first hitting time law for some jump-diffusion processes ...docs.isfa.fr/labo/2009.5(wp2105).pdf ·...
TRANSCRIPT
First hitting time law for some jump-diffusionprocesses : existence of a density
- Laure COUTIN (Université de Toulouse, IMT)- Diana Dorobantu (Université Lyon 1, Laboratoire SAF)
2009.5 (WP 2105)
Laboratoire SAF – 50 Avenue Tony Garnier - 69366 Lyon cedex 07 http://www.isfa.fr/la_recherche
First hitting time law for some jump-diffusion
processes : existence of a density∗
Laure Coutin
IMT, University of Toulouse
118 route de Narbonne, 31062, Toulouse
Diana Dorobantu
Université de Lyon, Université Lyon 1, Laboratoire de Science Actuarielle et Financière,
ISFA, 50 avenue Tony Garnier, F-69007, Lyon
Abstract. Let (Xt, t ≥ 0) be a diffusion process with jumps, sum of a Brownian motionwith drift and a compound Poisson process. We consider τx the first hitting time of a fixed levelx > 0 by (Xt, t ≥ 0). We prove that the law of τx has a density (defective when E(X1) < 0)with respect to the Lebesgue measure.
Keywords : Lévy process, jump-diffusion process, hitting time law.
∗The authors thank M. Pontier and Ph. Carmona for their careful reading and G. Letac for his helpfull
comments
1
1 Introduction
The main purpose of this paper is to show that the first hitting time distribution associated witha jump-diffusion process (sum of a Brownian motion with drift and an independent compoundPoisson process) has a density with respect to the Lebesgue measure.
Let (Xt, t ≥ 0) be a càd-làg process started at 0 and τx the first hitting time of level x > 0by X.
Lévy, in [16], computes the law of τx when X is a Brownian motion with drift. This resultis extended by Alili, Patie and Petersen [1] or Leblanc [13] to the case where X is an Ornstein-Uhlenbeck process. The case where X is a Bessel process was studied by Borodin and Salminenin [4].
Some results are also available when the process X has jumps. The first results are obtainedby Zolotarev [23] and Borokov [5] when X is a spectrally negative Lévy process. They give thelaw of τx . Moreover, if Xt has the probability density with respect to the Lebesgue measurep(t, x), then the law of τx has the density with repect to the Lebesgue measure f(t, x), wherexf(t, x) = tp(t, x). In this case where X has only negative jumps, Xτx = x almost surely.
If X is a spectrally positive Lévy process, Doney [7] gives an explicit formula for the jointLaplace transform of τx and the overshoot Xτx − x. When X is a stable Lévy process, Peskir[17], and Bernyk, Dalang and Peskir [2] obtain an explicit formula for the hitting time density.
The case where X has signed jumps is more recently studied. In [9], the authors give the lawof τx when X is the sum of a decreasing Lévy process and an independent compound processwith exponential jump sizes. This result is extended by Kou et Wang in [11] to the case ofa diffusion process with jumps where the jump sizes follow a double exponential law. Theycompute the Laplace transform of τx and derive an expression for the density of τx. For a moregeneral jump-diffusion process, Roynette, Vallois and Volpi [20] show that the Laplace transformof (τx, x − Xτx−, Xτx − x) is solution of some kind of random integral.
Doney and Kyprianou [8] studied the problem for general Lévy processes. They give thequintuple law of (Gτx−, τx − Gτx−, Xτx − x, x − Xτx−, x − Xτx−) where Xt = sups≤t Xs andGt = sup{s < t, Xs = Xs}.
Results are also available for some Lévy processes without Gaussian component, see Lefèvre,Loisel and Picard [14, 15, 18, 19]. Blanchet [3] considers a process satisfying the followingstochastic equation : dSt = St−(µdt + σ1φ(t)=0dWt + φ1φ(t)=φdNt), t ≤ T where T is a finite
horizon, µ ∈ R, σ > 0, φ(.) is a function taking two values 0 or φ, W is a Brownian motion, Nis a Poisson process with intensity 1
φ2 1φ(t)=φ and N is the compensated Poisson process.
The aim of our paper is to add to these studies the law of a first hitting time by a Lévyprocess which is the sum of a Brownian motion with drift and a compound Poisson process. Wedo not limit our study to a particular distribution of the jumps sizes.
This paper is organized as follows : Section 2 contains the main result (Theorem 2.1) whichgives the first hitting time law by a jump Lévy process. The following two sections (Section2.1 and Section 2.2) are dedicated to the proof of Theorem 2.1. In these sections we computethe derivative at t = 0 (Section 2.1) and at t > 0 (Section 2.2) of the hitting time distributionfunction. Section 3 contains the proofs of some useful results.
2
2 Hitting time law
Let m ∈ R, (Wt, t ≥ 0) be a standard Brownian motion, (Nt, t ≥ 0) be a Poisson processwith constant positive intensity a and (Yi, i ∈ N
∗) be a sequence of independent identicallydistributed random variables with distribution function FY . We suppose that the followingσ−fields σ(Yi, i ∈ N
∗), σ(Nt, t ≥ 0) and σ(Wt, t ≥ 0) are independent. Let (Tn, n ∈ N∗) be
the sequence of the jump times of the process N .
Let X be the process defined by
Xt = mt + Wt +
Nt∑
i=1
Yi, t ≥ 0. (1)
Let τx be the first hitting time of level x > 0 by the stochastic process (Xt, t ≥ 0) :τx = inf{u > 0 : Xu ≥ x}.
The main result of this paper is the following theorem. It gives the law of τx.
Theorem 2.1 If there exists β > 0 such that E(
eβ|Y1|)
< ∞, then the distribution function of τx
has a right derivative at 0 and is differentiable at every point of ]0, ∞[. The derivative, denotedf(., x), is equal to
f(0, x) =a
2
(
2 − FY (x) − FY (x−))
+a
4
(
FY (x) − FY (x−))
and for every t > 0 f(t, x) = aE (1τx>t(1 − FY ) (x − Xt)) + E
(
1τx>TNtf
(
t − TNt , x − XTNt
))
where
f(u, z) =| z |√2πu3
exp
[
−(z − mu)2
2u
]
1]0,∞[(u), u ∈ R, z ∈ R.
Furthermore, P(τx = ∞) = 0 if and only if m + aE(Y1) ≥ 0.
The proof of Theorem 2.1 is given in Sections 2.1 and 2.2.
Let X be the m drifted Brownian motion defined by
Xt = mt + Wt, for all t ≥ 0,
and τz be the first hitting time of z > 0 by X. Then, according to [10], f(., z) is the derivativeof the distribution function of τz (see Section 3 for more details).
Let (Ft)t≥0 be the completed natural filtration generated by the processes (Wt, t ≥ 0),(Nt, t ≥ 0) and the random variables (Yi, i ∈ N
∗).
Remark 2.2 This result is already known in the following cases :When X has no positive jumps, the law of τx is given in Theorem 46.4 page 348 of [21].When X is a stable Lévy process, with no negative jumps, the law of τx is given in [2].When X is a jump-diffusion where the jump sizes follow a double exponential law, the result isgiven in [11].
3
2.1 Existence of the right derivative at t = 0
In this section we prove the first part of Theorem 2.1. Our goal is to show that the distributionfunction of τx (i.e. t 7→ P(τx ≤ t)) has a right derivative at 0 and we compute this derivative.
For this purpose we split the probability P(τx ≤ h) according to the values of Nh :
P(τx ≤ h) = P(τx ≤ h, Nh = 0) + P(τx ≤ h, Nh = 1) + P(τx ≤ h, Nh ≥ 2).
Note that P(τx ≤ h, Nh ≥ 2) ≤ P(Nh ≥ 2) = 1 − e−ah − ahe−ah, thus
limh→0P(τx ≤ h, Nh ≥ 2)
h= 0.
The proof of the first part of Theorem 2.1 will be complete when the following lemma andproposition are shown :
Lemma 2.3 The term P(τx≤h, Nh=0)h converges to 0 when h goes to 0.
Proposition 2.4 If there exists β > 0 such that E(
eβ|Y1|)
< ∞, then for every x > 0 the termP(τx≤h, Nh=1)
h converges to a2 (2 − FY (x) − FY (x−)) + a
4 (FY (x) − FY (x−)) when h goes to 0.
Proof of Lemma 2.3.The essential observation is that on the set {ω : Nh(ω) = 0}, the processes (Xt, 0 ≤ t ≤ h) and(Xt, 0 ≤ t ≤ h) are equal , and P-almost surely τx ∧ h = τx ∧ h. Since τx is independent of N ,then
P(τx ≤ h, Nh = 0)
h=
e−ahP(τx ≤ h)
h.
The law of τx has a C∞ density (possibly defective) with respect to the Lebesgue measure, null
on ] −∞, 0]. Thus the limit of P(τx≤h, Nh=0)h exists and is equal to 0 when h goes to 0. 2
To prove Proposition 2.4, we use the same type of arguments as in [20] (for the proof ofTheorem 2.4). In [20], the authors compute the joint Laplace transform of (a passage time of aLévy process, overshoot) as solution of an integral equation.Proof of Proposition 2.4.We split the probability P(τx ≤ h, Nh = 1) into three parts according to the relative position ofτx and T1, the first jump time of the Poisson process N :
P(τx ≤ h, Nh = 1) = P(τx < T1, Nh = 1) + P(τx = T1, Nh = 1) + P(T1 < τx ≤ h, Nh = 1).(2)
Step 1 : We prove that the contribution to the limit limh→0+ h−1P(τx ≤ h, Nh = 1) of the firstterm on the right hand side of (2) is null.
Since the processes (Xt, 0 ≤ t < T1) and (Xt, 0 ≤ t < T1) are equal, then on the set{ω, τx(ω) < T1(ω)}, the stopping time τx and τx are equal. Thus
P(τx < T1, Nh = 1) = P(τx < T1 ≤ h, Nh = 1) ≤ P(τx ≤ h).
4
The law of τx has a C∞ density (possibly defective) with respect to the Lebesgue measure,
null on ] −∞, 0]. Thus P(τx<T1, Nh=1)h converges to 0 when h goes to 0.
Step 2 : We prove that the contribution to the limit limh→0+ h−1P(τx ≤ h, Nh = 1) of thesecond term on the right hand side of (2) is a
2 (2 − FY (x) − FY (x−)).
Note that
P(τx = T1, Nh = 1) = P(τx > T1, XT1 + Y1 ≥ x, T1 ≤ h < T2).
Here, for every n ∈ N∗, Tn = S1 + ... + Sn where (Si, i ≥ 1) is a sequence of independent
identically distributed random variables with exponential distribution with parameter a. Usingthe independence between (Si, i ≥ 1) and (Y1, X, τx) we get :
P(τx = T1, Nh = 1) =
∫ h
0ae−as1
∫ ∞
h−s1
ae−as2E
(
1τx>s11Y1≥x−Xs1
)
ds2ds1
=ae−ah
∫ h
0E
(
1τx>s1Y1≥x−Xs
)
ds.
Integrating with respect to Y1, we obtain :
P(τx = T1, Nh = 1) = ae−ah
∫ h
0E
(
(1 − FY )((x − Xs)−)
)
ds−ae−ah
∫ h
0E
(
1τx≤s(1 − FY )((x − Xs)−)
)
ds.
On the one hand, since FY is a càdlàg bounded function and X is a Brownian motion with drift,we get
lims→0E
(
FY ((x − Xs)−)
)
=FY (x) + FY (x−)
2.
On the other hand, since the distribution function of τx is differentiable, then
lims→0E
(
1τx≤s(1 − FY )((x − Xs)−)
)
= 0.
We deduce that
limh→0P(τx = T1, Nh = 1)
h=
a
2
(
2 − FY (x) − FY (x−))
.
Step 3 : We prove that the contribution to the limit limh→0+ h−1P(τx ≤ h, Nh = 1) of thethird term on the right hand side of (2) is a
4 (FY (x) − FY (x−)).
To this end, we state the following lemma :
Lemma 2.5 The term P(T1<τx≤h, Nh=1)h converges to a
4 (FY (x) − FY (x−)) when h goes to 0.
Proof Note that
P(T1 < τx ≤ h, Nh = 1) = P(T1 < τx ≤ h, T1 ≤ h < T2)
and T2 = T1 + S2 ◦ θT1 where θ is the translation operator. Here S2 is a random variable withexponential distribution with parameter a, independent of (T1, (Wt, t ≥ 0), (Yi, i ∈ N
∗)).
5
Moreover, on {ω : T1(ω) < τx(ω) ≤ h < T2(ω)}, Xs = XT1 + Xs−T1 ◦ θT1 when T1 < s ≤ hand τx = T1 + τx−XT1
◦ θT1 .
Strong Markov Property at the (Ft, t ≥ 0)-stopping time T1 gives :
P(T1 < τx ≤ h, Nh = 1) = E
(
1τx>T11h≥T1ET1
(
1τx−XT1≤h−T11h−T1<S2
))
where ET1(.) is E(. | FT1).
Integrating with respect to S2, we obtain :
P(T1 < τx ≤ h, Nh = 1) = E
(
1τx>T11h≥T1e−a(h−T1)
ET1
(
1τx−XT1≤h−T1
))
.
Remark that {ω : τx(ω) > T1(ω)} = {ω : τx(ω) > T1(ω)} ∩ {ω : XT1(ω) < x}. Consequently
P(T1 < τx ≤ h, Nh = 1) = − E
(
1τx≤T1≤h1XT1<xe−a(h−T1)
ET1
(
1τx−XT1≤h−T1
))
+ E
(
1h≥T11XT1<xe−a(h−T1)
ET1
(
1τx−XT1≤h−T1
))
.
Since the distribution function of τx has a null derivative at 0, then
limh→01
hE
(
1τx≤T1≤h1XT1<xe−a(h−T1)
ET1
(
1τx−XT1≤h−T1
))
= 0.
It remains to show that
limh→0+
G(h)
h=
a
4[F (x) − F (x−)] (3)
whereG(h) = E
(
1h≥T11XT1<xe−a(h−T1)
ET1
(
1τx−XT1≤h−T1
))
.
Integrating with respect to T1 and then using the fact that f(., z) is the derivative of the distri-bution function of τz, we get :
G(h) =
∫ h
0ae−as
E
[
1Xs+Y1<xe−a(h−s)E
s(1τx−Xs−Y1
≤h−s)]
ds
=ae−ah
∫ h
0
∫ h−s
0E
[
1Xs+Y1<xf(u, x − Xs − Y1)]
duds.
Since Xs = ms + Ws, we may apply Lemma 3.1 to µ = x − ms − Y1 and σ =√
s, then
G(h) =ae−ah
√2π
∫ h
0
∫ h−s
0E
[
e− (x−m(u+s)−Y1)2
2(u+s)
(
x − Y1
(u + s)3/2+
G√
s√u(u + s)
)
+
]
duds.
We make the following change of variable r = u + s.
G(h) =ae−ah
√2π
∫ h
0
∫ h
sE
[
e−(x−mr−Y1)2
2r
(
x − Y1
r3/2+
G√
s
r√
r − s
)
+
]
drds.
6
Firstly, we apply Fubini’s Theorem and secondly, we make the following change of variable v = sr .
G(h) =ae−ah
√2π
∫ h
0
∫ r
0E
[
e−(x−mr−Y1)2
2r
(
x − Y1
r3/2+
G√
s
r√
r − s
)
+
]
dsdr
=ae−ah
√2π
∫ h
0g(r)dr,
where
g(r) =
∫ 1
0E
[
e−(x−mr−Y1)2
2r
(
x − Y1√r
+G√
v√1 − v
)
+
]
dv.
But,
limr→0+
e−(x−mr−Y1)2
2r
(
x − Y1√r
+G√
v√1 − v
)
+
=
√v√
1 − vG+1x=Y1 ,
and
sup0≤r≤1
e−(x−mr−Y1)2
2r
(
x − Y1√r
+G√
v√1 − v
)
+
≤ supz≥0
ze−z2
2 + |m| +√
v√1 − v
|G|.
Then from Lebesgue’s Dominated Convergence Theorem we obtain :
limr→0g(r) = P(Y1 = x)E(G+)
∫ 1
0
√v√
1 − vdv =
√2π
4P (Y1 = x) .
We deduce the identity (3), i.e.
limh→0G(h)
h=
a
4P (Y1 = x) ,
which achieves the proof of Lemma 2.5 2
Propostion 2.4 is a consequence of the Steps 1 to 3. 2
2.2 Existence of the derivative at t > 0
Our task is now to show that the distribution function of τx is differentiable on R∗+ and to
compute its derivative. For this purpose we split the probability P(t < τx ≤ t + h) according tothe values of Nt+h − Nt :
P(t < τx ≤ t + h) =P(t < τx ≤ t + h, Nt+h − Nt = 0) + P(t < τx ≤ t + h, Nt+h − Nt = 1)
+P(t < τx ≤ t + h, Nt+h − Nt ≥ 2). (4)
The third term on the right hand side of (4) is upper bounded by
P(t < τx ≤ t + h, Nt+h − Nt ≥ 2) ≤ P(Nt+h − Nt ≥ 2) = 1 − e−ah − ahe−ah.
7
Therefore limh→0P(t<τx≤t+h, Nt+h−Nt≥2)
h = 0.
Let us study the second term on the right hand side of (4). Markov Property at t gives :
P(t < τx ≤ t + h, Nt+h − Nt = 1) = E(1τx>tPt(τx−Xt ≤ h, Nh = 1)),
where Pt(.) = P(.|Ft).
In virtue of Lemma 2.4,P
t(τx−Xt
≤h, Nh=1)
h converges to
a
2
[
2 − FY (x − Xt) − FY
(
(x − Xt)−)]
+a
4
[
FY (x − Xt) − FY
(
(x − Xt)−)]
and is upper bounded by P(Nh=1)h = ae−ah ≤ a. Dominated Convergence Theorem gives :
limh→0P(t < τx ≤ t + h, Nt+h − Nt = 1)
h= aE (1τx>t(1 − FY )(x − Xt))+
3a
4E (1τx>t∆FY (x − Xt))
where ∆FY (z) = FY (z) − FY (z−).
However the jumps set of FY (the distribution function of Y ) is countable and X has a density(cf. Proposition 3.12 page 90 of [6]). Thus E (1τx>t∆FY (x − Xt)) = 0 : indeed0 ≤ E (1τx>t∆FY (x − Xt)) ≤ E(1Y1=x−Xt) = E(1Xt=x−Y1) = 0. Therefore
limh→0P(t < τx ≤ t + h, Nt+h − Nt = 1)
h= aE (1τx>t(1 − FY ) (x − Xt)) .
The proof of the second part of Theorem 2.1 will be complete when the following propositionis shown :
Proposition 2.6 If there exists β > 0 such that E(
eβ|Y1|)
< ∞, then
limh→0P(t < τx ≤ t + h, Nt+h − Nt = 0)
h= E
(
1τx>TNtf
(
t − TNt , x − XTNt
))
,
where f is the function introduced in Theorem 2.1.
Proof Since TNt is not a stopping time, we can not apply Strong Markov Property. We split
P(t < τ ≤ t + h, Nt+h − Nt = 0) =∞
∑
k=0
P(t < τx ≤ t + h, Nt+h = Nt = k)
=P(t < τx ≤ t + h < T1) +∞
∑
k=1
P (t < τx ≤ t + h, Tk < t < t + h < Tk+1) .
On the set {(ω, t) : Tk(ω) < t}, we have Xt(ω) = XTk(ω) + Xt−Tk
◦ θTk(ω), hence on the set
{ω, τx(ω) > Tk(ω)}, τx = Tk + τx−XTk◦ θTk
. Strong Markov Property at the stopping time Tk
gives
P(t < τ ≤ t + h, Nt+h − Nt = 0) =e−a(t+h)P(t < τx ≤ t + h)
+∞
∑
k=1
E
(
1Tk<t1τx>TkE
Tk
(
1t−Tk<τx−XTk≤t+h−Tk
1t+h−Tk<Sk+1
))
.
8
On the set {(ω, t) : τz(ω) ≤ t < Sk+1(ω)}, we have τz = τz for every z < 0. Therefore
P(t < τx ≤ t + h, Nt+h − Nt = 0) =e−a(t+h)P(t < τx ≤ t + h)
+∞
∑
k=1
E
(
1Tk<t1τx>Tke−a(t+h−Tk)
ETk
(
1t−Tk<τx−XTk≤t+h−Tk
))
.
The FTk-conditional law of τx−XTk
has the density (possibly defective) f(., x − XTk), thus
P(t < τx ≤ t + h, Nt+h − Nt = 0) =e−a(t+h)P(t < τx ≤ t + h)
+∞
∑
k=1
E
(
1Tk<t1τx>Tke−a(t+h−Tk)
∫ t+h−Tk
t−Tk
f(u, x − XTk)du
)
.
Let us point out that e−a(t−Tk) = ETk
(
1Tk+1>t
)
then
P(t < τx ≤ t + h, Nt+h − Nt = 0) =e−ah
∫ t+h
tE (10≤t<T1) f(u, x)du
+ e−ah∞
∑
k=1
∫ t+h
tE
(
1Tk≤t<Tk+11τx>Tk
f(u − Tk, x − XTk))
du,
or shortly
P(t < τx ≤ t + h, Nt+h − Nt = 0) = e−ah
∫ t+h
tE
(
1TNt<τx f(u − TNt , x − XTNt
))
du. (5)
Since f is continuous with respect to u, then
limu→t+
1TNt<τx f(u − TNt , x − XTNt
) = 1TNt<τx f(t − TNt , x − XTNt
),
From Lemma 3.2, Propositions 3.5 and 3.6, f(u − TN−t, x − XTNt) is dominated uniformly in u
by a integrable random variable
f(u − TN−t, x − XTNt) ≤ cε,M (t − TNt)
1−ε
[
1
(x − XTNt)4ε
+ exp
(
2m
M(x − XTNt
)
)
]
.
Here 0 < ε < 1/4 and M > max(1, 2|m|β−1), (where E(eβ|Y |) < +∞, ) and cε,M , is a con-stant defined in Lemma 3.2 depending only on ε and M . Then, using Lebesgue’s DominatedConvergence Theorem in equation (5) we obtain
limh→0P(t < τx ≤ t + h, Nt+h − Nt = 1)
h= E
(
1τx>TNtf(t − TNt , x − XTNt
))
.
2
Using Proposition 2.6, the limit limh→0P(t<τx≤t+h)
h exists and is equal to
aE (1τx>t(1 − FY )(x − Xt)) + E
(
1τx>TNtf(t − TNt , x − XTNt
))
.
The following lemma allows to conclude the proof of Theorem 2.1. The results are known,for example in [15] and [22] but for sake of completeness we give a proof.
9
Lemma 2.7 For all x > 0, the stopping time τx is finite almost surely if and only if m+aE(Y1) ≥0.
Proof Remark thatP(τx = ∞) = P(supt≥0Xt < x).
Thanks to Theorem 7.2 page 183 of [12] which is a consequence of Strong Law of Large Numbers,
if m + aE(Y1) > 0, then limt→∞Xt = +∞ and
if m + aE(Y1) = 0, then limsupt→∞Xt = −liminft→∞Xt = ∞.
Therefore (see Exercise 39.11 page 271 of [21]), if m + aE(Y1) ≥ 0, then supt≥0Xt = +∞. Thisproves the first part of the lemma.
Conversely, let us suppose that m + aE(Y1) < 0. Then limt→∞ Xt = −∞, and accordingto Theorem 48.1 page 363 of [21], supt≥0 Xt < ∞. Assume that there exists x0 > 0, such thatP(τx0 < +∞) = 1. Then from all x, such that x ≤ x0 we have P(τx < +∞) = 1.
Now we use recurrence reasoning. Assume that for n ≥ 1 and for all x ≤ nx0, we haveP(τx < +∞) = 1. Let x such that nx0 < x ≤ (n + 1)x0, then using Strong Markov Property,
P(τx < +∞) = E(1τnx0<+∞1Xτnx0≥x) + E(1τnx0<+∞1Xτnx0
<xPτnx0 (τx−Xτnx0
< ∞)).
Using the recurrence hypothesis, since x − Xτnx0< x0, almost surely P
τnx0 (τx−Xτnx0< ∞) = 1
and then
P(τx < +∞) = 1.
We have proved that if there exists x0 > 0 such that P(τx0 < ∞) = 1, then for all x > 0,P(τx < ∞) = P(supt≥0 Xt ≥ x) = 1. This contradicts the fact that supt≥0 Xt < ∞. Then, ifm + aE(Y1) < 0, for all x > 0, P(τx < ∞) < 1. 2
3 Appendix
A Brownian motion with drift is a process
Xt = mt + Wt, t ≥ 0
with m ∈ R.
Let z > 0 and τz be the passage time defined by τz = inf{t ≥ 0 : Xt ≥ z}. By (5.12) page197 of [10], τz has the following law on R+ :
f(u, z)du + P(τz = ∞)δ∞(du)
where
f(u, z) =| z |√2πu3
exp
[
−(z − mu)2
2u
]
1]0,∞[(u), u ∈ R, and P(τz = ∞) = 1 − emz−|mz|. (6)
For a fixed z, the function f(., z) and all its derivatives admit 0 as limit at 0+. The functionf admits an extension (denoted f) of class C∞ on R, defined by f(u, z) = 0 for u ≤ 0. Moreoverit checks the two following lemmas :
10
Lemma 3.1 Let G be a Gaussian random variable N (0, 1) and let µ ∈ R, σ ∈ R+. Then for
every u ∈ R
E[f(u, µ + σG)1µ+σG>0] =1√2π
E
[
e− (µ−mu)2
2(σ2+u)
(
µ + σ2m
(σ2 + u)3/2+
σG√u(σ2 + u)
)
+
]
where (x)+ = max(x, 0).
Proof Using the probability density function of G and the definition of f (see (6)), we get :
E[f(u, µ + σG)1µ+σG>0] =1√
2πu3
∫
R
1√2π
|µ + σg|e−(µ+σg−mu)2
2u e−g2
2 1µ+σg>0dg.
Since |µ + σg|1µ+σg>0 = (µ + σg)1µ+σg>0 = (µ + σg)+, then
E[f(u, µ + σG)1µ+σG>0] =1√
2πu3
∫
R
1√2π
(µ + σg)+e−(µ+σg−mu)2
2u e−g2
2 dg.
A simple computation shows that (µ+σg−mu)2
u + g2 = σ2+uu
(
g + σ(µ−mu)σ2+u
)2+ (µ−mu)2
σ2+u. Therefore
E[f(u, µ + σG)1µ+σG>0] =e− (µ−mu)2
2(σ2+u)
√2πu3
∫
R
1√2π
(µ + σg)+e−σ2+u
2u
(
g+σ(µ−mu)
σ2+u
)2
dg.
Making the following change of variable x =√
σ2+uu
(
g + σ(µ−mu)σ2+u
)
, we conclude the proof
E[f(u, µ + σG)1µ+σG>0] =e− (µ−mu)2
2(σ2+u)
√2π
∫
R
1√2π
(
µ2 + σ2m
(σ2 + u)3/2+
σx√u(σ2 + u)
)
+
e−x2
2 dx.
2
Lemma 3.2 Let f : R × R → R+ be the function defined by
f(u, z) =
{
|z|√2πu3
exp[− (z−mu)2
2u ] si u > 0
0 si u ≤ 0,
where m ∈ R. Then for every ε > 0 and M ≥ 1, there exists a constant cε,M > 0 such that
f(u, z) ≤ cε,Mu−1+ε
[
1
| z |4ε+ exp
(
2mz
M
)]
.
Proof Let ε > 0 and M ≥ 1 be fixed. Remark that it is enough to prove that there exists aconstant cε,M > 0 such that
f(u, z) ≤ cε,Mu−1+ε 1
| z |2εexp
(mz
M
)
.
11
Then we conclude the proof using the inequality x1x2 ≤ x21+x2
22 for x1 = 1
|z|2ε and x2 = exp(
mzM
)
.
We now seek to upper bound the quotient f(u,z)
u−1+ε 1|z|2ε exp[mz
M], z < 0, u ∈ R+.
Since exp[− (z−mu)2
2u ] ≤ exp[− (z−mu)2
2uM ], then
f(u, z)
u−1+ε 1|z|2ε exp[mz
M ]≤
|z|√2πu3
exp[− (z−mu)2
2uM ]
u−1+ε 1|z|2ε exp[mz
M ]
=1√2π
(
z2
u
)
12+ε
exp
[
− z2
2uM
]
exp
[
−m2u
2M
]
≤M12+ε2ε
√π
(
z2
2uM
)
12+ε
exp
[
− z2
2uM
]
.
Since the function x 7→ x12+εe−x is continuous, null at 0 and at +∞, then it is bounded on R+.
Hence there exists cε > 0 such that x12+εe−x ≤ cε for any x ∈ R+. We apply this inequality to
x = z2
2uM , so ( z2
2uM )12+ε exp[− z2
2uM ] ≤ cε and the proof is complete. 2
Lemma 3.3 Let G be a Gaussian random variable N (0, 1), µ ∈ R, σ > 0 and 0 < α < 1. Thenthere exists two constants k1,α > 0, k2,α > 0, depending only on α, such that
E(|µ + σG|−α) ≤ k1,α
σ|µ|1−α + k2,ασ−α.
Proof Using the probability density function of G, we get :
E(|µ + σG|−α) =
∫ ∞
−µ
σ
1√2π
(µ + σx)−αe−x2
2 µx +
∫ −µ
σ
−∞
1√2π
(−µ − σx)−αe−x2
2 dx.
Integration by parts gives :
E(|µ + σG|−α) =
[
1√2π
e−x2
2(µ + σx)1−α
σ(1 − α)
]∞
−µ
σ
+1
σ(1 − α)
∫ ∞
−µ
σ
1√2π
(µ + σx)1−αxe−x2
2 dx
+
[
1√2π
e−x2
2(−µ − σx)1−α
−σ(1 − α)
]−µ
σ
−∞− 1
σ(1 − α)
∫ −µ
σ
−∞
1√2π
(−µ − σx)1−αxe−x2
2 dx
=1
σ(1 − α)
∫
R
1√2π
|µ + σx|1−αe−x2
2
(
x1x>−µ
σ− x1x≤−µ
σ
)
dx
=1
σ(1 − α)E
[
|µ + σG|1−αG(1G>−µ
σ− 1G≤−µ
σ)]
.
However for every µ ∈ R and σ > 0, we have
|G(1G>−µ
σ− 1G≤−µ
σ)| ≤ G − 2G1G≤0 = |G|.
Consequently E(|µ + σG|−α) ≤ 1σ(1−α)E(|G||µ + σG|1−α).
The inequality |µ + σG|1−α ≤ (|µ| + σ|G|)1−α ≤ |µ|1−α + σ1−α|G|1−α concludes the proof. 2
12
Lemma 3.4 For every t > 0 and (α, γ) ∈] − 1,∞[2, the following two series
∞∑
i≥1
E (1t>Ti(t − Ti)
αT γi ) zi
∞∑
i≥1
E (1t>Ti(t − Ti)
αT γi ) izi,
have an infinite radius of convergence.
Proof It is enough to prove that the first series has an infinite radius of convergence.
Note that for i ≥ 1, Ti admits as density the function u 7→ ai
(i−1)!ui−1e−au, thus
E (1t>Ti(t − Ti)
αT γi ) =
ai
(i − 1)!
∫ t
0e−au(t − u)αuγ+i−1du
≤ ai
(i − 1)!
∫ t
0(t − u)αuγ+i−1du
=ai
(i − 1)!tγ+i+α Γ(γ + i)Γ(α + 1)
Γ(γ + i + α + 1)
and the conclusion holds. 2
As a consequence of Lemma 3.4, we have the following two propositions which are used toprove Theorem 2.1.
Proposition 3.5 Assume that there exists β > 0 such that E(
eβ|Y1|)
< ∞. For every t > 0,ε > 0 and M ≥ 1 such that |2m
M | ≤ β, the random variable (t − TNt)−1+ε exp 2m
M (x − XTNt)
P-integrable.
Proof Note that
(t − TNt)−1+ε exp
2m
M(x − XTNt
) = t−1+ε exp2mx
M1t<T1 +
∞∑
i=1
1Ti≤t<Ti+1(t − Ti)−1+ε exp
2m
M(x − XTi
),
≤ t−1+ε exp2mx
M+
∞∑
i=1
1t>Ti(t − Ti)
−1+ε exp
[
2m
M(x − XTi
)
]
.
Let σ(Ti) be the σ-field generated by Ti, i > 0.Conditioning by Ti, we obtain that
E
[
exp
[
2m
M(x − XTi
)
]
|σ(Ti)
]
= e2mM
xe−2m2
MTi+
2m2
M2 TiE
(
e−2mM
Y1
)i.
Since M ≥ 1, then e−2m2
MTi+
2m2
M2 Ti ≤ 1, therefore
E
[
exp
[
2m
M(x − XTi
)
]
|σ(Ti)
]
≤ e2mM
xE
(
e−2mM
Y1
)i.
13
Consequently
E
[
(t − TNt)−1+ε exp
[
2m
M(x − XTNt
)
]]
≤ e2mM
x∞
∑
i=1
E[
1t>Ti(t − Ti)
−1+ε]
E
(
e−2mM
Y1
)i.
By the choice of M , E(e−2m
Mσ2 Y1) < E(eβ|Y1|) < ∞. We use Lemma 3.4, and the conclusion holds.2
Another useful result for the proof of Theorem 2.1 is the following :
Proposition 3.6 Assume that there exists β > 0 such that E(
eβ|Y1|)
< ∞. For every t > 0 and0 < ε < 1
4 the random variable (t − TNt)−1+ε|XTNt
− x|−4ε is P-integrable.
Proof Note that
(t − TNt)−1+ε|XTNt
− x|−4ε = t−1+ε|x|−4ε1t<T1 +
∞∑
i=1
1Ti≤t<Ti+1(t − Ti)−1+ε|XTi
− x|−4ε,
≤ |x|−4εt−1+ε +∞
∑
i=1
1Ti≤t(t − Ti)−1+ε|XTi
− x|−4ε.
We apply Lemma 3.3 to α = 4ε, G =WTi√
Ti, µ = mTi +
∑ij=1 Yj − x and σ =
√Ti. There
exists k1,ε > 0, k2,ε > 0 such that
E[
|XTi− x|−4ε|σ(Ti, Yj , j ≤ i)
]
≤ k1,ε√Ti
|mTi +
i∑
j=1
Yj − x|1−4ε + k2,εT−4εi .
Let us use the inequality x1−α1 ≤ 1 + x1, 0 < α < 1, x1 ≥ 0 for x1 = |mTi +
∑ij=1 Yj − x| and
α = 4ε :
E[
|XTi− x|−4ε|σ(Ti, Yj , j ≤ i)
]
≤ k1,εT− 1
2i + k1,εT
− 12
i |mTi +i
∑
j=1
Yj − x| + k2,εT−4εi .
Since |mTi +∑i
j=1 Yj − x| ≤ |m|Ti +∑i
j=1 |Yj | + |x|, then
E
[
1Ti<t(t − Ti)
−1+ε
|XTi− x|−4ε
]
≤(k1,ε + |x|)E[
1Ti<t(t − Ti)−1+εT
− 12
i
]
+ k1,ε|m|E[
1Ti<t(t − Ti)−1+εT
12
i
]
+k1,εiE(|Y1|)E[
1Ti<t(t − Ti)−1+εT
−12
i
]
+ k2,εE[
1Ti<t(t − Ti)−1+εT−4ε
i
]
.
We conclude the proof using Lemma 3.4. 2
14
References
[1] L. Alili, P. Patie, J.L. Pedersen, Representations of the First Hitting Time Density of anOrnstein-Uhlenbeck Process, Stochastic Models 21, 2005, pp. 967-980.
[2] V. Bernyk, R. C. Dalang, G. Peskir, The law of the supremum of stable Lévy processes withno negative jumps, Ann. Probab. Vol 36, Number 5, 2008, pp. 177-1789.
[3] C. Blanchet, Processus à sauts et risque de défaut, Ph.D. Thesis, University of d’Evry-Vald’Essonne, 2001.
[4] A. Borodin, P. Salminen, Handbook of Brownian Motion. Facts and Formulae, Birkhäuser,1996.
[5] A. A. Borokov, On the first passage time for one class of processes with independent incre-ments, Theor. Prob. Appl. 10, 1964, pp. 331-334.
[6] R. Cont, P. Tankov, Financial Modelling with Jump Processes, Chapman & Hall/CRCFinancial Mathematics Series, 2004.
[7] R.A. Doney, Hitting probabilities for spectrally positive Lévy processes, J. London, Math,Soc. (2) 44, 1991, pp. 556-576.
[8] R.A. Doney, A.E. Kyprianou, Overshoots and undershoots of Lévy processes, Ann. Appl.Probab. 16, 2005, pp. 91-106.
[9] M. Dozzi, P. Vallois, Level crossing times for certain processes without positive jumps,Bulletin des Sciences Mathématiques, 121, 1997, pp. 355-376.
[10] I. Karatzas, S.E. Shreve, Brownian Motion and Stochastic Calculus, Second Edition,Springer-Verlag, New-York, 1991.
[11] S.G. Kou, H. Wang, First passage times of a jump diffusion process, Adv. Appl. Prob. 35,2003, pp. 504-531.
[12] A. E. Kyprianou, Introductory Lectures on Fluctuations of Lévy Processes with Applica-tions, Springer-Verlag Berlin Heidelberg, 2006.
[13] B. Leblanc, Modélisation de la volatilité d’un actif financier et applications, Ph.D. Thesis,University of Paris VII, 1997.
[14] C. Lefèvre, S. Loisel, On finite-time ruin probabilities for classical risk models, ScandinavianActuarial Journal, Vol. 1, 2008, pp. 41-60.
[15] C. Lefèvre, S. Loisel, Finite-Time Horizon Ruin Probabilities for Independent or DependentClaim Amounts, Working paper WP2044, Cahiers de recherche de l’Isfa, 2008.
[16] P. Lévy, Processus stochastiques et mouvement brownien, Gauthier-Villars, 1948.
[17] G. Peskir, The law of the hitting times to points by a stable Lévy process with no-negativejumps, Research Report No. 15, Probability and Statistics Group School of Mathematics,The University of Manchester, 2007.
15
[18] P. Picard, C. Lefèvre, The probability of ruin in finite time with discrete claim size distri-bution, Scand. Actuar. J. (1), 1997, pp. 58-69.
[19] P. Picard, C. Lefèvre, The moments of ruin time in the classical risk model with discreteclaim size distribution, Insurance Math. Econom. 23 (2), 1998, pp. 157-172.
[20] B. Roynette, P. Vallois, A. Volpi, Asymptotic behavior of the hitting time, overshoot andundershoot for some Lévy processes, ESAIM PS, Vol. 12, 2008, pp. 58-93.
[21] K.I. Sato, Lévy Processes and Infinitely Divisible Distributions, Cambridge University Press: Cambridge, UK, 1999.
[22] A. Volpi, Etude asymptotique de temps de ruine et de l’overshoot, Ph.D. Thesis, Universityof Nancy 1, 2003.
[23] V. M. Zolotarev, The first passage time of a level and the behavior at infinity for a class ofprocesses with independent increments, Theor. Prob. Appl., 9, 1964, pp. 653-664.
16