Perpetual Integrals for Levy Processes

0.11 MB·English

by  Leif Doering

PERPETUAL INTEGRALS FOR LE´VY PROCESSES
LEIFDO¨RINGANDANDREASE.KYPRIANOU
5
1 Abstract. Given a L´evy 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 spatiallystationaryL´evy 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 mainresultofthepresentarticleisacharacterizationoffinitenessforperpetualintegralsofL´evy

1 processes:
v
Theorem 1. Suppose that ξ is a L´evy 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 L´evy 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 L´evy 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 L´evy 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.
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Leif Do¨ring,DepartementMathematik,ETH Zu¨rich,Ra¨mistrasse 10,8092Zu¨rich,Switzerland
AndreasE.Kyprianou,DepartmentofMathematicalSciences,Universityof Bath,ClavertonDown,Bath
BA2 7AY,United Kingdom