First passage time law for some jump-diffusion processes : existence of a density

0.23 MB·English

by  Laure Coutin

First hitting time law for some jump-di(cid:27)usion
∗
pro
esses : existen
e of a density
9
0
0
† ‡
2 Laure Coutin and Diana Dorobantu

r
p
A

0
1
(X , t 0)
t
Abstra
t. Let ≥ be a di(cid:27)usion pro
ess with jumps, sum of a Brownian motion
] τ
x
R with drift and a
ompound Poisson pro
ess. We
onsider the (cid:28)rst hitting time ofEa (cid:28)xed level
x > 0 (X , t 0) τ (X ) < 0)
t x 1
P by ≥ . We prove that the law of has a density (defe
tive when
.
h with respe
t to the Lebesgue measure.
t
a
m
Keywords : Lévy pro
ess, jump-di(cid:27)usion pro
ess, hitting time law.
[

1
v
1 Introdu
tion
9
6
6
The main purpose of this paper isto show that the (cid:28)rst hitting time distribution asso
iated with
1
. a jump-di(cid:27)usion pro
ess (sum of a Brownian motion with drift and an independent
ompound
4
0 Poisson pro
ess) has a density with respe
t to the Lebesgue measure.
9 (Xt, t 0) 0 τx x > 0
Let ≥ be a
àd-làg pro
ess started at and the (cid:28)rst hitting time of level
0 X
: by .
v
τ X
i x
X Lévy, in [16℄,
omputes the law of when is a Brownian motion with drift. This result
X
r is extended by Alili, Patie and PeterXsen [1℄ or Leblan
[13℄ to the
ase where is an Ornstein-
a
Uhlenbe
k pro
ess. The
ase where is a Bessel pro
ess was studied by Borodin and Salminen
in [4℄.
X
Some results are also available when the pro
ess has jumps. The (cid:28)rst results are obtained
X
by Zolotarev [23℄ and Borokov [5℄ when is a spe
trally negative Lévy pro
ess. They give the
τ X
x t
law of . Moreover, if has the probability density with respe
t to the Lebesgue measure
p(t,x), τ f(t,x)
x
then the law of has the density with repe
t to the Lebesgue measure , where
xf(t,x)= tp(t,x). X X = x
In this
ase where has only negative jumps, τx almost surely.
X
If is a spe
trally positive Lévy pro
ess, Doney [7℄ gives an expli
it formula for the joint
τ X x. X
Lapla
e transform of x and the overshoot τx − When is a stable Lévy pro
ess, Peskir
[17℄, and Bernyk, Dalang and Peskir [2℄ obtain an expli
it formula for the hitting time density.
∗
The authors thank M. Pontier and Ph. Carmona for their
areful reading and G. Leta
for his helpfull

om†ments
‡IMT, Universityof Toulouse, Fran
e, laure.
outinmath.univ-toulouse.fr
ISFA, Universityof Lyon, Fran
e, diana.dorobantuadm.univ-lyon1.fr
1
X
The
ase where has signed jumps is more re
ently studied. In [9℄, the authors give the law
τ X
x
of when is the sum of a de
reasing Lévy pro
ess and an independent
ompound pro
ess
with exponential jump sizes. This result is extended by Kou et Wang in [11℄ to the
ase of
a di(cid:27)usion pro
ess with jumps where the jump sizes follow a double exponential law. They
τ τ
x x

ompute the Lapla
e transform of and derive an expression for the density of . For a more
general jump-di(cid:27)usion pro
ess, Roynette, Vallois and Volpi [20℄ show that the Lapla
e transform
(τ ,x X ,X x)
of x − τx− τx − is solution of some kind of random integral.
Doney and Ky(pG¯rian,oτu [8℄G¯studi,eXd the pxr,oxblemXfor,gxenerXa¯l Lé)vy pro
eXs¯ses=. sTuhpey gXive the
qG¯uin=tuspulpe sla<wto,f X¯ τx=−Xx −. τx− τx − − τx− − τx− where t s≤t s and
t s s
{ }
Results are also available for some Lévy pro
esses without Gaussian
omponent, see Lefèvre,
Loisel and Pi
ard [14d,S15,=1S8, 1(9µ℄d.t+Blσa1n
het [d3W℄
o+nsφid1ers a dpNr˜o
)e,sst satTisfying thTe following
sto
hasti
equation : t t− φ˜(t)=0 t φ˜(t)=φ t ≤ where is a (cid:28)nite
µ R σ > 0 φ˜(.) 0 φ W N
horizon, ∈ , , is a fun
1t1ion taking twN˜o values or , is a Brownian motion,
is a Poisson pro
ess with intensity φ2 φ˜(t)=φ and is the
ompensated Poisson pro
ess.
The aim of our paper is to add to these studies the law of a (cid:28)rst hitting time by a Lévy
pro
ess whi
h is the sum of a Brownian motion with drift and a
ompound Poisson pro
ess. We
do not limit our study to a parti
ular distribution of the jumps sizes.
This paper is organized as follows : Se
tion 2
ontains the main result (Theorem 2.1) whi
h
gives the (cid:28)rst hitting time law by a jump Lévy pro
ess. The following two se
tions (Se
tion
2.1 and Se
tion 2.2) are dedi
ated to the proof of Theorem 2.1. In these se
tions we
ompute
t = 0 t > 0
the derivative at (Se
tion 2.1) and at (Se
tion 2.2) of the hitting time distribution
fun
tion. Se
tion 3
ontains the proofs of some useful results.
2 Hitting time law
R
m (W , t 0) (N , t 0)
t t
Let ∈ , ≥ be a standard BrowniaNn motion, ≥ be a Poisson pro
ess
a (Yi, i ∗)
with
onstant positive intensity and ∈ be a sequen
e of independent identi
ally
F
Y
distributed randomNvariables with distribution fun
tion . We suppose that the folNlowing
σ σ(Yi, i ∗) σ(Nt, t 0) σ(Wt, t 0) (Tn, n ∗)
−(cid:28)elds ∈ , ≥ and ≥ are independent. Let ∈ be
N
the sequen
e of the jump times of the pro
ess .
X
Let be the pro
ess de(cid:28)ned by
Nt
X = mt+W + Y , t 0.
t t i
≥ (1)
i=1
X
τ x > 0 (X , t 0)
x t
Let be the (cid:28)rst hitting time of level by the sto
hasti
pro
ess ≥ :
τ =inf u> 0 : X x
x u
{ ≥ }.
τ
x
The main result of this paper is the following theorem. It gives the law of .
2
β > 0 E eβY1 < τ
| | x
Theorem 2.1 If there exists su
h that ∞, then the distribution fun
tion of
0 ]0, [
has a right derivative at and is di(cid:27)erentiable a(cid:0)t ever(cid:1)y point of ∞ . The derivative, denoted
f(.,x)
, is equal to
a a
f(0,x) = 2 F (x) F (x ) + F (x) F (x )
Y Y − Y Y −
2 − − 4 −
t > 0 f(t,x)= aE(cid:0)(1 (1 F )(x X(cid:1) ))+(cid:0)E 1 f˜ t (cid:1)T ,x X
and for every τx>t − Y − t τx>TNt − Nt − TNt
(cid:16) (cid:16) (cid:17)(cid:17)
where z (z mu)2
f˜(u,z) = | | exp − 1 (u), u R, z R.
√2πu3 − 2u ]0,∞[ ∈ ∈
(cid:20) (cid:21)
P E
(τ = ) = 0 m+a (Y ) 0.
x 1
Furthermore, ∞ if and only if ≥
The proof of Theorem 2.1 is given in Se
tions 2.1 and 2.2.
X˜ m
Let be the drifted Brownian motion de(cid:28)ned by
X˜ = mt+W , t 0,
t t
for all ≥
τ˜ z > 0 X˜ f˜(.,z)
z
and be the (cid:28)rst hitting time of by . Then, a

ording to [10℄, is the derivative
τ˜
z
of the distribution fun
tion of (see Se
tion 3 for more details).
( ) (W , t 0)
t t 0 t
Let F ≥ be the
ompleted natural (cid:28)ltrNation generated by the pro
esses ≥ ,
(Nt, t 0) (Yi, i ∗)
≥ and the random variables ∈ .
Remark 2.2 This result is already known in the following
ases :
X τ
x
When has no positive jumps, the law of is given in Theorem 46.4 page 348 of [21℄.
X τ
x
When is a stable Lévy pro
ess, with no negative jumps, the law of is given in [2℄.
X
When is a jump-di(cid:27)usion where the jump sizes follow a double exponential law, the result is
given in [11℄.
t = 0
2.1 Existen
e of the right derivative at
In this se
tion we provePthe (cid:28)rst part of Theorem 2.1. Our goal is to show that the distribution
τ t (τ t) 0
x x
fun
tion of (i.e. 7→ ≤ ) has a right derivative at and we
ompute this derivative.
P
(τ h) N
x h
For this purpose we split the probability ≤ a

ording to the values of :
P P P P
(τ h) = (τ h, N = 0)+ (τ h, N = 1)+ (τ h, N 2).
x x h x h x h
≤ ≤ ≤ ≤ ≥
P(τ h, N 2) P(N 2) = 1 e ah ahe ah
x h h − −
Note that ≤ ≥ ≤ ≥ − − , thus
P
(τ h, N 2)
x h
lim ≤ ≥ = 0.
h 0
→ h
The proof of the (cid:28)rst part of Theorem 2.1 will be
omplete when the following lemma and
proposition are shown :
P
(τx≤h, Nh=0) 0 h 0
Lemma 2.3 The term h
onverges to when goes to .
3
β > 0 E eβY1 < x > 0
| |
PProposition 2.4 If there exists su
h that ∞, then for every the term
(τx≤hh, Nh=1)
onverges to a2 (2−FY(x)−FY(x−))(cid:0)+ a4 (F(cid:1)Y(x)−FY(x−)) when h goes to 0.
Proof of Lemma 2.3.
ω : N (ω) = 0 (X , 0 t h)
h t
T(X˜he,es0senttial ohb)servation is that Pon the set { τ h = τ˜}, thhe pro
essτ˜es ≤ ≤ aNnd
t x x x
≤ ≤ are equal , and -almost surely ∧ ∧ . Sin
e is independent of ,
then P(τ h, N = 0) e ahP(τ˜ h)
x h − x
≤ = ≤ .
h h
τ˜ C
x ∞
The law of has a densityP(possibly defe
tive) with respe
t to the Lebesgue measure, null
] ,0] (τx≤h, Nh=0) 0 h 0 2
on −∞ . Thus the limit of h exists and is equal to when goes to .
To prove Proposition 2.4, we use the same type of arguments as in [20℄ (for the proof of
Theorem 2.4). In [20℄, the authors
ompute the joint Lapla
e transform of (a passage time of a
Lévy pro
ess, overshoot) as solution of an integral equation.
Proof of Proposition 2.4P.
(τ h,N = 1)
x h
We split the probability ≤ into three parts a

ording to the relative position of
τ T , N :
x 1
and the (cid:28)rst jump time of the Poisson pro
ess
P P P P
(τ h, N = 1) = (τ < T , N = 1)+ (τ = T , N = 1)+ (T < τ h, N =1).
x h x 1 h x 1 h 1 x h
≤ ≤
(2)
lim h 1P(τ h, N = 1)
h 0+ − x h
Step 1 : We prove that the
ontribution to the limit → ≤ of the (cid:28)rst
term on the right hand side of (2) is null.
(X , 0 t < T ) (X˜ , 0 t < T )
t 1 t 1
Sin
e the pro
esses ≤ and ≤ are equal, then on the set
ω, τ (ω) < T (ω) , τ τ˜
x 1 x x
{ } the stopping time and are equal. Thus
P P P
(τ < T , N = 1) = (τ˜ < T h, N = 1) (τ˜ h).
x 1 h x 1 h x
≤ ≤ ≤
τ˜ C
x ∞
The law of has aP density (possibly defe
tive) with respe
t to the Lebesgue measure,
] ,0]. (τx<T1, Nh=1) 0 h 0
null on −∞ Thus h
onverges to whleimn goehs t1oP(.τ h, N = 1)
h 0+ − x h
Step 2 : We prove that the
ontribution toa(t2he lFim(itx) F→(x )). ≤ of the
se
ond term on the right hand side of (2) is 2 − Y − Y −
Note that
P(τ = T , N = 1) = P(τ˜ > T , X˜ +Y x, T h< T ).
x 1 h x 1 T1 1 ≥ 1 ≤ 2
N
n ∗ Tn = S1 +...+Sn (Si, i 1)
Here, for every ∈ , where ≥ is a sequen
e of independent
a
identi
ally distributed rand(oSm,viaria1b)les wi(tYh exX˜ponτ˜en)tial distribution with parameter . Using
i 1 x
the independen
e between ≥ and , , we get :
h
P(τ = T , N = 1) = ae as1 ∞ ae as2E 1 1 ds ds
x 1 h Z0 − Zh−s1 − (cid:16) τ˜x>s1 Y1≥x−X˜s1(cid:17) 2 1
h
=ae ah E 1 1 ds.
− τ˜x>s Y1 x X˜s
Z0 (cid:16) ≥ − (cid:17)
4
Y
1
Integrating with respe
t to , we obtain :
h h
P(τ = T , N = 1) = ae ah E (1 F )((x X˜ ) ) ds ae ah E 1 (1 F )((x X˜ ) ) ds.
x 1 h − Z0 (cid:16) − Y − s − (cid:17) − − Z0 (cid:16) τ˜x≤s − Y − s − (cid:17)
F X˜
Y
On the one hand, sin
e is a
àdlàg bounded fun
tion and is a Brownian motion with drift,
we get
F (x)+F (x )
lim E F ((x X˜ ) ) = Y Y − .
s 0 Y s −
→ − 2
(cid:16) (cid:17)
τ˜
x
On the other hand, sin
e the distribution fun
tion of is di(cid:27)erentiable, then
lim E 1 (1 F )((x X˜ ) ) = 0.
s→0 τ˜x≤s − Y − s −
(cid:16) (cid:17)
We dedu
e that
P
(τ = T , N = 1) a
x 1 h
lim = 2 F (x) F (x ) .
h 0 Y Y −
→ h 2 − −
(cid:0) (cid:1)
lim h 1P(τ h, N = 1)
h 0+ − x h
Step 3 : We prove that the
ontributionat(oFth(ex)limiFt (x →)). ≤ of the
third term on the right hand side of (2) is 4 Y − Y −
To this end, we state the following lemma :
P
Lemma 2.5 The term (T1<τx≤hh, Nh=1)
onverges to a4(FY(x)−FY(x−)) when h goes to 0.
Proof Note that
P P
(T <τ h, N = 1) = (T < τ h, T h< T )
1 x h 1 x 1 2
≤ ≤ ≤
T = T +S θ θ S
and 2 1 2 ◦ T1 where is the translation operator. Here 2 is a random vaNriable with
a (T1,(Wt, t 0),(Yi, i ∗)).
exponential distribution with parameter , independent of ≥ ∈
ω : T (ω) < τ (ω) h < T (ω) X = X +X˜ θ T < s h
Moreover, on { 1 x ≤ 2 }, s T1 s−T1 ◦ T1 when 1 ≤
τ = T +τ˜ θ
and x 1 x−XT1 ◦ T1.
( , t 0) T
t 1
Strong Markov Property at the F ≥ -stopping time gives :
P(T < τ h, N = 1) = E 1 1 ET1 1 1
1 x ≤ h τx>T1 h≥T1 τ˜x−XT1≤h−T1 h−T1<S2
(cid:16) (cid:16) (cid:17)(cid:17)
ET1(.) E(. )
where is | FT1 .
S
2
Integrating with respe
t to , we obtain :
P(T < τ h, N = 1) = E 1 1 e a(h T1)ET1 1 .
1 x ≤ h τx>T1 h≥T1 − − τ˜x−XT1≤h−T1
(cid:16) (cid:16) (cid:17)(cid:17)
ω : τ (ω)> T (ω) = ω :τ˜ (ω) > T (ω) ω :X < x .
Remark that { x 1 } { x 1 }∩{ T1(ω) } Consequently
P(T < τ h, N = 1) = E 1 1 e a(h T1)ET1 1
1 x ≤ h − τ˜x≤T1≤h XT1<x − − τ˜x−XT1≤h−T1
+E(cid:16)1 1 e a(h T1)ET1 1(cid:16) .(cid:17)(cid:17)
h≥T1 XT1<x − − τ˜x−XT1≤h−T1
(cid:16) (cid:16) (cid:17)(cid:17)
5
τ˜
x
Sin
e the distribution fun
tion of has a null derivative at 0, then
1
lim E 1 1 e a(h T1)ET1 1 = 0.
h→0h τ˜x≤T1≤h XT1<x − − τ˜x−XT1≤h−T1
(cid:16) (cid:16) (cid:17)(cid:17)
It remains to show that
G(h) a
lim = [F(x) F(x )]
−
h 0+ h 4 − (3)
→
where G(h) = E 1 1 e a(h T1)ET1 1 .
h≥T1 XT1<x − − τ˜x−XT1≤h−T1
(cid:16) (cid:16) (cid:17)(cid:17)
T f˜(.,z)
1
Integrating with respe
t to and then using the fa
t that is the derivative of the distri-
τ˜
z
bution fun
tion of , we get :
h
G(h) = ae asE 1 e a(h s)Es(1 ) ds
Z0 − h X˜s+Y1<x − − τ˜x−X˜s−Y1≤h−s i
h h s
=ae ah − E 1 f˜(u,x X˜ Y ) duds.
− X˜s+Y1<x − s− 1
Z0 Z0 h i
X˜ = ms+W µ = x ms Y σ = √s,
s s 1
Sin
e , we may apply Lemma 3.1 to − − and then
ae−ah h h−sE (x−m(u+s)−Y1)2 x Y1 G√s
G(h) = e− 2(u+s) − + duds.
√2π Z0 Z0 " (cid:18)(u+s)3/2 √u(u+s)(cid:19)+#
r = u+s
We make the following
hange of variable .
ae−ah h hE (x−mr−Y1)2 x Y1 G√s
G(h) = e− 2r − + drds.
√2π r3/2 r√r s
Z0 Zs (cid:20) (cid:18) − (cid:19)+(cid:21)
v = s
Firstly,weapplyFubini'sTheoremandse
ondly, wemakethefollowing
hangeofvariable r.
ae−ah h rE (x−mr−Y1)2 x Y1 G√s
G(h) = e− 2r − + dsdr
√2π r3/2 r√r s
Z0 Z0 (cid:20) (cid:18) − (cid:19)+(cid:21)
ae ah h
−
= g(r)dr,
√2π
Z0
where
1E (x−mr−Y1)2 x Y1 G√v
g(r) = e− 2r − + dv.
√r √1 v
Z0 (cid:20) (cid:18) − (cid:19)+(cid:21)
But,
rlim0+e−(x−m2rr−Y1)2 x√−rY1 + √G1√vv = √1√v vG+1x=Y1,
→ (cid:18) − (cid:19)+ −
and
(x−mr−Y1)2 x Y1 G√v z2 √v
sup e− 2r − + supze− 2 + m + G.
√r √1 v ≤ | | √1 v| |
0≤r≤1 (cid:18) − (cid:19)+ z≥0 −
6
Then from Lebesgue's Dominated Convergen
e Theorem we obtain :
1 √v √2π
P E P
lim g(r) = (Y = x) (G ) dv = (Y = x).
r 0 1 + 1
→ Z0 √1−v 4
We dedu
e the identity (3), i.e.
G(h) a
P
lim = (Y = x),
h 0 1
→ h 4
2
whi
h a
hieves the proof of Lemma 2.5
2
Propostion 2.4 is a
onsequen
e of the Steps 1 to 3.
t > 0
2.2 Existen
e of the derivative at
R
τx ∗+
Our task is now to show that the distribution fun
tion of isPdi(cid:27)erentiable on and to
(t < τ t+h)
x

ompute its derivative. For this purpose we split the probability ≤ a

ording to
N N
t+h t
the values of − :
P P P
(t < τ t+h) = (t <τ t+h, N N = 0)+ (t < τ t+h, N N = 1)
x x t+h t x t+h t
≤ ≤ − ≤ −
P
+ (t <τ t+h, N N 2).
x t+h t
≤ − ≥ (4)
The third term on the right hand side of (4) is upper bounded by
P(t < τ t+h, N N 2) P(N N 2) = 1 e ah ahe ah.
x t+h t t+h t − −
≤ − ≥ ≤ − ≥ − −
P
Therefore limh→0 (t<τx≤t+h,hNt+h−Nt≥2) =0.
t
Let us study the se
ond term on the right hand side of (4). Markov Property at gives :
P(t < τ t+h, N N = 1) = E(1 Pt(τ h, N =1)),
x ≤ t+h− t τx>t x−Xt ≤ h
Pt(.) =P(. )
t
where |F .
Pt
(τx−Xt≤h, Nh=1)
In virtue of Lemma 2.4, h
onverges to
a a
2 FY (x Xt) FY (x Xt)− + FY (x Xt) FY (x Xt)−
2 − − − − 4 − − −
(cid:2) P (cid:0) (cid:1)(cid:3) (cid:2) (cid:0) (cid:1)(cid:3)
(Nh=1) = ae ah a
and is upper bounded by h − ≤ . Dominated Convergen
e Theorem gives :
P
(t < τ t+h, N N = 1) 3a
lim x ≤ t+h − t = aE(1 (1 F )(x X ))+ E(1 ∆F (x X ))
h→0 h τx>t − Y − t 4 τx>t Y − t
∆F (z) = F (z) F (z )
Y Y Y −
where − .
F Y X
Y
Howeverthejumpssetof (thedistribEut(i1onfun∆
Ftion(xof X)i)s)
=ou0ntableand hasadensity
(0
f. EPr(o1posit∆ioFn 3(.x12 pXage))90 oEf([16℄). Thus)= Eτ(x1>t Y )−= 0t : indeed
≤ τx>t Y − t ≤ Y1=x−Xt Xt=x−Y1 . Therefore
P
(t < τ t+h, N N = 1)
lim x ≤ t+h − t = aE(1 (1 F )(x X )).
h→0 h τx>t − Y − t
The proof of the se
ond part of Theorem 2.1 will be
omplete when the following proposition
is shown :
7
β > 0 E eβY1 <
| |
Proposition 2.6 If there exists su
h that ∞, then
P (cid:0) (cid:1)
(t < τ t+h, N N = 0)
lim x ≤ t+h− t = E 1 f˜ t T ,x X ,
h→0 h τx>TNt − Nt − TNt
(cid:16) (cid:16) (cid:17)(cid:17)
f˜
where is the fun
tion introdu
ed in Theorem 2.1.
T
Proof Sin
e Nt is not a stopping time, we
an not apply Strong Markov Property. We split
P ∞ P
(t < τ t+h, N N = 0) = (t < τ t+h, N = N = k)
t+h t x t+h t
≤ − ≤
k=0
X
P ∞ P
= (t < τ˜ t+h < T )+ (t < τ t+h, T < t < t+h < T ).
x 1 x k k+1
≤ ≤
k=1
X
(ω,t) : T (ω) < t X (ω) = X (ω)+X θ (ω)
On the set { k }, we have t Tk t−Tk ◦ Tk , hen
e on the set
ω, τ (ω) > T (ω) , τ = T +τ θ . T
{ x k } x k x−XTk ◦ Tk Strong Markov Property at the stopping time k
gives
P(t < τ t+h, N N = 0) =e a(t+h)P(t < τ˜ t+h)
t+h t − x
≤ − ≤
+ ∞ E 1 1 ETk 1 1 .
Tk<t τx>Tk t−Tk<τ˜x−XTk≤t+h−Tk t+h−Tk<Sk+1
Xk=1 (cid:16) (cid:16) (cid:17)(cid:17)
(ω,t) :τ (ω) t < S (ω) τ = τ˜ z < 0
z k+1 z z
On the set { ≤ }, we have for every . Therefore
P(t < τ t+h, N N = 0) =e a(t+h)P(t < τ˜ t+h)
x t+h t − x
≤ − ≤
+ ∞ E 1Tk<t1τx>Tke−a(t+h−Tk)ETk 1t−Tk<τ˜x−XTk≤t+h−Tk .
Xk=1 (cid:16) (cid:16) (cid:17)(cid:17)
τ˜ f˜(.,x X )
The FTk-
onditional law of x−XTk has the density (possibly defe
tive) − Tk , thus
P(t < τ t+h, N N = 0) =e a(t+h)P(t < τ˜ t+h)
x t+h t − x
≤ − ≤
+ ∞ E 1Tk<t1τx>Tke−a(t+h−Tk) t+h−Tk f˜(u,x−XTk)du .
Xk=1 (cid:18) Zt−Tk (cid:19)
Let us point out that e−a(t−Tk) = ETk 1Tk+1>t then
(cid:0) t+h(cid:1)
P(t < τ t+h, N N = 0) =e ah E(1 )f˜(u,x)du
x ≤ t+h− t − Zt 0≤t<T1
t+h
+e ah ∞ E 1 1 f˜(u T ,x X ) du,
− Xk=1Zt (cid:16) Tk≤t<Tk+1 τx>Tk − k − Tk (cid:17)
or shortly
8
t+h
P(t < τ t+h, N N = 0) = e ah E 1 f˜(u T ,x X ) du.
x ≤ t+h− t − TNt<τx − Nt − TNt (5)
Zt (cid:16) (cid:17)
f˜ u,
Sin
e is
ontinuous with respe
t to then
lim 1 f˜(u T ,x X )= 1 f˜(t T ,x X ),
u t+ TNt<τx − Nt − TNt TNt<τx − Nt − TNt
→
f˜(u T ,x X ) u
From Lemma 3.2, Propositions 3.5 and 3.6, − N−t − TNt is dominated uniformly in
by a integrable random variable
1 2m
f˜(u T ,x X ) c (t T )1 ε +exp (x X ) .
− N−t − TNt ≤ ε,M − Nt − "(x−XTNt)4ε (cid:18)M − TNt (cid:19)#
0 < ε < 1/4 M > max(1,2m β 1), E(eβY ) < + ,) c ,
− | | ε,M
Here and | | (where ∞ and is a
on-
ε M
stant de(cid:28)ned in Lemma 3.2 depending only on and . Then, using Lebesgue's Dominated
Convergen
e Theorem in equation (5) we obtain
P
(t < τ t+h, N N = 1)
lim x ≤ t+h − t = E 1 f˜(t T ,x X ) .
h→0 h τx>TNt − Nt − TNt
(cid:16) (cid:17)
2
P
Using Proposition 2.6, the limit limh→0 (t<τxh≤t+h) exists and is equal to
aE(1 (1 F )(x X ))+E 1 f˜(t T ,x X ) .
τx>t − Y − t τx>TNt − Nt − TNt
(cid:16) (cid:17)
The following lemma allows to
on
lude the proof of Theorem 2.1. The results are known,
for example in [15℄ and [22℄ but for sake of
ompleteness we give a proof.
E
x > 0, τ m+a (Y )
x 1
Lemma 2.7 Forall thestoppingtime is(cid:28)nitealmostsurely ifandonlyif ≥
0.
Proof Remark that P P
(τ = )= (sup X < x).
x t 0 t
∞ ≥
Thanks toTheorem 7.2page 183of [12℄whi
h isa
onsequen
e of Strong Law of Large Numbers,
E
m+a (Y ) > 0, lim X = +
1 t t
if then →∞ ∞ and
E
m+a (Y ) = 0, limsup X = liminf X = .
1 t t t t
if then →∞ − →∞ ∞
E
m+a (Y ) 0 sup X = +
1 t 0 t
Therefore (see Exer
ise 39.11 page 271 of [21℄), if ≥ , then ≥ ∞. This
proves the (cid:28)rst part of the lemma.
E
m + a (Y ) < 0. lim X = ,
1 t t
Conversely, let us suppose that Then →∞ −∞ and a

ording
sup X < . x > 0,
tPo Theorem 48.1 page 363 of [21℄, t≥0 t ∞ Assume thaPt there exists 0 su
h that
(τ < + ) =1. x, x x (τ < + )= 1.
x0 ∞ Then from all su
h that ≤ 0 we have x ∞
9
n 1 x nx ,
0
P Now we use re
urren
e reasoning. Assume that for ≥ and for all ≤ we have
(τ < + )= 1. x nx < x (n+1)x ,
x 0 0
∞ Let su
h that ≤ then using Strong Markov Property,
P(τx < +∞)= E(1τnx0<+∞1Xτnx0≥x)+E(1τnx0<+∞1Xτnx0<xPτnx0(τx−Xτnx0 < ∞)).
Using the re
urren
e hypothesis, sin
e x−Xτnx0 < x0, almost surely Pτnx0(τx−Xτnx0 < ∞) = 1
and then
P
(τ <+ ) = 1.
x
∞
P
x > 0 (τ < ) = 1, x > 0,
PWe have provePd that if there exists 0 su
h that x0 ∞ then for all
(τ < ) = (sup X x) = 1. sup X < .
m+xaE(∞Y )< 0, t≥0x >t ≥0, P(τ < Th)i<s
1o.ntradi
ts the fa
t that t≥0 t ∞ Then,2if
1 x
for all ∞
3 Appendix
A Brownian motion with drift is a pro
ess
X˜ = mt+W , t 0
t t
≥
R
m
with ∈ .
z > 0 τ˜ τ˜ = inf t 0 : X˜ z .
z z t
Let and be the passage timRe de(cid:28)ned by { ≥ ≥ } By (5.12) page
τ˜
z +
197 of [10℄, has the following law on :
f˜(u,z)du+P(τ˜ = )δ (du)
z
∞ ∞
where
z (z mu)2
f˜(u,z) = √|2πu|3 exp − −2u 1]0,∞[(u), u∈ R, and P(τ˜z = ∞)= 1−emz−|mz|. (6)
(cid:20) (cid:21)
z f˜(.,z) 0 0+
f˜ For a (cid:28)xed , the fun
tion f˜ and all its derRivatives admitf˜(ua,sz)li=mi0t at u. T0he fun
tion
∞
admits an extension (denoted ) of
lass C on , de(cid:28)ned by for ≤ . Moreover
it
he
ks the two following lemmas :
G (0,1) µ R σ R+
Lemma 3R.1 Let be a Gaussian random variable N and let ∈ , ∈ . Then for
u
every ∈
1 (µ−mu)2 µ+σ2m σG
E[f˜(u,µ+σG)1µ+σG>0]= √2πE"e−2(σ2+u) (cid:18)(σ2+u)3/2 + √u(σ2+u)(cid:19)+#
(x) = max(x,0)
+
where .
10