1 Lévy Processes and Infinite Divisibility - Department of ...

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To see why, note that for k ≥ l, similar calculations to those in (4.2) show thatk∑∫||X (k) −X (l) || 2 = E[(X (k)T−X (l)T )2 ] = T λ n x 2 F n (dx)which tends to zero as k,l ↑ ∞ by the assumption (4.1). It follows that thereexists a martingale, say X = {X t : 0 ≤ t ≤ T} with respect to the filtration{F t : 0 ≤ t ≤ T} such that ||X − X (k) || → 0 as k ↑ ∞. Thanks to Doob’smaximal inequality (??), it follows moreover thatn=llim E( sup (X t −X (k)t ) 2 ) = 0. (4.3)k↑∞ 0≤t≤TFrom (4.3) it is implicit one-dimensional and hence finite-dimensional distributionsof X (k) converge to those of X. Consequently since the processes X (k) areall Lévy processes, for all 0 ≤ s ≤ t ≤ T,E(e iθ(Xt−Xs) ) = lim E(e iθ(X(k) t −X (k)s ) )k↑∞= lim E(e iθX(k) t−s )k↑∞= E(e iθXt−s )showing that X has stationary and independent increments. From Section 2.2we see that for all t ≥ 0{k∏∫}k∑E(e iθX(k) t) = E(e iθM(n) t) = exp − (1−e iθx +iθx) λ n F n (dx)n=1which converges to exp{−Ψ(θ)} as k ↑ ∞. Note in particular that the lastintegral converges thanks to the assumption (4.1).To show that X is a Lévy process we need only to deduce the paths of X areright continuous with left limits almost surely. However this is a direct resultof the fact that, if D[0,T] is the space of functions f : [0,T] → R which areright continuous with left limits, then D[0,T] is a closed space under the metricd(f,g) = sup t∈[0,T] |f(t)−g(t)| for f,g ∈ D[0,1]. See Exercise 8 for a proof ofthis fact.There is one outstanding issue that needs to be dealt with to complete theproof. In principle the limiting process X depends on T and we need to dismissthiseventuality. SupposeweindexX bythetimehorizonT, sayX T . Usingthatfor any two sequences of real numbers {a n } and {b n }, sup n a 2 n = (sup n |a n |) 2and sup n |a n +b n | ≤ sup n |a n |+sup n |b n |, we have together with an applicationof Minkowski’s inequality that, when T 1 ≤ T 2 ,E[sup(Xt T1−Xt T2) 2 ] 1/2 ≤ E[sup(Xt T1−X (k)t ) 2 ] 1/2 +E[sup(Xt T2−X (k)t ) 2 ] 1/2 .s≤T 1 s≤T 1 s≤T 1Hence taking limits as k ↑ ∞ we may appeal to (4.3) to deduce that the righthand side tends to 0. Consequently the two processesagree almost surely on thetime horizon [0,T 1 ] and we may comfortably say that the limit X does indeednot depend on the time horizon T.RRn=122
4.1 Proof of the Lévy-Itô decompositionWe are now in a position to prove the Lévy-Itô decomposition.Proof of Theorem 3.1. Recalling the decomposition (3.1) and the discussionthereafter, it suffices to prove the existence of the process X (3) and that it hasa countable number of discontinuities which are less than unity in magnitudeover each finite time horizon. However this follows immediately from Theorem4.1 when we take λ n = Π({x : 2 −(n+1) ≤ |x| < 2 −n }) andNote in particular thatF n (dx) = λ −1n Π(dx)| {x:2 −(n+1) ≤|x|
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Page 5 and 6: Theorem 1.2 (Lévy-Khintchine formu
Page 7 and 8: Using the fact that N has stationar
Page 9 and 10: AccordingtoTheorem1.2thereexistsaL
Page 11 and 12: where β ∈ [−1,1] η ∈ R and
Page 13 and 14: 0.20.10.0−0.10.0 0.2 0.4 0.6 0.8
Page 15 and 16: specific distributions, most of the
Page 17 and 18: The proof of existence of a Lévy p
Page 19 and 20: as n ↑ ∞ where ‖·‖ := 〈
Page 21: assume that the processes N (n) are
Page 25 and 26: With C (k,−) defined in the obvio
Page 27 and 28: ExercisesExercise 1 Using Definitio
Page 30 and 31: Exercise 8 Recall that D[0,1] is th
Page 32 and 33: and Γ (2) is a Gamma subordinator,
Page 34: [17] Lévy, P. (1924) Théorie des

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