Solution to selected problems.

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16. Let F t be natural filtration of B t satisfying usual hypothesis. By stationary increments property of Brownian motion and symmetry, W t = B 1−t − B 1 d = B1 − B 1−t d = B1−(1−t) = B t (21) This shows W t is Gaussian. W t has stationary increments because W t − W s = B 1−s − B 1−t d = Bt−s . Let G t be a natural filtration of W t . Then G t = σ(−(B 1 − B 1−s ) : 0 ≤ s ≤ t). By independent increments property of B t , G t is independent of F 1−t . For s > t, W s −W t = −(B 1−t −B 1−s ) ∈ F 1−t and hence independent of G t . 17. a) Fix ε > 0 and ω such that X·(ω) has a sample path which is right continuous, with left limits. Suppose there exists infinitely many jumps larger than ε at time {s n } n≥1 ∈ [0, t]. (If there are uncountably many such jumps, we can arbitrarily choose countably many of them.) Since [0, t] is compact, there exists a subsequence {s nk } k≥1 converging to a cluster point s ∗ ∈ [0, 1]. Clearly we can take further subsequence converging to s ∗ ∈ [0, 1] monotonically either from above or from below. To simplify notations, Suppose ∃{s n } n≥1 ↑ s ∗ . ( The other case {s n } n≥1 ↓ s ∗ is similar.) By left continuity of X t , there exists δ > 0 such that s ∈ (s ∗ − δ, s ∗ ) implies |X s − X s ∗ −| < ε/3 and |X s− − X s ∗ −| < ε/3. However for s n ∈ (s ∗ − δ, s ∗ ), |X sn − X s ∗ −| = |X sn − − X s ∗ − + △X sn | ≥ |△X sn | − |X sn − − X s ∗ −| > 2ε 3 This is a contradiction and the claim is shown. b) By a), for each n there is a finitely many jumps of size larger than 1/n. But J = {s ∈ [0, t] : |△X s | > 0} = ∪ ∞ n=1 {s ∈ [0, t] : |△X s| > 1/n}. We see that cardinality of J is countable. 18. By corollary to theorem 36 and theorem 37, we can immediately see that J ε and Z − J ε are Lévy processes. By Lévy -Khintchine formula (theorem 43), we can see that ψ J εψ Z−J ε = ψ Z . Thus J ε and Z − J ε are independent. ( For an alternative rigorous solution without Lévy -Khintchine formula, see a proof of theorem 36 .) 19. Let T n = inf{t > 0 : |X t | > n}. Then T n is a stopping time. Let S n = T n 1 {X0 ≤n}. We have {S n ≤ t} = {T n ≤ t, X 0 ≤ n} ∪ {X 0 > n} = {T n ≤ t} ∪ {X 0 > n} ∈ F t and S n is a stopping time. Since X is continuous, S n → ∞ and X S n 1 {Sn >0} ≤ n, ∀n ≥ 1. Therefore X is locally bounded. 20. Let H be an arbitrary unbounded random variable. (e.g. Normal) and let T be a positive random variable independent of H such that P (T ≥ t) > 0, ∀t > 0 and P (T < ∞) = 1. (e.g. Exponential). Define a process Z t = H1 {T ≥t} . Z t is a càdlàg process and adapted to its natural filtration with Z 0 = 0. Suppose there exists a sequence of stopping times T n ↑ ∞ such that Z Tn is bounded by some K n ∈ R. Observe that { H T n ≥ T > t Z Tn t = 0 otherwise Since Z T n is bounded by K n , P (Z T n > K n ) = P (H > K n )P (T n ≥ T > t) = 0. It follows that P (T n ≥ T > t) = 0, ∀n and hence P (T n ≤ T ) = 1. Moreover P (∩ n {T n ≤ T }) = P (T = ∞) = 1. This is a contradiction. (22) (23) 6
21. a) let a = (1 − t) −1 ( ∫ 1 t Y (s)ds) and Let M t = Y (ω)1 (0,t) (ω) + a1 [t,1) (ω). For arbitrary B ∈ B([0, 1]), {ω : M t ∈ B} = ((0, t) ∩ {Y ∈ B}) ∪ ({a ∈ B} ∩ (t, 1)). (0, t) ∩ {Y ∈ B} ⊂ (0, t) and hence in F t . {a ∈ B} ∩ (t, 1) is either (t, 1) or ∅ depending on B and in either case in F t . Therefore M t is adapted. Pick A ∈ F t . Suppose A ⊂ (0, t). Then clearly E(M t : A) = E(Y : A). Suppose A ⊃ (t, 1). Then ( ∫ 1 1 ) E(M t : A) = E(Y : A ∩ (0, t)) + E Y (s)ds : (t, 1) 1 − t t (24) =E(Y : A ∩ (0, t)) + ∫ 1 t Y (s)ds = E(Y : A ∩ (0, t)) + E(Y : (t, 1)) = E(Y : A). Therefore for ∀A ∈ F t , E(M t : A) = E(Y : A) and hence M t = E(Y |F t ) a.s. b) By simple calculation EY 2 = 1/(1 − 2α) < ∞ and Y ∈ L 2 ⊂ L 1 . It is also straightforward to compute M t = (1 − t) −1 ∫ 1 t Y (s)ds = (1 − α)−1 Y (t) for ω ∈ (t, 1). c) From b), M t (ω) = Y (ω)1 (0,t) (ω) + 1/(1 − α) −1 Y (t)1 (t,1) (ω). Fix ω ∈ (0, 1). Since 0 < α < 1/2, Y/(1 − α) > Y and Y is a increasing on (0, 1). Therefore, ( ) ( ) Y (t) sup M t = sup ∨ sup Y (ω) = Y (ω) Y (ω) ∨ Y (ω) = (25) 0 ε} ⊂ (0, ε] for all ε > 0. Then T (ω) ≤ ε on (ε, 1) for all ε > 0 and it follows that T ≡ 0. This is contradiction. Therefore there exists ε 0 such that {T > ε 0 } ⊃ (ε 0 , 1). Fix ε ∈ (0, ε 0 ). Then {T > ε} ⊃ {T > ε 0 } ⊃ (ε 0 , 1). On the other hand, since T is a stopping time, {T > ε} ⊂ (0, ε] or {T > ε} ⊃ (ε, 1). Combining these observations, we know that {T > ε} ⊃ (ε, 1) for all ε ∈ (0, ε 0 ). ∀ε ∈ (0, ε 0 ), there exists δ > 0 such that ε − δ > 0 and {T > ε − δ} ⊃ (ε − δ, 1). Especially T (ε) > ε − δ. Taking a limit of δ ↓ 0, we observe T (ε) ≥ ε on (0, ε 0 ). c) Using the notation in b), T (ω) ≥ ω on (0, ε 0 ) and hence M T (ω) = ω −1/2 on (0, ε 0 ). Therefore, EMT 2 ≥ E(1/ω : (0, ε 0)) = ∫ ε 0 0 ω−1 dω = ∞. Since this is true for all stopping times not identically equal to 0, M cannot be locally a L 2 martingale. d) By a), |M t (ω)| ≤ ω −1/2 ∨ 2 and M has bounded path for each ω. If M T 1 {T >0} were a bounded random variable, then M T would be bounded as well since M 0 = 2. However, from c) M T /∈ L 2 unless T ≡ 0 a.s. and hence M T 1 {T >0} is unbounded. 23. Let M be a positive local martingale and {T n } be its fundamental sequence. Then for t ≥ s ≥ 0, E(Mt Tn 1 {Tn >0}|F s ) = M T n s 1 {Tn >0}. By applying Fatou’s lemma, E(M t |F s ) ≤ M s a.s. Therefore 7
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: where Z t ′ = ∫ R xN t(·, dx),
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 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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