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

The term “Lévy process” honours the work of the French mathematicianPaul Lévy who, although not alone in his contribution, played an instrumentalroleinbringingtogetheranunderstandingandcharacterisationofprocesseswithstationary independent increments. In earlier literature, Lévy processes can befound under a number of different names. In the 1940s, Lévy himself referred tothem as a sub-class of processus additifs (additive processes), that is processeswith independent increments. For the most part however, research literaturethrough the 1960s and 1970s refers to Lévy processes simply as processes withstationary independent increments. One sees a change in language through the1980sandbythe1990stheuseofthe term“Lévyprocess”hadbecomestandard.From Definition 1.1 alone it is difficult to see just how rich a class of processesthe class of Lévy processes forms. De Finetti [4] introduced the notionof an infinitely divisible distribution and showed that they have an intimaterelationship with Lévy processes. This relationship gives a reasonably good impressionof how varied the class of Lévy processes really is. To this end, let usnow devote a little time to discussing infinitely divisible distributions.Definition 1.2 We say that a real-valued random variable Θ has an infinitelydivisible distribution if for each n = 1,2,... there exist a sequence of i.i.d. randomvariables Θ 1,n ,...,Θ n,n such thatΘ d = Θ 1,n +···+Θ n,nwhere = d is equality in distribution. Alternatively, we could have expressed thisrelation in terms of probability laws. That is to say, the law µ of a real-valuedrandom variable is infinitely divisible if for each n = 1,2,... there exists anotherlaw µ n of a real valued random variable such that µ = µ ∗nn . (Here µ ∗ n denotesthe n-fold convolution of µ n ).In view of the above definition, one way to establish whether a given randomvariable has an infinitely divisible distribution is via its characteristic exponent.Suppose that Θ has characteristic exponent Ψ(u) := −logE(e iuΘ ) for all u ∈ R.Then Θ has an infinitely divisible distribution if for all n ≥ 1 there exists acharacteristic exponent of a probability distribution, say Ψ n , such that Ψ(u) =nΨ n (u) for all u ∈ R.The full extent to which we may characteriseinfinitely divisible distributionsis described by the characteristic exponent Ψ and an expression known as theLévy–Khintchine formula. 2Theorem 1.1 (Lévy–Khintchine formula) A probability law µ of a realvaluedrandom variable is infinitely divisible with characteristic exponent Ψ,∫e iθx µ(dx) = e −Ψ(θ) for θ ∈ R,R2 Note, although it is a trivial fact, it is always worth reminding oneself of that one worksin general with the Fourier transform of a measure as opposed to the Laplace transform onaccount of the fact that the latter may not exist.3

if and only if there exists a triple (a,σ,Π), where a ∈ R, σ ≥ 0 and Π is ameasure concentrated on R\{0} satisfying ∫ R( ) 1∧x2Π(dx) < ∞, such thatΨ(θ) = iaθ + 1 ∫2 σ2 θ 2 + (1−e iθx +iθx1 (|x| 0, X t is arandom variable belonging to the class of infinitely divisible distributions. Thisfollows from the fact that for any n = 1,2,...,X t = X t/n +(X 2t/n −X t/n )+···+(X t −X (n−1)t/n ) (1.1)together with the fact that X has stationary independent increments. Supposenow that we define for all θ ∈ R, t ≥ 0,Ψ t (θ) = −logE ( e iθXt)then using (1.1) twice we have for any two positive integers m,n thatand hence for any rational t > 0,mΨ 1 (θ) = Ψ m (θ) = nΨ m/n (θ)Ψ t (θ) = tΨ 1 (θ). (1.2)Iftisanirrationalnumber,thenwecanchooseadecreasingsequenceofrationals{t n : n ≥ 1} such that t n ↓ t as n tends to infinity. Almost sure right continuityof X implies right continuity of exp{−Ψ t (θ)} (by dominated convergence) andhence (1.2) holds for all t ≥ 0.In conclusion, any Lévy process has the property that for all t ≥ 0E ( e iθXt) = e −tΨ(θ) ,where Ψ(θ) := Ψ 1 (θ) is the characteristic exponent of X 1 , which has an infinitelydivisible distribution.Definition 1.3 In the sequel we shall also refer to Ψ(θ) as the characteristicexponent of the Lévy process.It is now clear that each Lévy process can be associated with an infinitelydivisible distribution. What is not clear is whether given an infinitely divisibledistribution, one may construct a Lévy process X, such that X 1 has that distribution.This latter issue is affirmed by the following theorem which gives theLévy–Khintchine formula for Lévy processes.4

Theorem 1.2 (Lévy–Khintchine formula for Lévy processes) Suppose thata ∈ R, σ ≥ 0 and Π is a measure concentrated on R\{0} such that ∫ R (1 ∧x 2 )Π(dx) < ∞. From this triple define for each θ ∈ R,Ψ(θ) = iaθ + 1 ∫2 σ2 θ 2 + (1−e iθx +iθx1 (|x| 0 consider a probability distribution µ λ which is concentratedon k = 0,1,2... such that µ λ ({k}) = e −λ λ k /k!. That is to say the Poisson5

distribution. An easy calculation reveals that∑e iθk µ λ ({k}) = e −λ(1−eiθ )k≥0=[e − λ n (1−eiθ ) ] n.The right-hand side is the characteristic function of the sum of n independentPoisson processes, each of which with parameter λ/n. In the Lévy–Khintchinedecompositionwe seethat a = σ = 0and Π = λδ 1 , the Diracmeasuresupportedon {1}.Recall that a Poisson process, {N t : n ≥ 0}, is a Lévy process with distributionat time t > 0, which is Poisson with parameter λt. From the abovecalculations we haveE(e iθNt ) = e −λt(1−eiθ )and hence its characteristic exponent is given by Ψ(θ) = λ(1−e iθ ) for θ ∈ R.2.2 Compound Poisson ProcessesSuppose now that N is a Poisson random variable with parameter λ > 0 andthat {ξ i : i ≥ 1} is an sequence of i.i.d. random variables (independent of N)with common law F having no atom at zero. By first conditioning on N, wehave for θ ∈ R,E(e iθ∑ Ni=1 ξi ) = ∑ E(e iθ∑ ni=1 ξi )e −λλnn!n≥0= ∑ (∫ ) ne iθx F(dx) e −λλnn≥0 R n!= e −λ∫ R (1−eiθx )F(dx) . (2.1)Note we use the convention here and later that for any n = 0,1,2,...n∑= 0.n+1We see from (2.1) that distributions of the form ∑ Ni=1 ξ i are infinitely divisiblewith triple a = −λ ∫ 0

Using the fact that N has stationary independent increments together with themutual independence of the random variables {ξ i : i ≥ 1}, for 0 ≤ s < t < ∞,by writingX t = X s +∑N ti=N s+1it is clear that X t is the sum of X s and an independent copy of X t−s . Rightcontinuity and left limits of the process N also ensure right continuity and leftlimits of X. Thus compound Poisson processes are Lévy processes. From thecalculations in the previous paragraph, for each t ≥ 0 we may substitute N t forthe variable N to discover that the Lévy–Khintchine formula for a compoundPoisson process takes the form Ψ(θ) = λ ∫ R (1−eiθx )F(dx). Note in particularthat the Lévy measure of a compound Poisson process is always finite with totalmass equal to the rate λ of the underlying process N.If a drift of rate c ∈ R is added to a compound Poisson process so that nowξ i∑N tX t = ξ i +ct, t ≥ 0i=1thenitisstraightforwardtoseethattheresultingprocessisagainaLévyprocess.The associated infinitely divisible distribution is nothing more than a shiftedcompound Poisson distribution with shift c. The Lévy-Khintchine exponent isgiven by∫Ψ(θ) = λ (1−e iθx )F(dx)−icθ.RIf further the shift is chosen to centre the compound Poisson distribution thenc = λ ∫ RxF(dx) and then∫Ψ(θ) = (1−e iθx +iθx)λF(dx).2.3 Linear Brownian MotionTake the probability lawRµ s,γ (dx) :=1√2πs2 e−(x−γ)2 /2s 2 dxsupported on R where γ ∈ R and s > 0; the well-known Gaussian distributionwith mean γ and variance s 2 . It is well known that∫Re iθx µ s,γ (dx) = e −1 2 s2 θ 2 +iθγ=[e −1 2 ( s √ n) 2 θ 2 +iθ γ n] n7

showing again that it is an infinitely divisible distribution, this time with a =−γ, σ = s and Π = 0.We immediately recognise the characteristic exponent Ψ(θ) = s 2 θ 2 /2−iθγas also that of a linear Brownian motion,X t := sB t +γt, t ≥ 0,where B = {B t : t ≥ 0} is a standard Brownian motion. It is a trivial exerciseto verify that X has stationary independent increments with continuous pathsas a consequence of the fact that B does.2.4 Gamma ProcessesFor α,β > 0 define the probability measureµ α,β (dx) = αβΓ(β) xβ−1 e −αx dxconcentrated on (0,∞); the gamma-(α,β) distribution. Note that when β = 1this is the exponential distribution. We have∫ ∞0e iθx µ α,β (dx) ==1(1−iθ/α) β[ ] n1(1−iθ/α) β/nandinfinite divisibility follows. Forthe Lévy–Khintchinedecompositionwehaveσ = 0 and Π(dx) = βx −1 e −αx dx, concentrated on (0,∞) and a = − ∫ 10 xΠ(dx).However this is not immediately obvious. The following lemma proves to beuseful in establishing the above triple (a,σ,Π). Its proof is Exercise 3.Lemma 2.1 (Frullani integral) For all α,β > 0 and z ∈ C such that Rz ≤ 0we have1(1−z/α) β = ∞e−∫ 0 (1−ezx )βx −1 e −αx dx .To see how this lemma helps note that the Lévy–Khintchine formula for agamma distribution takes the formΨ(θ) = β∫ ∞0(1−e iθx ) 1 x e−αx dx = βlog(1−iθ/α)for θ ∈ R. The choice of a in the Lévy–Khintchine formula is the necessaryquantity to cancel the term coming from iθ1 (|x|

AccordingtoTheorem1.2thereexistsaLévyprocesswhoseLévy–Khintchineformula is given by Ψ, the so-called gamma process.Suppose now that X = {X t : t ≥ 0} is a gamma process. Stationaryindependent increments tell us that for all 0 ≤ s < t < ∞, X t = X s + ˜X t−swhere ˜X t−s is an independent copy of X t−s . The fact that the latter is strictlypositive with probabilityone (on account ofit being gamma distributed) impliesthat X t > X s almost surely. Hence a gamma process is an example of a Lévyprocess with almost surely non-decreasing paths (in fact its paths are strictlyincreasing). Another example of a Lévy process with non-decreasing paths isa compound Poisson process where the jump distribution F is concentrated on(0,∞). Note however that a gamma process is not a compound Poisson processon two counts. Firstly, its Lévy measure has infinite total mass unlike the Lévymeasure of a compound Poisson process, which is necessarily finite (and equalto the arrival rate of jumps). Secondly, whilst a compound Poisson process withpositive jumps does have paths, which are almost surely non-decreasing, it doesnot have paths that are almost surely strictly increasing.Lévy processes whose paths are almost surely non-decreasing (or simplynon-decreasing for short) are called subordinators.2.5 Inverse Gaussian ProcessesSuppose as usual that B = {B t : t ≥ 0} is a standard Brownian motion. Definethe first passage timeτ s = inf{t > 0 : B t +bt > s}, (2.2)that is, the first time a Brownian motion with linear drift b > 0 crosses abovelevel s. Recall that τ s is a stopping time 3 . Moreover,since Brownianmotion hascontinuous paths we know that B τs +bτ s = s almost surely. From the StrongMarkov Property it is known that {B τs+t+b(τ s +t)−s : t ≥ 0} is equal in lawto B and hence for all 0 ≤ s < t,τ t = τ s +˜τ t−s ,where ˜τ t−s is an independent copy of τ t−s . This shows that the process τ :={τ t : t ≥ 0} has stationary independent increments. Continuity of the paths of{B t +bt : t ≥ 0} ensures that τ has right continuous paths. Further, it is clearthat τ has almost surely non-decreasing paths, which guarantees its paths haveleft limits as well as being yet another example of a subordinator. According toits definition as a sequence of first passage times, τ is also the almost sure rightinverse of the path of the graph of {B t +bt : t ≥ 0} in the sense of (2.2). Fromthis τ earns its name as the inverse Gaussian process.3 We assume that the reader is familiar with the notion of a stopping time for a Markovprocess. By definition, the random time τ is a stopping time with respect to the filtration{G t : t ≥ 0} if for all t ≥ 0,{τ ≤ t} ∈ G t.9

According to the discussion following Theorem 1.1 it is now immediate thatfor each fixed s > 0, the random variable τ s is infinitely divisible. Its characteristicexponent takes the formΨ(θ) = s( √ −2iθ+b 2 −b)for all θ ∈ R and corresponds to a triple a = −2sb −1∫ b0 (2π)−1/2 e −y2 /2 dy, σ = 0and1 xΠ(dx) = s√ 2πx3 e−b2 2 dxconcentrated on (0,∞). The law of τ s can also be computed explicitly asµ s (dx) =s√2πx3 esb e −1 2 (s2 x −1 +b 2 x)for x > 0. The proof of these facts forms Exercise 6.2.6 Stable ProcessesStable processes are the class of Lévy processes whose characteristic exponentscorrespondto those of stable distributions. Stable distributions were introducedby [17, 18] as a third example of infinitely divisible distributions after Gaussianand Poisson distributions. A random variable, Y, is said to have a stable distributionif for all n ≥ 1 it observes the distributional equalityY 1 +···+Y nd= an Y +b n , (2.3)where Y 1 ,...,Y n are independent copies of Y, a n > 0 and b n ∈ R. By subtractingb n /n from each of the terms on the left-hand side of (2.3) one seesin particular that this definition implies that any stable random variable is infinitelydivisible. It turns out that necessarily a n = n 1/α for α ∈ (0,2]; see [3],Sect.VI.1. In that case we refer to the parameter α as the index. A smallerclass of distributions are the strictly stable distributions. A random variable Yis said to have a strictly stable distribution if it observes (2.3) but with b n = 0.In that case, we necessarily haveY 1 +···+Y nd= n 1/α Y. (2.4)The case α = 2 corresponds to zero mean Gaussian random variables and isexcluded in the remainder of the discussion as it has essentially been dealt within Sect.2.3.Stable random variables observing the relation (2.3) for α ∈ (0,1) ∪ (1,2)have characteristic exponents of the formΨ(θ) = c|θ| α (1−iβtan πα 2sgnθ)+iθη, (2.5)where β ∈ [−1,1], η ∈ R and c > 0. Stable random variables observing therelation (2.3) for α = 1, have characteristic exponents of the formΨ(θ) = c|θ|(1+iβ 2 sgnθlog|θ|)+iθη, (2.6)π10

where β ∈ [−1,1] η ∈ R and c > 0. Here we work with the definition ofthe sign function sgnθ = 1 (θ>0) − 1 (θ 0, α ∈ (0,2), β ∈ [−1,1] andη ∈ R Theorem 1.2 tells us that there exists a Lévy process, with characteristicexponent given by (2.5) or (2.6) according to this choice of parameters. Further,from the definition of its characteristic exponent it is clear that at each fixedtime the α-stable process will have distribution S(ct,α,β,η).2.7 Other ExamplesThere are many more known examples of infinitely divisible distributions (andhence Lévy processes). Of the many known proofs of infinitely divisibility for11

2.52.01.51.00.50.00.0 0.2 0.4 0.6 0.8 1.0Figure 1: A sample path of a Poisson process; Ψ(θ) = λ(1−e iθ ) where λ is thejump rate.0.80.60.40.20.0−0.2−0.40.0 0.2 0.4 0.6 0.8 1.0Figure 2: A sample path of a compound Poisson process; Ψ(θ) = λ ∫ R (1 −e iθx )F(dx) where λ is the jump rate and F is the common distribution of thejumps.12

0.20.10.0−0.10.0 0.2 0.4 0.6 0.8 1.0Figure 3: A sample path of a Brownian motion; Ψ(θ) = θ 2 /2.0.40.30.20.10.0−0.10.0 0.2 0.4 0.6 0.8 1.0Figure 4: A sample path of the independent sum of a Brownian motion and acompound Poisson process; Ψ(θ) = θ 2 /2+ ∫ R (1−eiθx )F(dx).13

0.20.0−0.2−0.40.0 0.2 0.4 0.6 0.8 1.0Figure 5: A sample path of a variance gamma processes. The latter has characteristicexponent given by Ψ(θ) = βlog(1−iθc/α+β 2 θ 2 /2α) where c ∈ R andβ > 0.0.60.40.20.0−0.2−0.4−0.60.0 0.2 0.4 0.6 0.8 1.0Figure 6: A sample path of a normal inverse Gaussian process; Ψ(θ) =δ( √ α 2 −(β +iθ) 2 − √ α 2 −β 2 ) where α,δ > 0, |β| < α.14

specific distributions, most of them are non-trivial, often requiring intimateknowledge of special functions. A brief list of such distributions might includegeneralised inverse Gaussian (see [5] and [14]), truncated stable (see [30], [8],[15], [1] and [2]), generalised hyperbolic (see [7]), Meixner (see [24]), Pareto (see[25] and [28]), F-distributions (see [10]), Gumbel (see [13] and [26]), Weibull(see [13] and [25]), lognormal (see [29]) and Student t-distribution (see [6] and[9]). See also the book of Steutel [27]Despite being able to identify a large number of infinitely divisible distributionsand hence associated Lévy processes, it is not clear at this point what thepaths of Lévy processes look like. The task of giving a mathematically preciseaccount of this lies ahead in the next section. In the meantime let us make thefollowing informal remarks concerning paths of Lévy processes.Exercise 1 shows that a linear combination of a finite number of independentLévy processes is again a Lévy process. It turns out that one may consider anyLévy process as an independent sum of a Brownian motion with drift and acountable number of independent compound Poisson processes with differentjump rates, jump distributions and drifts. The superposition occurs in sucha way that the resulting path remains almost surely finite at all times and,for each ε > 0, the process experiences at most a countably infinite numberof jumps of magnitude ε or less with probability one and an almost surelyfinite number of jumps of magnitude greater than ε, over all fixed finite timeintervals. If in the latter description there is always an almost surely finitenumber of jumps over each fixed time interval then it is necessary and sufficientthat one has the linear independent combination of a Brownian motion withdrift and a compound Poisson process. Depending on the underlying structureof the jumps and the presence of a Brownian motion in the described linearcombination, a Lévy process will either have paths of bounded variation on allfinite time intervals or paths of unbounded variation on all finite time intervals.WeincludesixcomputersimulationstogivearoughsenseofhowthepathsofLévyprocesseslook. Figs.1and2depictthepathsofPoissonprocessandacompoundPoissonprocess, respectively. Figs.3 and 4 show the paths of a Brownianmotionandtheindependent sumofaBrownianmotionandacompoundPoissonprocess, respectively. Finally Figs.5 and 6 show the paths of a variance gammaprocess and a normal inverse Gaussian processes. Both are pure jump processes(no Brownian component as described above). Variance gamma processes arediscussed in more detail later normal inverse Gaussian processes are Lévy processeswhose jump measure is given by Π(dx) = (δα/π|x|)exp{βx}K 1 (α|x|)dxfor x ∈ R where α,δ > 0, β ≤ |α| and K 1 (x) is the modified Bessel function ofthe third kind with index 1 (the precise definition of the latter is not worth thedetail at this moment in the text). Both experience an infinite number of jumpsover a finite time horizon. However, variance gamma processes have paths ofbounded variation whereas normal inverse Gaussian processes have paths ofunbounded variation. The reader should be warned however that computersimulations ultimately can only depict a finite number of jumps in any givenpath. All figures were very kindly produced by Antonis Papapantoleon.15

3 The Lévy-Itô decomposition intuitivelyOneofourmainobjectivesfortheremainderofthistextistoestablisharigorousunderstanding of the structure of the paths of Lévy processes. The way we shalldo this is to prove the assertion in Theorem 1.2 that given any characteristicexponent, Ψ, belonging to an infinitely divisible distribution, there exists a Lévyprocesswith the same characteristicexponent. This will be done by establishingthe so-called Lévy–Itô decomposition which describes the structure of a generalLévy process in terms of three independent auxiliary Lévy processes, each withdifferent types of path behaviour.According to Theorem 1.1, any characteristic exponent Ψ belonging to aninfinitely divisible distribution can be written, after some simple reorganisation,in the formΨ(θ) ={iaθ + 1 }2 σ2 θ 2{ ∫}+ Π(R\(−1,1)) (1−e iθx Π(dx))+|x|≥1{ ∫(1−e iθx +iθx)Π(dx)0

The proof of existence of a Lévy process with characteristic exponent givenby (3.1) thus boils down to showing the existence of a Lévy process, X (3) , whosecharacteristic exponent is given by Ψ (3) . Noting that∫(1−e iθx +iθx)Π(dx)0

4 The Lévy-Itô decomposition rigorouslyFix a time horizon T > 0. Let us assume that (Ω,F,{F t : t ∈ [0,T]},P) is afiltered probability space in which the filtration {F t : t ≥ 0} is complete withrespecttothe nullsetsofP andrightcontinuousin thesensethat F t = ⋂ s>t F s.Definition 4.1 Fix T > 0. Define M 2 T = M2 T (Ω,F,{F t : t ∈ [0,T]},P) tobe the space of real valued, zero mean, almost surely right-continuous, squareintegrable P-martingales with respect to the given filtration over the finite timeperiod [0,T].One luxury that follows from the assumptions on {F t : t ≥ 0} is that anyzero mean square integrable martingalewith respect to this filtration has a rightcontinuous modification 4 which is also a member of M 2 T .If we quotient out the equivalent classes of versions 5 of each martingale, itis straightforward to deduce that M 2 T is a vector space over the real numberswith zero element M t = 0 for all t ∈ [0,T] and all ω ∈ Ω. In fact, as we shallshortly see, M 2 T is a Hilbert space6 with respect to the inner product〈M,N〉 = E(M T N T ),where M,N ∈ M 2 T . It is left to the reader to verify the fact that 〈·,·〉 forms aninner product. The only mild technical difficulty in this verification is showingthat for M ∈ M 2 T , 〈M,M〉 = 0 implies that M = 0, the zero element. Notehowever that if 〈M,M〉 = 0 then by Doob’s Maximal Inequality, which saysthat for M ∈ M 2 T ,( )E sup Ms20≤s≤T≤ 4E ( M 2 T),wehavethatsup 0≤t≤T |M t | = 0almostsurely. ItfollowsnecessarilythatM t = 0for all t ∈ [0,T] with probability one.As alluded to above, we can show without too much difficulty that M 2 T isa Hilbert space. To do that we are required to show that given {M (n) : n =1,2,...} is a Cauchy sequence of martingales taken from M 2 T then there existsan M ∈ M 2 T such that ∥ ∥∥M−M∥(n) ∥ → 04 Recall that M ′ = {M t ′ : t ∈ [0,T]} is a modification of M if for every t ≥ 0, we haveP(M t ′ = Mt) = 1.5 Recall that M ′ = {M t ′ : t ∈ [0,T]} is a version of M if it is defined on the same probabilityspace and {∃t ∈ [0,T] : M t ′ ≠ Mt} is measurable with zero probability. Note that if M′ is amodification of M then it is not necessarily a version of M. However, it is obviously the casethat if M ′ is a version of M then it also fulfils the requirement of being a modification.6 Recall that 〈·,·〉 : L ×L → R is an inner product on a vector space L over the reals if itsatisfies the following properties for f,g ∈ L and a,b ∈ R; (i) 〈af +bg,h〉 = a〈f,h〉+b〈g,h〉for all h ∈ L, (ii) 〈f,g〉 = 〈g,f〉, (iii) 〈f,f〉 ≥ 0 and (iv) 〈f,f〉 = 0 if and only if f = 0.For each f ∈ L let ||f|| = 〈f,f〉 1/2 . The pair (L,〈·,·〉) are said to form a Hilbert space if allsequences, {f n : n = 1,2,...} in L that satisfy ||f n−f m|| → 0 as m,n → ∞, so called Cauchysequences, have a limit which exists in L.18

as n ↑ ∞ where ‖·‖ := 〈·,·〉 1/2 . To this end let us assume that the sequence ofprocesses { 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 of random variables {M (k)T: k ≥ 1} is a Cauchysequence in the Hilbert space of zero mean, square integrable random 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 ofE(M T |F ∗ t ) for t ∈ [0,T]and note that by definition∥∥M (n) −M∥ → 0as n tends to infinity. Clearly it is an Ft ∗ -adapted process and 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 and we see that M2 Tis indeed a Hilbert space.Having reminded ourselvesof some relevant properties of the space of squareintegrable martingales, let us consider a special class of martingale within thelatter class that are also Lévy processes and which are key to the proof of theLévy-Itô decomposition.Henceforth, we shall suppose that {ξ i : i ≥ 1} is a sequence of i.i.d. randomvariables with common law F (which does not assign mass to the origin) andthat 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 and right continuous.19

(ii) If moreover ∫ R x2 F(dx) < ∞ then M is a square integrable martingale suchthat∫E(Mt 2 ) = λt x 2 F(dx).Proof. For the first part, note that M has stationary and independent increments(in fact, as seen earlier, it is a Lévy process). Hence if we defineF t = σ(M s : s ≤ t) then for t ≥ s ≥ 0E(M t |F s ) = M s +E(M t −M s |F s ) = M s +E(M t−s ).Hence it suffices to prove that for all u ≥ 0, E(M u ) = 0. However the latter isevident since, by a classical computation (left to the reader) which utilizes theindependence of N and {ξ i : i ≥ 0},E(Nt∑i=1) ∫ξ i = λtE(ξ 1 ) = λt xF(dx).RNote that this computation also shows that E|M t | < ∞ for each t ≥ 0.For the second part, using the independence and distributional properties ifN and {ξ i : i ≥ 1} we have that⎡E(Mt 2 ) = E⎣as required.= E(Nt(Nt∫= λt∫= λt∫= λt∑i=1i=1ξ i) 2⎤⎦−λ 2 t 2 (∫i=1 j=1RR) 2xF(dx)⎛ ⎞∑ ∑N t∑N t(∫ξ)+E2 ⎝ 1 {j≠i} ξ i ξ j⎠−λ 2 t 2RRR(∫x 2 F(dx)+E(Nt 2 −N t)x 2 F(dx)+λ 2 t 2 (∫x 2 F(dx)RxF(dx)RR) 2−λ 2 t 2 (∫) 2 (∫x 2 F(dx) −λ 2 t 2) 2xF(dx)xF(dx)RxF(dx)RIf we recall the intuition in the discussion following (3.4) then our interest inthe type of martingales described in Lemma 4.1 (which are also Lévy processes)as far as the Lévy-Itô decomposition is concerned is tantamount to understandinghow to superimpose a countably infinite number of such processes togetherso that the resulting sum convergesin an appropriate sense. Precisely this pointis tackled in the next theorem.We need tointroduce somenotationfirst. Supposethat foreachn = 1,2,3,..the process N (n) = {N (n)twe understand the process N (n)t: t ≥ 0} is a Poisson process with rate λ n ≥ 0. Here= 0 for all t ≥ 0 if λ n = 0. We shall also20) 2) 2

assume that the processes N (n) are mutually independent. Moreover, for eachn = 1,2,3,... let us introduce the i.i.d. sequences {ξ (n)i : i = 1,2,3,...} (whichare themselves mutually independent sequences) with common distribution F nwhich does not assign mass to the origin and which satisfies ∫ x 2 F n (dx) < ∞.Associated with each pair (λ n ,F n ) is the square integrable martingale describedin Lemma 4.1 which we denote by M (n) = {M (n)t : t ≥ 0}. Suppose that wedenote by {F (n)t : t ≥ 0} as the natural filtration generated by the processM (n) ,then we can put all processes {M (n) : n ≥ 1} on the same probability space andtake them as martingales with respect to the common filtrationF t := σ( ⋃ n≥1F (n)t )which we may also assume without loss of generality is taken in its completedand right continuous form 8 .Theorem 4.1 If∑∫λ nn≥1Rx 2 F n (dx) < ∞ (4.1)then there exists a Lévy process X = {X t : t ≥ 0} defined on the same probabilityspace as the processes {M (n) : n ≥ 1} which is also a square integrablemartingale and whose characteristic exponent is given by∫Ψ(θ) =R(1−e iθx +iθx) ∑ λ n F n (dx)n≥1for all θ ∈ R such that the for each fixed T > 0,⎡ ( ) 2⎤k∑lim E ⎣sup X t − M (n) ⎦t = 0.k↑∞t≤TProof. First note that by linearity of conditional expectation, ∑ kn=1 M(k) is asquareintegrablemartingale. Infact, since, byindependenceandthemartingaleproperty, for i ≠ j we have E(M (i)t ) = E(M (i)t )E(M (j)t ) = 0, it follows that⎡( k∑E⎣n=1M (n)t) 2⎤⎦ =k∑n=1t M (j)n=1[E (M (n)t ) 2] = tk∑∫λ nn=1Rx 2 F n (dx) < ∞ (4.2)where the last equality follows by the assumption (4.1).Fix T > 0. We now claim that the sequence {X (k) : k ≥ 1} is a Cauchysequence with respect to ||·|| where X (k) = {X (k)t : 0 ≤ t ≤ T} andX (k)t =k∑n=1M (n)t .8 That is to say, if F t is not right continuous then we may work instead with F ∗ t = ⋂ s>t Fs.21

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|

5 Path variationThe proofofthe Lévy-Itôdecompositionrevealsalittle more than the decompositionin Theorem 3.1. It provides us the necessary point of view to characterizeprecisely the almost sure path variation of the process.Suppose that X is any Lévy process whose characteristic exponent we shallalways refer to in the form (3.5). Moreover we shall talk about X (1) , X (2)and X (3) as the three components of the Lévy-Itô decomposition mentioned inTheorem 3.1.If σ > 0, since it is known that Brownian motion has unbounded variationand since the continuous and discontinuous elements of the path of X makeindependent contributions to its variation, the latter must also have unboudnedvariation. In the case that σ = 0 however, the Lévy process may have eitherbounded or unbounded variation. Since the process X (2) is a compound Poissonprocess,whichobviouslyhaspathsofboundedvariation,the variationofX boilsdown to the variation of X (3) in this case.Recall that we may see the process X (3) as the uniform L 2 -limit over finitetime horizons of a sequence of partial sums of compound Poisson martingales= ∑ kn=0 M(n) t for t ≥ 0. Suppose in the aforementioned analysis we splitthe compound Poissonmartingalesinto twosub-procesescontainingthe positiveand negative jumps respectively. To be more precise, defineX (k)tandN t∑(n)M (n,+)t =N t∑(n)M (n,−)t =ξ (n)i 1 (n) (ξ i >0) −tλ ni=1|ξ (n)i|1 (n) (ξ i 0)is increasingin k and therefore has a limit almost surely (which may be infinite).Since X (k,+)t converges in distribution to a non-degenerate random variable, itfollows that∫ ∞∑limk↑∞ C(k,+) t < ∞ a.s. ⇔ x λ n F n (dx) < ∞(0,∞)24n=0

With C (k,−) defined in the obvious way it also follows thatand hence sincek∑∫limk↑∞ C(k,−) t < ∞ a.s. ⇔(−∞,0)N t∑(n)n=0 i=1ξ (n)i∫= X (k)t +tRxk∑n=0|x|∞∑λ n F n (dx) < ∞n=0λ n F n (dx) = C (k,+)t−C (k,−)t ,the left hand side above is almost surely absolutely convergent if and only if∫R|x|∞∑λ n F n (dx) < ∞.n=0The latter integral test is tantamount to saying∫|x|Π(dx) < ∞. (5.1)(−1,1)The importance of the previous calculations is that it is straightforward tocheck that the variation of X (3) over [0,t] is at least C (∞,+)t + C (∞,−)t whichexplodes if and only if (5.1) fails and, when it is well defined, is equal toC (∞,+)t∫+C (∞,−)t +tRxk∑λ n F n (dx)n=0which is almost surely finite if and only if (5.1) holds.In conclusion we have the following result.Lemma 5.1 Any Lévy process has paths of bounded variation if and only if∫σ = 0 and |x|Π(dx) < ∞.(−1,1)In the case that X has bounded variation, we may express the process X (3) inthe, now meaningful, wayX (3)t = ∑ ∫s 1 (|∆Xs|

Indeed the calculations abovehave shown that the sum ∑ s≤t ∆X s is necessarilyabsolutely convergent almost surely.Note that montone functions necessarily have bounded variation and hencea Lévy process is a subordinator only if it has paths of bounded variation. Inthat case inspecting (5.2) tells us the following.Lemma 5.2 A Lévy process is a subordinator if and only if it has paths ofbounded variation, Π(−∞,0) = 0 and∫δ := −a− xΠ(dx) ≥ 0.(0,1)In that case the characteristic exponent may be written∫Ψ(θ) = −iδθ+ (1−e iθx )Π(dx).(0,∞)26

ExercisesExercise 1 Using Definition 1.1, show that the sum oftwo (orindeed any finitenumber of) independent Lévy processes is again a Lévy process.Exercise 2 Suppose that S = {S n : n ≥ 0} is any random walk 9 and Γ p is anindependent random variable with a geometric distribution on {0,1,2,...} withparameter p.(i) Show that Γ p is infinitely divisible.(ii) Show that S Γp is infinitely divisible.Exercise 3 [ProofofLemma2.1]InthisexercisewederivetheFrullaniidentity.(i) Show for any function f such that f ′ exists and is continuous and f(0) andf(∞) are finite, that∫ ∞(f(ax)−f(bx)bdx = (f(0)−f(∞))log ,xa)0where b > a > 0.(ii) By choosing f(x) = e −x , a = α > 0 and b = α−z where z < 0, show that1(1−z/α) β = e−∫ ∞0 (1−ezx ) β x e−αx dxand hence by analytic extension show that the above identity is still validfor all z ∈ C such that Rz ≤ 0.Exercise 4 Establishing formulae (2.5) and (2.6) from the Lévy measure givenin (2.7) is the result of a series of technical manipulations of special integrals.In this exercise we work through them. In the following text we will use thegamma function Γ(z), defined byΓ(z) =∫ ∞0t z−1 e −t dtfor z > 0. Note the gamma function can also be analytically extended so thatit is also defined on R\{0,−1,−2,...} (see [16]). Whilst the specific definitionof the gamma function for negative numbers will not play an important role inthis exercise, the following two facts that can be derived from it will. For z ∈R\{0,−1,−2,...} the gamma function observes the recursion Γ(1+z) = zΓ(z)and Γ(1/2) = √ π.9 Recall that S = {S n : n ≥ 0} is a random walk if S 0 = 0 and for n = 1,2,3,... theincrements S n −S n−1 are independent and identically distributed.27

(i) Suppose that 0 < α < 1. Prove that for u > 0,∫ ∞0(e −ur −1)r −α−1 dr = Γ(−α)u αand show that the same equality is valid when −u is replaced by anycomplex number w ≠ 0 with Rw ≤ 0. Conclude by considering w = i that∫ ∞0(1−e ir )r −α−1 dr = −Γ(−α)e −iπα/2(E.1)as well as the complex conjugate of both sides being equal. Deduce (2.5)by considering the integral∫ ∞0(1−e iξθr )r −α−1 drfor ξ = ±1 and θ ∈ R. Note that you will have to take a = η −∫R x1 (|x|

Exercise 8 Recall that D[0,1] is the space of functions f : [0,1] → R whichare right continuous with left limits.(i) Define the norm ||f|| = sup x∈[0,1] |f(x)|. Use the triangle inequality todeduce that D[0,1] is closed under uniform convergence with respect tothe norm ||·||. That is to say, show that if {f n : n ≥ 1} is a sequence inD[0,1] and f : [0,1] → R such that lim n↑∞ ||f n −f|| = 0 then f ∈ D[0,1].(ii) Suppose that f ∈ D[0,1] and let ∆ = {t ∈ [0,1] : |f(t)−f(t−)| ̸= 0} (theset of discontinuity points). Show that ∆ is countable if ∆ c is countablefor all c > 0 where ∆ c = {t ∈ [0,1] : |f(t)−f(t−)| > c}. Next fix c > 0.By supposing for contradiction that ∆ c has an accumulation point, say x,show that the existence of either a left or right limit at x fails as it wouldimply that there is no left or right limit of f at x. Deduce that ∆ c andhence ∆ is countable.Exercise 9 The explicit construction of a Lévy process given in the Lévy–Itôdecomposition begs the question as to whether one may construct examples ofdeterministic functions which have similar properties to those of the paths ofLévyprocesses. The objective of this exercise is to do preciselythat. The readeris warned however, that this is purely an analytical exercise and one should notnecessarily think of the paths of Lévy processes as being entirely similar to thefunctions constructed below in all respects.(i) Let us recall the definition of the Cantor function which we shall use toconstruct a deterministic function which has bounded variation and thatis right continuous with left limits and whose points of discontinuity aredense in its domain. Take the interval C 0 := [0,1] and perform the followingiteration. For n ≥ 0 define C n as the union of intervals whichremain when removing the middle third of each of the intervals whichmake up C n−1 . The Cantor set C is the limiting object, ⋂ n≥0 C n and canbe described byC = {x ∈ [0,1] : x = ∑ k≥1α k3 k such that α k ∈ {0,2} for each k ≥ 1}.One sees then that the Cantor set is simply the points in [0,1] whichomits numbers whose tertiary expansion contain the digit 1. To describethe Cantor function, for each x ∈ [0,1] let j(x) be the smallest j for whichα j = 1 in the tertiary expansion of ∑ k≥1 α k/3 k of x. If x ∈ C thenj(x) = ∞ and otherwise if x ∈ [0,1]\C then 1 ≤ j(x) < ∞. The Cantorfunction is defined as followsf(x) = 1j(x)−12 + ∑ α ifor x ∈ [0,1].j(x) 2i+1 i=1Now consider the function g : [0,1] → [0,1] given by g(x) = f −1 (x)−axfor a ∈ R. Here we understand f −1 (x) = inf{θ : f(θ) > x}. Note that g is30

monotone if and only if a ≤ 0. Show that g has only positive jumps andthe value of x for which g jumps form a dense set in [0,1]. Show furtherthat g has bounded variation on [0,1].(ii) Now let us construct an example of a deterministic function which hasunbounded variation, whosepointsofdiscontinuityaredense initsdomainand that is right continuous with left limits. Denote by Q 2 the dyadicrationals. Consider a function J : [0,∞) → R as follows. For all x ≥ 0which are not in Q 2 , set J(x) = 0. It remains to assign a value for eachx = a/2 n where a = 1,3,5,... (even values of a cancel). Let{ 2J(a/2 n −nif a = 1,5,9,...) =−2 −n if a = 3,7,11,...and definef(x) =∑s∈[0,x]∩Q 2J(s).Show that f is uniformly bounded on [0,1], is right continuous with leftlimits and has unbounded variation over [0,1].Exercise 10 Show that any Lévy process of bounded variation may be writtenas the difference of two independent subordinators.Exercise 11 Suppose that X = {X t : t ≥ 0} is a Lévy process with characteristicexponent Ψ and τ = {τ s : s ≥ 0} is an independent subordinator withcharacteristic exponent Ξ. Show that Y = {X τs : s ≥ 0} is again a Lévy processwith characteristic exponent Ξ◦iΨ.Exercise 12 This exercise gives another explicit example of a Lévy process;the variance gamma process, introduced by [21] for modelling financial data.(i) Suppose that Γ = {Γ t : t ≥ 0} is a gamma subordinator with parametersα,β and that B = {B t : t ≥ 0} is an independent standard Brownianmotion. Show that for c ∈ R and σ > 0, the variance gamma processX t := cΓ t +σB Γt , t ≥ 0is a Lévy process with characteristic exponentΨ(θ) = βlog(1−i θcα + σ2 θ 22α ).(ii) Show that the variance gamma process is equal in law to the Lévy processΓ (1) −Γ (2) = {Γ (1)t −Γ (2)t : t ≥ 0},where Γ (1) is a Gamma subordinator with parameters(√ ) −11α (1) c=24α 2 + 1 σ 22 α + 1 cand β (1) = β2α31

and Γ (2) is a Gamma subordinator, independent of Γ (1) , with parametersα (2) =(√14c 2α 2 + 1 2) −1α − 1 cand β (2) = β.2ασ 2Exercise 13 Suppose that d is an integer greater than one. Choose a ∈ R dand let Π be a measure concentrated on R d \{0} satisfying∫R d (1∧|x| 2 )Π(dx)where |·| is the standard Euclidian norm. Show that it is possible to construct ad-dimensional process X = {X t : t ≥ 0} on a probability space (Ω,F,P) havingthe following properties.(i) The paths of X are right continuous with left limits P-almost surely in thesense that foreacht ≥ 0, P(lim s↓t X s = X t ) = 1andP(lim s↑t X s exists) =1.(ii) P(X 0 = 0) = 1, the zero vector in R d .(iii) For 0 ≤ s ≤ t, X t −X s is independent of σ(X u : u ≤ s).(iv) For 0 ≤ s ≤ t, X t −X s is equal in distribution to X t−s .(v) For any t ≥ 0 and θ ∈ R d ,andE(e iθ·Xt ) = e −Ψ(θ)tΨ(θ) = ia·θ+ 1 2 θ ·Aθ + ∫R d (1−e iθ·x +i(θ ·x)1 (|x|

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