Ruin probability in the delayed renewal risk model perturbed by a diffusion process

Abstract

In this paper, we consider the risk model perturbed by an independent diffusion process with a time delay in the arrival of the first claim. We derive the distribution of the counting process of the risk process, the integro-differential equations of the ruin probabilities and generalize its defective renewal equations. With claim amounts following exponential and mixed exponential distributions, explicit expressions and asymptotic properties of the ruin probabilities are derived. Numerical illustrations of the ruin probabilities are proposed when claim amounts are exponentially and mixed exponentially distributed. We extend the results to the case of the delayed renewal risk model with exchangeable risks which captures the possible dependence between inter-arrival times of the claims and derive the associated ruin probabilities.

Keywords

Ruin theory delay Poisson risk process renewal equation convolution formula diffusion process mixed exponential distribution

1. Introduction

The ordinary renewal risk model, usually referred to as the Sparre Andersen model, is often used to model the insurer’s surplus process. See [11,15,22], and references therein. This in turn is built on the classical compound Poisson process. Here, it is assumed that inter-claims times are exponentially distributed, claim sizes and inter-claims times are independent. Due to certain inadequacies of the classical compound risk model, the Gerber-Shiu function was introduced. This function allowed the inter-claims times to follow an arbitrary distribution. As a result, ruin probabilities and many ruin related quantities could be analytically studied. See [5,10,16]. As an extension to their work [9]. introduced a discounted penalty function with respect to the time of ruin, the surplus immediately before the ruin and the deficit at ruin to analyse these ruin related quantities in a unified manner. Subsequently the Gerber-Shiu discounted penalty function for the compound Poisson risk model has been extensively studied. See [16,19]. In the classical compound Poisson risk model, the stationary and independent increment assumptions on the surplus process play an important role. Sometimes, insurance claims may be delayed due to various reasons, thus classical assumptions are very restrictive in some applications.

In the literature the Gerber-Shiu function satisfies a defective renewal equation in the ordinary Sparre Andersen model, and one can use mathematical tools to solve the defective renewal equations. Because the assumption of independence and identical distribution for the claim inter time arrival may not be realistic, extensions to that model are made in many papers. Thus, the Gerber-Shiu function in the delayed renewal model has also been considered where the first inter claim time is assumed to follow a (possibly) different density than the common density of the subsequent inter-claims times as is discussed in [1,21,23]. Generalizations of the Gerber-Shiu function have also been studied, for example [6]. A Gerber-Shiu function that has been generalized to a more general cost function is considered in [4] and the Gerber-Shiu function at absolute ruin is considered in [3]. There has also been a considerable amount of research on models where the claim size and the inter-claims times are assumed to be dependent, preceding authors are [2,24,25].

Dufresne and Gerber [7] extended the classical model of collective risk theory by adding a diffusion process to the compound Poisson model. They showed that the probabilities of ruin (by oscillation or by a claim) satisfy certain defective renewal equations and derived the convolution formula for the probability of ruin and interpreted it in terms of the record highs of the aggregate loss process. [19] in their paper consider the surplus process of the classical continuous time risk model assuming independent diffusion (Wiener)process. They generalized the defective renewal equation for the expected discounted function of a penalty at the time of ruin in [9]. Wang [20] worked on a decomposition of the ruin probability for the risk process perturbed by diffusion. Gao and Wu [8] worked on the Gerber–Shiu discounted penalty function in a risk model with two types of delayed-claims and random income. They developed a new delayed model with random premium income and two types of by-claims, and then derived an integral system of equations for the Gerber–Shiu discounted penalty function and explicit solution of the Laplace transform of the discounted penalty function. They proved that the discounted penalty function satisfies a defective renewal equation and obtained an explicit result of the ruin probability under the exponential distribution. Zhang and Yang [27] worked on Gerber–Shiu analysis in a perturbed risk model with dependence between claim sizes and inter-claim times, where they also consider that the compound Poisson risk model is perturbed by a Brownian motion. Lee and Willmot [13] worked on the moments of the time to ruin in dependent Sparre Andersen models with emphasis on Coxian inter-claim times. The objective of this paper is to consider the risk model introduced by [7] and extends some results by taking into account the delay in the arrival of the first claim arrival time, and then extend the results to the case of the delayed renewal risk model with exchangeable risks and derive the associated ruin probabilities.

The paper is structured as follows. In Section 2, the risk model is presented. In Sections 3 and 4, we derive the distribution of the counting process and the equation for the Lundberg adjustment coefficient and its solution. Sections 5–8 deal with the integro-differential equations, the defective equations, their solutions and illustrations. Section 9 deals with the risk model with exchangeable risks and its numerical illustration followed by the conclusion.

2. Risk model

We consider the following compound aggregate delay Poisson risk process perturbed by an independent diffusion process. The claim inter-arrival times are assumed to be independent with exponential distribution with time delay in the arrival of the first claim. The surplus process at time t is: $\begin{eqnarray}\displaystyle U_{t}=u+ct+{\sigma}B_{t}-\displaystyle \mathop{\sum }_{i=1}^{N_{t}^{d}}X_{i},\quad \forall t\geq 0 & & \displaystyle\end{eqnarray}$ (1) where u is the initial surplus and c the rate at which premiums are received per unit of time. $N_{t}^{d}$ denote the delayed renewal process and the ordinary renewal process is denoted by N _t and 𝜎 represents the volatility that accounts for the perturbation of the diffusion process.

The hypotheses of the model are summarized in the following:

•

The time arrival of the first claim V ₁ is exponentially distributed with parameter 𝜆₁, i.e V ₁ ∼ Exp (𝜆₁);

•

Claim inter-arrival time $\{V_{i}\}_{i\geq 2}$ are independent and identically distributed (iid) with V _i ∼ V ∼ Exp (𝜆), ∀i ≥ 2;

•

$\{X_{i}\}_{i=1}^{\infty }$ are iid and distributed as the generic X;

•

$\{X_{i}\}_{i=1}^{\infty }$ and $\{V_{i}\}_{i\geq 1}$ are mutually independent;

•

the standard Brownian motion (B _t)_t≥0 is independent to the claim process.

if a random variable W is exponentially distributed, i.e, W ∼ Exp (𝜆) then its probability density function (p.d.f) is defined as f _W(t) = 𝜆e ^−𝜆t; its cumulative distribution function (c.d.f) is defined as F _W(t) = 1 − e ^−𝜆t, t ≥ 0 and the Laplace transform of f _W is $\tilde{f}_{W}(s)=E[e^{-sX}]=\left(\frac{{\lambda}}{{\lambda}+s}\right)$ .

3. Distribution of the counting process

In this section, we derive the distribution of the counting process of the delayed renewal process when the inter arrival times of the claims are assumed to be exponentially distributed, and check it is consistent with the distribution in the ordinary renewal process.

TheoremA .
Assuming that the arrival time of the first claim is exponentially distributed with parameter 𝜆₁ (i.e V ₁ ∼ Exp (𝜆₁)) and the inter arrival time of the others claims is exponentially distributed with parameter 𝜆 (i.e V ∼ Exp (𝜆₂)) the p.d.f of the counting process satisfies the following: $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-23.0pt}P[N_{t}^{d}=0]=e^{-{\lambda}_{1}t},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-23.0pt}P[N_{t}^{d}=n]=e^{-{\lambda}_{1}t}{\displaystyle \frac{{\lambda}^{n-1}{\lambda}_{1}}{({\lambda}-{\lambda}_{1})^{n}}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-13.0pt}\times ∼\left[1-e^{-({\lambda}-{\lambda}_{1})t}\displaystyle \mathop{\sum }_{k=0}^{n-1}{\displaystyle \frac{(({\lambda}-{\lambda}_{1})t)^{k}}{k!}}\right],\quad t\geq 0,n\geq 1.\nonumber\end{eqnarray}$

TheoremB .
Denote T _n = V ₁ + V ₂ + ⋯ + V _n, n ≥ 0 with T ₀ = 0. Since $\{N_{t}=n\}=\{T_{n}\leq t\lt T_{n+1}\}$ we have: $\begin{eqnarray}\displaystyle \hspace{-20.0pt}\text{} & \text{} & \displaystyle P[N_{t}^{d}=0]=P[t\lt V_{1}]=\int _{t}^{\infty }{\lambda}_{1}e^{-{\lambda}_{1}x}dx=e^{-{\lambda}_{1}t}\nonumber\end{eqnarray}$ $\begin{eqnarray}\displaystyle P[N_{t}^{d}=n]\text{} & =\text{} & \displaystyle \int _{0}^{t}f_{T_{1},N_{t}^{d}}(x,n)dx\nonumber\\ \displaystyle \text{} & =\text{} & \displaystyle \int _{0}^{t}f_{T_{1}|N_{t}^{d}}(x|n)P[N_{t}^{d}=n]dx.\end{eqnarray}$ (2)

From [14] we get $\begin{eqnarray}\displaystyle f_{T_{1}|N_{t}^{d}}(x|n)={\displaystyle \frac{f_{T_{1}}(x)P[N_{t-x}=n-1]}{P[N_{t}^{d}=n]}} & & \displaystyle\end{eqnarray}$ (3) and substituting (3) in (2) we have $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}P[N_{t}^{d}=n]=\int _{0}^{t}f_{T_{1}|N_{t}^{d}}(x|n)P[N_{t}^{d}=n]dx\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{5.0pt}=\int _{0}^{t}{\displaystyle \frac{f_{T_{1}}(x)P[N_{t-x}=n-1]}{P[N_{t}^{d}=n]}}P[N_{t}^{d}=n]dx\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{5.0pt}=\int _{0}^{t}P[N_{t-x}=n-1]f_{T_{1}}(x)dx.\nonumber\end{eqnarray}$ $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}P[N_{t}^{d}=n]=e^{-{\lambda}_{1}t}{\displaystyle \frac{{\lambda}^{(n-1)}{\lambda}_{1}}{(n-1)!}}\int _{0}^{t}z^{n-1}e^{-z({\lambda}-{\lambda}_{1})}dz\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}=e^{-{\lambda}_{1}t}{\displaystyle \frac{{\lambda}^{n-1}{\lambda}_{1}}{({\lambda}-{\lambda}_{1})^{n}}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-15.0pt}\times ∼\left[1-e^{-({\lambda}-{\lambda}_{1})t}\displaystyle \mathop{\sum }_{k=0}^{n-1}{\displaystyle \frac{(({\lambda}-{\lambda}_{1})t)^{k}}{k!}}\right],\quad t\geq 0,n\geq 1.\nonumber\end{eqnarray}$

TheoremC .
Using Lebesgue’s theorem, for ${\lambda}\rightarrow {\lambda}_{1},I_{n-1}=\int _{0}^{t}z^{n-1}e^{-z{\alpha}}dz\rightarrow \frac{t^{n}}{n}$ , we get the result for the ordinary renewal Poisson process i.e $P[N_{t}=n]=\frac{({\lambda}t)^{n}e^{-{\lambda}t}}{n!}$ .

4. Lundberg-type equation

Our goal is to determine the explicit expression for the ruin probability in the delayed risk model perturbed by a diffusion process. We introduce in this section a generalized version of the Lundberg equation for the risk process, and then we analyse the number of its roots. We need these roots to derive the defective renewal equation for the ruin probability.

Define the ruin probability by $\begin{eqnarray}\displaystyle m_{d}(u)=E\left[I({\tau}<\infty )|_{U(0)=u}\right], & & \displaystyle \nonumber\end{eqnarray}$ which is the first time the surplus level falls bellow zero. The time of ruin 𝜏 is: $\begin{eqnarray}\displaystyle {\tau}=\left\{\begin{array}{@{}l@{}}\inf \{t>0:U_{t}<0\}\\ \infty ∼\text{if }∼U_{t}\geq 0\end{array}\right.\quad \forall ∼t>0. & & \displaystyle \nonumber\end{eqnarray}$ To guarantee that ruin will not almost surely occur(i.e the ultimate ruin probability P[𝜏 < ∞|_U
₀=u] is not almost sure equal to one), the premium rate c is such that the $\begin{eqnarray}\displaystyle E[cV_{k}-X_{k}]>0,\quad k=1,2,\ldots . & & \displaystyle \nonumber\end{eqnarray}$

The delayed ruin probability m _d can be decomposed as m _d(u) = 𝜙_d(u) + 𝜓_d(u) where $\begin{eqnarray}\displaystyle {\phi}_{d}(u)=E[I({\tau}<\infty ,U_{{\tau}}<0)|_{U(0)=u}], & & \displaystyle \nonumber\end{eqnarray}$ and $\begin{eqnarray}\displaystyle {\psi}_{d}(u)=E[I({\tau}<\infty ,U_{{\tau}}=0)|_{U(0)=u}]. & & \displaystyle \nonumber\end{eqnarray}$ This decomposition can be explained by the fact that if the ruin occurs it can be caused by the oscillation or by the claim. Let’s denote by 𝜙, 𝜓 the ruin probability caused claim and by oscillation in the ordinary risk model.

Let’s consider the sequence of the surplus at the n ^th claim $\{U_{n}\}_{n=1}^{\infty }$ such that $\begin{eqnarray}\displaystyle U_{T_{n}}=u+cT_{n}+{\sigma}B_{T_{n}}-\displaystyle \mathop{\sum }_{k=1}^{N_{T_{n}}}X_{k}, & & \displaystyle \nonumber\end{eqnarray}$ where $T_{n}=\sum _{k=1}^{n}V_{k}$ is the time of the occurrence of the n − th claim. Setting the surplus immediately after the nth claim $U_{n}=u+cT_{n}+{\sigma}B_{T_{n}}-\sum _{k=1}^{n}X_{k}$ , with the properties of the Brownian motion (independent increment and stationarity) we get $\begin{eqnarray}\displaystyle U_{n}=^{d}u+\displaystyle \mathop{\sum }_{k=1}^{n}(cV_{k}+{\sigma}B_{V_{k}}-X_{k}). & & \displaystyle \nonumber\end{eqnarray}$ To proceed, we first determine s such that $\{e^{sU_{n}}\}_{n=0}^{\infty }$ is a martingale. Let’s Z _n = e ^{sU
_n}, therefore $\begin{eqnarray}\displaystyle Z_{n+1}=Z_{n}e^{s(cV_{n+1}+{\sigma}B_{V_{n+1}}-X_{n+1})}. & & \displaystyle \nonumber\end{eqnarray}$ Assuming that the adaptability and integrability conditions are met, the martingale condition is satisfied if $\begin{eqnarray}\displaystyle E[e^{s(cV_{n+1}+{\sigma}B_{V_{n+1}}-X_{n+1})}]=1. & & \displaystyle \nonumber\end{eqnarray}$ Let L (s) = E[e ^{−𝛿V +s (cV +𝜎B
_V−X)}]. The generalized Lundberg equation associated with the risk model in ((1)), n ≥ 2 is given by the following ordinary Lundberg equation, $\begin{eqnarray}\displaystyle L(s)=1, & & \displaystyle\end{eqnarray}$ (4) where $\begin{eqnarray}\displaystyle L(s)\text{} & =\text{} & \displaystyle \int _{0}^{\infty }\int _{0}^{\infty }E[e^{s(ct+{\sigma}B_{t}-x)}]f_{(X,V)(x,t)dxdt}\nonumber\\ \displaystyle \text{} & =\text{} & \displaystyle \int _{0}^{\infty }\int _{0}^{\infty }e^{s(ct-x)+{\displaystyle \frac{s^{2}{\sigma}^{2}}{2}}t}f_{(X,V)(x,t)dxdt}\nonumber\\ \displaystyle \text{} & =\text{} & \displaystyle \tilde{f}_{X}(s)\tilde{f}_{V}\left(-cs-{\displaystyle \frac{s^{2}{\sigma}^{2}}{2}}\right)\nonumber\\ \displaystyle \text{} & =\text{} & \displaystyle {\displaystyle \frac{{\lambda}\tilde{f}_{X}(s)}{{\lambda}-cs-{\displaystyle \frac{s^{2}{\sigma}^{2}}{2}}}}.\nonumber\end{eqnarray}$ Equation ((4)) has exactly one null root at 𝜌 = 0 in the right complex plane. See [19].

5. Integro-differential equation

In this section we derive the integro-differential equations satisfied by the ruin probability caused by the claims 𝜙_d and the ruin probability caused the perturbations 𝜓_d.

TheoremD .
Under the assumptions of the delayed and perturbed risk model defined in Section (2), the ruin probabilities satisfy the following equations.
The ruin probability 𝜙_d caused by claim satisfies the following differential equation. $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-28.0pt}{\lambda}_{1}{\phi}_{d}(u)=c{\phi}_{d}^{\prime }(u)+{\displaystyle \frac{{\sigma}^{2}}{2}}{\phi}_{d}^{^{\prime \prime }}(u)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-25.0pt}+{\lambda}_{1}\left[\int _{0}^{u}{\phi}(u-x)f_{X}(x)dx+\int _{u}^{\infty }f_{X}(x)dx\right].\end{eqnarray}$ (5)

The ruin probability 𝜓_d caused by the oscillation satisfies the following differential equation $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\lambda}_{1}{\psi}_{d}(u)=c{\psi}_{d}^{\prime }(u)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-10.0pt}+∼{\displaystyle \frac{{\sigma}^{2}}{2}}{\psi}_{d}^{^{\prime \prime }}(u)+{\lambda}_{1}\displaystyle \int _{0}^{u}{\psi}(u-x)f_{X}(x)dx.\end{eqnarray}$ (6)

TheoremE .
We consider a time interval and condition on the fact that the claim may occur or not to get the proof.

1. Let the consider the infinitesimal time interval [0, h], h > 0. where $\lim _{h\rightarrow 0}\frac{o(h)}{h}=0$ , By conditioning on the fact that a claim may occur or not and conditioning on the first claim in the time interval [0, h] we derive the following equation: $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\phi}_{d}(u)=E[I({\tau}\lt \infty ,U_{{\tau}}\lt 0)|_{U(0)=u}]\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-15.0pt}=∼P[N_{h}^{d}=0]E[{\phi}_{d}(u+ch+{\sigma}B_{h})]\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-15.0pt}+∼(1-P[N_{h}^{d}=0])\left\{\displaystyle \int _{0}^{h}E\left[\int _{0}^{u+ct+{\sigma}B_{t}}\right.\right.\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-15.0pt}\times ∼{\phi}(u+ct+{\sigma}B_{t}-x)f_{X}(x)dx\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-15.0pt}+∼\left.\left.\displaystyle \int _{u+ct+{\sigma}B_{t}}^{\infty }f_{X}(x)dx\right]f_{T_{1}|_{N_{h}^{d}=1}}(t)dt\right\}+o(h).\end{eqnarray}$ (7) We have the following approximating result: $\begin{eqnarray}\displaystyle P[N_{h}^{d}=0]\text{} & =\text{} & \displaystyle e^{-{\lambda}_{1}h}=1-{\lambda}_{1}h,P[N_{h}^{d}\geq 1]\nonumber\\ \displaystyle \text{} & =\text{} & \displaystyle 1-P[N_{h}^{d}=0]={\lambda}_{1}h+o(h).\end{eqnarray}$ (8)

Now let ${\phi}_{d}(u+ct+{\sigma}B_{t})={\mathcal{H}}(t,B_{t}).$ By the Ito’s lemma we have $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}d[{\phi}_{d}(u+ct+{\sigma}B_{t})]={\displaystyle \frac{\partial {\mathcal{H}}}{\partial t}}(t,B_{t})dt\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-5.0pt}+∼{\displaystyle \frac{\partial {\mathcal{H}}}{\partial B_{t}}}(t,B_{t})dB_{t}+{\displaystyle \frac{1}{2}}{\displaystyle \frac{\partial ^{2}{\mathcal{H}}}{\partial B_{t}^{2}}}(t,B_{t})d\langle B\rangle _{t}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-15.0pt}=c{\phi}_{d}^{\prime }(u+ct+{\sigma}B_{t})dt+{\sigma}{\phi}_{d}^{\prime }(u+ct+{\sigma}B_{t})dB_{t}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-5.0pt}+∼{\displaystyle \frac{{\sigma}^{2}}{2}}{\phi}_{d}^{^{\prime \prime }}(u+ct+{\sigma}B_{t})dt.\nonumber\end{eqnarray}$ Or equivalently, $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\phi}_{d}(u+ct+{\sigma}B_{t})={\phi}_{d}(u)+\int _{0}^{t}\Bigg[c{\phi}_{d}^{\prime }(u+cv+{\sigma}B_{v})\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-10.0pt}+∼{\displaystyle \frac{{\sigma}^{2}}{2}}{\phi}_{d}^{^{\prime \prime }}(u+ct+{\sigma}B_{t})\Bigg]dv\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-10.0pt}+∼\int _{0}^{t}{\sigma}{\phi}_{d}^{\prime }(u+cv+{\sigma}B_{v})dB_{v},\nonumber\end{eqnarray}$ thus $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}E[{\phi}_{d}(u+ct+{\sigma}B_{t})]={\phi}_{d}(u)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +\displaystyle \int _{0}^{t}E\big[c{\phi}_{d}^{\prime }(u+cv+{\sigma}B_{v})\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +{\displaystyle \frac{{\sigma}^{2}}{2}}{\phi}_{d}^{^{\prime \prime }}(u+cv+{\sigma}B_{v})\big]dv.\nonumber\end{eqnarray}$ From $\begin{eqnarray}\displaystyle \lim _{h\rightarrow 0}{\displaystyle \frac{\int _{0}^{h}f(t)dt}{h}}=f(0), & & \displaystyle\end{eqnarray}$ (9) we have $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\lim _{t\rightarrow 0}{\displaystyle \frac{1}{t}}\int _{0}^{t}E\Bigg[c{\phi}_{d}^{\prime }(u+cv+{\sigma}B_{v})\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}+∼{\displaystyle \frac{{\sigma}^{2}}{2}}{\phi}_{d}^{^{\prime \prime }}(u+cv+{\sigma}B_{v})\Bigg]dv=c{\phi}_{d}^{\prime }(u)+{\displaystyle \frac{{\sigma}^{2}}{2}}{\phi}_{d}^{^{\prime \prime }}(u),\nonumber\end{eqnarray}$ or for t →0 $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-24.0pt}\int _{0}^{t}\hspace{-4.0pt}E\hspace{-2.0pt}\left[c{\phi}_{d}^{\prime }(u+cv+{\sigma}B_{v})+{\displaystyle \frac{{\sigma}^{2}}{2}}{\phi}_{d}^{^{\prime \prime }}(u+cv+{\sigma}B_{v})\right]\hspace{-2.0pt}dv\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-10.0pt}=(c{\phi}_{d}^{\prime }(u)+{\displaystyle \frac{{\sigma}^{2}}{2}}{\phi}_{d}^{^{\prime \prime }}(u))t+o(t).\nonumber\end{eqnarray}$ Thus $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}E[{\phi}_{d}(u+ch+{\sigma}B_{h})]={\phi}_{d}(u)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-5.0pt}+∼ch{\phi}_{d}^{\prime }(u)+{\displaystyle \frac{{\sigma}^{2}}{2}}h{\phi}_{d}^{^{\prime \prime }}(u)+o(h).\end{eqnarray}$ (10)

Using $f_{T_{1}|_{N_{h}=1}}(t)=\frac{1}{h}$ from [14] and by substituting ((8)), ((10)) in the expression of ((7)) gives: $\begin{eqnarray} \displaystyle \text{} & \text{} & \displaystyle \hspace{-24.0pt}{\phi}_{d}(u)={\phi}_{d}(u)+\left[c{\phi}_{d}^{\prime }(u)+{\displaystyle \frac{{\sigma}^{2}}{2}}{\phi}_{d}^{^{\prime \prime }}(u)\right]h-{\lambda}_{1}{\phi}_{d}(u)h\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}+∼{\lambda}_{1}h\hspace{-3.0pt}\displaystyle \int _{0}^{h}\hspace{-3.0pt}E\left[\int _{0}^{u+ct+{\sigma}B_{t}}\hspace{-2.0pt}{\phi}(u+ct+{\sigma}B_{t}-x)f_{X}(x)dx\right.\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}+∼\left.\displaystyle \int _{u+ct+{\sigma}B_{t}}^{\infty }f_{X}(x)dx\right]{\displaystyle \frac{1}{h}}+o(h)\nonumber \end{eqnarray}$ or equivalently, $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-24.0pt}{\lambda}_{1}{\phi}_{d}(u)=c{\phi}_{d}^{\prime }(u)+{\displaystyle \frac{{\sigma}^{2}}{2}}{\phi}_{d}^{^{\prime \prime }}(u)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}+∼{\lambda}_{1}\hspace{-3.0pt}\int _{0}^{h}\hspace{-3.0pt}E\Bigg[\int _{0}^{u+ct+{\sigma}B_{t}}\hspace{-3.0pt}{\phi}(u+ct+{\sigma}B_{t}-x)f_{X}(x)dx\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}+∼\int _{u+ct+{\sigma}B_{t}}^{\infty }f_{X}(x)dx\Bigg]{\displaystyle \frac{1}{h}}dt\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}+∼o(h).\end{eqnarray}$ (11) In Eq. ((11)), letting h →0 and using ((9)) gives the following differential equation: $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\lambda}_{1}{\phi}_{d}(u)=c{\phi}_{d}^{\prime }(u)+{\displaystyle \frac{{\sigma}^{2}}{2}}{\phi}_{d}^{^{\prime \prime }}(u)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}+∼{\lambda}_{1}\left[\int _{0}^{u}{\phi}(u-x)f_{X}(x)dx+\int _{u}^{\infty }f_{X}(x)dx\right].\end{eqnarray}$ (12) Note that for 𝜆₁ →𝜆 we find the equation in the ordinary renewal risk model: $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\lambda}{\phi}(u)=c{\phi}^{\prime }(u)+{\displaystyle \frac{{\sigma}^{2}}{2}}{\phi}^{^{\prime \prime }}(u)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}+∼{\lambda}\left[\int _{0}^{u}{\phi}(u-x)f_{X}(x)dx+\int _{u}^{\infty }f_{X}(x)dx\right],\end{eqnarray}$ (13) which is the result of [7]. 2- Using the same arguments as in the derivation of 𝜙_d we get $\begin{eqnarray} \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\psi}_{d}(u)={\psi}_{d}(u)+[c{\psi}_{d}^{\prime }(u)+{\displaystyle \frac{{\sigma}^{2}}{2}}{\psi}_{d}^{^{\prime \prime }}(u)]dt-{\lambda}_{1}{\psi}_{d}(u)h\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}+∼{\lambda}_{1}\int _{0}^{h}E\Bigg[\int _{0}^{u+ct+{\sigma}B_{t}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}{\psi}(u+ct+{\sigma}B_{t}-x)f_{X}(x)dx\Bigg]dt+o(h),\nonumber \end{eqnarray}$ and letting h →0 gives the following differential equation $\begin{eqnarray}\displaystyle \hspace{-18.0pt}{\lambda}_{1}{\psi}_{d}(u) & = & \displaystyle c{\psi}_{d}^{\prime }(u)+{\displaystyle \frac{{\sigma}^{2}}{2}}{\psi}_{d}^{^{\prime \prime }}(u)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +∼{\lambda}_{1}\int _{0}^{u}{\psi}(u-x)f_{X}(x)dx.\end{eqnarray}$ (14) For 𝜆₁ →𝜆 we get $\begin{eqnarray}\displaystyle \hspace{-18.0pt}{\lambda}{\psi}(u) & = & \displaystyle c{\psi}^{\prime }(u)+{\displaystyle \frac{{\sigma}^{2}}{2}}{\psi}^{^{\prime \prime }}(u)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +∼{\lambda}\int _{0}^{u}{\psi}(u-x)f_{X}(x)dx,\end{eqnarray}$ (15) which is the result of [7].

6. The defective renewal equation

In this section we prove that the ruin probability caused by claims and the ruin probability caused the oscillation both satisfy a defective renewal equation. Let’s define $q(u)=\int _{u}^{\infty }f_{X}(x)dx=P[X>u]$ and denote its Laplace transform by $\tilde{q}(s)=\int _{0}^{\infty }e^{-su}q(u)du$ .

TheoremF .
Under the condition of the risk model,
The ruin probability 𝜙_d caused by the claims satisfies the defective renewal equation: $\begin{eqnarray}\displaystyle {\phi}_{d}(u)=k(u)+\int _{0}^{u}{\phi}_{d}(u-x)g(x)dx. & & \displaystyle\end{eqnarray}$ (16)

The ruin probability 𝜓_d caused by the oscillation satisfies the defective renewal equation: $\begin{eqnarray}\displaystyle {\psi}_{d}(u)=h(u)+\int _{0}^{u}{\psi}_{d}(u-x)g(x)dx. & & \displaystyle\end{eqnarray}$ (17) The Laplace transform of k, g and h are given by: $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-24.0pt}\tilde{k}(s)={\displaystyle \frac{-2{\lambda}_{1}{\lambda}\tilde{f}_{X}(s)[\tilde{q}(s)-E(X)]}{{\sigma}^{2}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)\left({\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}\right)}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-20.0pt}+{\displaystyle \frac{{\lambda}_{1}[\tilde{q}(s)-\tilde{q}(r)-\tilde{{\phi}}(r)\tilde{f}_{X}(r)]}{{\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}}}\left(1-{\displaystyle \frac{2{\lambda}(1-\tilde{f}_{X}(s))}{{\sigma}^{2}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}}\right),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-24.0pt}\tilde{h}(s)={\displaystyle \frac{{\displaystyle \frac{{\sigma}^{2}}{2}}(r-s)-{\lambda}_{1}\tilde{{\psi}}(r)\tilde{f}_{X}(r)}{{\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}}}\left(1-{\displaystyle \frac{2{\lambda}(1-\tilde{f}_{X}(s))}{{\sigma}^{2}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}}\right)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-20.0pt}+{\displaystyle \frac{{\lambda}_{1}\tilde{f}_{X}(s)}{\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)\left({\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}\right)}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-24.0pt}\tilde{g}(s)={\displaystyle \frac{2{\lambda}(1-\tilde{f}_{X}(s))}{{\sigma}^{2}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}},\tilde{{\phi}}(r)={\displaystyle \frac{{\lambda}[\tilde{q}(r)-E(X)]}{{\lambda}-{\lambda}_{1}-{\lambda}\tilde{f}_{X}(r)}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-24.0pt}\tilde{{\psi}}(r)={\displaystyle \frac{-{\displaystyle \frac{{\sigma}^{2}}{2}}r}{{\lambda}-{\lambda}_{1}-{\lambda}\tilde{f}_{X}(r)}}.\nonumber\end{eqnarray}$ and r the positive root of ${\lambda}_{1}-cs-\frac{{\sigma}^{2}}{2}s^{2}=0$ .

Note that when 𝜆₁ →𝜆 we get the result (12) of [9].

TheoremG .
Taking the Laplace transform of (5), we have $\begin{eqnarray}\displaystyle {\lambda}_{1}\tilde{{\phi}}_{d}(s) & = & \displaystyle c[s\tilde{{\phi}}_{d}(s)-{\phi}_{d}(0)]\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +∼{\displaystyle \frac{{\sigma}^{2}}{2}}[s^{2}\tilde{{\phi}}_{d}(s)-s{\phi}_{d}(0)-{\phi}_{d}^{\prime }(0)]\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +∼{\lambda}_{1}\left[\tilde{{\phi}}(s)\tilde{f}_{X}(s)+\tilde{q}(s)\right],\nonumber\end{eqnarray}$ which gives, $\begin{eqnarray}\displaystyle \hspace{-18.0pt}\tilde{{\phi}}_{d}(s)={\displaystyle \frac{-c{\phi}_{d}(0)-{\displaystyle \frac{{\sigma}^{2}}{2}}[s{\phi}_{d}(0)+{\phi}_{d}^{\prime }(0)]+{\lambda}_{1}[\tilde{{\phi}}(s)\tilde{f}_{X}(s)+\tilde{q}(s)]}{{\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}}}. & & \displaystyle \nonumber\end{eqnarray}$

Substituting the positive root r of the equation 𝜆₁− $cs-\frac{{\sigma}^{2}}{2}s^{2}=0$ and 𝜙_d(0) = 0 in the equivalent expression $({\lambda}-cs-\frac{{\sigma}^{2}}{2}s^{2}-{\lambda}\tilde{f}_{X}(s))\tilde{{\phi}}(s)=-c{\phi}(0)-\frac{{\sigma}^{2}}{2}$ $[s{\phi}(0)+{\phi}^{\prime }(0)]+{\lambda}\tilde{q}(s)$ implies that ${\lambda}_{1}[\tilde{{\phi}}(r)\tilde{f}_{X}(r)+\tilde{q}(r)]=\frac{{\sigma}^{2}}{2}{\phi}_{d}^{\prime }(0)$ . Thus $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-24.0pt}\tilde{{\phi}}_{d}(s)={\displaystyle \frac{{\lambda}_{1}[\tilde{q}(s)+\tilde{{\phi}}(s)\tilde{f}_{X}(s)-\tilde{q}(r)-\tilde{{\phi}}(r)\tilde{f}_{X}(r)]}{{\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}}}.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$ (18) In the expression of $\tilde{{\phi}}$ given by the Laplace transform from the Eq. ((13)), by substituting (𝜌 = 0) the null root of the equation ${\lambda}-cs-\frac{{\sigma}^{2}}{2}s^{2}={\lambda}\tilde{f}_{X}(s)$ and 𝜙(0) = 0 implies ${\lambda}E(X)={\lambda}\tilde{q}(0)=\frac{{\sigma}^{2}}{2}{\phi}^{\prime }(0)$ . Thus we get $\begin{eqnarray}\displaystyle \tilde{{\phi}}(s) & = & \displaystyle {\displaystyle \frac{-c{\phi}(0)-{\displaystyle \frac{{\sigma}^{2}}{2}}[s{\phi}(0)+{\phi}^{\prime }(0)]+{\lambda}\tilde{q}(s)}{{\lambda}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}-{\lambda}\tilde{f}_{X}(s)}}\nonumber\\ \displaystyle & = & \displaystyle {\displaystyle \frac{{\lambda}\tilde{q}(s)-{\lambda}E(X)}{{\lambda}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}-{\lambda}\tilde{f}_{X}(s)}}.\end{eqnarray}$ (19) By substituting ((19)) in ((18)) leads to the following expression of the Laplace transform: $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\tilde{{\phi}}_{d}(s)={\displaystyle \frac{{\lambda}_{1}[\tilde{q}(s)-\tilde{q}(r)-\tilde{{\phi}}(r)\tilde{f}_{X}(r)]}{\left({\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}\right)}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-20.0pt}+∼{\displaystyle \frac{{\lambda}_{1}{\lambda}\tilde{f}_{X}(s)[\tilde{q}(s)-E(X)]}{\left({\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}\right)\left({\lambda}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}-{\lambda}\tilde{f}_{X}(s)\right)}}.\end{eqnarray}$ (20)

Equivalently from ((20)) we have $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\left({\lambda}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}-{\lambda}\tilde{f}_{X}(s)\right)\tilde{{\phi}}_{d}(s)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}={\displaystyle \frac{{\lambda}_{1}[\tilde{q}(s)-\tilde{q}(r)-\tilde{{\phi}}(r)\tilde{f}_{X}(r)]}{\left({\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}\right)}}\left({\lambda}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}-{\lambda}\tilde{f}_{X}(s)\right)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-10.0pt}+∼{\displaystyle \frac{{\lambda}_{1}\tilde{f}_{X}(s)[{\lambda}\tilde{q}(s)-{\lambda}E(X)]}{\left({\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}\right)}}.\end{eqnarray}$ (21) And from the null root 𝜌 = 0 of the Lundberg equation we have the following $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\lambda}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}-{\lambda}\tilde{f}_{X}(s)=-{\displaystyle \frac{{\sigma}^{2}}{2}}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-10.0pt}+∼{\lambda}(1-\tilde{f}_{X}(s)).\end{eqnarray}$ (22) Substituting ((22)) in the Eq. ((21)) we get: $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\left(-{\displaystyle \frac{{\sigma}^{2}}{2}}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)+{\lambda}(1-\tilde{f}_{X}(s))\right)\tilde{{\phi}}_{d}(s)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-10.0pt}={\displaystyle \frac{{\lambda}_{1}[\tilde{q}(s)-\tilde{q}(r)-\tilde{{\phi}}(r)\tilde{f}_{X}(r)]}{\left({\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}\right)}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \times \left(-{\displaystyle \frac{{\sigma}^{2}}{2}}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)+{\lambda}\left(1-\tilde{f}_{X}(s)\right)\right)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +∼{\displaystyle \frac{{\lambda}_{1}\tilde{f}_{X}(s)[{\lambda}\tilde{q}(s)-{\lambda}E(X)]}{\left({\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}\right)}}\nonumber\end{eqnarray}$ which is equivalent to $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\left(-1+{\displaystyle \frac{{\lambda}(1-\tilde{f}_{X}(s))}{{\displaystyle \frac{{\sigma}^{2}}{2}}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}}\right)\tilde{{\phi}}_{d}(s)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-15.0pt}={\displaystyle \frac{{\lambda}_{1}[\tilde{q}(s)-\tilde{q}(r)-\tilde{{\phi}}(r)\tilde{f}_{X}(r)]}{\left({\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}\right)}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \times ∼\left(-1+{\displaystyle \frac{{\lambda}(1-\tilde{f}_{X}(s))}{{\displaystyle \frac{{\sigma}^{2}}{2}}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}}\right)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +∼{\displaystyle \frac{{\lambda}_{1}\tilde{f}_{X}(s)[{\lambda}\tilde{q}(s)-{\lambda}E(X)]}{{\displaystyle \frac{{\sigma}^{2}}{2}}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)\left({\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}\right)}},\nonumber\end{eqnarray}$ Thus $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\tilde{{\phi}}_{d}(s)={\displaystyle \frac{2{\lambda}(1-\tilde{f}_{X}(s))}{{\sigma}^{2}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}}\tilde{{\phi}}_{d}(s)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}+∼{\displaystyle \frac{{\lambda}_{1}[\tilde{q}(s)-\tilde{q}(r)-\tilde{{\phi}}(r)\tilde{f}_{X}(r)]}{\left({\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}\right)}}\left(1-{\displaystyle \frac{2{\lambda}(1-\tilde{f}_{X}(s))}{{\sigma}^{2}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}}\right)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}+{\displaystyle \frac{2{\lambda}_{1}\tilde{f}_{X}(s)[{\lambda}E(X)-{\lambda}\tilde{q}(s)]}{{\sigma}^{2}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)\left({\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}\right)}}.\nonumber\end{eqnarray}$ Setting $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\tilde{k}(s)={\displaystyle \frac{2{\lambda}_{1}\tilde{f}_{X}(s)[{\lambda}E(X)-{\lambda}\tilde{q}(s)]}{{\sigma}^{2}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)\left({\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}\right)}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-15.0pt}+\left({\displaystyle \frac{{\lambda}_{1}[\tilde{q}(s)-\tilde{q}(r)-\tilde{{\phi}}(r)\tilde{f}_{X}(r)]}{{\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}}}\right)\left(1-{\displaystyle \frac{2{\lambda}(1-\tilde{f}_{X}(s))}{{\sigma}^{2}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}}\right),\nonumber\end{eqnarray}$ and $\begin{eqnarray}\displaystyle \tilde{g}(s)={\displaystyle \frac{2{\lambda}(1-\tilde{f}_{X}(s))}{{\sigma}^{2}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}}, & & \displaystyle \nonumber\end{eqnarray}$ we get $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \tilde{{\phi}}_{d}(s)=\tilde{k}(s)+\tilde{g}(s)\tilde{{\phi}}_{d}(s)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle {\phi}_{d}(u)=k(u)+\int _{0}^{u}{\phi}_{d}(u-x)g(x)dx.\nonumber\end{eqnarray}$ 2- In the same way, taking the Laplace transform of ((14)), ((15)) and introducing the roots 𝜌 = 0 and r we get the following: $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\tilde{{\psi}}_{d}(s)\left({\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}\right)={\lambda}_{1}\tilde{{\psi}}(s)\tilde{f}_{X}(s)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}-∼{\displaystyle \frac{{\sigma}^{2}}{2}}s{\psi}_{d}(0)-\left({\lambda}_{1}\tilde{{\psi}}(r)\tilde{f}_{X}(r)-{\displaystyle \frac{{\sigma}^{2}}{2}}r{\psi}_{d}(0)\right),\end{eqnarray}$ (23) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \tilde{{\psi}}(s)={\displaystyle \frac{-{\displaystyle \frac{{\sigma}^{2}}{2}}{\psi}(0)s}{{\lambda}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}-{\lambda}\tilde{f}_{X}(s)}},\end{eqnarray}$ (24) with $c{\psi}(0)+\frac{{\sigma}^{2}}{2}{\psi}^{\prime }(0)=-\frac{{\sigma}^{2}}{2}{\rho}{\psi}(0)=0$ and $c{\psi}_{d}(0)+{\displaystyle \frac{{\sigma}^{2}}{2}}{\psi}_{d}^{\prime }(0)={\lambda}_{1}\tilde{{\psi}}(r)\tilde{f}_{X}(r)-{\displaystyle \frac{{\sigma}^{2}}{2}}r{\psi}_{d}(0)$ . Since 𝜓(0) = 𝜓_d(0) =1, substituting ((24)) in ((23)) and using ((22)) we get, $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\tilde{{\psi}}_{d}(s)=\left({\displaystyle \frac{2{\lambda}(1-\tilde{f}_{X}(s))}{{\sigma}^{2}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}}\right)\tilde{{\psi}}_{d}(s)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-15.0pt}+\left({\displaystyle \frac{{\displaystyle \frac{{\sigma}^{2}}{2}}s-{\lambda}_{1}\tilde{{\psi}}(r)\tilde{f}_{X}(r)}{{\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}}}\right)\left(1-{\displaystyle \frac{2{\lambda}(1-\tilde{f}_{X}(s))}{{\sigma}^{2}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}}\right)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-15.0pt}+∼{\displaystyle \frac{{\lambda}_{1}\tilde{f}_{X}(s)s}{s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)\left({\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}\right)}}.\nonumber\end{eqnarray}$

Thus $\begin{eqnarray}\displaystyle \hspace{-22.0pt}\tilde{h}(s) & = & \displaystyle \left({\displaystyle \frac{{\displaystyle \frac{{\sigma}^{2}}{2}}(r-s)-{\lambda}_{1}\tilde{{\psi}}(r)\tilde{f}_{X}(r)}{{\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}}}\right)\left(1-{\displaystyle \frac{2{\lambda}(1-\tilde{f}_{X}(s))}{{\sigma}^{2}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}}\right)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +∼{\displaystyle \frac{{\lambda}_{1}\tilde{f}_{X}(s)}{\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)\left({\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}\right)}}.\nonumber\end{eqnarray}$

TheoremH .
If 𝜆₁ →𝜆 we get the result of [19] and the one of [9] with 𝛿 = 0. $\begin{eqnarray}\displaystyle \begin{array}{@{}l@{}}\lim _{{\lambda}_{1}\rightarrow {\lambda}}\tilde{k}(s)={\displaystyle \frac{2{\lambda}[E(X)-\tilde{q}(s)]}{{\sigma}^{2}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}},\\ \lim _{{\lambda}_{1}\rightarrow {\lambda}}\tilde{g}(s)={\displaystyle \frac{2{\lambda}(1-\tilde{f}_{X}(s))}{{\sigma}^{2}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}},\\ \lim _{{\lambda}_{1}\rightarrow {\lambda}}\tilde{h}(s)={\displaystyle \frac{1}{s+{\displaystyle \frac{2c}{{\sigma}^{2}}}}}.\end{array} & & \displaystyle\end{eqnarray}$ (25)

the reasons for the remark are that: for ${\lambda}_{1}\rightarrow {\lambda},\tilde{q}(s)-$ $\tilde{q}(r)-\tilde{{\phi}}(r)\tilde{f}_{X}(r)=\tilde{q}(s)-\tilde{q}(r)+\frac{{\lambda}\tilde{f}_{X}(r)[\tilde{q}(r)-E(X)]}{{\lambda}-{\lambda}_{1}-{\lambda}\tilde{f}_{X}(r)}\rightarrow \tilde{q}(s)-E(X)$ , thus $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-27.0pt}\lim _{{\lambda}_{1}\rightarrow {\lambda}}\tilde{k}(s)={\displaystyle \frac{2{\lambda}[\tilde{q}(s)-E(X)]}{{\sigma}^{2}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-21.0pt}\times ∼\left[{\displaystyle \frac{-{\lambda}\tilde{f}_{X}(s)}{({\lambda}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2})}}+{\displaystyle \frac{{\displaystyle \frac{{\sigma}^{2}}{2}}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}{({\lambda}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2})}}+{\displaystyle \frac{-{\lambda}(1-\tilde{f}_{X}(s))}{\left({\lambda}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}\right)}}\right].\nonumber\end{eqnarray}$ From the Eq. (22) we get $\begin{eqnarray}\displaystyle \displaystyle \lim _{{\lambda}_{1}\rightarrow {\lambda}}\tilde{k}(s)={\displaystyle \frac{2{\lambda}[E(X)-\tilde{q}(s)]}{{\sigma}^{2}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}}. & & \displaystyle \nonumber\end{eqnarray}$ We also note that $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\lim _{{\lambda}_{1}\rightarrow {\lambda}}\tilde{h}(s)=\left({\displaystyle \frac{-{\displaystyle \frac{{\sigma}^{2}}{2}}s}{{\lambda}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}}}\right)\left(1-{\displaystyle \frac{2{\lambda}(1-\tilde{f}_{X}(s))}{{\sigma}^{2}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}}\right)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-15.0pt}+{\displaystyle \frac{{\lambda}\tilde{f}_{X}(s)s}{s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)\left({\lambda}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}\right)}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}=\left({\displaystyle \frac{{\displaystyle \frac{{\sigma}^{2}}{2}}s}{{\displaystyle \frac{{\sigma}^{2}}{2}}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}}\right)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-15.0pt}\times ∼\left({\displaystyle \frac{{\lambda}\tilde{f}_{X}(s)}{({\lambda}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2})}}-{\displaystyle \frac{{\displaystyle \frac{{\sigma}^{2}}{2}}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}{{\lambda}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}}}+{\displaystyle \frac{2{\lambda}(1-\tilde{f}_{X}(s))}{{\lambda}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}}}\right)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}={\displaystyle \frac{1}{s+{\displaystyle \frac{2c}{{\sigma}^{2}}}}}\text{ from the equation (22).}\nonumber\end{eqnarray}$ Finally ${\psi}(u)=e^{-\frac{2c}{{\sigma}^{2}}u}+\int _{0}^{u}{\psi}(u-x)g(x)dx$ with g given in (26).
6.1. Expressions of g, k and h

With the following relation for all positive 𝜌, $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\displaystyle \frac{\tilde{q}({\rho})-\tilde{q}(s)}{s-{\rho}}}=\int _{0}^{\infty }\tilde{q}({\rho})e^{-sy}e^{{\rho}y}dy\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-5.0pt}-\int _{0}^{\infty }e^{-sy}\int _{0}^{y}e^{{\rho}(y-{\alpha})}q({\alpha})d{\alpha}dy\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-15.0pt}=\int _{0}^{\infty }\left(\int _{0}^{\infty }e^{-{\rho}{\alpha}}q({\alpha})d{\alpha}\right)e^{-sy}e^{{\rho}y}dy\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-5.0pt}-\int _{0}^{\infty }e^{-sy}\int _{0}^{y}e^{{\rho}(y-{\alpha})}q({\alpha})d{\alpha}dy\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-15.0pt}=\int _{0}^{\infty }e^{-sy}\left(\int _{0}^{\infty }e^{(y-{\alpha}){\rho}}q({\alpha})d{\alpha}\right)dy\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-5.0pt}-\int _{0}^{\infty }e^{-sy}\int _{0}^{y}e^{{\rho}(y-{\alpha})}q({\alpha})d{\alpha}dy\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-15.0pt}=\int _{0}^{\infty }e^{-sy}\left(\int _{y}^{\infty }e^{{\rho}(y-{\alpha})}q({\alpha})d{\alpha}\right)dy,\nonumber\end{eqnarray}$ if we set $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \tilde{A}(s)={\displaystyle \frac{2{\lambda}_{1}\tilde{f}_{X}(s)}{{\sigma}^{2}(s-r)\left(s+r+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \tilde{B}(s)={\displaystyle \frac{2{\lambda}[E(X)-\tilde{q}(s)]}{{\sigma}^{2}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \tilde{h}_{1}={\displaystyle \frac{\left[{\displaystyle \frac{{\sigma}^{2}}{2}}(r-s)-{\lambda}_{1}\tilde{{\psi}}(r)\tilde{f}_{X}(r)\right]}{\left({\lambda}_{1}-cs-{\displaystyle \frac{{\sigma}^{2}}{2}}s^{2}\right)}}\nonumber\end{eqnarray}$ $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\tilde{D}(s)={\displaystyle \frac{2{\lambda}_{1}(\tilde{q}(s)-\tilde{q}(r)-\tilde{{\phi}}(r)\tilde{f}_{X}(r))}{{\sigma}^{2}s\left(s+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}={\displaystyle \frac{2{\lambda}_{1}(\tilde{q}(r)-\tilde{q}(s))}{{\sigma}^{2}(s-r)\left(s+r+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-15.0pt}+{\displaystyle \frac{2{\lambda}_{1}\tilde{f}_{X}(r)}{{\sigma}^{2}(s-r)\left(s+r+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}}{\displaystyle \frac{{\lambda}(\tilde{q}(r)-E(X))}{{\lambda}-{\lambda}_{1}-{\lambda}\tilde{f}_{X}(r)}}.\nonumber\end{eqnarray}$ $\tilde{k}$ and $\tilde{h}$ can be re-expressed as $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \tilde{k}(s)=\tilde{A}(s)\tilde{B}(s)+\tilde{D}(s)(1-\tilde{g}(s)),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \tilde{h}(s)=\tilde{h}_{1}(s)(1-\tilde{g}(s))-{\displaystyle \frac{\tilde{A}(s)}{s+{\displaystyle \frac{2c}{{\sigma}^{2}}}}}.\nonumber\end{eqnarray}$ Then the expression of g, k, h are $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}g(x)={\displaystyle \frac{2{\lambda}}{{\sigma}^{2}}}\int _{0}^{x}e^{-\left({\rho}+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)(x-y)}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \times ∼\left(\int _{y}^{\infty }e^{{\rho}(y-{\alpha})}f_{X}({\alpha})d{\alpha}\right)dy,\end{eqnarray}$ (26) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}k(x)=\int _{0}^{x}A(x-y)B(y)dy+D(x)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle -\int _{0}^{x}g(x-y)D(y)dy,\end{eqnarray}$ (27) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}h(x)=h_{1}(x)-\int _{0}^{x}g(x-y)h_{1}(y)dy+D(x)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle -\int _{0}^{x}A(x-y)e^{-{\displaystyle \frac{2c}{{\sigma}^{2}}}y}dy,\end{eqnarray}$ (28) where $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}A(x)={\displaystyle \frac{2{\lambda}_{1}}{{\sigma}^{2}}}\int _{0}^{x}e^{-(r+{\displaystyle \frac{2c}{{\sigma}^{2}}})(x-{\alpha})}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \quad \times ∼\left(\int _{0}^{{\alpha}}e^{r({\alpha}-y)}f_{X}(y)dy\right)d{\alpha},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}B(x)={\displaystyle \frac{2{\lambda}}{{\sigma}^{2}}}\int _{0}^{x}e^{-\left({\rho}+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)(x-y)}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \quad \times ∼\left(\int _{y}^{\infty }e^{{\rho}(y-{\alpha})}q({\alpha})d{\alpha}\right)dy,\nonumber\end{eqnarray}$ $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-24.0pt}D(x)={\displaystyle \frac{2{\lambda}}{{\sigma}^{2}}}\left[\int _{0}^{x}e^{-\left(r+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)(x-y)}\hspace{-3.0pt}\left(\int _{y}^{\infty }e^{r(y-{\alpha})}q({\alpha})d{\alpha}\right)dy\right.\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}+\left.{\displaystyle \frac{{\lambda}\tilde{f}_{X}(r)(\tilde{q}(r)-\tilde{q}({\rho}))}{{\lambda}-{\lambda}_{1}-{\lambda}\tilde{f}_{X}(r)}}\int _{0}^{x}e^{r(x-y)}e^{-\left(r+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)y}dy\right],\nonumber\end{eqnarray}$ $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}h_{1}(x)=e^{-\left(r+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)x}+{\displaystyle \frac{2{\lambda}_{1}}{{\sigma}^{2}}}\tilde{{\psi}}(r)\tilde{f}_{X}(r)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \quad \times ∼\int _{0}^{x}e^{r(x-y)}e^{-{\displaystyle \frac{2c}{{\sigma}^{2}}}y}dy.\nonumber\end{eqnarray}$

7. Representation of the solution

The goal of this section is to derive the solution of the integro-differential equations. Let’s $G(x)=\frac{\int _{0}^{x}g(y)dy}{\int _{0}^{\infty }g(y)dy}$ be the the associated claim size c.d.f.

As discussed in detail in [26], Properties of the solution of the defective renewal equation depend heavily on those of the associated claim size distribution. The solutions of the Eqs (16) and (17) can be represented as follow, $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\phi}_{d}(u)=k(u)+{\displaystyle \frac{1}{1-{\pi}}}\int _{0}^{u}k(u-x)dQ(x),\end{eqnarray}$ (29) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\psi}_{d}(u)=h(u)+{\displaystyle \frac{1}{1-{\pi}}}\int _{0}^{u}h(u-x)dQ(x),\end{eqnarray}$ (30) Where $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\pi}=\int _{0}^{\infty }g(x)dx=\tilde{g}(0)={\displaystyle \frac{{\lambda}E(X)}{c}},\end{eqnarray}$ (31) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle Q(x)=\displaystyle \mathop{\sum }_{n=1}^{\infty }(1-{\pi}){\pi}^{n}G^{\ast n}(x),\end{eqnarray}$ (32) G ^∗n the n-fold convolution of independent random varaible with cdf G. We have 0 < 𝜋 <1 from the condition assuring that the ruin is not almost surely certain.

TheoremI .

Considering that the claims are exponentially distributed i.e X ∼ Exp (𝜃) then the expression of Q defined in (32) satisfies the following $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\displaystyle \frac{1}{1-{\pi}}}{\displaystyle \frac{d}{dx}}Q(x)={\displaystyle \frac{{\displaystyle \frac{2{\lambda}}{{\sigma}^{2}}}}{\sqrt{\left({\theta}-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)^{2}+{\displaystyle \frac{8{\lambda}}{{\sigma}^{2}}}}}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-5.0pt}(e^{-{\beta}x}-e^{-{\alpha}x}),\quad \forall x\geq 0,\end{eqnarray}$ (33) where $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\alpha}={\displaystyle \frac{{\theta}}{2}}+{\displaystyle \frac{c}{{\sigma}^{2}}}+{\displaystyle \frac{1}{2}}\sqrt{\left({\theta}-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)^{2}+{\displaystyle \frac{8{\lambda}}{{\sigma}^{2}}}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle {\beta}={\displaystyle \frac{{\theta}}{2}}+{\displaystyle \frac{c}{{\sigma}^{2}}}-{\displaystyle \frac{1}{2}}\sqrt{\left({\theta}-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)^{2}+{\displaystyle \frac{8{\lambda}}{{\sigma}^{2}}}}.\nonumber\end{eqnarray}$

TheoremJ .

The Laplace transform of (16), (17) we have $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \tilde{{\phi}}_{d}={\displaystyle \frac{\tilde{k}(s)}{1-\tilde{g}(s)}}=\tilde{k}(s)+{\displaystyle \frac{\tilde{g}(s)}{1-\tilde{g}(s)}}\tilde{k}(s),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \tilde{{\psi}}_{d}={\displaystyle \frac{\tilde{h}(s)}{1-\tilde{g}(s)}}=\tilde{h}(s)+{\displaystyle \frac{\tilde{g}(s)}{1-\tilde{g}(s)}}\tilde{h}(s).\nonumber\end{eqnarray}$ Let define R such that $\tilde{R}(s)=\frac{\tilde{g}(s)}{1-\tilde{g}(s)}$ . By identification, with the uniqueness Laplace transform of (29), (30) and taking the inverse of the Laplace transform, we can see that $\frac{1}{1-{\pi}}\frac{d}{dx}Q(x)=R(x)$ . By substituting X ∼ Exp (𝜃) We have that $\begin{eqnarray}\displaystyle {\displaystyle \frac{\tilde{g}(s)}{1-\tilde{g}(s)}}\text{} & =\text{} & \displaystyle {\displaystyle \frac{{\displaystyle \frac{2{\lambda}}{{\sigma}^{2}}}}{s^{2}+\left({\theta}+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)s+{\displaystyle \frac{2c{\theta}}{{\sigma}^{2}}}\left(1-{\displaystyle \frac{{\lambda}}{c{\theta}}}\right)}}\nonumber\\ \displaystyle \text{} & =\text{} & \displaystyle {\displaystyle \frac{2{\lambda}}{{\sigma}^{2}}}{\displaystyle \frac{1}{(s+{\alpha})(s+{\beta})}}.\nonumber\end{eqnarray}$ Since ${\Delta}=({\theta}+\frac{2c}{{\sigma}^{2}})^{2}-\frac{8c{\theta}}{{\sigma}^{2}}(1-\frac{{\lambda}}{c{\theta}})=({\theta}-\frac{2c}{{\sigma}^{2}})^{2}+\frac{8{\lambda}}{{\sigma}^{2}}>0$ thus there exists two negatives roots −𝛼, −𝛽 such that $\begin{eqnarray}\displaystyle \hspace{-16.0pt}{\displaystyle \frac{\tilde{g}(s)}{1-\tilde{g}(s)}} & = & \displaystyle {\displaystyle \frac{2{\lambda}}{{\sigma}^{2}}}{\displaystyle \frac{1}{(s+{\alpha})(s+{\beta})}}\nonumber\\ \displaystyle & = & \displaystyle -{\displaystyle \frac{{\displaystyle \frac{2{\lambda}}{{\sigma}^{2}}}}{\sqrt{\left({\theta}-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)^{2}+{\displaystyle \frac{8{\lambda}}{{\sigma}^{2}}}}}}\left[{\displaystyle \frac{1}{s+{\alpha}}}-{\displaystyle \frac{1}{s+{\beta}}}\right]\nonumber\end{eqnarray}$ with $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\alpha}={\displaystyle \frac{{\theta}}{2}}+{\displaystyle \frac{c}{{\sigma}^{2}}}+{\displaystyle \frac{1}{2}}\sqrt{\left({\theta}-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)^{2}+{\displaystyle \frac{8{\lambda}}{{\sigma}^{2}}}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle {\beta}={\displaystyle \frac{{\theta}}{2}}+{\displaystyle \frac{c}{{\sigma}^{2}}}-{\displaystyle \frac{1}{2}}\sqrt{\left({\theta}-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)^{2}+{\displaystyle \frac{8{\lambda}}{{\sigma}^{2}}}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle {\alpha}-{\beta}=\sqrt{\left({\theta}-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)^{2}+{\displaystyle \frac{8{\lambda}}{{\sigma}^{2}}}}.\nonumber\end{eqnarray}$ By taking inverse of the Laplace transform of $\tilde{R}$ we get: $\begin{eqnarray}\displaystyle R(x)={\displaystyle \frac{{\displaystyle \frac{2{\lambda}}{{\sigma}^{2}}}}{\sqrt{\left({\theta}-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)^{2}+{\displaystyle \frac{8{\lambda}}{{\sigma}^{2}}}}}}(e^{-{\beta}x}-e^{-{\alpha}x}), & & \displaystyle \nonumber\end{eqnarray}$ and notice that $\frac{1-{\pi}}{{\pi}}R$ is a density function.

The mixed-exponential distribution is a good approximation of the claim amounts. Useful computational formulas are available for the ruin probabilities, together with related quantities such as stop-loss moments and the distribution of the deficit at ruin when the claim amount distribution is of mixed-Exponential type (e.g., [12] and [21]). The class of exponential mixtures is preserved under a wide variety of risk-theoretic operations. In particular, the residual lifetime distribution, the equilibrium or integrated tail distribution, the aggregate claims distribution, and the conditional distribution of the deficit at ruin (given that ruin occurs) are all different mixtures of the same Erlangs (e.g., [16,26]). More generally, in the Sparre Andersen or renewal risk model with mixed-Erlang claim amounts, the ladder height distribution is a different mixture of the same Erlangs [22]. The following two theorems gives the formula of the ruin probability when the claim amounts are exponentially distributed and are a mixture of two Exponential distributions.

TheoremK .

Given that the claims sizes are exponentially distributed,

The delayed ruin probability 𝜙_d caused by the claims is: $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-24.0pt}{\phi}_{d}(u)={\displaystyle \frac{{\displaystyle \frac{2{\lambda}_{1}}{{\sigma}^{2}}}\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)}{\left[{\theta}+r\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)\right]\left[{\theta}-\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)\left(r+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)\right]}}e^{-\left(r+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)u}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +∼{\displaystyle \frac{-{\displaystyle \frac{2{\lambda}}{{\sigma}^{2}}}}{({\alpha}-{\beta})\left[{\theta}-{\alpha}\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)\right]}}e^{-{\alpha}u}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +∼{\displaystyle \frac{{\displaystyle \frac{2{\lambda}}{{\sigma}^{2}}}}{({\alpha}-{\beta})\left[{\theta}-{\beta}\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)\right]}}e^{-{\beta}u}.\nonumber\end{eqnarray}$

The delayed ruin probability 𝜓_d caused by the oscillation is: $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\psi}_{d}(u)={\displaystyle \frac{r\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}+{\displaystyle \frac{c{\theta}}{{\lambda}_{1}}}\right)}{\left({\theta}+r\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)\right)\left[1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}-{\displaystyle \frac{{\theta}r{\sigma}^{2}}{2{\lambda}_{1}}}\right]}}e^{-\left(r+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)u}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +∼{\displaystyle \frac{-{\theta}}{({\alpha}-{\beta})\left[1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}+{\displaystyle \frac{{\theta}}{{\lambda}_{1}}}\left(c-{\displaystyle \frac{{\sigma}^{2}}{2}}{\alpha}\right)\right]}}e^{-{\alpha}u}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +∼{\displaystyle \frac{{\theta}}{({\alpha}-{\beta})\left[1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}+{\displaystyle \frac{{\theta}}{{\lambda}_{1}}}\left(c-{\displaystyle \frac{{\sigma}^{2}}{2}}{\beta}\right)\right]}}e^{-{\beta}u}.\nonumber\end{eqnarray}$

The ruin probability in the delayed risk model perturbed by a diffusion if claims are exponentially distributed is therefore $\begin{eqnarray}\displaystyle m_{d}(u)={\phi}_{d}(u)+{\psi}_{d}(u) & & \displaystyle \nonumber\end{eqnarray}$ where 𝛼, 𝛽 are defined previously and $r=-\frac{c}{{\sigma}^{2}}+\sqrt{\frac{c^{2}}{{\sigma}^{4}}+\frac{2{\lambda}_{1}}{{\sigma}^{2}}}$ .

TheoremL .

The proof can be done using the Lemma 7.1 with the expression of h and k in ((26)) the representation of the solution, or using directly the expression of the Laplace transform and inverting the Laplace transform, and draw by the uniqueness the solution. By substituting the exponential distribution directly in the laplace transform of 𝜙_d and 𝜓_d we get $\begin{eqnarray}\displaystyle \tilde{{\phi}}_{d}={\displaystyle \frac{{\displaystyle \frac{-2{\lambda}_{1}^{2}}{{\sigma}^{2}}}\left\{\left(s-r+{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}r\right)\left[s^{2}+s({\theta}+{\displaystyle \frac{2c}{{\sigma}^{2}}})+{\displaystyle \frac{2c{\theta}}{{\sigma}^{2}}}\left(1-{\displaystyle \frac{{\lambda}}{c{\theta}}}\right)\right]+{\displaystyle \frac{2{\lambda}}{{\sigma}^{2}{\lambda}_{1}}}(r{\lambda}-r{\lambda}_{1}-{\lambda}_{1}{\theta})\right\}}{(r{\lambda}-r{\lambda}_{1}-{\lambda}_{1}{\theta})(s+{\theta})(s-r)\left(s+r+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)\left[s^{2}+s\left({\theta}+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)+{\displaystyle \frac{2c{\theta}}{{\sigma}^{2}}}\left(1-{\displaystyle \frac{{\lambda}}{c{\theta}}}\right)\right]}}\hspace{32.0pt} & & \displaystyle\end{eqnarray}$ (34) and can be written in the following way $\begin{eqnarray}\displaystyle \hspace{-14.0pt}\tilde{{\phi}}_{d}={\displaystyle \frac{a_{1}}{s+{\theta}}}+{\displaystyle \frac{a_{2}}{s-r}}+{\displaystyle \frac{a_{3}}{s+r+{\displaystyle \frac{2c}{{\sigma}^{2}}}}}+{\displaystyle \frac{a_{4}}{s+{\alpha}}}+{\displaystyle \frac{a_{5}}{s+{\beta}}} & & \displaystyle \nonumber\end{eqnarray}$ by drawing and rewriting the coefficients we get a ₁ = a ₂ = 0, $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}a_{3}={\displaystyle \frac{{\displaystyle \frac{2{\lambda}_{1}}{{\sigma}^{2}}}\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)}{\left[{\theta}+r\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)\right]\left[{\theta}-\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)(r+{\sigma}^{2})\right]}},\nonumber\end{eqnarray}$ $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}a_{4}={\displaystyle \frac{-{\displaystyle \frac{2{\lambda}}{{\sigma}^{2}}}}{({\alpha}-{\beta})\left[{\theta}-{\alpha}\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)\right]}},\nonumber\end{eqnarray}$ $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}a_{5}={\displaystyle \frac{{\displaystyle \frac{2{\lambda}}{{\sigma}^{2}}}}{({\alpha}-{\beta})\left[{\theta}-{\beta}\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)\right]}}.\nonumber\end{eqnarray}$ Thus by taking the inverse of the Laplace transform we get the result. In the same way $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-25.0pt}\tilde{{\psi}}_{d}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-20.0pt}={\displaystyle \frac{\left(s-{\displaystyle \frac{r^{2}\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)}{{\theta}+r\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)}}\right)\left[s^{2}+s\left({\theta}+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)+{\displaystyle \frac{2c{\theta}}{{\sigma}^{2}}}\left(1-{\displaystyle \frac{{\lambda}}{c{\theta}}}\right)\right]-{\displaystyle \frac{2{\lambda}_{1}}{{\sigma}^{2}}}}{(s-r)\left(s+r+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)\left[s^{2}+s\left({\theta}+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)+{\displaystyle \frac{2c{\theta}}{{\sigma}^{2}}}\left(1-{\displaystyle \frac{{\lambda}}{c{\theta}}}\right)\right]}}\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$ (35) and can also be written as

$\begin{eqnarray}\displaystyle \hspace{-22.0pt}\tilde{{\psi}}_{d}(s) & = & \displaystyle {\displaystyle \frac{\left({\displaystyle \frac{r\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}+{\displaystyle \frac{c{\theta}}{{\lambda}_{1}}}\right)}{\left({\theta}+r\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)\right)\left[1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}-{\displaystyle \frac{{\theta}r{\sigma}^{2}}{2{\lambda}_{1}}}\right]}}\right)}{s+r+{\displaystyle \frac{2c}{{\sigma}^{2}}}}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +∼{\displaystyle \frac{\left({\displaystyle \frac{-{\theta}}{({\alpha}-{\beta})\left[1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}+{\displaystyle \frac{{\theta}}{{\lambda}_{1}}}\left(c-{\displaystyle \frac{{\sigma}^{2}}{2}}{\alpha}\right)\right]}}\right)}{s+{\alpha}}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +∼{\displaystyle \frac{\left({\displaystyle \frac{{\theta}}{({\alpha}-{\beta})\left[1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}+{\displaystyle \frac{{\theta}}{{\lambda}_{1}}}\left(c-{\displaystyle \frac{{\sigma}^{2}}{2}}{\beta}\right)\right]}}\right)}{s+{\beta}}}.\nonumber\end{eqnarray}$

TheoremM .

Assuming that the arrival time of the first claim is exponentially distributed with parameter 𝜆₁, the inter arrival time of the others claims exponentially distributed with parameter 𝜆 and given that the claims are mixed exponentially distributed,

the ruin probability caused by the claim is $\begin{eqnarray}\displaystyle {\phi}_{d}(u)=a_{1}e^{-\left(r+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)u}+\displaystyle \mathop{\sum }_{i=1}^{3}a_{1+i}e^{-{\rho}_{i}u}. & & \displaystyle\end{eqnarray}$ (36)

the ruin probability caused by the oscillation is $\begin{eqnarray}\displaystyle {\psi}_{d}(u)=b_{1}e^{-(r+{\displaystyle \frac{2c}{{\sigma}^{2}}})u}+\displaystyle \mathop{\sum }_{i=1}^{3}b_{1+i}e^{-{\rho}_{i}u}. & & \displaystyle\end{eqnarray}$ (37)

The ruin probability in the delayed risk model perturbed by a diffusion if claims are mixed exponential is therefore m _d(u) = 𝜙_d(u) + 𝜓_d(u).

where −𝜌_i, i = 1, 2, 3 are the roots of the function $\begin{eqnarray}\displaystyle P_{3}(s) & = & \displaystyle \Bigg\{s^{3}+\left({\theta}_{1}+{\theta}_{2}+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)s^{2}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +∼\left[{\theta}_{1}{\theta}_{2}+{\displaystyle \frac{2c}{{\sigma}^{2}}}({\theta}_{1}+{\theta}_{2})-{\displaystyle \frac{2{\lambda}}{{\sigma}^{2}}}\right]s\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +∼{\displaystyle \frac{2c{\theta}_{1}{\theta}_{2}}{{\sigma}^{2}}}-{\displaystyle \frac{2{\lambda}}{{\sigma}^{2}}}[(1-p){\theta}_{1}+p{\theta}_{2}]\Bigg\},\nonumber\end{eqnarray}$ and the coefficients are given by: $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle a_{1}={\displaystyle \frac{{\displaystyle \frac{{\lambda}_{1}}{{\sigma}^{2}}}w_{1}\left(-r-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}{\left(r+{\displaystyle \frac{c}{{\sigma}^{2}}}\right)\left({\theta}_{1}-r-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)\left({\theta}_{2}-r-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)\left({\rho}_{1}-r-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)\left({\rho}_{2}-r-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)\left({\rho}_{3}-r-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle a_{1+i}={\displaystyle \frac{{\displaystyle \frac{2{\lambda}_{1}}{{\sigma}^{2}}}w_{1}(-{\rho}_{i})}{(r+{\rho}_{i})\left(r+{\displaystyle \frac{2c}{{\sigma}^{2}}}-{\rho}_{i}\right)({\theta}_{1}-{\rho}_{i})({\theta}_{2}-{\rho}_{i})\mathop{\prod }_{j=1,j\neq i}^{3}({\rho}_{j}-{\rho}_{i})}},\quad i=1,2,3,\nonumber\end{eqnarray}$ $\begin{eqnarray}\displaystyle b_{1}={\displaystyle \frac{w_{2}\left(-r-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}{\left(-2r-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)\left({\rho}_{1}-r-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)\left({\rho}_{2}-r-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)\left({\rho}_{3}-r-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)}},\hspace{38.0pt} & & \displaystyle\end{eqnarray}$ (38) $\begin{eqnarray}\displaystyle b_{1+i}={\displaystyle \frac{w_{2}(-{\rho}_{i})}{(-r-{\rho}_{i})(r+{\displaystyle \frac{2c}{{\sigma}^{2}}}-{\rho}_{i})\mathop{\prod }_{j=1,j\neq i}^{3}({\rho}_{j}-{\rho}_{i})}},\quad i=1,2,3. & & \displaystyle \nonumber\end{eqnarray}$ w ₁, w ₂ are defined in ((41)) and ((42)).

TheoremN .

In this proof we first by derive the explicit expression of the Laplace transform when claims are mixed exponential claim and express the ruin probability.

That is X|𝛩 = 𝜃 ∼ Exp (𝜃) where $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\Theta}=\left\{\begin{array}{@{}l@{}}{\theta}_{1}\text{ with probability }p\\ {\theta}_{2}\text{ with probability }1-p\end{array}\right.\quad \forall ∼t>0\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle P[X\leq x]=pP[X\leq x|{\Theta}={\theta}_{1}]\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \qquad +∼(1-p)P[X\leq x|{\Theta}={\theta}_{2}]\nonumber\end{eqnarray}$ then the density function $\begin{eqnarray}\displaystyle f_{X}(x)=pf_{X|{\Theta}={\theta}_{1}}(x)+(1-p)f_{X|{\Theta}={\theta}_{2}}(x) & & \displaystyle \nonumber\end{eqnarray}$ and $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\forall s\in \mathbb{C}\text{ s.th }Re(s)<0,\tilde{f}_{X}(s)=p\tilde{f}_{X|{\Theta}={\theta}_{1}}(s)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}+∼(1-p)\tilde{f}_{X|{\Theta}={\theta}_{2}}(s)=p{\displaystyle \frac{{\theta}_{1}}{s+{\theta}_{1}}}+(1-p){\displaystyle \frac{{\theta}_{2}}{s+{\theta}_{2}}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}q(u)=\int _{u}^{\infty }f_{X}(x)dx,\tilde{q}(s)={\displaystyle \frac{p}{s+{\theta}_{1}}}+{\displaystyle \frac{1-p}{s+{\theta}_{2}}}\nonumber\end{eqnarray}$ With this mixed distribution, the Laplace transform $\tilde{{\phi}}_{d}$ defined in (20) of the probability caused by a claim 𝜙_d and the Laplace transform $\tilde{{\psi}}_{d}$ defined in (23) of the probability caused by oscillation 𝜓_d become: $\begin{eqnarray}\displaystyle \hspace{-18.0pt}\tilde{{\phi}}_{d}(s)={\displaystyle \frac{-{\displaystyle \frac{2{\lambda}_{1}}{{\sigma}^{2}}}w_{1}(s)}{(s-r)\left(s+r+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)(s+{\theta}_{1})(s+{\theta}_{2})P_{3}(s)}}, & & \displaystyle\end{eqnarray}$ (39)

and $\begin{eqnarray}\displaystyle \tilde{{\psi}}_{d}(s)={\displaystyle \frac{w_{2}(s)}{(s-r)\left(s+r+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)P_{3}(s)}}, & & \displaystyle\end{eqnarray}$ (40) with $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}P_{3}(s)=\Bigg\{s^{3}+\left({\theta}_{1}+{\theta}_{2}+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)s^{2}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +∼\left[{\theta}_{1}{\theta}_{2}+{\displaystyle \frac{2c}{{\sigma}^{2}}}({\theta}_{1}+{\theta}_{2})-{\displaystyle \frac{2{\lambda}}{{\sigma}^{2}}}\right]s\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +∼{\displaystyle \frac{2c{\theta}_{1}{\theta}_{2}}{{\sigma}^{2}}}-{\displaystyle \frac{2{\lambda}}{{\sigma}^{2}}}[(1-p){\theta}_{1}+p{\theta}_{2}]\Bigg\},\nonumber\end{eqnarray}$ $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}w_{1}(s)=Q_{2}(s)P_{3}(s)-{\displaystyle \frac{2{\lambda}}{{\sigma}^{2}{\theta}_{1}{\theta}_{2}}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \times ∼\left[{\theta}_{1}{\theta}_{2}+s(p{\theta}_{1}+(1-p){\theta}_{2})\right]\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \times ∼[{\theta}_{1}{\theta}_{2}-(p{\theta}_{2}+(1-p){\theta}_{1})(s+{\theta}_{1}+{\theta}_{2})],\end{eqnarray}$ (41) and $\begin{eqnarray}\displaystyle Q_{2}(s) & = & \displaystyle -{\displaystyle \frac{(s-r)\{s(r+p{\theta}_{2}+(1-p){\theta}_{1})-{\theta}_{1}{\theta}_{2}+(p{\theta}_{2}+(1-p){\theta}_{1})(r+{\theta}_{1}+{\theta}_{2})\}}{(r+{\theta}_{1})(r+{\theta}_{2})}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle -∼{\displaystyle \frac{{\lambda}r\{{\theta}_{1}{\theta}_{2}-(p{\theta}_{2}+(1-p){\theta}_{1})[(r+{\theta}_{1}+{\theta}_{2})]\}\{{\theta}_{1}{\theta}_{2}+r[(1-p){\theta}_{2}+p{\theta}_{1}]\}}{{\theta}_{1}{\theta}_{2}\{-{\lambda}_{1}(r+{\theta}_{1})(r+{\theta}_{2})+{\lambda}r[r+(1-p){\theta}_{1}+p{\theta}_{2}]\}(r+{\theta}_{1})(r+{\theta}_{2})}}\{(s+{\theta}_{1})(s+{\theta}_{2})\},\nonumber\end{eqnarray}$ with $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}w_{2}(s)=\left[s-r-{\displaystyle \frac{{\lambda}_{1}r[{\theta}_{1}{\theta}_{2}+r(p{\theta}_{1}+(1-p){\theta}_{2})]}{-{\lambda}_{1}(r+{\theta}_{1})(r+{\theta}_{2})+{\lambda}r[+(1-p){\theta}_{1}+p{\theta}_{2}]}}\right]\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}P_{3}(s)-{\displaystyle \frac{2{\lambda}_{1}}{{\sigma}^{2}}}[{\theta}_{1}{\theta}_{2}+s(p{\theta}_{1}+(1-p){\theta}_{2})].\end{eqnarray}$ (42) If we set p =1, the expressions of Laplace transform (39) and (40) become the expressions in (34) and (35) of the case exponential claim.

The degree three polynomial function P ₃(s) has exactly three roots in the set of complex number $\mathbb{C}$ . The Eq. ((4)) is equivalent to $\begin{eqnarray}\displaystyle {\displaystyle \frac{-{\displaystyle \frac{{\sigma}^{2}}{2}}sP_{3}(s)}{(s+{\theta}_{1})(s+{\theta}_{2})}} & = & \displaystyle 0.\nonumber\end{eqnarray}$ By the Section 4, the Eq. ((4)) has exactly one root(null root) in the right complex plane, thus the other root are in the left complex plane. The coefficient of the degree three polynomial function P ₃(s) are positive thus one root of P ₃(s) at least is negative say −𝜌₁ and the other −𝜌_i, i = 2,3 are also negative or conjugate such that Re (−𝜌_i) < 0, i = 2,3. We notice that w ₁(−𝜃₁) = w ₁(−𝜃₂) = w ₁(r) = 0 and w ₂(r) = 0. The Eqs ((39)) and ((40)) can the be written as $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\tilde{{\phi}}_{d}(s)={\displaystyle \frac{a_{1}}{s+r+{\displaystyle \frac{2c}{{\sigma}^{2}}}}}+{\displaystyle \frac{a_{2}}{s+{\rho}_{1}}}+{\displaystyle \frac{a_{3}}{s+{\rho}_{2}}}+{\displaystyle \frac{a_{4}}{s+{\rho}_{3}}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\tilde{{\psi}}_{d}(s)={\displaystyle \frac{b_{1}}{s+r+{\displaystyle \frac{2c}{{\sigma}^{2}}}}}+{\displaystyle \frac{b_{2}}{s+{\rho}_{1}}}+{\displaystyle \frac{b_{3}}{s+{\rho}_{2}}}+{\displaystyle \frac{b_{4}}{s+{\rho}_{3}}},\nonumber\end{eqnarray}$ where the coeffients a _i, b _i, i = 1, …,4 are defined in ((38)). By taking the inverse of the Laplace transforms, we obtain the result.

7.1. Asymptotic behaviour of the ruin probability

In this section we analyse the behaviour of the ruin probability in the delayed risk model perturbed by a diffusion when claims are exponentially distributed. We first consider the case the initial surplus is infinity then let the volatility tends into zero and obtained the ruin probability in the ordinary risk model with delayed in the arrival of the first claims. And by letting 𝜆₁ tends to 𝜆, we compare the resulting ruin probability with ordinary case to see the consistency of the results. $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\displaystyle \lim _{{\sigma}\rightarrow 0}{\alpha}=+\infty ,\displaystyle \lim _{{\sigma}\rightarrow 0}r+{\displaystyle \frac{2c}{{\sigma}^{2}}}=+\infty ,\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\displaystyle \lim _{{\sigma}\rightarrow 0}{\beta}=\lim {\displaystyle \frac{2c{\theta}-2{\lambda}}{{\displaystyle \frac{{\sigma}^{2}{\theta}}{2}}+c+{\displaystyle \frac{1}{2}}\sqrt{({\sigma}^{2}{\theta}-2c)^{2}+8{\sigma}^{2}{\lambda}}}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle ={\theta}-{\displaystyle \frac{{\lambda}}{c}}.\nonumber\end{eqnarray}$ The ruin probability in the ordinary and perturbed risk model is given by the sum of $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\displaystyle \lim _{{\lambda}_{1}\rightarrow {\lambda}}{\phi}_{d}(u)={\displaystyle \frac{-2{\lambda}}{{\theta}({\alpha}-{\beta}){\sigma}^{2}}}e^{-{\alpha}u}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +∼{\displaystyle \frac{2{\lambda}}{{\theta}({\alpha}-{\beta}){\sigma}^{2}}}e^{-{\beta}u},\nonumber\end{eqnarray}$ and $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\displaystyle \lim _{{\lambda}_{1}\rightarrow {\lambda}}{\psi}_{d}(u)={\displaystyle \frac{-{\lambda}}{({\alpha}-{\beta})\left(c-{\displaystyle \frac{{\sigma}^{2}}{2}}{\alpha}\right)}}e^{-{\alpha}u}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +∼{\displaystyle \frac{{\lambda}}{({\alpha}-{\beta})\left(c-{\displaystyle \frac{{\sigma}^{2}}{2}}{\beta}\right)}}e^{-{\beta}u}.\nonumber\end{eqnarray}$ Now letting the volatility 𝜎 tends to zero, we obtain $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\lim _{{\sigma}\rightarrow 0}{\phi}_{d}(u)={\displaystyle \frac{{\displaystyle \frac{{\lambda}}{c}}}{{\theta}-\left({\theta}-{\displaystyle \frac{{\lambda}}{c}}\right)\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)}}e^{-\left({\theta}-{\displaystyle \frac{{\lambda}}{c}}\right)u},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\text{and }\lim _{{\sigma}\rightarrow 0}{\psi}_{d}(u)=0,\nonumber\end{eqnarray}$ which leads to the expression of the delayed ruin probability of the delayed ordinary risk model.

The ultimate ruin probability with infinity initial surplus is given by $\begin{eqnarray}\displaystyle \lim _{u\rightarrow \infty }m_{d}(u)=\lim _{u\rightarrow \infty }({\phi}_{d}(u)+{\psi}_{d}(u))=0. & & \displaystyle \nonumber\end{eqnarray}$ Note that if we let simultaneously 𝜆₁ →𝜆 and 𝜎 →0 we get the well known ruin probability in the ordinary Sparre Anderson risk model, i.e $\begin{eqnarray}\displaystyle \hspace{-22.0pt}\lim _{{\lambda}_{1}\rightarrow {\lambda},{\sigma}\rightarrow 0}m_{d}(u)=\lim _{{\lambda}_{1}\rightarrow {\lambda},{\sigma}\rightarrow 0}({\phi}_{d}(u))={\displaystyle \frac{{\lambda}}{c{\theta}}}e^{-\left({\theta}-{\displaystyle \frac{{\lambda}}{c}}\right)u}. & & \displaystyle \nonumber\end{eqnarray}$ And in the case the initial surplus is zero, We have $\begin{eqnarray}\displaystyle \lim _{{\lambda}_{1}\rightarrow {\lambda}}{\phi}_{d}(0)={\displaystyle \frac{-2{\lambda}}{({\alpha}-{\beta}){\theta}{\sigma}^{2}}}+{\displaystyle \frac{2{\lambda}}{({\alpha}-{\beta}){\theta}{\sigma}^{2}}}=0, & & \displaystyle \nonumber\end{eqnarray}$ and $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\lim _{{\lambda}_{1}\rightarrow {\lambda}}{\psi}_{d}(0)={\displaystyle \frac{{\lambda}}{{\alpha}-{\beta}}}\left({\displaystyle \frac{-1}{c-{\displaystyle \frac{{\sigma}^{2}}{2}}{\alpha}}}+{\displaystyle \frac{1}{c-{\displaystyle \frac{{\sigma}^{2}}{2}}{\beta}}}\right)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle ={\displaystyle \frac{-{\lambda}{\displaystyle \frac{{\sigma}^{2}}{2}}}{c^{2}-{\displaystyle \frac{{\sigma}^{2}}{2}}({\alpha}+{\beta})c+{\displaystyle \frac{{\sigma}^{4}}{4}}{\alpha}{\beta}}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle ={\displaystyle \frac{-{\lambda}{\displaystyle \frac{{\sigma}^{2}}{2}}}{c^{2}-{\displaystyle \frac{c{\theta}{\sigma}^{2}}{2}}-c^{2}+{\displaystyle \frac{{\sigma}^{2}}{2}}(c{\theta}-{\lambda})}}=1,\end{eqnarray}$ (43) thus $\begin{eqnarray}\displaystyle \lim _{{\lambda}_{1}\rightarrow {\lambda}}m_{d}(0)=\lim _{{\lambda}_{1}\rightarrow {\lambda}}{\phi}_{d}(0)+\lim _{{\lambda}_{1}\rightarrow {\lambda}}{\psi}_{d}(0)=1. & & \displaystyle \nonumber\end{eqnarray}$ More specifically, m _d(0) = 𝜙_d(0) + 𝜓_d(0) = 1 with 𝜙_d(0) = 0 and 𝜓_d(0) = 1.

8. Numerical illustration

In this section, the goal is to provide a numerical illustration of the ruin probability and from the obtained results, analyse the impact of changes in volatility on ruin probability and the impact of the initial surplus on the ruin probability.

8.1. Case claim amounts are exponentially distributed

For the illustration, we consider c = 1, 𝜃 = 1, 𝜆 = 0.75, 𝜆₁ = 0.55

Fig. 1.

Ruin probability caused by claims with fixed volatility.

Fig. 2.

Ruin probability caused by oscillation with fixed volatility.

As expected, when the initial capital increases, we notice that both probabilities i.e ruin probability caused by claims and ruin probability caused by the oscillation decrease, and for zero initial surplus the ruin probability caused by oscillation is one.

The graphics show that, as the volatility in the diffusion process is high the probability caused by oscillation is much more important and increases with respect to the volatility.

8.2. Case claim amounts are mixed exponentially distributed

We provide numerical example for the ruin probability in the mixed exponential distribution of claims.

Fig. 3.

Ruin probability caused by claim with fixed initial capital.

Fig. 4.

Ruin probability caused by oscillation with fixed initial capital.

Table 1

Numerical illustration in case of mixed exponential distributed

c = 3, 𝜃₁ = 2, 𝜃₂ = 3.5, p = 0.5, 𝜆₁ = 1.25, 𝜆 = 0.5, 𝜎 = 0.5

𝜙_d(u) = −0.1787e ^−24.409u + 0.1089e ^−1.904u + 0.0528e ^−3.409u + 0.0170e ^−24.186u

𝜓_d(u) = 1.2149e ^−24.409u + 0.01010e ^−1.904u + 0.00947e ^−3.409u − 0.2345e ^−24.186u

c = 2.75, 𝜃₁ = 5, 𝜃₂ = 0.75, p = 0.25, 𝜆₁ = 0.85, 𝜆 = 0.25, 𝜎 = 1

𝜙_d(u) = −1.3630e ^−5.793u + 0.26415e ^−0.6718u + 0.01328e ^−4.8348u + 1.08565e ^−5.7433u

𝜓_d(u) = 8.61613e ^−5.793u + 0.04124e ^−0.6718u + 0.45843e ^−4.8348u − 8.1158e ^−5.7433u

c = 3.75, 𝜃₁ = 2.5, 𝜃₂ = 3.5, p = 0.6, 𝜆₁ = 3.85, 𝜆 = 1.25, 𝜎 = 1.01

𝜙_d(u) = −0.55486e ^−8.2654u + 0.29765e ^−2.16928u + 0.05907e ^−3.33073u + 0.19814e ^−7.85219u

𝜓_d(u) = 1.92504e ^−8.2654u + 0.14965e ^−2.16928u + 0.06239e ^−3.33073u − 1.13709e ^−7.85219u

As in the exponential distribution case, we remark that the ruin probability caused by claims is zero and the ruin probability caused by oscillation is one when the initial is zero.

9. Delayed renewal risk model with exchangeable inter claim time

All this papers with the objective of determining the ruin probability for compound renewal sums, assumed that the distribution of the inter-occurrence times are independent. But in many applications, the assumption that the subsequent inter arrival times are independent turns out to be too restrictive and generalizations to dependent scenarios are called for. For instance, in car insurance, if there has been a long waiting time before a claim, the next inter-arrival time can be long, as well, because the policyholders are potentially “good drivers”; or it could be the other way around, where some policyholders only start to use their cars a long time after purchasing them, then claims would suddenly arrive more frequently after a long silence. On way to treat that dependence is the use of exchangeable risks. Let’s define the exchangeable risk model in this section. We assume that the inter claim inter time arrival V _i, i = 1, … are dependent such that V ₁|_𝛬₁ andV _i|_𝛬, i = 2, … are independent and distributed exponentially. $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\bar{F}_{(V_{1},V_{2},\ldots ,V_{n})}(x_{1},x_{2},\ldots ,x_{n})\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}=∼P[V_{1}>x_{1},V_{2}>x_{2},\ldots ,V_{n}>x_{n}]\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}=\int _{0}^{\infty }\hspace{-3.0pt}\int _{0}^{\infty }\hspace{-3.0pt}P[V_{1}>x_{1},V_{2}>x_{2},\ldots ,V_{n}>x_{n}|_{({\Lambda}_{1},{\Lambda})}]\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-5.0pt}f_{({\Lambda}_{1},{\Lambda})}({\lambda}_{1},{\lambda})d{\lambda}_{1}d{\lambda}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}=\int _{0}^{\infty }\int _{0}^{\infty }e^{-{\lambda}_{1}x_{1}-{\lambda}\displaystyle \mathop{\sum }_{i=2}^{n}x_{i}}f_{({\Lambda}_{1},{\Lambda})}({\lambda}_{1},{\lambda})d{\lambda}_{1}d{\lambda}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}=\tilde{f}_{({\Lambda}_{1},{\Lambda})}\left(x_{1},\displaystyle \mathop{\sum }_{i=2}^{n}x_{i}\right).\nonumber\end{eqnarray}$ If 𝛬₁ and 𝛬 are equal in distribution, we get the ordinary case. since $\begin{eqnarray}\displaystyle \bar{F}_{V_{1}}(x_{1})=\int _{0}^{\infty }e^{-{\lambda}_{1}x_{1}}f_{{\Lambda}_{1}}({\lambda}_{1})d{\lambda}_{1}=\tilde{f}_{{\Lambda}_{1}}(x_{1}) & & \displaystyle \nonumber\end{eqnarray}$ we have that $\begin{eqnarray}\displaystyle \tilde{f}_{{\Lambda}_{1}}^{-1}(\bar{F}_{V_{1}}(x_{1}))=x_{1}. & & \displaystyle \nonumber\end{eqnarray}$ Therefore $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\displaystyle \bar{F}_{(V_{1},V_{2},\ldots ,V_{n})}(x_{1},x_{2},\ldots ,x_{n})=\tilde{f}_{({\Lambda}_{1},{\Lambda})}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \times ∼\left(\tilde{f}_{{\Lambda}_{1}}^{-1}(\bar{F}_{V_{1}}(x_{1})),\displaystyle \mathop{\sum }_{i=2}^{n}\tilde{f}_{{\Lambda}}^{-1}(\bar{F}_{V_{i}}(x_{1}))\right).\nonumber\end{eqnarray}$ By the [18] theorem $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}\bar{F}_{(V_{1},V_{2},\ldots ,V_{n})}(x_{1},x_{2},\ldots ,x_{n})\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle =C(\bar{F}_{V_{1}}(x_{1}),\bar{F}_{V_{2}}(x_{2}),\ldots ,\bar{F}_{V_{n}}(x_{n})),\nonumber\end{eqnarray}$ Thus we obtain the following Archimedian copula, see [17], $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}C(u_{1},u_{2},\ldots ,u_{n})=\tilde{f}_{({\Lambda}_{1},{\Lambda})}\left(\tilde{f}_{{\Lambda}_{1}}^{-1}(u_{1}),\displaystyle \mathop{\sum }_{i=2}^{n}\tilde{f}_{{\Lambda}}^{-1}(u_{i})\right).\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$ (44)

TheoremO .

Considering that the mixing random variable 𝛬₁, 𝛬 are independent and geometrically distributed with parameters q ₁ and q, P[𝛬 = k] = q (1 − q)^k−1 we have $\begin{eqnarray}\displaystyle \tilde{f}_{({\Lambda}_{1},{\Lambda})}(s,t)\text{} & =\text{} & \displaystyle E(e^{-s{\Lambda}_{1}-t{\Lambda}})=E(e^{-s{\Lambda}_{1}})E(e^{-t{\Lambda}})\nonumber\\ \displaystyle \text{} & =\text{} & \displaystyle {\displaystyle \frac{q_{1}}{q_{1}-1+e^{s}}}{\displaystyle \frac{q}{q-1+e^{t}}},\nonumber\end{eqnarray}$ and $\begin{eqnarray}\displaystyle \tilde{f}_{{\Lambda}_{1}}(s)={\displaystyle \frac{q_{1}}{q_{1}-1+e^{s}}},\tilde{f}_{{\Lambda}}^{-1}(y)=ln\left(1-q-{\displaystyle \frac{q}{y}}\right). & & \displaystyle \nonumber\end{eqnarray}$ By (44) the copula of (V ₁, V ₂, …, V _n) is then $\begin{eqnarray}\displaystyle \hspace{-22.0pt}C(u_{1},u_{2},\ldots ,u_{n})=u_{1}{\displaystyle \frac{q}{q-1+\mathop{\prod }_{i=2}^{n}\left(1-q-{\displaystyle \frac{q}{u_{i}}}\right)}}. & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$ (45)

9.1. Expression of the ruin probability in the delayed and perturbed risk model with exchangeable inter claim time arrival

Under the assumption mentioned above in this section, If we condition on (𝛬₁, 𝛬) we get the get the conditional probability (caused by claims and by oscillation), 𝜙_d(𝜆₁, 𝜆, u) and 𝜓_d(𝜆₁, 𝜆, u). The unconditional probabilities ${\phi}_{d}^{\ast }$ and by oscillation ${\psi}_{d}^{\ast }$ is therefore give by:

If (𝛬₁, 𝛬) are discrete $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\phi}_{d}^{\ast }(u)=\displaystyle \mathop{\sum }_{{\lambda}_{1}=c{\theta}}^{\infty }\mathop{\sum }_{{\lambda}=c{\theta}}^{\infty }P[{\Lambda}_{1}={\lambda}_{1},{\Lambda}={\lambda}]\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-15.0pt}+\displaystyle \mathop{\sum }_{{\lambda}_{1}=0}^{c{\theta}}\displaystyle \mathop{\sum }_{{\lambda}=0}^{c{\theta}}{\phi}_{d}({\lambda}_{1},{\lambda},u)P[{\Lambda}_{1}={\lambda}_{1},{\Lambda}={\lambda}],\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\psi}_{d}^{\ast }(u)=\displaystyle \mathop{\sum }_{{\lambda}_{1}=c{\theta}}^{\infty }\displaystyle \mathop{\sum }_{{\lambda}=c{\theta}}^{\infty }P[{\Lambda}_{1}={\lambda}_{1},{\Lambda}={\lambda}]\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-15.0pt}+\displaystyle \mathop{\sum }_{{\lambda}_{1}=0}^{c{\theta}}\displaystyle \mathop{\sum }_{{\lambda}=0}^{c{\theta}}{\psi}_{d}({\lambda}_{1},{\lambda},u)P[{\Lambda}_{1}={\lambda}_{1},{\Lambda}={\lambda}].\nonumber\end{eqnarray}$

If (𝛬₁, 𝛬) are continuous $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\phi}_{d}^{\ast }(u)=\int _{c{\theta}}^{\infty }\int _{c{\theta}}^{\infty }f_{({\Lambda}_{1},{\Lambda})}({\lambda}_{1},{\lambda})d{\lambda}_{1}d{\lambda}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}+\int _{0}^{c{\theta}}\int _{0}^{c{\theta}}{\phi}_{d}({\lambda}_{1},{\lambda},u)f_{({\Lambda}_{1},{\Lambda})}({\lambda}_{1},{\lambda})d{\lambda}_{1}d{\lambda},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\psi}_{d}^{\ast }(u)=\int _{c{\theta}}^{\infty }\int _{c{\theta}}^{\infty }f_{({\Lambda}_{1},{\Lambda})}({\lambda}_{1},{\lambda})d{\lambda}_{1}d{\lambda}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}+\int _{0}^{c{\theta}}\int _{0}^{c{\theta}}{\psi}_{d}({\lambda}_{1},{\lambda},u)f_{({\Lambda}_{1},{\Lambda})}({\lambda}_{1},{\lambda})d{\lambda}_{1}d{\lambda}.\nonumber\end{eqnarray}$

Under the assumptions that the 𝛬₁ and 𝛬 are independent and geometrically distributed, and if condition on (𝛬₁, 𝛬) the inter time arrival are independent and exponentially distributed, since P[𝛬 ≥ n] = (1 − q)ⁿ⁻¹ the unconditional ruin probabilities are given by the following.

The ruin probability caused by claims with exchangeable risk is $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\phi}_{d}^{\ast }(u)=\displaystyle \mathop{\sum }_{{\lambda}_{1}=\lfloor c{\theta}\rfloor +1}^{\infty }\displaystyle \mathop{\sum }_{{\lambda}=\lfloor c{\theta}\rfloor +1}^{\infty }P[{\Lambda}_{1}={\lambda}_{1}]P[{\Lambda}={\lambda}]\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}+\displaystyle \mathop{\sum }_{{\lambda}_{1}=0}^{\lfloor c{\theta}\rfloor }\displaystyle \mathop{\sum }_{{\lambda}=0}^{\lfloor c{\theta}\rfloor }{\phi}_{d}({\lambda}_{1},{\lambda},u)P[{\Lambda}_{1}={\lambda}_{1}]P[{\Lambda}={\lambda}]\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}=(1-q_{1})^{\lfloor c{\theta}\rfloor }(1-q)^{\lfloor c{\theta}\rfloor }\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}+\displaystyle \mathop{\sum }_{{\lambda}_{1}=0}^{\lfloor c{\theta}\rfloor }\displaystyle \mathop{\sum }_{{\lambda}=0}^{\lfloor c{\theta}\rfloor }{\phi}_{d}({\lambda}_{1},{\lambda},u)P[{\Lambda}_{1}={\lambda}_{1}]P[{\Lambda}={\lambda}].\nonumber\end{eqnarray}$

The ruin probability caused by oscillations with exchangeable risk is $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\psi}_{d}^{\ast }(u)=(1-q_{1})^{\lfloor c{\theta}\rfloor }(1-q)^{\lfloor c{\theta}\rfloor }\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}+\displaystyle \mathop{\sum }_{{\lambda}_{1}=0}^{\lfloor c{\theta}\rfloor }\displaystyle \mathop{\sum }_{{\lambda}=0}^{\lfloor c{\theta}\rfloor }{\psi}_{d}({\lambda}_{1},{\lambda},u)P[{\Lambda}_{1}={\lambda}_{1}]P[{\Lambda}={\lambda}]\nonumber\end{eqnarray}$

where 𝜙_d(𝜆₁, 𝜆, u) = 𝜙_d(u) and 𝜓_d(𝜆₁, 𝜆, u) = 𝜓_d(u) defined in (K ). $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\phi}_{d}({\lambda}_{1},{\lambda},u)=p_{{\lambda}_{1},{\lambda}}e^{-\left(r_{{\lambda}_{1}}+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)u}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-5.0pt}+∼q_{{\lambda}_{1},{\lambda}}e^{-{\alpha}_{{\lambda}}u}+{\delta}_{{\lambda}_{1}{\lambda}}e^{-{\beta}_{{\lambda}}u},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\psi}_{d}({\lambda}_{1},{\lambda},u)=p_{{\lambda}_{1},{\lambda}}^{\prime }e^{-\left(r+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)u}+q_{{\lambda}_{1},{\lambda}}^{\prime }e^{-{\alpha}u}+{\delta}_{{\lambda}_{1}{\lambda}}^{\prime }e^{-{\beta}u},\nonumber\end{eqnarray}$ With $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}p_{{\lambda}_{1},{\lambda}}={\displaystyle \frac{{\displaystyle \frac{2{\lambda}_{1}}{{\sigma}^{2}}}(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}})}{\left[{\theta}+r(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}})\right]\left[{\theta}-\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)\left(r+{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)\right]}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}q_{{\lambda}_{1},{\lambda}}={\displaystyle \frac{-{\displaystyle \frac{2{\lambda}}{{\sigma}^{2}}}}{({\alpha}-{\beta})\left[{\theta}-{\alpha}\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)\right]}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\delta}_{{\lambda}_{1}{\lambda}}={\displaystyle \frac{{\displaystyle \frac{2{\lambda}}{{\sigma}^{2}}}}{({\alpha}-{\beta})\left[{\theta}-{\beta}\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)\right]}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}p_{{\lambda}_{1},{\lambda}}^{\prime }={\displaystyle \frac{r\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}+{\displaystyle \frac{c{\theta}}{{\lambda}_{1}}}\right)}{\left({\theta}+r\left(1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}\right)\right)\left[1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}-{\displaystyle \frac{{\theta}r{\sigma}^{2}}{2{\lambda}_{1}}}\right]}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}q_{{\lambda}_{1},{\lambda}}^{\prime }={\displaystyle \frac{-{\theta}}{({\alpha}-{\beta})\left[1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}+{\displaystyle \frac{{\theta}}{{\lambda}_{1}}}\left(c-{\displaystyle \frac{{\sigma}^{2}}{2}}{\alpha}\right)\right]}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\delta}_{{\lambda}_{1}{\lambda}}^{\prime }={\displaystyle \frac{{\theta}}{({\alpha}-{\beta})\left[1-{\displaystyle \frac{{\lambda}}{{\lambda}_{1}}}+{\displaystyle \frac{{\theta}}{{\lambda}_{1}}}\left(c-{\displaystyle \frac{{\sigma}^{2}}{2}}{\beta}\right)\right]}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\alpha}_{{\lambda}}={\displaystyle \frac{{\theta}}{2}}+{\displaystyle \frac{c}{{\sigma}^{2}}}+{\displaystyle \frac{1}{2}}\sqrt{\left({\theta}-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)^{2}+{\displaystyle \frac{8{\lambda}}{{\sigma}^{2}}}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}{\beta}_{{\lambda}}={\displaystyle \frac{{\theta}}{2}}+{\displaystyle \frac{c}{{\sigma}^{2}}}-{\displaystyle \frac{1}{2}}\sqrt{\left({\theta}-{\displaystyle \frac{2c}{{\sigma}^{2}}}\right)^{2}+{\displaystyle \frac{8{\lambda}}{{\sigma}^{2}}}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{-22.0pt}r_{{\lambda}_{1}}=-{\displaystyle \frac{c}{{\sigma}^{2}}}+\sqrt{{\displaystyle \frac{c^{2}}{{\sigma}^{4}}}+{\displaystyle \frac{2{\lambda}_{1}}{{\sigma}^{2}}}}.\nonumber\end{eqnarray}$

If 𝛬₁ and 𝜆 are continuous, since the integral expressions of the ruin probabilities in the delayed risk model with exchangeable inter claim time in are difficult to calculate, one can discretize the mixing random variables (𝛬₁ and 𝜆) then use the formulae in the discrete case.

9.1.1. Numerical computation of the ruin probability for risk model with exchangeable inter claim time

We provide a numerical example for the ruin probability in the risk model with exchangeable inter claim and exponential distribution of claims.

Table 2
Numerical illustration of ruin probabilities in risk model with exchangeable inter claim time

u c 𝜃 𝜎 q ₁ q ${\phi}_{d}^{\ast }(u)$ ${\psi}_{d}^{\ast }(u)$

0 1 1.5 0.25 0.9 0.9 0.01000000 0.82000000

1 1 1.5 0.25 0.9 0.9 0.33717541 0.02557218

2 1 1.5 0.25 0.9 0.9 0.21151701 0.01959137

3 1 1.5 0.25 0.9 0.9 0.13412028 0.01590761

4 1 1.5 0.25 0.9 0.9 0.08644935 0.01363867

5 1 1.5 0.25 0.9 0.9 0.05708742 0.01224116

1 1 1.5 0.25 0.9 0.1 0.04635282 0.01173024

1 1 1.5 0.25 0.9 0.2 0.08270565 0.01346048

1 1 1.5 0.22 0.9 0.3 0.11905847 0.01519073

1 1 1.5 0.25 0.9 0.5 0.19176412 0.01865121

1 1 1.5 0.25 0.9 0.7 0.26446976 0.02211170

1 1 1.5 0.25 0.9 0.9 0.33717541 0.02557218

3.25 2.5 2.15 0.05 0.75 0.25 0.1044902 0.05636531

3.25 2.5 2.15 0.25 0.75 0.25 0.1046333 0.05761663

3.25 2.5 2.15 0.5 0.75 0.25 0.1050191 0.06162310

3.25 2.5 2.15 0.75 0.75 0.25 0.1055117 0.06860135

3.25 2.5 2.15 1 0.75 0.25 0.1060099 0.07898944

u	c	𝜃	𝜎	q ₁	q	${\phi}_{d}^{\ast }(u)$	${\psi}_{d}^{\ast }(u)$
0	1	1.5	0.25	0.9	0.9	0.01000000	0.82000000
1	1	1.5	0.25	0.9	0.9	0.33717541	0.02557218
2	1	1.5	0.25	0.9	0.9	0.21151701	0.01959137
3	1	1.5	0.25	0.9	0.9	0.13412028	0.01590761
4	1	1.5	0.25	0.9	0.9	0.08644935	0.01363867
5	1	1.5	0.25	0.9	0.9	0.05708742	0.01224116
1	1	1.5	0.25	0.9	0.1	0.04635282	0.01173024
1	1	1.5	0.25	0.9	0.2	0.08270565	0.01346048
1	1	1.5	0.22	0.9	0.3	0.11905847	0.01519073
1	1	1.5	0.25	0.9	0.5	0.19176412	0.01865121
1	1	1.5	0.25	0.9	0.7	0.26446976	0.02211170
1	1	1.5	0.25	0.9	0.9	0.33717541	0.02557218
3.25	2.5	2.15	0.05	0.75	0.25	0.1044902	0.05636531
3.25	2.5	2.15	0.25	0.75	0.25	0.1046333	0.05761663
3.25	2.5	2.15	0.5	0.75	0.25	0.1050191	0.06162310
3.25	2.5	2.15	0.75	0.75	0.25	0.1055117	0.06860135
3.25	2.5	2.15	1	0.75	0.25	0.1060099	0.07898944

We can notice that the ruin probability (caused by claims and by oscillations) both decrease when the initial capital increases. And the probability caused by oscillations increases as the volatility increases. When the initial surplus is zero we see that the ruin probability caused by claims is not zero but close and the ruin probability caused by oscillations is also not one but close. This because of numerical computations and, conditionally on 𝛬₁, 𝛬 the probabilities are respectively 0, and 1, and by integrating upon the distribution of 𝛬₁, 𝛬 we get the obtained results.

10. Conclusion

Insurance and financial institutions deals with risks and so the models that account for the oscillation must capture the error and randomness that may arise in the premium or in the claims. By considering the surplus process, the integro- differential equations of ruin probabilities caused by claims and by oscillations are derived. By considering exponential claim distribution, analytical expression of the ruin probability is determined when there is a delayed in the time arrival of the first claim in the surplus model perturbed a diffusion process. The numerical illustrations confirm the expectancy and the ruin probability increases as the volatility is much more important and the ruin probability decreases as the initial capital increases. The expression of the ruin probability is derived in the risk model with exchangeable inter claim time. Further extensions can be made by using Erlang distribution for inter-claim time, using a jump or Lévy process; or considering the volatility as stochastic to account for the states of economical growth and recession in a regime switching economy.

Footnotes

Acknowledgements

The authors would like to thank the anonymous referees for constructive comments that improved the contents and presentation of this paper.

Author contributions

This research paper is done by Essodina Takouda in the framework of a doctoral programme under the supervision of Prof. Franck Adékambi.

Funding

This work was supported by the Global Excellence and Stature (GES) 4.0 scholarship of the University of Johannesburg (UJ).

Conflicts of interest

The authors declare no conflict of interest.

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