Solution to selected problems.

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Chapter 2. Semimartingales and Stochastic Integrals 1. Let x 0 ∈ R be a discontinuous point of f. Wlog, we can assume that f is a right continuous function with left limit and △f(x 0 ) = d > 0. Since inf t {B t = x 0 } < ∞ and due to Strong Markov property of B t , we can assume x 0 = 0. Almost every Brownian path does not have point of decrease (or increase) and it is a continuous process. So B· visit x 0 = 0 and changes its sign infinitely many times on [0, ɛ] for any ɛ > 0. Therefore, Y t = f(B t ) has infinitely many jumps on any compact interval almost surely. Therefore, ∑ (△Y s ) 2 = ∞ a.s. s≤t 2. By dominated convergence theorem, lim E Q[|X n − X| ∧ 1] = lim E P n→∞ n→∞ [ |X n − X| ∧ 1 · dQ dP ] = 0 |X n − X| ∧ 1 → 0 in L 1 (Q) implies that X n → X in Q-probability. 3. B t is a continuous local martingale by construction and by independence between X and Y , [B, B] t = α 2 [X, X] t + (1 − α 2 )[Y, Y ] t = t So by Lévy theorem, B is a standard 1-dimensional Brownian motion. [X, B] t = αt, [Y, B] t = √ 1 − α 2 t 4. Assume that f(0) = 0. Let M t = B f(t) and G t = F f(t) where B· is a one-dimensional standard Brownian motion and F t is a corresponding filtration. Then E[M t |G s ] = E [ B f(t) |F f(s) ] = Bf(s) = M s Therefore M t is a martingale w.r.t. G t . Since f is continuous and B t is a continuous process, M t is clearly continuous process. Finally, [M, M] t = [B, B] f(t) = f(t) If f(0) > 0, then we only need to add a constant process A t to B f(t) such that 2A t M 0 +A 2 t = −B 2 f(0) for each ω to get a desired result. 5. Since B is a continuous process with △B 0 = 0, M is also continuous. M is local martingale since B is a locally square integrable local martingale. [M, M] t = ∫ t 0 H 2 s ds = ∫ t 0 1ds = t So by Lévy ’s characterization theorem, M is also a Brownian motion. 12
6. Pick arbitrary t 0 > 1. Let Xt n = 1 (t0 − 1 ,∞)(t) for n ≥ 1, X t = 1 [t0 ,∞), Y t = 1 [t0 ,∞). Then n X n , X, Y are finite variation processes and Semimartingales. lim n Xt n = X t almost surely. But lim n [X n , Y ] t0 = 0 ≠ 1 = [X, Y ] t0 7. Observe that [X n , Z] = [H n , Y · Z] = H n · [Y, Z], [X, Z] = [H, Y · Z] = H · [Y, Z] and [Y, Z] is a semimartingale. Then by the continuity of stochastic integral, H n → H in ucp implies, H n · [Y, Z] → H · [Y, Z] and hence [X n , Z] → [XZ] in ucp. 8. By applying Ito’s formula, [f n (X) − f(X), Y ] t =[f n (X 0 ) − f(X 0 ), Y ] t + + 1 2 ∫ t 0 ∫ t (f” n (X s ) − f” n (X s ))d[[X, X], Y ] s =(f n (X 0 ) − f(X 0 ))Y 0 + 0 ∫ t 0 (f ′ n(X s ) − f ′ (X s ))d[X, Y ] s (f ′ n(X s ) − f ′ (X s ))d[X, Y ] s Note that [[X, X], Y ] ≡ 0 since X and Y are continuous semimartingales and especially [X, X] is a finite variation process. As n → ∞, (f n (X 0 )−f(X 0 ))Y 0 → 0 for arbitrary ω ∈ Ω. Since [X, Y ] has a path of finite variation on compacts, we can treat f n (X) · [X, Y ], f(X) · [X, Y ] as Lebesgue-Stieltjes integral computed path by path. So fix ω ∈ Ω. Then as n → ∞, sup |f n (B s ) − f(B s )| = sup |f n (x) − f(x)| → 0 0≤s≤t inf s≤t B s≤x≤sup s≤t B s since f n ′ → f uniformly on compacts. Therefore ∫ t ∣ ∣∣∣ ∣ (f n(X ′ s ) − f ′ (X s ))d[X, Y ] s ≤ sup |f n (B s ) − f(B s )| 0≤s≤t 0 ∫ t 0 d|[X, Y ]| s −→ 0 9. Let σ n be a sequence of random partition tending to identity. By theorem 22, ∑ ( ) ( ) [M, A] = M 0 A 0 + lim M T i+1 n − M Ti n A T i+1 n − A Ti n n→∞ i ∣ ≤ 0 + lim sup n→∞ i ∣ ∣M T i+1 n − M Ti n ∣ ∑ i ∣ ∣A T n i+1 − A T n i ∣ = 0 since M is a continuous process and ∑ ∣ i ∣A T i+1 n − A Ti n ∣ < ∞ by hypothesis. Similarly ∑ ( ) ( ) [A, A] = M0 2 + lim A T i+1 n − A Ti n A T i+1 n − A Ti n n→∞ Therefore, ≤ 0 + lim sup n→∞ i i ∣ ∣A T i+1 n − A Ti n ∣ ∑ i [X, X] = [M, M] + 2[M, A] + [A, A] = [M, M] ∣ ∣A T n i+1 − A T n i ∣ = 0 13
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: So that Y R ◦ θ R (ω) = { 1 R
Page 15 and 16: 15. If M 0 = 0 then by Theorem 42,
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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