PERPETUAL INTEGRALS FOR LE´VY PROCESSES LEIFDO¨RINGANDANDREASE.KYPRIANOU 5 1 Abstract. Given a Lévy process ξ we ask for necessary and sufficient conditions for almost sure ∞ 0 finiteness of the perpetual integral 0 f(ξs)ds, where f is a positive locally integrable function. If 2 µ=E[ξ1]∈(0,∞)andξ haslocaltRimesweprovethe0-1law ∞ n P f(ξs)ds<∞ ∈{0,1} a (cid:16)Z0 (cid:17) J withtheexactcharacterization ∞ ∞ 4 P f(ξs)ds<∞ =0 ⇐⇒ f(x)dx=∞. (cid:16)Z0 (cid:17) Z ] Theproof uses spatiallystationaryLévy processes, local timecalculations, Jeulin’s lemmaand the R Hewitt-Savage 0-1law. P . h The study of perpetual integrals f(X )ds with finite or infinite horizon for diffusion processes X t s a hasa longhistorypartiallybecausReoftheiruse inthe analysisofstochasticdifferentialequationsand m insurance, financial mathematics as the present value of a continuous stream of perpetuities. [ The mainresultofthepresentarticleisacharacterizationoffinitenessforperpetualintegralsofLévy 1 processes: v Theorem 1. Suppose that ξ is a Lévy process that has strictly positive mean µ<∞, local times and 5 4 is not a compound Poisson process. If f is a measurable locally integrable positive function, then the 6 following 0-1 law holds: 0 ∞ ∞ 0 (T1) P f(ξ )ds<∞ =1 ⇐⇒ f(x)dx<∞ s . (cid:16)Z0 (cid:17) Z 1 and 0 ∞ ∞ 5 (T2) P f(ξ )ds<∞ =0 ⇐⇒ f(x)dx=∞. 1 s : (cid:16)Z0 (cid:17) Z v Let us briefly compare the theorem with the existing literature: i X (i) If ξ is a Brownian motion with positive drift, then results were obtained through the Ray-Knight r theorem, Jeulin’s lemma and Khashminkii’s lemma by Salminen/Yor [11], [12]. a (ii) For spectrally negative ξ, i.e. ξ only jumps downwards, the equivalence was obtained in Khosh- nevisan/Salminen/Yor [12], see also Example 3.9 of Schilling/Voncracek [9]. The spectrally negative case also turns out to be easier in our proof. (iii) If f is (ultimately) decreasing, results for general Lévy processes have been proved in Erick- son/Maller[5].InthiscasetheresultstatedinTheorem1followseasilyfromthelawoflargenumbers by estimating 2µt>ξ > 1µt for t big enough. The very same argumentalso shows that for µ=+∞ t 2 the integral test (T) fails in general. For a result in this case we again refer to [5]. Remark 2. It is not clear whether or not the assumption ξ having local time plays a role. For (ultimately)decreasingf theexistenceoflocaltimeisclearlynotneeded,whereaswehavenoconjecture for general f. Proof of Theorem 1 Before going into the proof let us fix some notation and facts needed below. For more definitions and background we refer for instance to [1] or [8]. The law of ξ issued from x∈R will be denoted by Px, 1 2 LEIFDO¨RINGANDANDREASE.KYPRIANOU abbreviating P=P0, and the characteristic exponent is defined as Ψ(λ):=−logE exp(iλξ ) , λ∈R. 1 (cid:2) (cid:3) We recall from Theorem V.1 of [1] that ξ has local times L (x) if and only if t t≥0,x∈R (cid:0) (cid:1) ∞ 1 (1) R dr <∞. Z (cid:18)1+Ψ(r)(cid:19) −∞ This means that for any bounded measureable function f :R→[0,∞) the occupation time formula t f(ξ )ds= f(x)L (x)dx, t≥0, s t Z0 ZR holds almost surely. A consequence of (1) is also that points are non-polar. More precisely, a Theorem of Kesten and Bretagnolle states that P(τ <∞)> 0 for all x> 0 if τ = inf{t: ξ = x}, see for instance Theorem x x t 7.12 of [8]. Throughout we assume ξ is transient so that ε R( 1 )dr < ∞ and, consequently, (1) implies −ε ψ(r) ∞ R 1 dr <∞. But then Theorem II.16Rof [1] implies that the potential measure −∞ Ψ(r) R (cid:16) (cid:17) ∞ U(dx)= P ξ ∈dx ds s Z 0 (cid:0) (cid:1) has a bounded density u(x) with respect to the Lebesgue measure. We start with the easy direction of Theorem 1: Proof of Theorem 1, Sufficiency of Integral Test. Supposethat f(x)dx<∞.Since weassumethat R ξ is transient and has a local time we can use the existence andRboundedness of the potential density to obtain ∞ ∞ E f(ξ )ds = f(x) P(ξ ∈dx)ds= f(x)u(x)dx≤supu(x) f(x)dx<∞. s s hZ0 i ZR Z0 ZR x∈R ZR Since finiteness of the expectationimplies almostsure finiteness the sufficiency ofthe integraltest for almost sure finiteness of the perpetual integral is proved. (cid:3) For the reverse direction we use Jeulin’s lemma, here is a simple version: Lemma3. Suppose(Xx)x∈R arenon-negative,non-trivialandidenticallydistributedrandomvariables on some probability space (Ω,F,P), then P f(x)X dx<∞ =1 =⇒ f(x)dx<∞. x (cid:16)ZR (cid:17) ZR Proof. Since X are identically distributed, we may choose ε > 0 so that P(X > ε) = δ > 0 for all x x x∈R. Since f(x)X dx is almost surely finite, there is some N ∈N so that P(A )>1−δ/2 with R x N A ={ f(xR)X dx>N}. Hence, we have N R x R E X 1 ≥εP({X >ε}∩A )>εδ/2>0. x AN x N (cid:2) (cid:3) But then N ≥NP(A )≥E f(x)X dx1 = f(x)E[X 1 ]dx≥εδ/2 f(x)dx. N hZR x ANi ZR x AN ZR The proof is now complete. (cid:3) Note that there are different versions of Jeulin’s lemma (see for instance [10]). Most commonly, one refers to Jeulin’s lemma if P( f(x)X dx < ∞) > 0 but more assumptions on the X are posed. R x x Those extra assumptions on XRx are not satisfied in our setting but we can employ a 0-1 law that allows us to work with P( f(x)X dx<∞)=1. R x R PERPETUAL INTEGRALS FOR LE´VY PROCESSES 3 We would like to apply Jeulin’s lemma via the occupation time formula ∞ t ∞ ∞ f(ξ )ds= lim f(ξ )ds= lim f(x)L (x)dx= f(x)L (x)dx s s t ∞ Z0 t↑∞Z0 t↑∞Z0 Z0 withL (x):=lim L (x).TheargumentistoosimplisticbecausethedistributionofL (x)depends ∞ t↑∞ t ∞ on x. Only if ξ is spectrally negative the laws L (x) are independent of x by the strong Markov ∞ property.To makethis idea workweworkwith randomizedinitialconditionsinstead.In whatfollows we chose a particularly convenient initial distribution motivated by a result from fluctuation theory (see Lemma 3 of [2]). Our assumption E[ξ ]<∞ implies that 1 (2) P ξ −z ∈dy z=→⇒∞ρ(dy), Tz (cid:0) (cid:1) where T = inf{t ≥ 0 : ξ ≥ z} and ρ is a non-degenerate probability law, called the stationary z t overshoot distribution. The convergence in (2) is a consequence of the quintuple law of [8] and the distribution ρ can be written down explicitly. Since ρ is the stationary overshootdistribution we have Pρ ξ −a∈dy := Px ξ −a∈dy ρ(dx)=ρ(dy), ∀a>0, Ta Z Ta (cid:0) (cid:1) (cid:0) (cid:1) so that spatial stationarity holds due to the strong Markov property: under Pρ (3) (ξ(a)) :=(ξ −a) t t≥0 Ta+t t≥0 has law Pρ for all a>0. The stationarity property (3) will be the key to apply Jeulin’s lemma. Lemma 4. For any x>0 we have Pρ(L (x)∈dy)=Pρ(L (1)∈dy). ∞ ∞ Proof. Firstnote thatPρ(T <∞)=1 for all x>0 andL(x) only starts to increaseatsome time at x · or after T . Then x ∞ Pρ(L (x)∈dy) = Pz(L (x)∈dy)Pρ(ξ ∈dz) ∞ Z ∞ Tx 0 ∞ = Pz+x(L (x)∈dy)Pρ(ξ −x∈dz) Z ∞ Tx 0 ∞ = Pz+x(L (x)∈dy)ρ(dz) ∞ Z 0 ∞ = Pz(L (0)∈dy)ρ(dz), ∞ Z 0 using the strong Markov property, the spatial stationarity of Pρ and spatial homogeneity of ξ. Since the righthand side is independent of x the proof is complete. (cid:3) Next, we use the Hewitt-Savage 0-1 law (see [4]) in order to get the weak version of Jeulin’s lemma going. If X0,X1,... denotes a sequence of random variables taking values in some measurable space, thenaneventA∈σ(X0,X1,...)iscalledexchangeableifitisinvariantunderfinite permutations(i.e. only finitely many indices are changed) of the sequence X0,X1,.... The Hewitt-Savage 0-1 law states that any exchangeable event of an iid sequence has probability 0 or 1. Lemma 5. P ∞f(ξ )ds<∞ ∈{0,1}. 0 s (cid:0)R (cid:1) ∞ Proof. The idea of the proof is to write Λ := { f(ξ )ds < ∞} as an exchangeable event with 0 s respectto the iidincrements ofξ onintervals[n,nR+1]sothat P(Λ)∈{0,1}.LetD denote the RCLL functions w :[0,1]→R. If ξ is the given Lévy process, then define the increment processes as (ξn) =(ξ −ξ ) . t t∈[0,1] n+t n t∈[0,1] 4 LEIFDO¨RINGANDANDREASE.KYPRIANOU The Lévy property implies that the sequence ξ0,ξ1,... is iid on D. Furthermore, note that ξ can be reconstructed from the ξn through n−1 ξ =ξn + ξi ∀r∈[n,n+1). r r−n 1 Xi=0 Using that g : (w ) 7→ (w ) , g : (w,w′) 7→ (w +w′) and g : (w ) 7→ 1 t t∈[0,1) 1 t∈[0,1) 2 t∈[0,1) t t t∈[0,1) 3 t t∈[0,1) 1f(w )ds are measurable mappings, there are measurable mappings gn :Dn →R such that 0 s R 1 n−1 f ξn+ ξi dr =gn(ξ0,...,ξn). Z0 (cid:16) r Xi=0 1(cid:17) As a consequence we find that ∞ ∞ n+1 f(ξ )ds<∞ = f(ξ )ds<∞ s s nZ0 o nnX=0Zn o ∞ 1 n−1 = f ξn+ ξi dr <∞ nnX=0Z0 (cid:16) r Xi=0 1(cid:17) o ∞ = gn(ξ0,...,ξn)<∞ nnX=0 o ∈σ(ξ0,ξ1,...). Since clearly Λ is exchangeable for ξ0,ξ1,... the Hewitt-Savage 0-1 law implies the claim. (cid:3) Lemma 6. Suppose P( ∞f(ξ )ds<∞)=1, then Pρ( ∞f(ξ )ds<∞)=1. 0 s 0 s R R Proof. The statement is obvious if ξ is a subordinator, so we assume it is not. Next we show that Px( ∞f(ξ )ds < ∞) = 1 for any x > 0. To see this we use the strong Markov 0 s property at τ0 = inf{tR: ξt = 0} which is finite with positive probabillity since points in R are non-polar: ∞ ∞ Px f(ξ )ds<∞ ≥Px f(ξ )ds<∞,τ <∞ s s 0 (cid:16)Z0 (cid:17) (cid:16)Zτ0 (cid:17) ∞ =Px f(ξ −ξ )ds<∞,τ <∞ (cid:16)Z0 s+τ0 τ0 0 (cid:17) ∞ =P0 f(ξ )ds<∞ Px(τ <∞) s 0 (cid:16)Z0 (cid:17) >0. But then the 0-1 law of Lemma 5 implies that Px ∞f(ξ )ds<∞ =1. Finally, we obtain 0 s (cid:16)R (cid:17) ∞ ∞ Pρ f(ξ )ds<∞ = Px f(ξ )ds<∞ ρ(dx)= ρ(dx)=1 s s (cid:16)Z0 (cid:17) ZR (cid:16)Z0 (cid:17) ZR and the proof is complete. (cid:3) Now we are ready to prove the more delicate part of Theorem 1. Proof of Theorem 1, Necessity of Integral Test. Suppose P( ∞f(ξ )ds<∞)>0 which then implies 0 s P( ∞f(ξ )ds < ∞) = 1 by Lemma 5. Hence, Pρ( ∞f(ξR)ds < ∞) = 1 by Lemma 6. Using the 0 s 0 s occRupation time formula we get R ∞ t f(ξ )ds= lim f(ξ )ds= lim f(x)L (x)dx= f(x)L (x)dx Pρ-a.s. s s t ∞ Z0 t→∞Z0 t→∞ZR ZR In Lemma 4 we proved that L (x) is independent of x under Pρ so that Jeulin’s Lemma implies ∞ f(x)dx<∞. (cid:3) R R PERPETUAL INTEGRALS FOR LE´VY PROCESSES 5 Acknowledgement. The authors thank Jean Bertoin for several discussions on the topic. References [1] J.Bertoin:”Lévyprocesses.”CambridgeTracts inMathematics 121,CambridgeUniversityPress,1996. [2] J.Bertoin,M.Savov:”SomeapplicationsofdualityforLévyprocessesinahalf-line.”Bull.LondonMath.Soc.43, pp.97-110,2011. [3] D. Dufresne: ”The distribution of a perpetuity, with applications to risk theory and pension funding.” Scand. ActuarialJ.,pp.39-79,1990. [4] E.Hewitt, L.J. Savage: “Symmetricmeasures onCartesianproducts.”Trans.Amer.Math.Soc., 80, pp.470-501 (1955). [5] K.B. Erickson, R.A. Maller: ”Generalised Ornstein-Uhlenbeck processes and the convergence of Lévy integrals.” SéminairedeProbabilitésXXXVIII,2005. [6] D. Khoshnevisan, P. Salminen, M. Yor: ”A note on a.s. finiteness of perpetual integral functionals of diffusions.” Electr.Comm.Probab.,Vol.11,pp.108-117. [7] A. Kyprianou, J.C. Pardo, V. Rivero: ”Exact and asymptotic n-tuple laws at first and last passage.” Ann. Appl. Probab.Volume20,pp.522-564,2010. [8] A.Kyprianou:“Fluctuations ofLévyProcesseswithApplications.”Springer(2013). [9] R. Schilling, Z. Vondraˇcek: “Absolute continuity and singularity of probability measures induced by a purely discontinuous Girsanovtransformofastableprocess.”arXiv:1403.7364 [10] A. Matsumoto, K. Yano: ”On a zero-one law for the norm process of transient random walk.” Séminaire de ProbabilitésXLIII,pp.105-126, 2011. [11] P.Salminen,M.Yor:”PropertiesofperpetualintegralfunctionalsofBrownianmotionwithdrift.”Ann.I.H.P.,41, pp.335-347,2005. [12] P. Salminen, M.Yor:”Perpetual integral fuctionals as hitting and occupation times.”Electr.J. Probab., Vol.10, p.371-419,2005. Leif Do¨ring,DepartementMathematik,ETH Zu¨rich,Ra¨mistrasse 10,8092Zu¨rich,Switzerland AndreasE.Kyprianou,DepartmentofMathematicalSciences,Universityof Bath,ClavertonDown,Bath BA2 7AY,United Kingdom

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