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Abstract and Applied Analysis

Volume 2014 (2014), Article ID 747262, 10 pages

http://dx.doi.org/10.1155/2014/747262

## Exit Problems for Jump Processes Having Double-Sided Jumps with Rational Laplace Transforms

School of Mathematical Sciences, Qufu Normal University, Qufu, Shandong 273165, China

Received 5 December 2013; Accepted 18 December 2013; Published 2 February 2014

Academic Editor: Xinguang Zhang

Copyright © 2014 Yuzhen Wen and 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

We consider the two-sided first-exit problem for a jump process having jumps with rational Laplace transform. We derive the joint distribution of the first passage time to two-sided barriers and the value of process at the first passage time. As applications, we present explicit expressions of the dividend formulae for barrier strategy and threshold strategy.

#### 1. Introduction

Let be a jump process defined on the filtered probability space and where is the starting point of , is a Poisson process with rate , constant represents the drift, and the jump sizes are independent and identically distributed random variables whose probability density function is given by . Moreover, it is assumed that and are independent. Recently, one-sided and two-sided first-exit problems for processes with two-sided jumps have attracted a lot of attention in applied probability. For example, Perry et al. [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. Chi [4] discussed the first passage time to two barriers of a hyperexponential jump diffusion process. Closed form expressions are obtained in [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. [6], Lewis and Mordecki [7], Kuznetsov [8], Perry et al. [1], and the references therein. On the other hand, the jump processes with two-sided jumps have been studied widely in dividend problems. The problem of first-exit from a half-line is of fundamental interest with regard to the expected total discounted dividends up to ruin. See, for example, Chi [4], Chi and Lin [9], and Yin et al. [10].

Motivated by some related work mentioned above, we will assume is a sequence of iid random variables with a density given by

We assume that , , , , and , and , and that and for all .

Define to be the first-exit time of to two flat barriers and ; that is, and the infimum of an empty set is defined as .

The rest of the paper is organized as follows. In Section 2, we study the roots of the generalized Lundberg equation. Section 3 concentrates on deriving the joint distribution of the first-exit time and a nonnegative measurable function of the process value at the first-exit time to two flat barriers. In Section 4, we will study two kinds of dividend payment strategies by using the results in Section 3.

#### 2. Preliminary Results

The infinitesimal generator of the jump process (1) is given by for any continuously differentiable function and the Laplace exponent is a rational function, which has the following partial fraction decomposition: We see that has poles at and with the corresponding multiplicities and . Note that has poles (counting with multiplicity) in the half-plane and poles in the half-plane . We denote four marks: (i) is the number of poles of in the half-plane ,(ii) is the number of zeros of in the half-plane .

In [8], the author has given the solutions of the Cramér-Lundberg equation . However, for this simple Lévy process , we will give another simple proof for the zeros of this equation.

Lemma 1. *Assume that .*(1)*If , then and .*(2)*If , then and .*(3)*If , then and .*

*Proof. *We denote
Then and are analytic in . Define a half circle , , where , where is an arbitrary positive constant. Note that for and is bounded for . When is large enough,
on the boundary of the half circle in . On the imaginary axis, we observe that
Since has roots with negative real part, so the equation , has roots with negative real part according to Rouché’s theorem. Similarly, we can prove that , has roots with positive real part. The proofs of (2) and (3) are similar to the proof of (1).

*Remark 2. *In [8], the author states that it is very unlikely for the equation to have multiple solutions, so we assume all the roots of this equation are distinct. We denote and for notational simplicity.

#### 3. Distribution of the First-Exit Time to Two Flat Barriers

In this section, we consider the distribution of the first-exit time to two flat barriers. Let be positive integers. Suppose , , , , and . For , and , , define Note that and are matrices and is an matrix.

Theorem 3. *Consider any nonnegative measurable function such that
**
for , , , . For any and , one has the following.**(1) If , then
**
Here
**
are both -dimensional vectors, and satisfies
**
where is an -dimensional vector. Moreover, when is a nonsingular matrix, is the unique solution of (13); that is,
**(2) If , then
**
Here and are both -dimensional vectors, and satisfies
**
where is an -dimensional vector. Moreover, when is a nonsingular matrix, is the unique solution of (16); that is,
**(3) If , then
**
Here and are both -dimensional vectors, and satisfies
**
where is an -dimensional vector. Moreover, when is a nonsingular matrix, is the unique solution of (19); that is,
*

*Proof. *When the jump process exits the band , sometimes it hits the boundary exactly and sometimes it jumps across the boundaries when leaving. We introduce 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 .

It is easy to check that
is a zero-mean martingale for every and every . We apply the sampling theorem to the martingale and get ; that is,
Inserting the roots of in (22), we get that the right side of (22) becomes zero.

From Perry et al. [1], we know that for , , and , given , the overshoot is independent of and distributed with scale parameter and for , , and , given , the excess is independent of and distributed with scale parameter . Therefore, we know that, for and , given , the overshoot is independent of and distributed with scale parameter and for and , given , the excess is independent of and distributed with scale parameter .

If , by the law of total probability, we have
Inserting , and , , in (22), it follows that, for ,
and for
Then the vector satisfies and we have (11). If is nonsingular, we have .

The proof of (2) and (3) is similar and we omit their proofs.

Corollary 4. *If the matrix is nonsingular, then for any , , one has the following.**(1) If , then
**
where
**(2) If , then
**
where
**(3) If , then
**
where
*

Corollary 5. *For any , , one has the following.**(1) If , then
**
Here is a vector uniquely determined by the following linear system:
**
where is an -dimensional vector and is an nonsingular matrix:
**(2) If , then
**
Here is a vector uniquely determined by the following linear system:
**
where is an -dimensional vector and is an nonsingular matrix:
*

Corollary 6. *For any , , one has the following.**(1) If , then
**
Here is a vector uniquely determined by the following linear system:
**
where is an -dimensional vector and is an nonsingular matrix:
**(2) If , then
**
Here is a vector uniquely determined by the following linear system:
**
where is an -dimensional vector and is an nonsingular matrix:
*

*Example 7. *When , , and , , then the equation has real roots: , , and (). Let
Denote by
Then we have
where
We define as follows: let be obtained from by changing to in (i)If , then we have
where
(ii)If , then we have
where
(iii)If , then we have
where
(iv) If , , then we have
where
(v)If , , then we have
where
(vi)If , then we have
where

#### 4. Dividend Problems

In this section, we are going to derive dividend formulae for barrier strategy and threshold strategy, based on the results obtained in Section 3.

For barrier strategy, it is assumed that dividends are paid according to a barrier strategy . Such a strategy has a level of the barrier ; when the surplus exceeds the barrier, the excess is paid out immediately as the dividend. The modified surplus at time is given by where denote the aggregate dividends paid between time and time ; that is,

Let be the time of ruin for the modified surplus and the present value of all dividends until ruin, where is the discount factor.

Denote by the expectation of the discounted dividends until ruin, if the barrier strategy with parameter is applied; that is, Note that we have

Theorem 8. *Let and . Then for , one has the following.*

(1) for ,

(2) if , then

(3) if , then

*Proof. *It follows from the conditional memoryless property, the conditional independence, and the strong Markov property that, for , we have
Assertion (2) follows from (1) and , and assertion (3) follows from (1) and . This ends the proof.

For threshold strategy, we assume that dividends are paid according to the following strategy governed by parameters and . Whenever the modified surplus is below the threshold level , no dividends are paid. However, when the surplus is above this threshold level, dividends are paid at a constant rate that does not exceed the premium rate . We define the modified risk process by , where Let denote the present value of all dividends until the time of ruin , where with if for all . Here is the discount factor. Denote by the expected discounted value of dividend payments; that is,