Two-Sided First Exit Problem for Jump Diffusion Distribution Processes Having Jumps with a Mixture of Erlang ()
1. Introduction
Consider the following jump diffusion process
(1.1)
where the constant 
is the starting point of
, 
and 
represent the drift and the volatility of the diffusion part, respectively, 
is a standard Brownian motion with
, 
is a Poisson process with rate
, and the jumps sizes 
are assumed to be i.i.d. real valued random variables with common density
. Moreover, it is assumed that the random processes
, 
and random variables 
are mutually independent. In this paper we are interested in the density 
of following type
(1.2)
where
, 
, 
, 
and that
, 
for all
. Moreover,
Define 
to be the first exit time of 
to two flat barriers 
and 
, i.e.
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 component. For some related works, see Perry et al. [8], Cai and Kou [9], Lewis and Mordecki [10] and the references therein.
Motivated by works mentioned above, the main objective of this paper is to study the first exit time of the process (1.1) with jump density (1.2) from an interval and the overshoot over the boundary at the exit time. In Section 2, we study the roots of the generalized Lundberg 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
for any twice continuously differentiable function 
and the Lévy exponent of 
is given by
By analytic continuation, the function 
can be extended to the complex plane except at finitely many poles. In the following, we consider the resulting extension 
of
, i.e.,
Let us denote 
and
.
In [11], Kuznetsov has studied the roots of the equation
. However, for this particular Lévy process
, we will give another simple proof for the roots of this equation.
Lemma 2.1. For fix
, the generalized CramérLundberg equation
has 
complex roots 
with 
for 
and 
with 
for
.
Proof. Let
Firstly, we prove that for given
, 
has 
roots with negative real parts. Set
with
where 
is an arbitrary positive constant. Applying Rouchés theorem on the semi-circle
, consisting of the imaginary axis running from 
to 
and with radius 
running clockwise from 
to
. We let 
and denote by 
the limiting semi-circle. It is known that both 
and 
are analytic in
. We want to show that
Notice that 
for
, and
is bounded for
. Hence, for
,
on the boundary of the half circle in
. For
, we have 
(see Lewis and Mordecki [10]). On the other hand,
Thus we have
. Since
has 
roots with negative real parts, so equation 
has 
roots with negative real parts. Similarly, we can prove 
has 
roots with positive real parts.
In the rest of this paper, we assume all the roots of equation 
are distinct and denote 
, 
for notational simplicity, and denote 
(or 
in the sequel) representing the expectation (or probability) when 
starts from
. We denote a sequence of events
, 
= {
: 
crosses 
at time 
by the 
th phase of 
th positive jump whose parameter is 
},
= {
: 
crosses 
at time 
by the 
th phase of 
th negative jump whose parameter is 
}
for
, 
, 
, 
, 
and
.
Theorem 2.2. For any
, we have
(2.1)
(2.2)
Furthermore, conditional on 
, the stopping time 
is independent of the overshoot 
(the undershoot
). More precisely, for any
, we have
(2.3)
(2.4)
Proof. Firstly, we prove (2.1) and (2.3). It suffices to show
(2.5)
since (2.1) can be obtained by letting 
in (2.5) and then dividing both sides of the resulting equation by
. It is known that an Erlang(n) random variable can be expressed as an independent sum of 
exponential random variables with same parameters. Let 
the 
independent exponentially distributed random variables with parameter
. Denote by 
the arrival times of the Poisson process
, and let 
be the field generated by process
,
. It follows that
With
, we have
Thus we have
(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
, 
, 
, 
, 
, 
, we have
Corollary 2.4. For any
, we have
where
for
, 
, 
, 
.
Corollary 2.5. For
, 
, 
, 
, we have
Remark 2.6. When
, 
, (2.1) and (2.3) reduce to Equations (8) and (9) of Cai [3], respectively.
3. Main Results
In this section, we study the distribution of the first exit problem to two barriers. We first define three vectors:
where
Let
Define a matrix 
.
Theorem 3.1. Consider any nonnegative measurable function 
such that 
and 
for
, 
, 
,
. For any 
and
, we have
(3.1)
where 
satisfies
(3.2)
Moreover, when 
is a non-singular matrix, 
is the unique solution of (3.2), i.e.,
(3.3)
Proof. By the law of total probability, we have
It follows from Corollary 2.4, for
, 
, 
, 
, we have
Combining these equations, we get
The expressions for
, 
, 
and 
can be determined as follows. Let 
denote the set of functions 
such that 
is twice continuously differentiable and bounded for 
with 
and 
bounded for
. By applying Itô formula to the process
, we have that for 
and
,
where 
is a martingale with
. Note that we have 
as
.
For any
, we can easily obtain from the above equation that
where the last term of the above equation is a mean-0 martingale. This implies that
(3.4)
By simple calculation, the function 
with 
and 
satisfies 
for
. It follows from (3.4) that the process
is a martingale. Then
(3.5)
Setting 
for 
and 
for 
in (3.5), we have the following linear equations:
and
Then the vector 
satisfies 
and we have (3.1). If 
is non-singular, we have 
. This completes the proof.
Corollary 3.2. For any
, we have
(3.6)
where
and
Remark 3.3. When
, 
, (3.1) and (3.6) reduce to equation (6) and (15) of [4], respectively.
From Theorem 3.1, choosing 
to be
,
, 
, 
, 
, 
and 
respectively, we can obtain the following corollaries.
Corollary 3.4. 1) For any
, we have
(3.7)
where
is determined by the linear system
. Here
2) For any
, we have
(3.8)
where
is determined by the linear system
. Here
Corollary 3.5. 1) For 
and any
, 
, we have
(3.9)
where
is determined by the linear system
. Here
2) For 
and any
, 
, we have
(3.10)
where
is determined by the linear system
. Here
Note that the difference of 
and 
is exactly 
. Thus we obtain the following results.
Corollary 3.6. 1) For
, and for any
, we have
(3.11)
where
is determined by the linear system
. Here
2) For 
and any
, 
, we have
(3.12)
where
is determined by the linear system
. Here
To end the paper, we give an example.
Example 3.7. When
,
and
, the equation 
has 
real roots:
, 
, 
and 
. Let
Denote 
by
Then we have
where
We define 
(
, 
,
) and 
(
, 
,
) as follows: let 
(
, 
,
) be obtained from 
(
, 
,
) by changing 
to 
in 
(
,
); let 
(
, 
,
) be obtained from 
(
, 
,
) by changing 
to 
in 
(
, 
,
).
• If
, then we have
where
• 
• If
, then we have
where
• 
• If
, then we have
where
• 
• If
, 
, then we have
• 
• If
, 
, then we have
where
• 
• If
, then we have
where
• 
• If
, then we have
where
When
, we have
Therefore, we have
These results are all consistent with that of Theorem 3.1 of Kou and Wang [2] for the one-sided exit problem of the doubly exponential jump diffusion process.
NOTES
#Corresponding author.