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

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

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as n ↑ ∞ where ‖·‖ := 〈·,·〉 1/2 . To this end let us assume that the sequence <strong>of</strong>processes { M (n) : n = 1,2,... } is a Cauchy sequence, in other words,{ [ (E M (m)T) ]} 2 1/2−M (n)T→ 0 as m,n ↑ ∞.Necessarily then the sequence <strong>of</strong> r<strong>and</strong>om variables {M (k)T: k ≥ 1} is a Cauchysequence in the Hilbert space <strong>of</strong> zero mean, square integrable r<strong>and</strong>om variablesdefined on (Ω,F T ,P), say L 2 (Ω,F T ,P), endowed with the inner product〈M,N〉 = E(MN). Hence there exists a limiting variable, say M T inL 2 (Ω,F T ,P) satisfying{ [E (M (n)T−M T ) 2]} 1/2→ 0as n ↑ ∞. Define the martingale M to be the right continuous version 7 <strong>of</strong>E(M T |F ∗ t ) for t ∈ [0,T]<strong>and</strong> note that by definition∥∥M (n) −M∥ → 0as n tends to infinity. Clearly it is an Ft ∗ -adapted process <strong>and</strong> by Jensen’sinequalityE ( Mt2 )= E(E(M T |Ft ∗ ) 2)≤ E ( E ( MT|F 2 t∗ ))= E ( MT2 )which is finite. Hence Cauchy sequences converge in M 2 T <strong>and</strong> we see that M2 Tis indeed a Hilbert space.Having reminded ourselves<strong>of</strong> some relevant properties <strong>of</strong> the space <strong>of</strong> squareintegrable martingales, let us consider a special class <strong>of</strong> martingale within thelatter class that are also <strong>Lévy</strong> processes <strong>and</strong> which are key to the pro<strong>of</strong> <strong>of</strong> the<strong>Lévy</strong>-Itô decomposition.Henceforth, we shall suppose that {ξ i : i ≥ 1} is a sequence <strong>of</strong> i.i.d. r<strong>and</strong>omvariables with common law F (which does not assign mass to the origin) <strong>and</strong>that N = {N t : t ≥ 0} is a Poisson process with rate λ > 0.Lemma 4.1 Suppose that ∫ |x|F(dx) < ∞.R(i) The process M = {M t : t ≥ 0} where∑N t ∫M t := ξ i −λt xF(dx)Ri=1is a martingale with respect to its natural filtration.7 Here we use the fact that {F ∗ t: t ∈ [0,T]} is complete <strong>and</strong> right continuous.19

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