http://dx.doi.org/10.4236/am.2013.48153 Published Online August 2013 (http://www.scirp.org/journal/am)
Two-Sided First Exit Problem for Jump Diffusion
Processes Having Jumps with a Mixture of Erlang
Distribution
*
Yuzhen Wen#, Chuancun Yin
School of Mathematical Sciences, Qufu Normal University, Qufu, China Email: #[email protected], [email protected]
Received May 29, 2013; revised June 29, 2013; accepted July 7, 2013
Copyright © 2013 Yuzhen Wen, Chuancun Yin. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
ABSTRACT
In this paper, we consider the two-sided first exit problem for jump diffusion processes having jumps with rational Laplace transforms. We investigate the probabilistic property of conditional memorylessness, and drive the joint distri- bution of the first exit time from an interval and the overshoot over the boundary at the exit time.
Keywords: First Exit Time; Two-Sided Jumps; Jump Diffusion Process; Overshoot
1. Introduction
Consider the following jump diffusion process
1
, 0
Nt
t t k
k
X u ct W Y t
, (1.1) where the constant is the starting point of u
Xt 0t ,
and
c represent the drift and the volatility of the diffusion part, respectively,
Wt t0 is a standard0
Brownian motion with W0 ,
t 0 is a Poissont
N
process with rate , and the jumps si 1 2,
are assumed to be i.i.d. real valued random variables with common density . Moreover, it is assumed that the random processes
t t0 , t t 0 and randomvariables 1 2 are mutually independent. In this
paper we are interested in the density of following type
zes
Y Y,
p x
W
N
Y Y, ,p
1
0 1 1
1 ˆ
ˆ
ˆ 0 1 1
e 1 ! ˆ
ˆ e ,
1 !
j
j
j
j
i i
m J
x j
ij x
j i
i i m J
x j
ij x
j i
x
p x p I
i
x
p I
i
(1.2)
where J J m m, ,ˆ j,ˆj , pij,pˆij ,
Re
j 0 ,j and that
ˆRe 0 ij, ˆiˆj for all i j.
Moreover,
ˆ ˆ
1 1 1 1
ˆ 1.
mj mj
J J
ij ij
j i j i
p p
Define to be the first exit time of Xt to two flat
barriers a and b
a<b
, i.e.
inf t 0 :Xt borXt a .
Recently, one-sided and two-sided first exit problems for processes with two-sided jumps have attracted a lot of attentions in applied probability (see [1-7]). For example, Perry and Stadje [1] studied two-sided first exit time for processes with two-sided exponential jumps; Kou and Wang [2] studied the one-sided first passage times for a jump diffusion process with exponential positive and negative jumps. Cai [3] investigated the first passage time of a hyper-exponential jump diffusion process. Cai et al. [4] discussed the first passage time to two barriers of a hyper-exponential jump diffusion process. Closed form expressions are obtained in Kadankova and Veraverbeke [5] for the integral transforms of the joint distribution of the first exit time from an interval and the value of the overshoot through boundaries at the exit time for the Poisson process with an exponential compo- nent. For some related works, see Perry et al. [8], Cai and Kou [9], Lewis and Mordecki [10] and the references therein.
*This work was supported by the National Natural Science Foundation
of China (No. 11171179) and Natural Science Foundation of Shandong Province (No. ZR2010AQ015).
#Corresponding author.
and the overshoot over the boundary at the exit time. In Section 2, we study the roots of the generalized Lund- berg equation and conditional memory lessness. The main results of this paper are given in Section 3.
2. Preliminary Results
It is easy to see that the infinitesimal generator of is given by
Xt t0
2
1 2
d
L x x c x
u x y u x p y y
for any twice continuously differentiable function
x and the Lévy exponent of
Xt t0 is given by
2 2 ˆ ˆ
1 1 1 1
1 1
ln e 2
ˆ ˆ
1 . ˆ
zXt
i i
mj mj
J J
ij j ij j
i i
j i j i
j j
g z E z zc
t
p p
z z
By analytic continuation, the function g z
can be extended to the complex plane except at finitely many poles. In the following, we consider the resulting extension G z
of g z
, i.e.,
ˆ ˆ 2 2
1 1 1 1
ˆ ˆ 1
1 , .
2 ˆ
i i
mj mj
J J
ij j ij j
i i
j i j i
j j
p p
G z z zc z C
z z
Let us denote J1 j j
M
m and Mˆ
Jjˆ1mˆj.In [11], Kuznetsov has studied the roots of the equation G z
. However, for this particular Lévy process X , we will give another simple proof for the roots of this equation.Lemma 2.1. For fix 0, the generalized Cramér- Lundberg equation
G z has M Mˆ 2
1 , 2 ,
complex roots
1
, M
1, 2, , 1 i M
with for
and
Re i 0
ˆ1 2 1
ˆ
,ˆ , ,ˆM
, ,
ˆ 1
with
for i M .
ˆi
0Re 1, 2
Proof. Let
2 21
1 , ,
2
G z z zc z
ˆ ˆ 2
1 1 1 1
ˆ ˆ
, .
ˆ
i i
mj mj
J J
ij j ij j
i i
j i j i
j j
p p
G z z
z z
Firstly, we prove that for given 0, G z
hasˆ 1
M roots with negative real parts. Set
: ,
r
C z z r zC with r max1 j Jˆ
ˆj ,where is an arbitrary positive constant. Applying Rouchés theorem on the semi-circle r , consisting of
the imaginary axis running from to and with radius
C ir
ir
r running clockwise from ir to . We let and denote by the limiting semi-circle. It is
ir
r C
known that both and
are analytic in . We want to show that
ˆ ˆ
1
1 ˆ
j
m J
j
j z G z
2
z G z
ˆ ˆ1 ˆ
j
m J
j
j
C
ˆ ˆ
ˆ ˆ
1 2
1 1
ˆ j ˆ j , .
J m J m
j j
j j
z G z z G z z C
Notice that G z1
for Re
z , and
ˆ ˆ2 1 1 1 1
ˆ ˆ
p p
j j
i i
ij j ij j
J m J m
i i
j i j i
G z
isbounded for Re
z . Hence, for Re
z ,
ˆ ˆ
ˆ ˆ
1 2
1 1
ˆ j ˆ j
J m J m
j j
j j
z G z z G z
on the boundary of the half circle in C. For aR, we have G2
ia (see Lewis and Mordecki [10]). Onthe other hand,
2 21 1
1
i Re i
2
G a G a a .
Thus we have G1
ia G2
ia . Since
ˆ ˆ
1
1 ˆ
j
m J
j
j z G z
has Mˆ 1 roots with negative real parts, so equation G z
has Mˆ 1 roots with negative real parts. Similarly, we can prove
,G z 0 has M1 roots with positive real parts.
In the rest of this paper, we assume all the roots of equation G(z)= are distinct and denote
i ,M 1
i i 1, 2,
ˆ
,ˆ ,M 1
i ˆi i1, 2, u
E
for notational simplicity, and denote (or in the sequel) representing the expectation (or probability) when
u
P
t
X starts from . We denote a sequence of events
u
0 :
K X b , G0
:X a
jil
K = {: X crosses b at time by the th phase of th positive jump whose parameter is
l
j i l = {
G : X crosses a at time by the
th phase of th negative jump whose parameter is l
ˆ i
j
for , ,
}
1, 2, ,
j J j 1, 2, , Jˆ i1, , mj, ˆ
1, , j
i m, l1, , i and l1, , i
0
x
. Theorem 2.2. For any , we have
0
e !
j
h i l
j
, x u
jil h
x
P X b x K
h
(2.1)
ˆ0 ˆ
e
! .
j
h i l
j x
u
j i l h
x
P X a x G
h
(2.2)Furthermore, conditional on
: Kjil
, the stopping time
: Gj i l
is independent ofthe overshoot Xb (the undershoot X a). More precisely, for any x0, we have
,
, ,
u
jil
u u
jil jil
P t X b x K
P t K P X b x K
(2.3)
,
, .
u
j i l
u u
j i l j i l
P t X a x G
P t G P X a x G
Proof. Firstly, we prove (2.1) and (2.3). It suffices to show
0
, ,
e ,
!
j
u
jil h
i l
j x u
jil h
P t X b x K
x
P t K
h
,(2.5)
since (2.1) can be obtained by letting in (2.5) and then dividing both sides of the resulting equation by
t
u jil
P K . It is known that an Erlang(n) random variable can be expressed as an independent sum of exponential random variables with same parameters. Let n
, 2, , n
, 1
i
V i the independent exponentially distributed random variables with parameter i
n
. Denote by 1 2 the arrival times of the Poisson process ,
and let
, ,
T T N
,0
t Xs s t
F be the field generated by process Xs, 0 s t. It follows that
1 1
, ,
, ,
jil
jil
u
K
u
n K
n n
P t X b x I
P T t X b x I P
n.
(2.4) With Uu
t u ct W t
, we have
0
0
0
max ,
1 2
max ,
, ,
max , , ,
, ,
, ,
jil
n jil
n
n jil n
s n s Tn
jil n
s n s Tn
u
n n K
u
s T n K
s T
u
T K T n
X b T t
u
u n n K T n
X b T t
P P T t X b x I
P X b X b x T t I
E P X b x I T I
E P U T Y Y Y b x I T I
F
F
0
0
1
max ,
1
1 1 1 1
max ,
1 1 =1 =1
, ,
, ,
jil n
s n s Tn
n
s n s Tn
n u
n u n k K T n
X b T t
k
i n l n l
u
ij k u n k k l u n j k T n
X b T t
k l k k k k
E P Y b x U T Y I T I
p E P V b x U T Y V V b U T Y V T I
F
F
0
e , .
!
j
jil
h i l
j x u
n K
h
x
P T t I
h
Thus we have
0
, ,
e ,
!
jil
j
jil
u
K h
i l
j x u
K h
P t X b x I
x
P t I
h
(2.2) and (2.4) can be obtained similarly. This completes the proof.
The following results are immediate consequences of Theorem 2.2.
. Corollary 2.3. For j1, 2, , J , j 1, 2, , Jˆ ,
1, , j
i m , i 1, , mˆj, , , we
have 1, ,
1
e ,
i l
X b j
u
jil
j
E I K
0,
1
ˆ
e , 0
ˆ
i l
X a j
u
j i l
j
E I G
.
Corollary 2.4. For any x0, we have
1
0
e , !
j
h k
j x
u
jk h
x
P X b x K
h
1
ˆ0
ˆ
e , !
j
h k
j x
u
j k h
x
P X a x G
h
, ,
u
jk
u u
jk jk
P t X b x K
P t K P X b x K
,
, ,
u
j k
u u
j k
P t X a x G
P t G P X a x G
j k
where
1 1 2 j j 1,
jk jk j k jm m k
K K K K
1 1 2 j j 1
j k j k j k j m m k
G G G G
for j1, 2, , J , 1, 2, ,k mj, j 1, 2, , Jˆ, ˆ
1, 2, , j
k m.
Corollary 2.5. For j1, 2, , J , 1, 2, ,k mj, ,
ˆ 1, 2, ,
j J k 1,2, , mˆj, we have
e ,
k
X b j
u
jk
j
E I K
0,
ˆe ,
ˆ k
X a j
u
j k
j
E I G
0.
Remark 2.6. When , , (2.1) and (2.3) reduce to Equations (8) and (9) of Cai [3], respectively.
1 i
m nj1
3. Main Results
In this section, we study the distribution of the first exit problem to two barriers. We first define three vectors:
ˆ
T
ˆ
11 11 ˆ
, , , , , J, , , , , , ,
J
u u u d d d
jk Jm j k Jm
f f b f f f f a f f f
ˆ
0 0 ˆ ˆ
π e , , e , e , , e
JmJ JmJ
u u u u
K K G
E I E I E I E I
,
G
u
e1u b,e2u b, ,eM 1u b,eˆ1u a,eˆ2u a, ,eˆMˆ 1u a
.
where
1 0
e
d , 1, , , 1, , , 1 !
j
k y
j j
u
jk j
y
f f y b y j J k m
k
1 ˆ
0 ˆ ˆ e ˆ
ˆ d , 1, , , 1, , . 1 !
j
k y
j j
d
j k j
y
f f y a y j J k m
k
Let
ˆ1 ˆ2 ˆˆ 1
1 1,1, ,1, , , , M ,
b a
b a b a
C e e e 2
e1 ,e2 , ,eM 1 ,1,1, ,1
a b a b a b
C ,
ˆ
1 1
ˆ
1 1
ˆ ˆ
ˆ
1 1 1 1
ˆ ˆ
1 1 1 ˆ 1
e e
ˆ ˆ
, 1, 2, ,
e e
ˆ ˆ
M
i i i M i
i i
r b a
r b a
i i i i
i i M i i M
i
m m r b a m r b a m
i i i i
m m
i i M i i M
,
A i J
1 1
1 1
ˆ
1 1 1 1
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ
1 1
1 1
ˆ ˆ ˆ ˆ
e e
ˆ ˆ ˆ ˆ ˆ ˆ
ˆ , 1, 2, ,
ˆ ˆ ˆ ˆ
e e
ˆ ˆ ˆ ˆ
ˆ ˆ
M
i i
M
i i
i i
a b a b
i i i i
i i M i i M
i
m m
a b m a b m
i i i i
m m
i i M
i i M
B i
. J
Define a matrix
.
Tˆ
1 1 2 1
T T T T T T
J J
D C A A C B B
Theorem 3.1. Consider any nonnegative measurable function f such that fjku and
d j k
f for
, 1, 2,
,
j J k1, , mj, j 1,,Jˆ, k 1, , mˆj.
For any 0 and
,b , we have
X
u a e
u
E f πf, (3.1)
where π satisfies
πD. (3.2) Moreover, when D is a non-singular matrix, is the unique solution of (3.2), i.e.,
1
π D. (3.3) Proof. By the law of total probability, we have
00
0
0
1 1 1
ˆ ˆ
1 1 1
1 1
ˆ
1
e e e
e e
e e
e
j
jil
j
j i l
j
jk
m
J i
u u u
K K
j i l
m
J i
u u
G G
j i l
m J
u u
K K
j k
J u
G
j k
E f X E f X I E f X I
E f X I E f X I
E f X I E f X I
E f X I
ˆ
1
e .
j
j k
m u
G
E f X I
It follows from Corollary 2.4, for j1, 2, , J, 1, , j
k m , j 1, , Jˆ, k 1,,mˆj, we have
0 0
e e
u u
K K
E f X I E I f b ,
,
0 0
e e
u u
G G
E f X I E I f a
1
0 e
e
e d ,
1 !
jk
j jk
u
K
k y
j j
u K
E f X I
y
E I f y b y
k
1 ˆ
0
e
ˆ ˆ e
e d .
1 !
j k
j j k
u
G
k y
j j
u G
E f X I
y
E I f y a y
k
.
u jk
d j k
Combining these equations, we get
00
1 1 ˆ ˆ
1 1
e
e e
e e
j
j
j k
u
m J
u u
K K jk
j k
m J
u u
G G
j k
E f X
E I f b E I f
E I f a E I f
The expressions for u e 0 ,
K
E I Eu e IKjk
,
0
e u
G
E I and Eu e
j k
G
I
can be determined as follows. Let denote the set of functions g R: R such that g u
is twice continuously differentiable and bounded for a<u<b with g u
and g u
bounded for a u b. By applying Itô formula to the process Xt, we have that for t0 and g ,
t
0t
s d t,g X g u
Lg X sMhave aXsb as s.
For any 0, we can easily obtain from the above equation that
t t
0 e
e d
t s s
0 e
t
d ,
s s t
g X
g u
Lg X g X s
M, 0.
where the last term of the above equation is a mean-0 martingale. This implies that
0
e
e d
t u
t
t
u s
s s
E g X
g u E Lg X g X s t
(3.4) By simple calculation, the function
e xg x with
G and C satisfies Lg
Xs g X
s 0 for a x b. It follows from (3.4) that the process
e t Xt :
t
0 is a martingale. Then
0
0
0
1
1 1 1
1 ˆ
ˆ
ˆ
1 1 1
1
e e e e e
ˆ
e e e e
ˆ
e e
j
jil
j
j i l
i l
m
J i
j
X u b u e b
K K
j i l j
i l
m
J i
j
u a u a
G G
j i l j
J
u b
K
j
E E I E I
E I E I
E I
u u
0
1
ˆ ˆ
ˆ
1 1
e e
ˆ
e e e e .
ˆ
j
jk
j
j k
k m
j
u b
K
k j
k m
J
j
u a u a
G G
j k j
E I
E I E I
(3.5)
Setting i for i1, 2, , M1 and ˆi for i1, 2, , Mˆ 1 in (3.5), we have the following linear
equations:
0
0
1 1
ˆ ˆ
1 1
e e e e e
ˆ
e e e e ,
ˆ
j
i i i
jk
j
i i
j k
k m
J
j
u u X u b u
K K
j k j i
k m
J
j
a a
u u
G G
j k j i
E E I E I
E I E I
eib
and
0
0
ˆ ˆ ˆ ˆ
1 1
ˆ ˆ
ˆ ˆ ˆ
1 1
e e e e e e
ˆ
ˆ
e e e e .
ˆ ˆ
j
i i i i
jk
j
i i
j k
k m
J
j
u u X u b u b
K K
j k j i
k m
J
j
a a
u u
G G
j k j i
E E I E I
E I E I
Then the vector satisfies π π D
u and we have (3.1). If is non-singular, we have. This completes the proof. D
1u D
π
Corollary 3.2. For any
min ˆ1, ˆ2 , , min 1, 2 , , Jˆ
e ,
ˆ
, J
, we
have
ˆ
1 1
ˆ
1 1
e ei j
M M
u a u b
X
u b
i j
i j
E e
(3.6)where
T
ˆ
1, 2, , M 1, , , ,1 2 M 1 ,
D J and
1 1 T ˆ
ˆ
1 1 1
ˆ
1 1 1
ˆ ˆ
1, , , , , , , ,e , e , , e .
ˆ ˆ
J
m
m m
a b a b a b
J J J
J J J
J
Remark 3.3. When , , (3.1) and (3.6) reduce to equation (6) and (15) of [4], respectively.
1 i
m nj1
From Theorem 3.1, choosing
X a0 I
, IX b y , IX a y , IXb, IXa and eX respectively, we can obtain the following co-
rollaries.
f X to be IX b0
,
Corollary 3.4. 1) For any 0, we have
ˆ
1 1
ˆ 0
1 1
0
ˆ ˆ
e e e
1
j i
M M
a u b u
u
i j
X b
i j
u a
E I a u
u b
b, (3.7)where
T
ˆ
1 2 1 1 2 1
ˆ : ˆ ˆ, , , ˆM , , , ,ˆ ˆ ˆM
is determined by the linear system DˆI1. Here
T
ˆ 1 1,1, ,1,0,0, ,01 M M 2.
I
2) For any 0, we have
ˆ
1 1 ˆ
0
1 1
1
ˆ ˆ
ˆ ˆ
e e e
0
j i
M M
a u b u
u
i j
X a
i j
u a
E I b u
u b
> > ,a (3.8)where
1 2 1 1 2 ˆ 1
ˆ ˆ ˆ ˆ ˆ ˆ ˆ
ˆT : ˆ ˆ, , , ˆ , , , ,ˆ ˆ ˆ
M M
is determined by the linear system Dˆˆ I2. Here
T
ˆ 2 0,0, ,0,1,1, ,11 M M 2 .
I
Corollary 3.5. 1) For f X
IX a y and any 0, y0, we have
ˆ
1 1
ˆ
1 1
0 ,
e e e
1
j i
M M
a u b u
u
i j
X b y
i j
u b b y u a
E I a u b
u b y
or
,
(3.9)where
T
ˆ 1, 2, , M 1, , , ,1 2 M 1
is determined by the linear system D =I1
. Here
1
1 1
1 1
1 T
1
0 0
ˆ
1 2
0,e , , e , ,e , , e ,0, ,0 ;
! !
J
J J
t
t m
m
J
y y
y y
t t
M M
y y
I
t t
2) For f X
IXa<y and any 0
, y0, we have
ˆ
1 1
ˆ <
1 1
1
e e e
0 ,
j i
M M
a u b u
u
i j
X a y
i j
u a y
E I b u a
u a y a u b
, or
(3.10)where
T
ˆ 1, 2, , M 1, , , ,1 2 M 1
is determined by the linear system D I2. Here
ˆ 1
ˆ ˆ
1 1
ˆ 1 ˆ 1
ˆ
ˆ ˆ
ˆ 1 ˆ
T 2
0 0
ˆ
1 2
ˆ ˆ
0, ,0,0,e , , e , ,e , , e .
! !
J
J J
t
t m
m
y J y
y y
t t
M M
y y
I
t t
Note that the difference of u e 0
X b
E I
e 0
u
X a
E I
and e >0
u
X b
E I
is
exactly
e <0
u
X a
E I
e u
X b
E I
E e IX a
u
. Thus we obtain the following results. Corollary 3.6. 1) For f X
IX b , and for any
0, we have
ˆ
1 1 ˆ
1 1
0 o
e e e
1
j i
M M
a u b u
u
i j
X b
i j
u b u a
E I a u b
u b
r
,
I
(3.11)
where
T
ˆ 1, 2, , M 1, , , ,1 2 M 1
is determined by the linear system D 1. Here
T
ˆ
1 1,0, ,0,0, ,01 M M 2 .
I
2) For f X
IX a and any
0, y0, we have
ˆ
1 1
ˆ ˆ 1
1 1
0 o
e e e
1
M M
a u b u
u i M
i j
X a
i j
u a u b
E I b u a
u a
r
,
(3.12)where
T
ˆ 1, , ,2 M 1, , , ,1 2 M 1
is determined by the linear system D I2. Here
T
ˆ
2 0, 0, , 0,1, 0, , 01 M M 2.
I
To end the paper, we give an example. Example 3.7. When Jm1 Jˆ mˆ11,
1x ˆ ˆeˆ1x
p I
11 1e x 0 11 1 x 0 p x p I and11 ˆ11 1
p p , the equation G z
has real roots: 41
, 2, ˆ1 and ˆ2
. Let
ˆ2 ˆ1ˆ10112
1 2
1 2
1 2
1 2
ˆ ˆ
ˆ ˆ
1 1 1 1
1 1 1 2 1 1 1 2
1 1 1 1
1 1 1 2 1 1 1 1
1 1 e e
e e
ˆ ˆ
.
e e 1 1
ˆe ˆe ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ
b a b a
b a b a
a b a b
a b a b
D
Denote 1 by D
11 21 31 41
12 22 32 42
1
13 23 33 43
14 24 34 44
1
.
d d d d
d d d d
D
d d d d
D
d d d d
Then we have
1 2 ˆ1 ˆ2
1 2 3 4
e
e e e e
u
u b u b u a u a
E f X
,
where
11 21 11 31 41 11
1
12 22 11 32 42 11
2
13 23 11 33 43 11
3
14 24 11 34 44 11
4
,
,
,
,
u d
u d
u d
u d
d f b d f d f a d f D
d f b d f d f a d f D
d f b d f d f a d f D
d f b d f d f a d f D
