Solution to selected problems.

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10. X 2 is P -semimartingale and hence Q-semimartingale by Theorem 2. By Corollary of theorem 15, (X − · X) Q is indistinguishable form (X − · X) P . Then by definition of quadric variation, [X, X] P = X 2 − (X − · X) P = X 2 − (X − · X) Q = [X, X] Q up to evanescent set. 11. a) Λ = [−2, 1] is a closed set and B has a continuous path, by Theorem 4 in chapter 1, T (ω) = inf t : B t /∈ Λ is a stopping time. b) M t is a uniformly integrable martingale since M t = E[B T |F t ]. Clearly M is continuous. Clearly N is a continuous martingale as well. By Theorem 23, [M, M] t = [B, B] T t = t ∧ T and [N, N] t = [−B, −B] T t = t ∧ T . Thus [M, M] = [N, N]. However P (M t > 1) = 0 ≠ P (N t > 1). M and N does not have the same law. 12. Fix t ∈ R + and ω ∈ Ω. WLOG we can assume that X(ω) has a càdlàg path and ∑ s≤t |△X s(ω)| < ∞. Then on [0, t], continuous part of X is bounded by continuity and jump part of X is bounded by hypothesis. So {X s } s≤t is bounded. Let K ⊂ R be K = [inf s≤t X s (ω), sup s≤t X s (ω)]. Then f, f ′ , f” is bounded on K. Since ∑ s≤t {f(X s) − f(X s− ) − f ′ (X s− )△X s } is an absolute convergent series, (see proof of Ito’s formula (Theorem 32), it suffices to show that ∑ s≤t {f ′ (X s− )△X s } < ∞. By hypothesis, ∑ ∣ s≤t ∣f ′ (X s− )△X s ∣ ∣ ≤ sup x∈K |f ′ (x)| ∑ s≤t |△X s | < ∞ Since this is true for all t ∈ R + and almost all ω ∈ Ω, the claim is shown. 13. By definition, stochastic integral is a continuous linear mapping J X : S ucp → D ucp . (Section 4). By continuity, H n → H under ucp topology implies H n · X → H · X under ucp topology. 14. Let Â = 1 + A to simplify notations. Fix ω ∈ Ω and let Â−1 · X = Z. Then Z ∞ (ω) < ∞ by hypothesis and Â · Z = X. Applying integration by parts to X and then device both sides by Â yields X t Â = Z t − 1 Â ∫ t 0 Z s dÂs Since Z ∞ < ∞ exists, for any ɛ > 0, there exists τ such that |Z t − Z ∞ | < ε for all t ≥ τ. Then for t > τ, ∫ 1 t Z s dÂs = 1 ∫ τ Z s dÂs + 1 ∫ t Â t − (Z s − Z ∞ )dÂs + Z Âτ ∞ Â t Â t Â t 0 0 τ Let’s evaluate right hand side. As t → ∞, the first term goes to 0 while the last term converges to Z ∞ by hypothesis. By construction of τ, the second term is smaller than ε(A t − A τ )/A t , which converges to ε. Since this is true for all ε > 0, 1 lim t→∞ Â ∫ t 0 Z s dÂs = Z ∞ Thus lim t→∞ X t /Ât = 0. 14 Â t
15. If M 0 = 0 then by Theorem 42, there exists a Brownian motion B such that M t = B [M,M]t . By the law of iterated logarithm, lim t→∞ M t B [M,M]t B τ = lim = lim [M, M] t t→∞ [M, M] t τ→∞ τ = 0 If M 0 ≠ 0, Let X t = M t − M 0 . Then [X, X] t = [M, M] t − M 2 0 . So [X, X] t → ∞ and M t [M, M] t = X t + M 0 [X, X] t + M 2 0 → 0 17. Since integral is taken over [t − 1/n, t] and Y is adapted, Xt n is also an adapted process. For t > s ≥ 1/n such that |t − s| ≤ 1/n, ∫ t ∫ t−1/n |Xt n − Xs n | = n Y τ dτ − Y τ dτ ≤ 2n(t − s) ∣ s s−1/n ∣ sup |Y τ | s− 1 n ≤τ≤t Since X is constant on [0, t] and [1, ∞), let π be a partition of [t, 1] and M = sup s− 1 n ≤τ≤t |Y τ | < ∞. Then for each n, total variation of X n is finite since sup |Xt n i+1 − Xt n i | ≤ 2nM ∑ π π (t i+1 − t i ) = 2nM(1 − t) Therefore X n is a finite variation process and in particular a semimartingale. By the fundamental theorem of calculus and continuity of Y , Y s = lim n→∞ n ∫ s s−1/n Y sdu = lim n→∞ Xt n for each t > 0. At t = 0, lim n X n = 0 = Y 0 . So X n → Y for all t ≥ 0. However Y need not be a semimartingale since Y t = (B t ) 1/3 where B t is a standard Brownian motion satisfies all requirements but not a semimartingale. 18. B t has a continuous path almost surely. Fix ω such that B t (ω) is continuous. Then lim n→∞ Xt n (ω) = B t (ω) by Exercise 17. By Theorem 37, the solution of dZt n = Zs−dX n s n , Z 0 = 1 is = exp (X t n − 1 ) 2 [Xn , X n ] t = exp(Xt n ), Z n t since X n is a finite variation process. On the other hand, Z t = exp(B t − t/2) and lim n→∞ Zn t = exp(B t ) ≠ Z t 19. A n is a clearly càdlàg and adapted. By periodicity and symmetry of triangular function, ∫ π 2 0 |dA n s | = 1 n ∫ π 2 Since total variation is finite, A n t lim sup n→∞ 0≤t≤ π 2 0 |d sin(ns)| = 1 n · n ∫ π 2n 0 d(sin(ns)) = ∫ π 2 0 d(sin(s)) = 1 is a semimartingale. On the other hand, |A n t | = lim n→∞ 1 n = 0 15
Page 1 and 2: Solution to selected problems. Chap
Page 3 and 4: 10. (a) Let N t = ∑ i 1 i (N i t
Page 5 and 6: where Z t ′ = ∫ R xN t(·, dx),
Page 7 and 8: 21. a) let a = (1 − t) −1 ( ∫
Page 9 and 10: For this, let t 0 = 0 and t n+1 =
Page 11 and 12: So that Y R ◦ θ R (ω) = { 1 R
Page 13: 6. Pick arbitrary t 0 > 1. Let Xt n
Page 17 and 18: 22. Claim that for all integer k
Page 19 and 20: Chapter 3. Semimartingales and Deco
Page 21 and 22: lemma Let X t be a càdlàg adapted
Page 23 and 24: 20. Let {T n } be an increasing seq
Page 25: 31. Let T be an arbitrary F µ stop

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