Solution to selected problems.

More documents

Recommendations

Info

( By hypothesis E [A, A] 1/2 ∞ ) 1 2 < ∞. By BDG inequality and the fact that M is a bounded martingale, E([M, M] 1/2 ) ≤ c 1 E[M ∗ ∞] < ∞ for some positive constant c 1 . This complete the proof. 24. Since we assume that the usual hypothesis holds throughout this book (see page 3), let Ft 0 = σ{T ∧ s : s ≤ t} and redefine F t by F t = ∩ ɛ Ft+ɛ 0 ∨ N . Since {T < t} = {T ∧ t < t} ∈ F t , T is F-stopping time. Let G = {G t } be a smallest filtration such that T is a stopping time, that is a natural filtration of the process X t = 1 {T ≤t} . Then G ⊂ F. For the converse, observe that {T ∧ s ∈ B} = ({T ≤ s} ∩ {T ∈ B}) ∪ ({T > s} ∩ {s ∈ B}) ∈ F s since {s ∈ B} is ∅ or Ω, {T ∈ B} ∈ F T and in particular {T ≤ s}∩{T ∈ B} ∈ F s and {T > s} ∈ F s . Therefore for all t, T ∧s, (∀s ≤ t) is G t measurable. Hence Ft 0 ⊂ G t . This shows that G ⊂ F. (Note: we assume that G satisfies usual hypothesis as well. ) 25. Recall that the {(ω, t) : △A t (ω) ≠ 0} is a subset of a union of disjoint predictable times and in particular we can assume that a jump time of predictable process is a predictable time. (See the discussion in the solution of exercise 12). For any predictable time T such that E[△Z T |F T − ] = 0, △A T = E[△A T |F T − ] = E[△M T |F T − ] = 0 a.s. (62) 26. Without loss of generality, we can assume that A 0 = 0 since E[△A 0 |F 0− ] = E[△A 0 |F 0 ] = △A 0 = 0. For any finite valued stopping time S, E[A S ] ≤ E[A ∞ ] since A is an increasing process. Observe that A ∞ ∈ L 1 because A is a process with integrable variation. Therefore A {A S } S is uniformly integrable and A is in class (D). Applying Theorem 11 (Doob-Meyer decomposition) to −A, we see that M = A − Ã is a uniformly integrable martingale. Then 0 = E[△M T |F T − ] = E[△A T |F T − ] − E[△ÃT |F T − ] = 0 − E[△ÃT |F T − ]. a.s. (63) Therefore Ã is continuous at time T . 27. Assume A is continuous. Consider an arbitrary increasing sequence of stopping time {T n } ↑ T where T is a finite stopping time. M is a uniformly integrable martingale by theorem 11 and the hypothesis that Z is a supermartingale of class D. Then ∞ > EZ T = −EA T and in particular A T ∈ L 1 . Since A T ≥ A Tn for each n. Therefore by Doob’s optional sampling theorem, Lebesgue’s dominated convergence theorem and continuity of A yields, lim n E[Z T − Z Tn ] = lim n E[M T − N Tn ] − lim n E[A T − A Tn ] = −E[lim n (A T − A Tn )] = 0. (64) Therefore Z is regular. Conversely suppose that Z is regular and assume that A is not continuous at time T . Since A is predictable, so is A − and △A. In particular, T is a predictable time. Then there exists an announcing sequence {T n } ↑ T . Since Z is regular, 0 = lim n E[Z T − Z Tn ] = lim E[M T − M Tn ] − lim E[A T − A Tn ] = E[△A T ]. (65) Since A is an increasing process and △A T ≥ 0. Therefore △A T = 0 a.s. This is a contradiction. Thus A is continuous. 24
31. Let T be an arbitrary F µ stopping time and Λ = {ω : X T (ω) ≠ X T − (ω)}. Then by Meyer’s theorem, T = T Λ ∧ T Λ c where T Λ is totally inaccessible time and TΛ c is predictable time. By continuity of X, Λ = ∅ and T Λ = ∞. Therefore T = T Λ c. It follows that all stopping times are predictable and there is no totally inaccessible stopping time. 32. By exercise 31, the standard Brownian space supports only predictable time since Brownian motion is clearly a strong Markov Feller process. Since O = σ([S, T [: S, T are stopping time and S ≤ T ) and P = σ([S, T [: S, T are predictable times and S ≤ T ), if all stopping times are predictable O = P. 35. E(M t ) ≥ 0 and E(M t ) = exp [B τ t − 1/2(t ∧ τ)] ≤ e. So it is a bounded local martingale and hence martingale. If E(−M) is a uniformly integrable martingale, there exists E(−M ∞ ) such that E[E(−M ∞ )] = 1. By the law of iterated logarithm, exp(B τ − 1/2τ)1 {τ=∞} = 0 a.s. Then [ ( E[E(−M ∞ )] = E exp −1 − τ ) ] 1 2 {τ
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 and 14: 6. Pick arbitrary t 0 > 1. Let Xt n
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: 20. Let {T n } be an increasing seq

Solution to selected problems.

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?