formula for PM. In that case lim u ηu u 2 convergence as in the rst point. = d, where d = b y νdy is the drift. Use the dominated 3. Show that the characteristic exponent η is bounded if and only if X is a compound Poisson process. If you assume that η is bounded, show that e ηu = cosuyνdy and that e ty2 /2 νdy = 2πt e λ2 /2t e ηλdλ supe ηλ. λ Deduce further that X has bounded variations in the sense of the preceding point. Exercise 2 First round : subordinators A non-decreasing Lévy process with values in + is called a subordinator.. Show that a real valued Lévy process X = X t : t is a subordinator if its characteristic exponent has the form ηu = ibu + e iyu νdy, where b and the Lévy measure has the support in + and satises y νdy, or equivalently y νdy <. 2 + y Moreover X t = tb + s t X s. Use the positivity and the monotonicity to prove that X contains no Brownian part. Then use the existence of all moments of the third term in the Lévy-Itô decomposition of X to obtain 2.
2 2. Show that for a subordinator where the Laplace exponent φλ = ηiλ = bλ + E e λxt = e tφλ, 3 Analyse the analytic continuation of u E e iuxt. e λy νdy A subordinator X is called one-sided stable process if for each a there corresponds a constant ca such that ax t and X t ca have the same law. a Show that c in this denition is continuous ans satises the equation caã = cacã. Then deduce that ca = a α, with some α > the index. b Deduce further that φa = caφ and hence E e λxt = e t r λα, r > the rate. 5 By using the concavity of φ deduce that α,. c Prove or assume that for α,, e λy Γ α dy = λ α, y+α α where here Γ is the Euler's gamma function. Deduce that the stable subordinators with index α and rate r have the Laplace exponent 4 with Lévy measure νdy = αr Γ α. 6 y+α d An example : the stable subordinator with index 2 and rate have the probability density t f Xt s := 2 s 3/2 e t2 /4s, s. π Indeed, set g t λ := E e λxt = e λs f Xt sds, prove that g tλ = t/2 λg t λ, that g t = and deduce that g t λ = e tλ/2. Exercise 3 ising scale : transience and recurrence Let X = X t : t be a Lévy process with the characteristic exponent η. Denote by P t the associated semigroup, given by P t fx = fx + ypx t dy, with f a non-negative measurable function. ecall that the resolvent λ is given, for measurable f by λ fx = e λt P t fxdt = E x e λt fx t dt.. Let τ an exponential random time with parameter λ. Show that E fx τ = λ λ f. 2
3 2. Denote the Fourier transform of a function g L by Fgu = e i u x gxdx attention, not the same denition as in the course. Show that for every f L L, FP t fu = exp { tη u } Ffu, and F λ fu = λ + η u Ffu, 7 where t and λ >. Moreover if A denotes the generator of P t and D its domain, show that if f D and Af L, then FAfu = η uffu. 3. Let us introduce a familiy of measures called the potential measures {Ux, : x } given, for B B, by Ux, B = P x X t Bdt = E x {Xt B}dt [, ]. If T B = inf{t : X t B} denotes the rst entrance time into B, show that Ux, B = E x {Xt B}dt = Uy, BP x X TB dy, 8 where B is the closure of B. T B 4. We say that the process X is transient if the for every compact set K, Ux, K <, x, or equivalent if U, K < since Ux, K = U, K {x}. Here we denoted by B B = {x x : x B, x B }. We say that a process is recurrent if U, B = for every open ball B centered in. We want to prove that the process is either transient or recurrent. a Suppose that ε > such that U, B <, where B = ε, ε, and let B = [ ε 3, ε 3]. Justify the following relations Ux, B sup y B Uy, B = sup y B U, B {y} U, B B U, B <. Use 8 for the rst inequality. b Deduce that for every y, Ux, {y} + B < and then Ux, K < for any compact K. 5. Test of transience : the Lévy process X is transient i for some r > small enough lim sup λ r r e B λ + ηu du <. 9 a For r > arbitrary small consider f = [ r,r] [ r,r] the convolution. Clearly it can be proved seen that f 2r [ 2r,2r] is continuous non-negative with support [ 2r, 2r] and also that { [2/u sinru] 2 if u Ffu = 4r 2 otherwise 3
4 is a bounded continuous non-negative function. Show that for λ >, λ f = [ 2 ] 2e 2π u sinru du. λ + η u Use Fourier inversion, 7 and the fact that λ f is a real number. b Deduce that and latter quantity is innite whenever and then X is recurrent. 2rU, [ 2r, 2r] lim λ λ f, lim sup λ c Conversely, assume that for r >, Set r r e 2r lim sup e λ 2r λ + ηu λ + ηu du = du <. { [2/x sinrx] 2 if x gx = fu = 4r 2 otherwise having its Fourier transform Fgu = 2π [ r,r] [ r,r]. Deduce un expression of λ g. One can use the same argument as in the previous point. d Prove that U, [ π 3r, π 3r ] <. For instance one can prove and use that gx r2, when x π 3r. Conclude that X is transient. Exercise 4 Final step : pathwise uniqueness Let X = X t : t be a one-dimensional symmetric stable with index α, 2 having its Lévy measure given by νdz = z α on and its generator Lfx = [fx + z fx { z } zf x] z α dz, for C 2 functions.. Set X n t = s t X s { Xs n}. Show that X n is a Lévy process and give its Lévy measure, then show that X n is a square integrable martingale. 2. Let H t be a bounded predictable process. Show that Zt n := t H sdxs n is a square integrable martingale. 3. Set U n t = X t X n t. Show that E t H s dus n CE X s { Xs >n}, for some constant C. Show that the right hand side of the latter inequality is nite and tends to, as n. 4. Deduce that the process Z t = t H sdx s is a martingale. 4 s t
5 5. Let f be a C 2 b function with bounded rst and second derivatives and set Ks, z := fz s + H s z fz s f Z s H s z. Justify each of following equalities = fz + where M t = fz t = fz + t t f Z s dz s + f Z s dz s + t t t with intensity measure dtνdz. 6. Prove that for each m, V m t = f Z s dz s + s t[fz s fz s f Z s Z s ] t Ks, znds, dz = fz + M t + t Ks, zdsνdz, Ks, zñds, dz. Here N is the PM associated to X z m Ks, zñds, dz is a martingale and that M t is the limit of martingales t f Z s dz s + Vt m. One can use the fact that for each k > m, Vt k Vt m is a martingale and show that t E Ks, z Nds, dz + dsνdz C m α, m< z for some constant C not depending on m. 7. Show that, if H s, we have t Ks, zdsνdz = H s α LfZ s. One should come back to the expression of K, perform the change of variable w = H s z and recall the expression of L. Conclude that even if H s =, fz t = fz + M t + 8. We study the uniqueness of the following SDE t where F is supposed bounded continuous such that H s α LfZ s ds. dy t = F Y t dx t, Y = y, 2 F x F y ρ x y, x, y, 3 with ρ : [, a non-decreasing continuous function, ρ =. More precisely we try to prove that if dx =, 4 ρx α then the solution of the SDE 2 is pathwise unique. + a Let Y and Y 2 be any two solutions of 2 set Z t = Yt Yt 2 and H t = F Yt F Y 2 Deduce that Z t = t H sdx s. b Dene A t = t H s α ds. Justify that A t < use the fact that F is bounded. t. 5
6 c Let a n such that a n a n+ ρx α dx = n. For each n let h n be a non-negative C 2 function with the support in [a n+, a n ], whose integral is and with h n x 2/nρx α. Why is this possible? d Denote by p t x, y the transition density for X t, that is the density of P x X t dy. Fix λ >, let r λ x = e λt p t x, dt and λ fx = fyr λ x ydy. Who is λ here? It can be proved that r λ x is bounded and is continuous in x admitted. Furthermore r λ x < r λ, if x. Finally, set f n = λ h n x. Show that if h n is C 2, then f n is C 2. One can interchanges dierentiation and integration and uses translation invariance. e Show that Lf n = L λ h n = λ λ h n h n = λf n h n. f Use Itô's product formula and to deduce E e λat f n Z t f n = E t g Conclude from the preceding two points that f n E e λat f n Z t = E e λas H s α Lf n Z s ds E t t e λas H s α h n Z s ds. e λas λ H s α f n Z s ds. Show, by using the properties of h n and the fact that H s ρz s, that the right hand side of the latter equality is less that 2t/n hence tends to, as n. h Justify that h n ydy δ weakly, as n and that f n x r λ x as n. Show that r λ E e λat r λ Z t =. i Conclude that PZ t = = for each t hence Z is identically. Exercice Bonus Last word : another approach to pathwise uniqueness Consider the same SDE 2 as in 8 of the Exercise 4 and make the same assumptions on F and ρ. ecall that N is the PM associated to X with intensity measure dtνdz.. Explain why the equation 2 can be written as Y t = y + t z t F X s zñds, dz + z > F X s znds, dz Consider again the sequence a n such that a n a n+ ρx α dx = n and also h n non-negative C 2 even functions with the support in [a n+, a n ], whose integral is and with h n x 2/nρx α. Set ux = x α and u n = u h n. Justify that u n s ux, as n. 3. In this question we prove that Lu n = cu n with c a constant independent of n. Set u ɛ x = uxe ɛ x and u ɛ n = u ɛ h n. The functions u ɛ n belongs to S. a Use Exercise 3 point 2, with same notations, and show FLu ɛ nξ = cα ξ α Fu ɛ nξ. b Show that Fu ɛ nξ = c α [ ɛ iξ α + ɛ iξ α] Fh n ξ. c For ξ compute lim ɛ [ ɛ iξ α + ɛ ξ α] and deduce Lu n = lim ɛ Lu ɛ n = c u n. 6
7 4. Denote again Y and Y 2 two solutions of 2 and set Z t = Yt Yt 2. Show that + t u n Z t u n = t F Y s F Y 2 s α Lu n Z s ds [ u n Zs + F Y s F Y 2 s z u n Z s ]Ñds, dz. 5. Show that F x F y α Lu n x, y cρ x y α h n x y c/n. 6. Set, for k, T k = inf{t > : Z t > k}. Deduce that [ ] [ t T k E u n Z t Tk u n + E Conclude that E [ Z t Tk α ] = and then then Z t = a.s. c ] n ds. 7
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