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

Authors: Yuzhen Wen; Chuancun Yin
Journal: Abstract and Applied Analysis
 (2014)

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## 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.


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## Body

## 1. Introduction

LetX={X(t):t≥0} be a jump process defined on the filtered probability space (Ω,ℱ,𝔽=ℱt:t≥0,P) and
(1)X(t)=x+μt+∑k=1N(t)‍Yk≡x+μt+S(t),t≥0,
where x is the starting point of X(t), {Nt;t≥0} is a Poisson process with rate λ, constant μ∈R represents the drift, and the jump sizes {Yk;k≥1} are independent and identically distributed random variables whose probability density function is given by p(y). Moreover, it is assumed that Nt and Yk 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{Yk:k=1,2,…} is a sequence of iid random variables with a density p(x) given by
(2)p(x)=∑j=1J‍∑i=1mj‍pijρjixi-1(i-1)!e-ρjx1{x>0}+∑j=1J^‍∑i=1m^j‍p^ijρ^ji|x|i-1(i-1)!eρ^jx1{x<0}.We assume thatJ,J^∈N,  mj,  m^j∈N,  pij,p^ij∈R+,  and ∑j=1J‍∑i=1mj‍pij+∑j=1J^‍∑i=1m^j‍p^ij=1,  Re(ρj)>0  and  Re(ρ^j)>0,   and that  ρi≠ρj  and ρ^i≠ρ^j for all i≠j.Defineτ to be the first-exit time of Xt to two flat barriers a and b  (a<b); that is,
(3)τ≔inf{t≥0:Xt≥borXt≤a},
and the infimum of an empty set is defined as +∞.The rest of the paper is organized as follows. In Section2, 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
(4)(Lu)(x)=μu′(x)+λ∫-∞∞‍[u(x+y)-u(x)]p(y)dy
for any continuously differentiable function u(x) and the Laplace exponent G(z)=lnE[exp(zX1)] is a rational function, which has the following partial fraction decomposition:
(5)G(z)=zμ+λ(∑j=1J‍∑i=1mj‍pij(ρj)i(ρj-z)i+∑j=1J^‍∑i=1m^j‍p^ij(ρ^j)i(ρ^j+z)i-1),22222222222222222222222222222222222z∈C.
We see that G(z) has poles at ρj(1≤j≤J) and -ρ^j(1≤j≤J^) with the corresponding multiplicities mj and m^j. Note that G(z) has ∑j=1J‍mj poles (counting with multiplicity) in the half-plane Re(z)>0 and ∑j=1J^‍m^j poles in the half-plane Re(z)<0. We denote four marks: (i)
M{M^} is the number of poles of G(z) in the half-plane Re(z)>0{Re(z)<0},(ii)
K{K^} is the number of zeros of G(z)-θ in the half-plane Re(z)>0{Re(z)<0}.In [8], the author has given the solutions of the Cramér-Lundberg equation G(z)=θ. However, for this simple Lévy process X, we will give another simple proof for the zeros of this equation.Lemma 1.
Assume thatθ>0.(1)
Ifμ>0, then K=M+1 and K^=M^.(2)
Ifμ<0, then K=M and K^=M^+1.(3)
Ifμ=0, then K=M and K^=M^.Proof.
We denote(6)G1(z)=zμ-λ-θ,z∈C,G2(z)=λ(∑j=1J‍∑i=1mj‍pij(ρj)i(ρj-z)i+∑j=1J^‍∑i=1m^j‍p^ij(ρ^j)i(ρ^j+z)i),2222222222222222222222222222z∈C.
Then (∏j=1J^‍(ρ^j+z)m^j)G1(z) and (∏j=1J^‍(ρ^j+z)m^j)G2(z) are analytic in C-. Define a half circle |x|=r, x∈C-, where r>ɛ+max1≤j≤J^{|ρ^j|}, where ɛ is an arbitrary positive constant. Note that |G1(z)|→∞ for Re(z)→-∞ and |G2(z)|≤λ∑j=1J‍∑i=1mj‍(|pij||ρj|i/εi)+λ∑j=1J^‍∑i=1m^j‍(|p^ij||ρ^j|i/εi) is bounded for Re(z)→-∞. When r is large enough,
(7)|(∏j=1J^‍(ρ^j+z)m^j)G1(z)|>|(∏j=1J^‍(ρ^j+z)m^j)G2(z)|
on the boundary of the half circle in C-. On the imaginary axis, we observe that
(8)|G1(ia)|=|λ+θ+iaμ|>λ=|G2(ia)|,a∈R.
Since (∏j=1J^‍(ρ^j+z)m^j)G1(z) has M^ roots with negative real part, so the equation G(z)=θ,θ>0, has M^+1 roots with negative real part according to Rouché’s theorem. Similarly, we can prove that G(z)=θ, θ>0 has M 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 G(z)=θ to have multiple solutions, so we assume all the roots of this equation are distinct. We denote γi(θ)=γi(i=1,2,…,M+1) and  γ^i(θ)=γ^i(i=1,2,…,M^+1) 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. Letk,J,mj,J^,m^j′ be positive integers. Suppose I1k=(1,1,…,1)1×k, k=M+1, M^+1, E1M^=(er^1(b-a),…,er^M^(b-a))1×M^, and E1M=(er1(b-a),…,erM(b-a))1×M. For i=M, M+1 and j=M^, M^+1, define
(9)A11i=(ρ1ρ1-γ1⋯ρ1ρ1-γi⋮⋱⋮(ρ1ρ1-γ1)m1⋯(ρ1ρ1-γi)m1⋮⋱⋮ρJρJ-γ1⋯ρJρJ-γi⋮⋱⋮(ρJρJ-γ1)mJ⋯(ρJρJ-γi)mJ),A12j=(er^1(b-a)ρ1ρ1-γ^1⋯er^M^(b-a)ρ1ρ1-γ^j⋮⋱⋮er^1(b-a)ρ1m1(ρ1-γ^1)m1⋯er^M^(b-a)ρ1m1(ρ1-γ^j)m1⋮⋱⋮er^1(b-a)ρJρJ-γ^1⋯er^M^(b-a)ρJρJ-γ^j⋮⋱⋮er^1(b-a)ρJmJ(ρJ-γ^1)mJ⋯er^M^(b-a)ρJmJ(ρJ-γ^j)mJ)A21i=(eγ1(a-b)ρ^1ρ^1+γ1⋯eγM+1(a-b)ρ^1ρ^1+γi⋮⋱⋮eγ1(a-b)ρ^1m^1(ρ^1+γ1)m^1⋯eγM+1(a-b)ρ^1m^1(ρ^1+γi)m^1⋮⋱⋮eγ1(a-b)ρ^J^ρ^J^+γ1⋯eγM+1(a-b)ρ^J^ρ^J^+γi⋮⋱⋮eγ1(a-b)ρ^J^m^J^(ρ^J^+γ1)m^J^⋯eγM+1(a-b)ρ^J^m^J^(ρ^J^+γi)m^J^),A22j=(ρ^1ρ^1+γ^1⋯ρ^1ρ^1+γ^j⋮⋱⋮(ρ^1ρ^1+γ^1)m^1⋯(ρ^1ρ^1+γ^j)m^1⋮⋱⋮ρ^J^ρ^J^+γ^1⋯ρ^J^ρ^J^+γ^j⋮⋱⋮(ρ^J^ρ^J^+γ^1)m^J^⋯(ρ^J^ρ^J^+γ^j)m^J^)A1=(I1M+1E1×M^A11M+1A12M^A21M+1A22M^),A2=(A11MA12M^+1E1MI1M^+1A21MA22M^+1),A3=(A11MA12M^A21MA22M^).
Note that A1 and A2 are (M+M^+1)×(M+M^+1) matrices and A3 is an (M+M^)×(M+M^) matrix.Theorem 3.
Consider any nonnegative measurable functionf such that
(10)fjku=∫0∞‍f(y+b)(ρjy)k-1ρje-ρjy(k-1)!dy<∞,fj′k′d=∫-∞0‍f(y+a)(-ρ^jy)k′-1ρ^jeρ^jy(k′-1)!dy<∞
for j=1,2,…,J, k=1,…,mj, j′=1,…,J^, k′=1,…,m^j′. For any θ>0 and u∈(a,b), one has the following.
(1) Ifμ>0, then
(11)Eu[e-θτf(Xτ)]=π1f1.
Here
(12)π1=([e-θτ1GJ^,m^J^]Eu[e-θτ1K0],Eu[e-θτ1K11],…,Eu[e-θτ1KJ,mJ],Eu[e-θτ1G11],…,Eu[e-θτ1GJ^,m^J^])T,f1=(f(b),f11u,…,fjku,…,fJmJu,f11d,…,fj′k′d,…,fJ^m^J^d)T
are both M+M^+1-dimensional vectors, and π1 satisfies
(13)π1A1=ϖ1,
where ϖ1=(eγ1(u-b),eγ2(u-b),…,eγM+1(u-b),eγ^1(u-a),eγ^2(u-a),…,eγ^M^(u-a)) is an M+M^+1-dimensional vector. Moreover, when A1 is a nonsingular matrix, π1 is the unique solution of (13); that is,
(14)π1=ϖ1·A1-1.
(2) Ifμ<0, then
(15)Eu[e-θτf(Xτ)]=π2f2.
Here π2=([e-θτ1GJ^,m^J^]Eu[e-θτ1K11],…,Eu[e-θτ1KJ,mJ],Eu[e-θτ1G0],Eu[e-θτ1G11],…,Eu[e-θτ1GJ^,m^J^])T and f2=(f11u,…,fjku,…,fJmJu,f(a),f11d,…,fj′k′d,…,fJ^m^J^d)T are both M+M^+1-dimensional vectors, and π2 satisfies
(16)π2A2=ϖ2,
where ϖ2=(eγ1(u-b),eγ2(u-b),…,eγM(u-b),eγ^1(u-a),eγ^2(u-a),…,eγ^M^+1(u-a)) is an M+M^+1-dimensional vector. Moreover, when A2 is a nonsingular matrix, π2 is the unique solution of (16); that is,
(17)π2=ϖ2·A2-1.
(3) Ifμ=0, then
(18)Eu[e-θτf(Xτ)]=π3f3.
Here π3=(Eu[e-θτ1K11],…,Eu[e-θτ1KJ,mJ],Eu[e-θτ1G11],…,Eu[e-θτ1GJ^,m^J^])T and f3=(f11u,…,fjku,…,fJmJu,f11d,…,fj′k′d,…,fJ^m^J^d)T are both M+M^-dimensional vectors, and π3 satisfies
(19)π3A3=ϖ3,
where ϖ3=(eγ1(u-b),eγ2(u-b),…,eγM(u-b),eγ^1(u-a),eγ^2(u-a),…,eγ^M^(u-a)) is an M+M^-dimensional vector. Moreover, when A3 is a nonsingular matrix, π3 is the unique solution of (19); that is,
(20)π3=ϖ3·A3-1.Proof.
When the jump processX exits the band (a,b), sometimes it hits the boundary exactly and sometimes it jumps across the boundaries when leaving. We introduce a sequence of events:K0≔{ω:Xτ=b},  G0≔{ω:Xτ=a},Kjil≔{ω:X(·) crosses b at time τ by the lth phase of ith positive jump whose parameter is ρj},Gj′i′l′≔{ω:X(·) crosses a at time τ by the l′th phase of i′th negative jump whose parameter is ρ^j′}Kjk=Kjk1∪Kj(k+1)2∪⋯∪Kjmj(mj-k+1),Gj′k′=Gj′k′1∪Gj′(k′+1)2∪⋯∪Gj′mj′(mj′-k′+1)forj=1,2,…,J,  j′=1,2,…,J^,  i=1,…,mi,  i′=1,…,m^i,  l=1,…,i, and l′=1,…,i′.
It is easy to check that(21)Mt=e-θt+βXt-eβX0-(G(z)-θ)∫0t‍e-θs+βXsds,t≥0
is a zero-mean martingale for every θ>0 and every β∈C. We apply the sampling theorem to the martingale M=(Mt∣t≥0) and get Eu[Mτ]=Eu[M0]=0; that is,
(22)Eu[e-θτ+βXτ]-eβX0=(G(z)-θ)Eu[∫0τ‍e-θs+βXsds].
Inserting the roots of G(z)=θ in (22), we get that the right side of (22) becomes zero.
From Perry et al. [1], we know that for l=1,2,…,i,  i=1,2,…,mj, and j=1,2,…,J, given Kjil, the overshoot Xτ-b is independent of τ and Erlang(i+j-l) distributed with scale parameter ρj and for l′=1,…,i′,  i′=1,…,m^i, and j′=1,2,…,J^, given Gj′i′l′, the excess a-Xτ is independent of τ and Erlang(i′+j′-l′) distributed with scale parameter ρ^j′. Therefore, we know that, for j=1,2,…,J and k=1,2,…,mj, given Kjk, the overshoot Xτ-b is independent of τ and Erlang(k) distributed with scale parameter ρj and for j′=1,2,…,J^ and k′=1,2,…,m^j′, given Gj′k′, the excess a-Xτ is independent of τ and Erlang(k′) distributed with scale parameter ρ^j′.
Ifμ>0, by the law of total probability, we have
(23)Eu[e-θτf(Xτ)]=Eu[e-θτf(Xτ)1K0]+∑j=1J‍∑i=1mj‍∑l=1i‍Eu[e-θτf(Xτ)1Kjil]+∑j′=1J^‍∑i′=1m^j′‍∑l′=1i′‍Eu[e-θτf(Xτ)1Gj′i′l′]=Eu[e-θτf(Xτ)1K0]+∑j=1J‍∑k=1mj‍Eu[e-θτf(Xτ)1Kjk]+∑j′=1J^‍∑k′=1m^j′‍Eu[e-θτf(Xτ)1Gj′k′]=Eu[e-θτ1K0]f(b)+∑j=1J‍∑k=1mj‍Eu[e-θτ1Kjk]fjku+∑j′=1J^‍∑k′=1m^j′‍Eu[e-θτ1Gj′k′]fj′k′d.
Inserting γi,  i=1,2,…,M+1 and  γ^i,  i=1,2,…,M^,  in (22), it follows that, for i=1,2,…,M+1,
(24)eγiu=Eu[e-θτ+γiXτ]=Eu[e-θτ1K0]eγib+∑j=1J‍∑k=1mj‍Eu[e-θτ1Kjk](ρjρj-γi)keγib+∑j′=1J^‍∑k′=1m^j′‍Eu[e-ρ^τ1Gj′k′](ρ^j′ρ^j′+γi)k′eγia
and for i=1,2,…,M^(25)eγ^iu=Eu[e-θτ+γ^iXτ]=Eu[e-θτ1K0]eγ^ib+∑j=1J‍∑k=1mj‍Eu[e-θτ1Kjk](ρjρj-γ^i)keγ^ib+∑j′=1J^‍∑k′=1m^j′‍Eu[e-ρ^τ1Gj′k′](ρ^j′ρ^j′+γ^i)k′eγ^ia.
Then the vector π satisfies π1·A1=ϖ1 and we have (11). If A1 is nonsingular, we have π1=ϖ1A1-1.
The proof of (2) and (3) is similar and we omit their proofs.Corollary 4.
If the matrixD is nonsingular, then for any δ∈(-min(|ρ^1|,|ρ^2|,…,|ρ^J^|), min(|ρ1|,|ρ2|,…,|ρJ|)), one has the following.
(1) Ifμ>0, then
(26)Eu[e-θτ+δXτ]=eδb·(∑i=1M+1‍ωieγi(u-b)+∑j=1M^‍νjeγ^j(u-a)),222222222222222222222222222222222222a<u<b,
where
(27)A1·(ω1,ω2,…,ωM+1,ν1,ν2,…,νM^)T=J1(δ),J1(δ)=((ρ^1ρ^1+δ)m^11,(ρ1ρ1-δ)m1,…,ρ1ρ1-δ,…,(ρJρJ-δ)mJ,…,ρJρJ-δ,(ρ^1ρ^1+δ)m^1eδ(a-b),…,ρ^J^ρ^J^+δeδ(a-b))T.
(2) Ifμ<0, then
(28)Eu[e-θτ+δXτ]=eδb·(∑i=1M‍ω¯ieγi(u-b)+∑j=1M^+1‍ν¯jeγ^j(u-a)),222222222222222222222222222222222222a<u<b,
where
(29)A2·(ω¯1,ω¯2,…,ω¯M,ν¯1,ν¯2,…,ν¯M^+1)T=J2(δ),J2(δ)=((ρ1ρ1-δ)m1,…,ρ1ρ1-δ,…,(ρJρJ-δ)mJ,…,ρJρJ-δ,eδ(a-b),(ρ^1ρ^1+δ)m^1eδ(a-b),…,ρ^J^ρ^J^+δeδ(a-b))T.
(3) Ifμ=0, then
(30)Eu[e-θτ+δXτ]=eδb·(∑i=1M+1‍ω~ieγi(u-b)+∑j=1M^+1‍ν~jeγ^j(u-a)),222222222222222222222222222222222222a<u<b,
where
(31)A3·(ω~1,ω~2,…,ω~M+1,ν~1,ν~2,…,ν~M^+1)T=J3(δ),J3(δ)=((ρ^1ρ^1+δ)m^1(ρ1ρ1-δ)m1,…,ρ1ρ1-δ,…,(ρJρJ-δ)mJ,…,ρJρJ-δ,(ρ^1ρ^1+δ)m^1eδ(a-b),…,ρ^J^ρ^J^+δeδ(a-b))T.Corollary 5.
For anyθ>0, u,b∈R, one has the following.
(1) Ifμ>0, then
(32)Eu[e-θτbf(Xτb)]={f(u),u≥b;∑l=1M+1‍ωleγl(u-b),u<b.
Here ω=(ω1,ω2,…,ωM+1)T is a vector uniquely determined by the following linear system:
(33)D1ω=f¯1,
where f¯1=(f(b),f11u,…,f1m1u,…,fJ1u,…,fJmJu) is an (M+1)-dimensional vector and D1 is an (M+1)×(M+1) nonsingular matrix:
(34)D1=(11⋯1ρ1ρ1-γ1ρ1ρ1-γ2⋯ρ1ρ1-γM+1⋮⋮⋮⋮(ρ1ρ1-γ1)m1(ρ1ρ1-γ2)m1⋯(ρ1ρ1-γM+1)m1⋮⋮⋮⋮ρJρJ-γ1ρJρJ-γ2⋯ρJρJ-γM+1⋮⋮⋮⋮(ρJρJ-γ1)mJ(ρJρJ-γ2)mJ⋯(ρJρJ-γM+1)mJ)
(2) Ifμ≤0, then
(35)Eu[e-θτbf(Xτb)]={f(u),u≥b;∑l=1M‍ω-leγl(u-b),u<b.
Here ω¯=(ω¯1,ω¯2,…,ω¯M)T is a vector uniquely determined by the following linear system:
(36)D2ω¯=f¯2,
where f¯2=(f11u,…,f1m1u,…,fJ1u,…,fJmJu) is an M-dimensional vector and D2 is an M×M nonsingular matrix:
(37)D2=(ρ1ρ1-γ1ρ1ρ1-γ2⋯ρ1ρ1-γM+1⋮⋮⋮⋮(ρ1ρ1-γ1)m1(ρ1ρ1-γ2)m1⋯(ρ1ρ1-γM+1)m1⋮⋮⋮⋮ρJρJ-γ1ρJρJ-γ2⋯ρJρJ-γM+1⋮⋮⋮⋮(ρJρJ-γ1)mJ(ρJρJ-γ2)mJ⋯(ρJρJ-γM+1)mJ).Corollary 6.
For anyθ>0, u,a∈R, one has the following.
(1) Ifμ>0, then
(38)Eu[e-θτaf(Xτa)]={f(u),u≤a;∑l=1M^‍ωleγl(u-a),u>a.
Here ν=(ν1,ν2,…,νM^)T is a vector uniquely determined by the following linear system:
(39)C1ν=f~1,
where f~1=(f11d,…,f1m^1d,…,fJ^1d,…,fJ^m^J^d) is an M^-dimensional vector and C1 is an (M^)×(M^) nonsingular matrix:
(40)C1=(ρ^1ρ^1+γ^1ρ^1ρ^1+γ^2⋯ρ^1ρ^1+γ^M^⋮⋮⋮⋮(ρ^1ρ^1+γ^1)m^1(ρ^1ρ^1+γ^2)m^1⋯(ρ^1ρ^1+γ^M^)m^1⋮⋮⋮⋮ρ^J^ρ^J^+γ^1ρ^J^ρ^J^+γ^2⋯ρ^J^ρ^J^+γ^M^⋮⋮⋮⋮(ρ^J^ρ^J^-γ^1)m^J^(ρ^J^ρ^J^+γ^2)m^J^⋯(ρ^J^ρ^J^+γ^M^)m^J^).
(2) Ifμ≤0, then
(41)Eu[e-θτaf(Xτa)]={f(u),u≤a;∑l=1M^+1‍ω-lev-l(u-a),u<a.
Here ν¯=(ν¯1,ν¯2,…,ν¯M^+1)T is a vector uniquely determined by the following linear system:
(42)C2ν¯=f~2,
where f~2=(f(a),f11d,…,f1m^1d,…,fJ^1d,…,fJ^m^J^d) is an M^+1-dimensional vector and C2 is an (M^+1)×(M^+1) nonsingular matrix:
(43)C2=(11⋯1ρ^1ρ^1+γ^1ρ^1ρ^1+γ^2⋯ρ^1ρ^1+γ^M^+1⋮⋮⋮⋮(ρ^1ρ^1+γ^1)m^1(ρ^1ρ^1+γ^2)m^1⋯(ρ^1ρ^1+γ^M^+1)m^1⋮⋮⋮⋮ρ^J^ρ^J^+γ^1ρ^J^ρ^J^+γ^2⋯ρ^J^ρ^J^+γ^M^+1⋮⋮⋮⋮(ρ^J^ρ^J^-γ^1)m^J^(ρ^J^ρ^J^+γ^2)m^J^⋯(ρ^J^ρ^J^+γ^M^+1)m^J^).Example 7.
WhenJ=m1=J^=m^1=1,  p(x)=p11ρ1e-ρ1xI(x>0)+p^11ρ^1eρ^1xI(x<0), and p11+p^11=1, μ>0, then the equation G(z)=θ has 3 real roots: γ1,  γ2, and γ^1 (-∞<γ^1<0<γ1<γ2<∞). Let
(44)A1=(11eγ^1(b-a)ρ1ρ1-γ1ρ1ρ1-γ2ρ1eγ^1(b-a)ρ1-γ^1ρ^1eγ1(a-b)ρ^1+γ1ρ^1eγ2(a-b)ρ^1+γ2ρ^1ρ^1+γ^1).
Denote A1-1 by
(45)A1-1=1|A1|(a11-a21a31-a12a22-a32a13-a23a33).
Then we have
(46)Eu[e-θτf(Xτ)]=ω1eγ1(u-b)+ω2eγ2(u-b)+ω3eγ^1(u-a),
where
(47)ω1=a11f(b)-a21f11u+a31f11d|A1|,ω2=-a12f(b)+a22f11u-a32f11d|A1|,ω3=a13f(b)-a23f11u+a33f11d|A1|,f11u=∫0∞‍f(y+b)ρ1e-ρ1ydy,f11d=∫-∞0‍f(y+a)ρ^1eρ^1ydy,|A1|=ρ1ρ^1(γ2-γ1)(ρ1-γ1)(ρ1-γ2)(ρ^1+γ^1)+ρ1ρ^1(γ1-γ^1)e(γ2-γ^1)(a-b)(ρ1-γ1)(ρ1-γ^1)(ρ^1+γ2)+ρ1ρ^1(γ1-γ^2)e(γ1-γ^1)(a-b)(ρ1-γ2)((ρ1-γ^1))(ρ^1+γ1),a11=ρ1ρ^1(ρ1-γ2)(ρ^1+γ^1)-ρ1ρ^1e(γ2-γ^1)(a-b)(ρ1-γ^1)(ρ^1+γ2),a21=ρ^1ρ^1+γ^1-ρ^1e(γ2-γ^1)(a-b)ρ^1+γ2,a13=ρ1ρ^1eγ2(a-b)(ρ1-γ1)(ρ^1+γ2)-ρ1ρ^1eγ1(a-b)(ρ1-γ2)(ρ^1+γ1),a23=ρ1eγ2(a-b)ρ^1+γ2-ρ^1eγ1(a-b)ρ^1+γ1,a31=ρ(γ^1-γ2)eγ^(b-a)(ρ1-γ^1)(ρ^1-γ2),a33=ρ(γ1-γ2)(ρ1-γ1)(ρ^1-γ2).
We define a12(a22,a32) as follows: let a12(a22,a32) be obtained from a11(a21,a31) by changing γ2 to γ1 in a11(a21,a31).(i)
Iff(Xτ)=eδXτ, then we have
(48)Eu[e-θτ+δXτ]=eδb(ω1′eγ1(u-b)+ω2′eγ2(u-b)+ω3′eγ^1(u-a)),a<u<b,
where
(49)f(b)=eδb,f11u=eδbρ1ρ1-δ,f11d=eδaρ^1ρ^1+δ,ω1′=a11-a21(ρ1/(ρ1-δ))+a31(ρ^1eδ(a-b)/(ρ^1+δ))|A1|,ω2′=-a12+a22(ρ1/(ρ1-δ))-a32(ρ^1eδ(a-b)/(ρ^1+δ))|A1|,ω3′=a13-a23(ρ1/(ρ1-δ))+a33(ρ^1eδ(a-b)/(ρ^1+δ))|A1|.(ii)
Iff(Xτ)=I(Xτ-b≥0), then we have
(50)Eu[e-θτI(Xτ-b≥0)]=ω^1eγ1(u-b)+ω^2eγ2(u-b)+ω^3eγ^1(u-a),a<u<b,
where
(51)f(b)=1,f11u=1,f11d=0,ω^1=a11-a21|A1|,ω^2=-a12+a22|A1|,ω^3=a13-a23|A1|.(iii)
Iff(Xτ)=I(Xτ-a≤0), then we have
(52)Eu[e-θτI(Xτ-a≤0)]=ω^^1eγ1(u-b)+ω^^2eγ2(u-b)+ω^3eγ^1(u-a),a<u<b,
where
(53)f(b)=0,f11u=0,f11d=1,ω^^1=a31|A1|,ω^^2=-a32|A1|,ω^^3=a33|A1|.(iv)
Iff(Xτ)=I(Xτ-b>y), y>0, then we have
(54)Eu[e-θτI(Xτ-b>y)]=ωˇˇ1eγ1(u-b)+ωˇˇ2eγ2(u-b)+ωˇˇ3eγ^1(u-a),a<u<b,
where
(55)f(b)=0,f11u=e-ρ1y,f11d=0,ωˇ1=-a21e-ρ1y|A1|,ωˇ2=a22e-ρ1y|A1|,ωˇ3=-a23e-ρ1y|A1|.(v)
Iff(Xτ)=I(Xτ-a<-y), y>0, then we have
(56)Eu[e-θτI(Xτ-a<-y)]=ωˇˇ1eγ1(u-b)+ωˇˇ2eγ2(u-b)+ωˇˇ3eγ^1(u-a),a<u<b,
where
(57)f(b)=0,f11u=0,f11d=e-ρ^1y,ωˇˇ1=-a31e-ρ^1y|A1|,ωˇˇ2=a32e-ρ^1y|A1|,ωˇˇ3=-a33e-ρ^1y|A1|.(vi)
Iff(Xτ)=I(Xτ=b), then we have
(58)Eu[e-θτI(Xτ=b)]=ω~1eγ1(u-b)+ω~2eγ2(u-b)+ω~3eγ^1(u-a),a<u<b,
where
(59)f(b)=1,f11u=0,f11d=0,ω~1=a11|A1|,ω~2=-a12|A1|,ω~3=a13|A1|.


## 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 Section3.For barrier strategy, it is assumed that dividends are paid according to a barrier strategyb. Such a strategy has a level of the barrier b>0; when the surplus exceeds the barrier, the excess is paid out immediately as the dividend. The modified surplus at time t is given by
(60)U(t)=X(t)-D(t),t≥0,
where D(t) denote the aggregate dividends paid between time 0 and time t; that is,
(61)D(t)=(max0≤s≤tX(s)-b)+.Let(62)T=inf{t≥0:U(t)<0}
be the time of ruin for the modified surplus U and
(63)D=∫0T‍e-θtdD(t)
the present value of all dividends until ruin, where θ>0 is the discount factor.Denote byV(u;b) the expectation of the discounted dividends until ruin, if the barrier strategy with parameter b is applied; that is,
(64)V(u;b)=E[D∣X(0)=u],0≤u≤b.
Note that we have
(65)V(u;b)=u-b+V(b;b),u>b.Theorem 8.
Letτb+=inf{t≥0:X(t)≥b} and τ=T∧τb+. Then for θ>0, one has the following.(1) for0<u<b,
(66)V(u;b)=∑j=1J‍∑i=1mj‍mj+1-iηjEu[e-θτb+1Kji]+V(b,b)(∑j=1J‍∑i=1mj‍Eu[e-θτb+1K0]22222222222+∑j=1J‍∑i=1mj‍Eu[e-θτb+1Kji]);(2) ifμ>0, then
(67)V(b;b)=∑j=1J∑i=1mj((mj+1-i)/ηj)(Eu[e-θτb+1Kji])′|u=b(Eu[e-θτb+1K0]+∑j=1J∑i=1mjEu[e-θτb+1Kji])′|u=b;(3) ifμ≤0, then
(68)V(b;b)=∑j=1J∑i=1mj((mj+1-i)/ηj)Eb[e-θτb+1Kji]1-∑j=1J∑i=1mjEu[e-θτb+1Kji].Proof.
It follows from the conditional memoryless property, the conditional independence, and the strong Markov property that, for0≤u≤b, we have
(69)V(u;b)=Eu[∫0τ‍e-θtdD(t)]=Eu[∫0τ‍e-θtdD(t)1{τb+<T}]=Eu[(Xτ-b+V(b;b))e-θτb+1{τb+<T}1{Xτ>b}]+Eu[∫τb+T‍e-θtdD(t)1{τb+<T}1{Xτ=b}]=∑j=1J‍∑i=1mj‍Eu[(Xτb+-b)e-θτb+1Kji]+V(b;b)∑j=1J‍∑i=1mj‍Eu[e-θτb+1Kji]+Eu[(∫0T‍e-θtdD(t)·θτb+∣ℱτb+X)e-θτb+1{τb+<T}1{Xτ=b}E×(∫0T‍e-θtdD(t)·θτb+∣ℱτb+X)]=∑j=1J‍∑i=1mj‍mj+1-iηjEu[e-θτb+1Kji]+V(b;b)∑j=1J‍∑i=1mj‍Eu[e-θτb+1Kji]+Eu[(∫0T‍e-θtdD(t)·θτb+)e-θτb+1{τb+<τu}1{Xτ=b}EXτb+×(∫0T‍e-θtdD(t)·θτb+)]=∑j=1J‍∑i=1mj‍mj+1-iηiEu[e-θτb+1Kji]+V(b,b)(∑j=1J‍∑i=1mj‍Eu[e-θτb+1K0]222222222222+∑j=1J‍∑i=1mj‍Eu[e-θτb+1Kji]).
Assertion (2) follows from (1) and V′(b;b)=1, and assertion (3) follows from (1) and P(K0)=0. This ends the proof.For threshold strategy, we assume that dividends are paid according to the following strategy governed by parametersb>0 and α>0. Whenever the modified surplus is below the threshold level b, no dividends are paid. However, when the surplus is above this threshold level, dividends are paid at a constant rate α that does not exceed the premium rate μ>0. We define the modified risk process Ub={Ub(t):t≥0} by Ub(t)=X(t)-Db(t), where
(70)Db(t)=α∫0t‍1(Ub(s)>b)ds.
Let Db denote the present value of all dividends until the time of ruin T~2,
(71)Db=α∫0T~2‍e-θtdDb(t),
where T~2=inf{t>0:Ub(t)<0} with T~2=∞ if Ub(t)=0 for all t≥0. Here θ>0 is the discount factor. Denote by V(u,b) the expected discounted value of dividend payments; that is, V(u,b)=E[Db∣Ub(0)=u].Similar to the proof of Theorem8, we have the following result.Theorem 9.
Letτb-=inf{t≥0:X(t)≤b}, τb+=inf{t≥0:X(t)≥b}, and τ=T∧τb+. Then, for θ>0, one has the following.
(1) foru>b,
(72)V(u;b)=αθ-αθEu[e-θτb-1F0]-αθ∑j=1J^‍∑i=1m^j‍Eu[e-θτb-1Fji]+V(b,b)Eu[e-θτb-1F0]+∑j=1J^‍∑i=1m^j‍Eu[e-θτb-1Fji]fji;
(2) forb>u>0, if μ>0, then
(73)V(u;b)=α∑j=1J‍∑i=1mj‍mj+1-iηjEu[e-θτb+1Kji]+V(b,b)Eu[e-θτb+1K0]+V(b,b)∑j=1J‍∑i=1mj‍Eu[e-θτb+1Kji];
(3) forb>u>0, if μ≤0, then
(74)V(u;b)=α∑j=1J‍∑i=1mj‍mj+1-iηjEu[e-θτb+1Kji]+V(b,b)∑j=1J‍∑i=1mj‍Eu[e-θτb+1Kji],
where F0={ω:Ub(τb-)=b},  Fji={ω:Ub(τb-)<b,-YNτb-~Erlang(ρ^j,i)},  j=1,2,…,J^,  i=1,2,…,m^j,  and  fji=∫-∞0‍V(y+b;b)((-ρ^jy)i-1ρ^jeρ^jy/(i-1)!)dy.Remark 10.
Ifmj=1,  j=1,2,…,J,  m^j=1, and  j=1,2,…,J^, that is, the process X has the hyperexponential downward jumps and upward jumps, then we have the known results (Theorems 3.2 and 3.4, of Yin et al. [10]).



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*Source: 747262-2014-02-02.xml*