Solution to selected problems.

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29. Let M t be a Lévy process and local martingale. M t has a representation of the form ∫ ∫ M t = B t + x [N t (·, dx) − tν(dx)] + αt + xN t (·, dx) (34) {|x|≤1} {|x|>1} First two terms are martingales. Therefore WLOG, we can assume that ∫ M t = αt + xN t (·, dx) (35) {|x|>1} M t has only finitely many jumps on each interval [0, t] by exercise 17 (a). Let {T n } n≥1 be a sequence of jump times of M t . Then P (T n < ∞) = 1, ∀n and T n ↑ ∞. We can express M t by a sum of compound Poisson process and a drift term (See Example in P.33): M t = ∞∑ U i 1 {t≥Ti } − αt. (36) i=1 Since M t is local martingale by hypothesis, M is a martingale if and only if M t ∈ L 1 and E[M t ] = 0, ∀t. There exists a fundamental sequence {S n } such that M S n is a martingale, ∀n. WLOG, we can assume that M S n is uniformly integrable. If U 1 ∈ L 1 , M t ∈ L 1 for every α and M t becomes martingale with α ∗ = EN t EU 1 /t. Furthermore, for any other α, M t can’t be local martingale since for any s<strong>to</strong>pping time T , Mt T = L T t + (α ∗ − α)(t ∧ T ) where L t is a martingale given by α = α ∗ and Mt T can’t be a martingale if α ≠ α ∗ . It follows that if M t is only a local martingale, U 1 /∈ L 1 . By exercise 24, since M T 1− t Therefore, F T1 − = σ(M T 1− , T 1 ) ∨ N = σ(T 1 ) ∨ N (37) = M t 1 {t 0 : Z t ≥ z}. Then T z is a s<strong>to</strong>pping time and P (T z < ∞) = 1. Let’s define a coordinate map ω(t) = Z t (ω). Let R = inf{s < t : Z s ≥ z}. We let { { 1 s ≤ t , ω(t − s) < z − y Y s (ω) = , Y ′ 1 s ≤ t , ω(t − s) > z + y s(ω) = (40) 0 otherwise 0 otherwise 10
So that Y R ◦ θ R (ω) = { 1 R ≤ t , Z t < z − y 0 otherwise , Y ′ R ◦ θ R (ω) = { 1 R ≤ t , Z t > z + y 0 otherwise (41) Strong Markov property implies that on {R < ∞} = {T z ≤ t} = {S t ≥ z}, E 0 (Y R ◦ θ R |F R ) = E ZR Y R , E 0 (Y ′ R ◦ θ R |F R ) = E ZR Y ′ R (42) Since Z R ≥ z and Z is symmetric, E a Y s = E a 1 {Zt−s z+y} = E a Y ′ s, ∀a ≥ z, s < t. (43) By taking expectation, P 0 (T z ≤ t, Z t < z − y) = E(E 0 (Y R ◦ θ R |F R ) : R < ∞) = E(E ZR Y R : R < ∞) ≤E(E ZR Y ′ R : R < ∞) = E(E 0 (Y ′ R ◦ θ R |F R ) : R < ∞) = P 0 (T z ≤ t, Z t > z + y) =P (Z t > z + y) 31. We define T z , R, ω(t) as in exercise 30. We let { 1 s ≤ t , ω(t − s) ≥ z Y s (ω) = 0 otherwise (44) (45) Since Z R ≥ z and Z is symmetric, E a Y s = E a 1 {Zt−s ≥z} ≥ 1 2 ∀a ≥ z, s < t. (46) By the same reasoning as in exercise 30, taking expectation yields P 0 (Z t ≥ z) = P 0 (T z ≤ t, Z t ≥ z) = E(E 0 (Y R ◦ θ R |F R ) : R < ∞) = E(E ZR Y R : R < ∞) ≥E( 1 2 : R < ∞) = 1 2 P (R < ∞) = 1 2 P (S t ≥ z) (47) 11
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: For this, let t 0 = 0 and t n+1 =
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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