# On the Expected Discounted Penalty Function for

Authors: Huiming Zhu; Ya Huang; Xiangqun Yang; Jieming Zhou
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

We focus on the expected discounted penalty function of a compound Poisson risk model with random incomes and potentially delayed claims. It is assumed that each main claim will produce a byclaim
with a certain probability and the occurrence of the byclaim may be delayed depending on associated main claim amount. In addition, the premium number process is assumed as a Poisson process. We derive the integral equation satisfied by the expected discounted penalty function. Given that the premium size is exponentially distributed, the explicit expression for the Laplace transform of the expected discounted penalty function is derived. Finally, for the exponential claim sizes, we present the explicit formula for the expected discounted penalty function.


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## Body

## 1. Introduction

In the classical risk theory, assumption of independence among claims is an important condition to the study of risk models. However, in many practical situations, the assumption is often inconsistent with the operation of insurance companies. In reality, claims may be time-correlated for various reasons, and it is important to study risk model which is able to depict this phenomenon. Since the work by Waters and Papatriandafylou [1], many researchers have studied various kinds of dependencies among claim amounts and claim numbers, such as Gerber [2], Shiu [3], Dickson [4], Willmot [5], and Ambagaspitiya [6, 7]. Among others, in the case of the compound binomial model, Yuen and Guo [8] consider a specific dependence structure between the claim sizes and interclaim times. Under their assumption, each claim causes a byclaim but the occurrence of the byclaim may be delayed. Further, based on the same model, Xiao and Guo [9] investigate the joint distribution of the surplus immediately prior to ruin and the deficit at ruin.Note that the risk model referred above is based on the assumption that the probability of delay of each byclaim is constant and independent of claim amounts. Albrecher and Boxma [10] consider a generalization of the classical risk model to a dependent setting where the distribution of the time between two claim occurrences depends on the previous claim size. Motivated by the idea, Zou and Xie [11] introduce a risk model with an interesting dependence structure between the amount of main claim and the occurrence of byclaim. It is a natural extension for the delayed claims model due to the fact that the bigger the claim amount for main claim (such as car damage) is, the greater odds of the byclaim (such as injury) would be delayed in the actual practice with insurer. Based on the structure, we consider an improved payment mode named potentially delayed claims where the main claim induces a byclaim with a certain probability. The improvement is inspired from a series of examples similar to the case referred above, in which main claim does not induce byclaim with probability 1.Because the insurance company may have lump sums of income, we apply potentially delayed claims to the compound Poisson risk model in the presence of random incomes. Since Boucherie et al. [12] described the random incomes by adding a compound Poisson process with positive jumps to the classical risk model, many authors have studied similar topics. Boikov [13] studies ruin problem of a risk model with stochastic premium process. Bao [14] considers a risk model, in which the premium is a Poisson process instead of a linear function of time. Labbé et al. [15] consider a risk model where the stochastic incomes follow a compound Poisson process and research the case when the premiums have Erlang distributions in more depth. Hao and Yang [16] analyze the expected discounted penalty function of a compound Poisson risk model with random incomes and delayed claims. Yu [17] also studies the expected discounted penalty function in a Markov regime-switching risk model with random income.In this paper, we aim at the expected discounted penalty function of an extensive risk model with random incomes and potentially delayed claims. This paper generalizes the model of Hao and Yang [16]. Based on the extensive model, we obtain explicit expression of the expected discounted penalty function, while [16] derives defective renewal equations of it only. When the main claim induces a byclaim with probability 1 and the byclaim is delayed with a constant probability, the results in this paper will reduce to them in [16]. So [16] can be seen as a special case of this paper. In addition, Zou and Xie [11] derive the probability of ruin in the risk model with delayed claims, but this paper obtains the expected discounted penalty function which contains the probability of ruin. If we define the expected discounted penalty function with the same expression as ruin probability and assume that the premium is a linear function of time, we can get the same results as [11].The rest of this paper is structured as follows. In Section2, we introduce the compound Poisson risk model with random incomes and potentially delayed claims. In Section 3, we derive an integral equation for the expected discounted penalty function and obtain explicit expression of its Laplace transform when the premium income is exponentially distributed. The defective renewal equation satisfied by the expected discounted penalty function is studied in Section 4. Section 5 obtains explicit result for the expected discounted penalty function with positive initial surplus when the claim amounts from both classes are exponentially distributed. Section 6 concludes the paper.


## 2. Model

Now, we can show the extensive risk model with random incomes and potentially delayed claims in mathematics. On the one hand, we denote the aggregate premium incomes at timet by SX(t) which is a compound Poisson process, and {N1(t):t≥0} is the corresponding Poisson income number process with parameter λ1. The premium incomes amounts {Xi}i≥1 are assumed to be independent and identically distributed (i.i.d.) positive random variables with common distribution FX, probability function fX, and mean μX. So we get SX(t)=∑j=1N1(t)‍Xj. On the other hand, we consider a continuous time model which involves two types of insurance claims, namely, the main claims and the byclaims. Let the aggregate main claims process be a compound Poisson process and let {N2(t):t≥0} be the corresponding Poisson claim number process with parameter λ2. Its jump times are denoted by {Vi}i≥1 with V0=0. The main claim amounts {Yi}i≥1 and the byclaim amounts {Zi}i≥1 are assumed to be independent and identically distributed (i.i.d.) positive random variables with common distribution FY and FZ, respectively. Moreover, they are independent and their means are denoted by μY and μZ. Then the surplus process of the risk model is defined as
(1)U(t)=u+∑j=1N1(t)‍Xj-∑i=1N2(t)‍Yi-R(t),
where U(0)=u is the initial capital and R(t) is the sum of all byclaims Zi that occurred before time t. We assume that N1(t) and N2(t) are mutually independent.With the assumption of potentially delayed claims, the claim occurrence process is to be of the following type: there will be a main claimYi at every epoch Vi of the Poisson process and the main claim Yi will induce a byclaim Zi with probability q. If the main claim amount Yi induces a byclaim Zi and the main claim amount Yi is less than a threshold Mi, the byclaim Zi and its associated main claim Yi occur simultaneously; otherwise, the occurrence of the byclaim Zi is delayed to Vi+1 and main claim Yi+1 occurs simultaneously. From Zou and Xie [18], we know that
(2)E[∑j=1N1(t)‍Xj-∑i=1N2(t)‍Yi-R(t)]=λ1tμX-[(1-e-λ2t)λ2tμY+λ2tqμZmmmmmmmmm-qP(Y1≥M1)μZ(1-e-λ2t)].Therefore, we further assume that(3)λ1μX>λ2(μY+qμZ).
This assumption ensures that the safety loading is positive.LetT≜inf⁡{t≥0:U(t)<0} be the time of ruin with T=∞ if U(t)≥0 for all t≥0. The ruin probability is defined by Ψ(u)≜P(T<∞∣U(0)=u), u≥0. The expected discounted penalty function is of the following form:
(4)ϕ(u)≜E[e-δTω(U(T-),|U(T)|)I(T<∞)∣U(0)=u],mmmmmmmmmmmmmmmmmmmmmmmmnu≥0,
where δ≥0 is a constant and ω(x1,x2) is a nonnegative measurable function defined on [0,∞)×[0,∞). I(A) is the indicator function of event A. |U(T)| is the deficit at ruin and U(T-) is the surplus immediately prior to ruin.


## 3. The Expected Discounted Penalty Function of the Exponential Premium Income

To handle the surplus process (1), we consider a slight change in the risk model. Instead of having one main claim Y1 and a byclaim Z1 with probability P(Y1<M1) at the first epoch V1, another byclaim Z is added at the first epoch V1; that is, byclaim Z and main claim Y1 occur at V1 simultaneously. Hence, the corresponding surplus process U1(t) of this auxiliary risk model is defined as
(5)U1(t)=u+∑j=1N1(t)‍Xj-∑i=1N2(t)‍Yi-R(t)-Z,
where Z denotes the other byclaim amount, U1(0)=u. Assume that Z and {Zi}i≥1 are i.i.d. positive random variables.The expected discounted penalty function for this auxiliary risk model is denoted byϕ1(u) which is useful to derive ϕ(u).Obviously there will be a main claimY1 at the first epoch V1. Let W1 be the time for the first premium. The first claim can be or cannot be earlier than the first premium. If it is, there are three situations.(1)
The main claim does not induce a byclaim; then the surplus process gets renewed except for the initial value. The probability of this event is1-q.(2)
The main claim induces a byclaimZ1 and the main claim size Y1<M1; then the byclaim Z1 also occurs at the first epoch V1; the surplus process U(t) will renew itself with different initial reserve. The probability of this event is qP(Y1<M1).(3)
The main claim induces a byclaimZ1 and the main claim size Y1≥M1; then the occurrence of the byclaim Z1 will be delayed to V2; that is, the delayed byclaim Z1 and the main claim Y2 occur simultaneously. In this case, U(t) will not renew itself but transfer to the auxiliary model described in the paragraph above. The probability of this event is qP(Y1≥M1).Conditioning on the time of the first event, we have(6)ϕ(u)=∫0∞‍λ1e-(λ1+λ2+δ)tdt∫0∞‍ϕ(u+x)dFX(x)+∫0∞‍λ2e-(λ1+λ2+δ)tdt(1-q)×[∫0u‍ϕ(u-y)dFY(y)+∫u∞‍ω(u,y-u)dFY(y)]+∫0∞‍λ2e-(λ1+λ2+δ)tdtq×[∬0<y+z<u(1-FM(y))ϕ(u-y-z)dFY(y)mmmmm×dFZ(z)mmmmm+∬y+z>u(1-FM(y))mmmmmmmimmm×∬0<y+z<uω(u,y+z-u)dFY(y)dFZ(z)]+∫0∞‍λ2e-(λ1+λ2+δ)tdtq×[∫0u‍FM(y)ϕ1(u-y)dFY(y)mm+∫u∞‍FM(y)ω(u,y-u)dFY(y)]=λ1λ1+λ2+δA(u)+λ2(1-q)λ1+λ2+δ×[∫0u‍ϕ(u-y)dFY(y)+ω1(u)]+λ2qλ1+λ2+δ×[∬0<y+z<u(1-FM(y))ϕ(u-y-z)mmmmmmm×dFY(y)dFZ(z)+ω2(u)∬0<y+z<u]+λ2qλ1+λ2+δ[∫0u‍FM(y)ϕ1(u-y)mmmmmmmmm×dFY(y)+ω3(u)∫0u‍],
where
(7)A(u)=∫0∞‍ϕ(u+x)dFX(x),ω1(u)=∫u∞‍ω(u,y-u)dFY(y),ω2(u)=∬y+z>u(1-FM(y))×ω(u,y+z-u)dFY(y)dFZ(z),ω3(u)=∫u∞‍FM(y)ω(u,y-u)dFY(y).
Similarly, for the expected discounted penalty function ϕ1(u) of the auxiliary risk model, we have
(8)ϕ1(u)=∫0∞‍λ1e-(λ1+λ2+δ)tdt∫0∞‍ϕ1(u+x)dFX(x)+∫0∞‍λ2e-(λ1+λ2+δ)tdt(1-q)×[∫0u‍ϕ(u-y)dFY*FZ(y)+∫u∞‍ω(u,y-u)dFY*FZ(y)]+∫0∞‍λ2e-(λ1+λ2+δ)tdtq×[∬0<y+z<u(1-FM(y))ϕ(u-y-z)mmmmmmm×dFY(y)dFZ*FZ(z)m+∬y+z>u(1-FM(y))m∬0<y+z<u∫‍×ω(u,y+z-u)dFY(y)dFZ*FZ(z)]+∫0∞‍λ2e-(λ1+λ2+δ)tdtq×[∬0<y+z<uFM(y)ϕ1(u-y-z)mmmm×dFY(y)dFZ(z)mmmm+∬y+z>uFM(y)mmmmmmmmm∬0<y+z<u∫‍×ω(u,y+z-u)dFY(y)dFZ(z)]=λ1λ1+λ2+δA1(u)+λ2(1-q)λ1+λ2+δ×[∫0u‍ϕ(u-y)dFY*FZ(y)+ω4(u)]+λ2qλ1+λ2+δ×[∬0<y+z<u(1-FM(y))ϕ(u-y-z)∬0<y+z<u∫‍×dFY(y)dFZ*FZ(z)+ω5(u)∬0<y+z<u]+λ2qλ1+λ2+δ[∬0<y+z<ummmmmmmmmmm×FM(y)ϕ1(u-y-z)mmmmmmmmmmm×∬0<y+z<udFY(y)dFZ(z)+ω6(u)],
where
(9)A1(u)=∫0∞‍ϕ1(u+x)dFX(x),ω4(u)=∫u∞‍ω(u,y-u)dFY*FZ(y),ω5(u)=∬y+z>u(1-FM(y))×ω(u,y+z-u)dFY(y)dFZ*FZ(z),ω6(u)=∬y+z>uFM(y)×ω(u,y+z-u)dFY(y)dFZ(z).In the following, we will give the Laplace transforms of theϕ(u) and ϕ1(u).Letχ1(y)=FM(y)FY′(y) and χ2(y)=(1-FM(y))FY′(y). For Re(s)≥0, we define
(10)χ^1(s)≜∫0∞‍e-syχ1(y)dy=E[exp⁡(-sY)I(Y≥M)]=∫0∞‍e-syFM(y)dFY(y),χ^2(s)≜∫0∞‍e-syχ2(y)dy=E[exp⁡(-sY)I(Y<M)]=∫0∞‍e-sy(1-FM(y))dFY(y),b^1(s)≜∫0∞‍e-sydFY(y),b^2(s)≜∫0∞‍e-sydFZ(y),b^3(s)≜∫0∞‍e-sydFY*FZ(y),b^4(s)≜∫0∞‍e-sydFZ*FZ(y).Taking the Laplace transforms of (6) and (8), we obtain
(11)ϕ^(s)=λ1λ1+λ2+δA^(s)+λ2(1-q)λ1+λ2+δ×[ϕ^(s)b^1(s)+ω^1(s)]+λ2qλ1+λ2+δ×[ϕ^(s)χ^2(s)b^2(s)+ϕ^1(s)χ^1(s)+ω^2(s)+ω^3(s)],(12)ϕ^1(s)=λ1λ1+λ2+δA^1(s)+λ2(1-q)λ1+λ2+δ×[ϕ^(s)b^3(s)+ω^4(s)]+λ2qλ1+λ2+δ×[ϕ^(s)χ^2(s)b^4(s)mmm+ϕ^1(s)χ^1(s)b^2(s)+ω^5(s)+ω^6(s)].Now we introduce the Dickson-Hipp operatorTr studied in Dickson and Hipp [19]. Define
(13)Trh(x)=∫x∞‍e-r(y-x)h(y)dy,
where h(x) is a real-valued function and r is a complex number. As in Li and Garrido [20], we find Trh(0)=h^(r). For distinct r and s,
(14)TrTsh(x)=TsTrh(x)=Trh(x)-Tsh(x)s-r.
If r=s,
(15)TrTsh(x)=∫x∞‍(y-x)e-r(y-x)h(y)dy.Suppose that the premium incomesXj are exponentially distributed; that is,
(16)FX(x)=1-e-x/μX.According to the definition and properties of the Dickson-Hipp operator, we take the Laplace transform ofA(u) and A1(u); then
(17)A^(s)=ϕ^(s)-ϕ^(1/μX)1-sμX,(18)A^1(s)=ϕ^1(s)-ϕ^1(1/μX)1-sμX.Plugging (17) and (18) into (11) and (12), respectively, and then making some simplifications, we obtain
(19)ϕ^(s)=((1-sμX)λ1+λ2+δmmm×[χ3(s)ω^(s)+λ2q(1-sμX)χ^1(s)λ1+λ2+δω^*(s)]mmm-l1(s)[χ3(s)ω^(s)+λ2q(1-sμX)χ^1(s)λ1+λ2+δω^*(s)])×((1-sμX-λ1λ1+λ2+δ)2mmnm-(1-sμX)λ2b^1(s)(1-q+qb^2(s))λ1+λ2+δmmmm×(1-sμX-λ1λ1+λ2+δ)2(1-sμX-λ1λ1+λ2+δ))-1,(20)ϕ^1(s)=((1-sμX)λ1+λ2+δmmmm×[χ4(s)ω^*(s)mmmnm+λ2(1-sμX)((1-q)b^3(s)+qχ^2(s)b^4(s))λ1+λ2+δmmnmm×ω^(s)χ4]-l2(s)(1-sμX)λ1+λ2+δ)mmmm×((1-sμX-λ1λ1+λ2+δ)2-(1-sμX)mmmmmmm×λ2b^1(s)(1-q+qb^2(s))λ1+λ2+δmmmmmmm×(1-sμX-λ1λ1+λ2+δ)2(1-sμX-λ1λ1+λ2+δ))-1,
where
(21)ω^(s)=λ2[(1-q)ω^1(s)+q(ω^2(s)+ω^3(s))],ω^*(s)=λ2[(1-q)ω^4(s)+q(ω^5(s)+ω^6(s))],χ3(s)=1-sμX-λ1λ1+λ2+δ-(1-sμX)λ2qχ^1(s)b^2(s)λ1+λ2+δ,χ4(s)=1-sμX-λ1λ1+λ2+δ-(1-sμX)×λ2[(1-q)b^1(s)+qχ^2(s)b^2(s)]λ1+λ2+δ,l1(s)=λ1χ3(s)λ1+λ2+δϕ^(1μX)+λ1λ2q(1-sμX)χ^1(s)(λ1+λ2+δ)2ϕ^1(1μX),l2(s)=λ1χ4(s)λ1+λ2+δϕ^1(1μX)+λ1λ2(1-sμX)[(1-q)b^3(s)+qχ^2(s)b^4(s)](λ1+λ2+δ)2×ϕ^(1μX).To solveϕ^(s) and ϕ^1(s) in (19) and (20), we need to find ϕ^(1/μX) and ϕ^1(1/μX). Here we will first consider the zeros of the denominators of (19) and (20) or equally the zeros of the following equation:
(22)(1-sμX-λ1λ1+λ2+δ)mmmm×[1λ2b^1(s)(1-q+qb^2(s))λ1+λ2+δ-sμX-λ1λ1+λ2+δmmmmnm-λ2b^1(s)(1-q+qb^2(s))λ1+λ2+δ(1-sμX)]=0.Lemma 1.
Forδ>0, the denominators of (19) and (20) have exactly two distinct positive real roots, say, ρ1(δ) and ρ2(δ)=(λ2+δ)/(λ1+λ2+δ)μX. Further, ρ1(δ) and ρ2(δ) are the only roots on the right half of the complex plane.Proof.
To prove Lemma1, it is equal to show that (22) has exactly two roots in the right half complex plan. Firstly, (22) can be rewritten as
(23)(1-sμX-λ1λ1+λ2+δ)l(s)=0,
where
(24)l(s)=(1-sμX)×[1-λ2b^1(s)(1-q+qb^2(s))λ1+λ2+δ]-λ1λ1+λ2+δ.
Forδ>0, s≥0, it is easy to check that l(0)=δ/(λ1+λ2+δ)>0 and lim⁡s→+∞l(s)=-∞. And
(25)l′(s)=-μX[1-λ2b^1(s)(1-q+qb^2(s))λ1+λ2+δ]-(1-sμX)×λ2[b^1′(s)(1-q+qb^2(s))+b^1(s)qb^2′(s)]λ1+λ2+δ<-μX[1-λ2b^1(s)(1-q+qb^2(s))λ1+λ2+δ]-λ2[b^1′(s)(1-q+qb^2(s))+b^1(s)qb^2′(s)]λ1+λ2+δ<-μX[1-λ2b^1(s)(1-q+qb^2(s))λ1+λ2+δ]+λ2[μY(1-q+qb^2(s))+b^1(s)qμZ]λ1+λ2+δ<-μXλ1+δλ1+λ2+δ+λ2(μY+qμZ)λ1+λ2+δ<λ2(μY+qμZ)-λ1μXλ1+λ2+δ<0,
which implies l(s) is a strictly decreasing function of s. So l(s)=0 has exactly one positive real root, say, ρ1(δ). Obviously, ρ1(δ) is also one positive real root of (22). Note that (λ2+δ)/(λ1+λ2+δ)μX is another positive real root of (22), say, ρ2(δ) and l((λ2+δ)/(λ1+λ2+δ)μX)≠0. That means ρ1(δ)≠ρ2(δ), so we conclude that (22) has exactly two distinct positive real roots, say ρ1(δ) and ρ2(δ).
Now, we prove thatρ1(δ) is the exactly one positive real root of equation l(s)=0 on the right half of the complex plane. When s is on the half-circle, |z|=r(r>0) and Re(z)≥0 on the complex plane, for r sufficiently large,
(26)|1-sμX-λ1λ1+λ2+δ|>1-λ1λ1+λ2+δ>λ2λ1+λ2+δ>λ2|b^1(s)(1-q+qb^2(s))(1-sμX)|λ1+λ2+δ,
while for s on the imaginary axis, Re(z)=0, the last inequality is true as well. That is to say, on the boundary of the contour enclosed by the half-circle and the imaginary axis,
(27)|1-sμX-λ1λ1+λ2+δ|>λ2|b^1(s)[1-q+qb^2(s)](1-sμX)|λ1+λ2+δ.
Then we conclude, by Rouché’s theorem, that on the right half of the complex plane, the number of roots of the equation l(s)=0 equals the number of roots of the equation 1-sμX-λ1/(λ1+λ2+δ)=0. Furthermore, the latter has exactly one root on the right half of the complex plane. It follows that l(s)=0 has exactly one positive real root, say, ρ1(δ), on the right half of the complex plane. It is easy to see that ρ2(δ)=(λ2+δ)/(λ1+λ2+δ)μX is the exactly one positive real root of equation 1-sμX-λ1/(λ1+λ2+δ)=0 on the right half of the complex plane. It follows from all of the above that (22) has exactly two distinct positive real roots ρ1(δ) and ρ2(δ) on the right half of the complex plane. Hence, the lemma is proved.Remark 2.
From Klimenok [21], we know that lim⁡δ→0+ρ1(δ)=0. Thereafter, we denote them by ρ1, ρ2 for simplicity.
Sinceϕ^(s) is finite for all s with Re(s)>0, we know ρi, i=1,2, should also be zeros of the numerator in (19); that is,
(28)(1-ρiμX)λ1+λ2+δ×[χ3(ρi)ω^(ρi)+λ2q(1-ρiμX)χ^1(ρi)λ1+λ2+δω^*(ρi)]=l1(ρi),i=1,2.
By solving these linear equations, we get(29)ϕ^(1μX)=(1-ρ1μX)(1-ρ2μX)m×(λ1[(1-ρ1μX)χ^1(ρ1)χ3(ρ2)mmmmm-(1-ρ2μX)χ^1(ρ2)χ3(ρ1)])-1m·[λ2qχ^1(ρ1)χ^1(ρ2)λ1+λ2+δχ^1(ρ1)χ3(ρ2)ω^(ρ2)-χ^1(ρ2)χ3(ρ1)ω^(ρ1)mmm+λ2qχ^1(ρ1)χ^1(ρ2)λ1+λ2+δmmm·((1-ρ2μX)ω^*(ρ2)mmmmm-(1-ρ1μX)ω^*(ρ1))λ2qχ^1(ρ1)χ^1(ρ2)λ1+λ2+δ],ϕ^1(1μX)=((λ1+λ2+δ)χ3(ρ1)ω^(ρ1)mnm+λ2q(1-ρ1μX)χ^1(ρ1)ω^*(ρ1))m×(λ1λ2qχ^1(ρ1))-1m-λ1(λ1+λ2+δ)χ3(ρ1)λ1λ2q(1-ρ1μX)χ^1(ρ1)ϕ^(1μX).Then the explicit expression for theϕ^(s) and ϕ^1(s) can be obtained by (19) and (20), respectively.


## 4. The Defective Renewal Equation for the Expected Discounted Penalty Function

In this section, we study the defective renewal equation satisfied by the expected discounted penalty function. Note that (19) can be rewritten as
(30)ϕ^(s)=f^1(s)+f^2(s)h^1(s)-h^2(s),
where
(31)f^1(s)=-λ1ϕ^(1/μX)λ1+λ2+δ(1-sμX-λ1λ1+λ2+δ),f^2(s)=(1-sμX)λ1+λ2+δ×[χ3(s)ω^(s)+λ2q(1-sμX)χ^1(s)λ1+λ2+δω^*(s)]+(λ1λ2q(1-sμX)χ^1(s)[b^2(s)ϕ^(1μX)-ϕ^1(1μX)]×[b^2(s)ϕ^(1μX)-ϕ^1(1μX)])×((λ1+λ2+δ)2)-1,h^1(s)=(1-sμX-λ1λ1+λ2+δ)2,h^2(s)=(1-sμX)×λ2b^1(s)(1-q+qb^2(s))λ1+λ2+δ×(1-sμX-λ1λ1+λ2+δ).Lemma 3.
The Laplace transform of the expected discounted penalty functionϕ^(s) satisfies
(32)ϕ^(s)=TsTρ1Tρ2h2(0)μX2ϕ^(s)+TsTρ1Tρ2f2(0)μX2.Proof.
Sinceϕ^(s) is analytic for all s with Re(s)≥0, we know ρi, i=1,2, are zeros of the numerator in (30). It means f^1(ρi)=-f^2(ρi), i=1,2. Because f^1(s) is a polynomial of degree 1, using Lagrange interpolating theorem, we obtain
(33)f^1(s)=f^1(ρ1)s-ρ2ρ1-ρ2+f^1(ρ2)s-ρ1ρ2-ρ1=-f^2(ρ1)s-ρ2ρ1-ρ2-f^2(ρ2)s-ρ1ρ2-ρ1.
It yields
(34)f^1(s)+f^2(s)=(-f^2(ρ1)(s-ρ2)+f^2(ρ2)(s-ρ1)+(s-ρ2)f^2(s)-(s-ρ1)f^2(s))×(ρ1-ρ2)-1=((s-ρ2)[f^2(s)-f^2(ρ1)]-(s-ρ1)[f^2(s)-f^2(ρ2)])×(ρ1-ρ2)-1=(s-ρ1)(s-ρ2)×TsTρ2f2(0)-TsTρ1f2(0)ρ1-ρ2=(s-ρ1)(s-ρ2)TsTρ2Tρ1f2(0).
The denominator of (30) can be dealt with in a similar way. From Lemma 1, we know that h^1(ρi)=h^2(ρi), i=1,2. Because h^1(s) is a polynomial of degree 2, using Lagrange interpolating theorem, we obtain
(35)h^1(s)=h^1(0)(s-ρ1)(s-ρ2)ρ1ρ2+s(h^1(ρ1)ρ1s-ρ2ρ1-ρ2+h^1(ρ2)ρ2s-ρ1ρ2-ρ1)=h^1(0)(s-ρ1)(s-ρ2)ρ1ρ2+s(h^2(ρ1)ρ1s-ρ2ρ1-ρ2+h^2(ρ2)ρ2s-ρ1ρ2-ρ1)=h^1(0)(s-ρ1)(s-ρ2)ρ1ρ2+h^2(ρ1)s-ρ2ρ1-ρ2+h^2(ρ2)s-ρ1ρ2-ρ1+(s-ρ1)(s-ρ2)×(h^2(ρ1)ρ11ρ1-ρ2+h^2(ρ2)ρ21ρ2-ρ1).
Then using Property 6 of the Dickson-Hipp operator given in Li and Garrido [20], we have
(36)h^1(s)-h^2(s)=h^1(0)(s-ρ1)(s-ρ2)ρ1ρ2+h^2(ρ1)s-ρ2ρ1-ρ2+h^2(ρ2)s-ρ1ρ2-ρ1+(s-ρ1)(s-ρ2)×(h^2(ρ1)ρ11ρ1-ρ2+h^2(ρ2)ρ21ρ2-ρ1)-h^2(s)=(s-ρ1)(s-ρ2)×(h^1(0)ρ1ρ2+h^1(ρ1)ρ1(ρ1-ρ2)mmmm-h^1(ρ2)ρ2(ρ1-ρ2)+h^2(ρ1)(ρ1-ρ2)(s-ρ1)mmmm-h^2(ρ2)(ρ1-ρ2)(s-ρ2)-h^2(s)(s-ρ1)(s-ρ2))=(s-ρ1)(s-ρ2)×(T0Tρ1Tρ2h1(0)-TsTρ1Tρ2h2(0)).
It is easy to check thatT0Tρ1Tρ2h1(0)=μX2 which makes (36) become
(37)h^1(s)-h^2(s)=(s-ρ1)(s-ρ2)(μX2-TsTρ1Tρ2h2(0)).
Invoking (34) and (37) into (30), we could obtain
(38)ϕ^(s)=TsTρ2Tρ1f2(0)μX2-TsTρ1Tρ2h2(0),
which leads to (32). This completes the proof.Now, we are ready to derive the defective renewal equation forϕ(u).Theorem 4.
ϕ
(
u
) satisfies the following integral equation:
(39)ϕ(u)=∫0u‍ϕ(u-y)Tρ1Tρ2h2(y)μX2dy+Tρ1Tρ2f2(u)μX2.Proof.
Equation (39) follows easily from the inverse Laplace transform in (32). We could like to point out that (39) is also a defective renewal equation. This can be verified by showing that
(40)T0Tρ1Tρ2h2(0)μX2<1.
Forδ>0, putting s=0 in (36), it follows that
(41)T0Tρ1Tρ2h2(0)μX2=1-h^1(0)-h^2(0)μX2ρ1ρ2=1-δ(λ2+δ)μX2ρ1ρ2(λ1+λ2+δ)2<1.
Now we consider the caseδ=0. Setting s=ρ1(δ) in the denominators of (19) and (20), we have
(42)1-ρ1(δ)μX-λ1λ1+λ2+δ-λ2b^1(ρ1(δ))[1-q+qb^2(ρ1(δ))]λ1+λ2+δ×(1-ρ1(δ)μX)=0.
Differentiating both sides of this equation with respect toδ and then setting δ=0, we have
(43)ρ1′(0)=1λ1μX-λ2(μY+qμZ)>0.
Then taking the limitδ→0 in (41) and using L’Hôspital’s rule, we obtain
(44)T0Tρ1Tρ2h2(0)μX2=1-1μX2ρ2(0)(λ1+λ2+δ)2×lim⁡δ→0+(λ2+δ)δρ1(0)=1-λ2μX2ρ1′(0)ρ2(0)(λ1+λ2)2<1.
Thus, (39) is a defective renewal equation, and the proof is complete.Remark 5.
Whenq≡1 and P(Y1≥M1)=θ, then χ^1(s)=θb^1(s), χ^2(s)=(1-θ)b^1(s). In the case, each main claim induces a byclaim, and its associated byclaim occurs simultaneously with probability 1-θ, or the occurrence of the byclaim may be delayed with probability θ. Actually, the risk model given by (1) will be the compound Poisson risk model with delayed claims and random incomes studied by Hao and Yang [16]. Then, by some simple calculations, we can find that (39) in Theorem 4 is consistent with (4.6) in [16].Remark 6.
The explicit analytic solution to the defective renewal (39) can be obtained by compound geometric distribution (see Lin and Willmot [22]).


## 5. Explicit Results for Exponential Claim Size Distributions

We now consider the case where both claim sizes are exponentially distributed, that is, distribution functionsFY~Exp(ν) and FZ~Exp(ω), where ν=1/μY and ω=1/μZ. Then we have
(45)b^1(s)=νν+s,b^2(s)=ωω+s,b^3(s)=νω(ν+s)(ω+s),b^4(s)=ω2(ω+s)2.For the special caseFM~Exp(μ), we obtain
(46)χ^2(s)=∫0∞‍e-sye-μydFY(y)=b^1(s+μ),χ^1(s)=b^1(s)-b^1(s+μ).
So we have
(47)χ^2(s)=νν+s+μ,χ^1(s)=νν+s-νν+s+μ=νμ(ν+s)(ν+s+μ).LetB^1(s)≜χ^1(s)b^2(s)ω^(s), B^2(s)≜χ^1(s)ω^*(s); then f^2(s) can be written as
(48)f^2(s)=(1-sμX)λ1+λ2+δ×(1-sμX-λ1λ1+λ2+δ)ω^(s)-λ2q(1-sμX)2(λ1+λ2+δ)2[B^1(s)-B^2(s)]+λ1λ2q(λ1+λ2+δ)2×[ϕ^(1μX)ωνμ(1-sμX)(ω+s)(ν+s)(ν+s+μ)mmm-ϕ^1(1μX)νμ(1-sμX)(ν+s)(ν+s+μ)].
So we can derive
(49)TsTρ2Tρ1f2(0)=μXλ1+λ2+δ×[(1-μXρ1)TsTρ1ω(0)+μXTsω(0)]-λ12λ2q(λ1+λ2+δ)4×[TsTρ2Tρ1B1(0)-TsTρ2Tρ1B2(0)]-λ2qμX[2-μX(ρ1+ρ2)](λ1+λ2+δ)2×[TsTρ1B1(0)-TsTρ1B2(0)]-λ2qμX2(λ1+λ2+δ)2[TsB1(0)-TsB2(0)]+λ1λ2q(λ1+λ2+δ)2ϕ^(1μX)·[ωνμ[s2+(a2+ρ1)s+ρ12+a2ρ1+a1]g(s)g(ρ1)XXYλ1(λ1+λ2+δ)m×ωνμ[c0s2+(c0a2+c1)s+c0a1+c1a2+c2]g(s)g(ρ1)g(ρ2)mωνμ[s2+(a2+ρ1)s+ρ12+a2ρ1+a1]g(s)g(ρ1)XXY+μXωνμ[s2+(a2+ρ1)s+ρ12+a2ρ1+a1]g(s)g(ρ1)]-λ1λ2q(λ1+λ2+δ)2ϕ^1(1μX)·[λ1(λ1+λ2+δ)·(νμ[(2ν+μ+s)(2ν+μ+ρ1+ρ2)mmmmmmmi-ν(ν+μ)+ρ1ρ2])×((ν+s)(ν+μ+s)(ν+ρ1)mi×(ν+μ+ρ1)(ν+ρ2)(ν+μ+ρ2))-1+μXνμ(2ν+μ+ρ1+s)(ν+s)(ν+μ+s)(ν+ρ1)(ν+μ+ρ1)],
where
(50)a0=ων(ν+μ),a1=ων+(ω+ν)(ν+μ),a2=ω+2ν+μ,c0=a1+a2(ρ1+ρ2)+ρ12+ρ1ρ2+ρ22,c1=-a0+ρ1ρ2(a2+ρ1+ρ2),c2=-a0(ρ1+ρ2)-a1ρ1ρ2+ρ12ρ22,g(x)=(ω+x)(ν+x)(ν+μ+x).From (32) and (37), we know that
(51)ϕ^(s)=(s-ρ1)(s-ρ2)TsTρ2Tρ1f2(0)h^1(s)-h^2(s).
It turns out that (51) can be transformed to another expression by multiplying both denominator and numerator by g(s):
(52)ϕ^(s)=g(s)(s-ρ1)(s-ρ2)TsTρ2Tρ1f2(0)g(s)(h^1(s)-h^2(s)).
The common denominator of (52), denoted by D5(s), is a polynomial of degree 5 with the leading coefficient μX2, given by
(53)D5(s)=(ω+s)(ν+s)(ν+μ+s)×(1-sμX-λ1λ1+λ2+δ)2-(ν+μ+s)(1-sμX)×λ2ν((1-q)(ω+s)+qω)λ1+λ2+δ×(1-sμX-λ1λ1+λ2+δ).
Obviously, D5(s) has five roots on the complex plane and all the complex roots are in conjugate pairs. Noting that s=ρ1, s=ρ2, and s=-(ν+μ) are three roots, we have
(54)D5(s)=μX2(s-ρ1)(s-ρ2)(s+ν+μ)(s+R1)(s+R2).Note also that allRis have positive real parts, since, otherwise, they also are roots of (22) which is a contradiction to the conclusion of Lemma 1.DenoteR0=ν+μ. Furthermore, if R0, R1, and R2 are distinct, we obtain, by partial fractions, that
(55)as2+bs+c(s+R0)(s+R1)(s+R2)=r1(a,b,c)s+R0+r2(a,b,c)s+R1+r3(a,b,c)s+R2,(s+ω)(s+ν)(s+R1)(s+R2)=1+h1s+R1+h2s+R2,
where
(56)r1(a,b,c)=aR02-bR0+c(R1-R0)(R2-R0),r2(a,b,c)=aR12-bR1+c(R0-R1)(R2-R1),r3(a,b,c)=aR22-bR2+c(R0-R2)(R1-R2),h1=(ν-R1)(ω-R1)R2-R1,h2=(ν-R2)(ω-R2)R1-R2.
Then (52) can be simplified to
(57)ϕ^(s)=1μX2(1+∑i=12‍his+Ri)×{μXλ1+λ2+δλ2qμX2(λ1+λ2+δ)2×[(1-μXρ1)TsTρ1ω(0)+μXTsω(0)]-λ12λ2q(λ1+λ2+δ)4×[TsTρ2Tρ1B1(0)-TsTρ2Tρ1B2(0)]-λ2qμX[2-μX(ρ1+ρ2)](λ1+λ2+δ)2×[TsTρ1B1(0)-TsTρ1B2(0)]-λ2qμX2(λ1+λ2+δ)2[TsB1(0)-TsB2(0)]}+λ1λ2qμX2(λ1+λ2+δ)2ϕ^(1μX)·[∑j=02‍λ1ωνμ(λ1+λ2+δ)g(ρ1)g(ρ2)m×∑j=02‍rj(c0,c0a2+c1,c0a1+c1a2+c2)s+Rjm+μXωνμg(ρ1)∑j=02‍rj(1,a2+ρ1,ρ12+a2ρ1+a1)s+Rj∑j=02‍]-λ1λ2qμX2(λ1+λ2+δ)2ϕ^1(1μX)·[∑j=02‍∑j=02‍(λ1νμ)×((λ1+λ2+δ)(ν+ρ1)×(ν+μ+ρ1)(ν+ρ2)×(ν+μ+ρ2))-1×∑j=02‍rj(d1,d2,d3)s+Rj+μXνμ(ν+ρ1)(ν+μ+ρ1)×∑j=02‍rj(1,2ν+μ+ω+ρ1,ω(2ν+μ+ρ1))s+Rj∑j=02‍],
where
(58)d1=2ν+μ+ρ1+ρ2,d2=(2ν+μ+ω)d1-ν(ν+μ)+ρ1ρ2,d3=ω(d2-ωd1).Taking the inverse Laplace transforms, we can derive explicit expressions forϕ(u):
(59)ϕ(u)=1μX2Λ(u)+1μX2∑i=12‍hie-Riu*_Λ(u)+λ1λ2qμX2(λ1+λ2+δ)2ϕ^(1μX)·[∑j=02‍λ1ωνμ(λ1+λ2+δ)g(ρ1)g(ρ2)m×∑j=02‍rj(c0,c0a2+c1,c0a1+c1a2+c2)e-Rjum+μXωνμg(ρ1)∑j=02‍rj(1,a2+ρ1,ρ12+a2ρ1+a1)e-Rju]-λ1λ2qμX2(λ1+λ2+δ)2ϕ^1(1μX)·[∑j=02‍(λ1νμ)×((λ1+λ2+δ)(ν+ρ1)mmmmmmmmmmm×(ν+μ+ρ1)(ν+ρ2)(ν+μ+ρ2))m×∑j=02‍rj(d1,d2,d3)e-Rjum+μXνμ(ν+ρ1)(ν+μ+ρ1)m×∑j=02‍rj(1,2ν+μ+ω+ρ1,ω(2ν+μ+ρ1))e-Rju],
where *_ denotes the operation of convolution which is different from the distribution functions convolution:
(60)Λ(u)=μXλ1+λ2+δ×[(1-μXρ1)Tρ1ω(u)+μXω(u)]-λ12λ2q(λ1+λ2+δ)4×[Tρ2Tρ1B1(u)-Tρ2Tρ1B2(u)]-λ2qμX[2-μX(ρ1+ρ2)](λ1+λ2+δ)2×[Tρ1B1(u)-Tρ1B2(u)]-λ2qμX2(λ1+λ2+δ)2×[B1(u)-B2(u)].Equation (59) is the explicit expression for ϕ(u) with the case where both the claim sizes are exponentially distributed.


## 6. Conclusions

We have generalized the results in [11, 16]. It is assumed that the premium income process is a compound Poisson process; moreover, every main claim will produce a byclaim with a certain probability and the occurrence of the byclaim may be delayed depending on associated main claim amount. We not only derive the integral equation satisfied by the expected discounted penalty function, but also obtain the explicit expression for the Laplace transform of the expected discounted penalty function when the premium size is exponentially distributed. Finally, for the exponential claim sizes, we present the explicit formula for the expected discounted penalty function.



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*Source: 717269-2014-03-05.xml*