Laplace Transformation of the Ruin Time for a Risk Model with a Parisian Implementation Delay

Abstract

In this paper, we apply the concept of the Parisian implementation delay to dividend payments and assume that the claim amount paid $Z_i$ by the insurance company for the ith time follows an exponential distribution. We give the Laplace transformation of the ruin time for a risk model with a Parisian implementation delay and give display expressions using the scale function.

1. Introduction

Generally, in articles and books on risk theory, it is customary to make deterministic decisions. For example, the determination of the time of bankruptcy is traditionally defined as the moment when the company’s earnings process reaches the level of 0 for the first time, and the company declares bankruptcy. On the other hand, in dividend payments, the determination of the dividend payment time is traditionally defined as the payment of dividends when the company’s earnings process exceeds a specific threshold of b (>0). However, from a practical perspective, this decision-making approach is unreasonable because there may be a time difference between the company making decisions and their specific implementation. Generally, companies need to continuously observe their current operational (financial) situation before making further decisions. In response to this issue, some scholars have proposed the Parisian implementation delay concept, which allows for a certain delay in the execution of decisions, which is more realistic.

In recent years, Parisian bankruptcy has been of great concern in bankruptcy theory, which means that bankruptcy will only be declared when the surplus process is below the 0 level for more than a delay time length r (≥0). The Parisian bankruptcy time is defined as: ${\kappa }_r^U=inf\{t>0:t-g_t^U>r\},$ where the constant r represents the delay time. The delay time here is also known as the Parisian implementation delay [1,2]. The Parisian implementation delay has been applied to two different aspects, namely bankruptcy and dividend payment decisions. For more detailed theoretical results, interested readers can refer to [3,4,5,6,7,8,9,10,11,12].

From a practical perspective, we can consider extending the classic concept of bankruptcy to Parisian bankruptcy. Ref. [5] first proposed the concept of Parisian implementation delay in insurance risk theory and defined the Parisian bankruptcy moment; that is, bankruptcy is declared only when the surplus process continues below the 0 level for a specified length of time r (>0). That is to say, if a company can continue operating at a negative earnings level within a specified period of time, it is considered a normal operating enterprise without having declared bankruptcy. In this regard, the Parisian bankruptcy time is a fairer bankruptcy calculation than the classical bankruptcy time. Ref. [7] provided the Laplace transformation of the Parisian ruin time for the classical composite Poisson risk process. For more research on Parisian bankruptcy issues, see [9,13,14].

On the other hand, the Parisian implementation delay concept can be applied to dividend payments. Under this strategy, dividends are only distributed when the earnings process continues above a constant dividend barrier b (>0) for a given period of time r (>0). That is to say, even if the company’s earnings exceed the dividend barrier b (>0), but later due to poor operations, the earnings level falls below the barrier within the given time r (>0), dividends will not be distributed. In contrast, if the enterprise has been operating well and has been above the dividend barrier for a sufficient period of time, then the surplus exceeding the barrier will be paid in a lump sum dividend payment. This dividend payment strategy is more realistic. Ref. [7] studied the application of the Parisian implementation delay to dividends in a composite Poisson risk model. In [8], the Parisian implementation delayed dividend problem in the Brownian motion risk model was studied. In addition, in [3], the author considered the dividend problem of a dual risk model with a Parisian implementation delay and provided an expression for the Laplace transform at the bankruptcy time. The dividend payment strategy considering the Parisian implementation delay will make the model more practical. Especially for companies, it is meaningful for their actual operations to make dividend payments after maintaining financial health for a long time. In this article, we consider a dual risk model with a Parisian implementation delay in dividends. For more detailed theoretical results, interested readers can refer to [15,16,17,18,19,20,21,22].

The innovation of this paper lies in applying the concept of Parisian implementation delay to dividend payments; we give the Laplace transformation of the ruin time for a risk model with a Parisian implementation delay and give display expressions using the scale function.

The remainder of the article is organized as follows. Section 2 discusses the risk model with a Parisian implementation delay in dividends and provides some preparatory knowledge. Section 3 presents the main results and proofs of this paper. Section 4 gives some discussions and conclusions.

2. Preliminaries and Lemmas

In this article, we discuss the dual risk model, which is

$$\begin{array}{c}\hfill U\left(t\right)=u-ct+\sum _{i=1}^{N\left(t\right)}{Z}_i,t\ge 0,\end{array}$$

(1)

where $u=U\left(0\right)\ge 0$ represents the initial funds of the insurance company, $c(>0)$ represents the expenses or expenses of the insurance company per unit time, $Z_i(i\ge 1)$ represents the ith income value of the insurance company, $N\left(t\right)$ represents the number of income occurrences of the insurance company within the time period of $[0,t]$, and $\{N_t,t\ge 0\}$ is a Poisson process with an intensity of $\lambda$. We assume that $\left\{N\right(t),t\ge 0\}$ and ${\left\{Z_i\right\}}_{i=1,2,\dots }$ are independent of each other.

The dual risk model can be applied in reality to risk models that require regular investment, such as life insurance annuities, as well as to enterprises that require continuous investment and have variable returns, such as research companies.

Generally, we study dual risk models that have a certain dividend strategy, such as a barrier dividend strategy or a threshold dividend strategy. In this article, we discuss the risk model with a Parisian implementation delay in dividends.

Assuming there is a dividend barrier b (>0), for a general dividend strategy, as long as the earnings process $U\left(t\right)$ exceeds the dividend barrier b, dividends will be immediately distributed. The Parisian implementation delay in dividends refers to the fact that dividends are only distributed when the earnings process $U\left(t\right)$ continues above the dividend barrier b for a period of time r (>0), where r is a definite delay time. The dividend value is equal to the portion of the surplus value that exceeds the dividend barrier b when the delay period ends; but, it is not equal to the surplus value at the moment when it exceeds b.

According to the Parisian implementation delay in dividends strategy, we further define the modified surplus process $\{U_b\left(t\right),t\ge 0\}$ considering dividend payments,

$$\begin{array}{c}\hfill U_b\left(t\right)=U\left(t\right)-L_t,t\ge 0.\end{array}$$

(2)

where $U_b\left(0\right)=U\left(0\right)=u\ge 0$ is defined as the initial capital, and $L_t$ represents the cumulative dividend value at time t.

We will follow the traditional definition of bankruptcy, which is to declare bankruptcy immediately when the modified surplus process $\{U_b\left(t\right),t\ge 0\}$ reaches the 0 level. It is not difficult to know that the surplus value before bankruptcy and the deficit at the time of bankruptcy are both equal to 0.

Definition 1.

In the modified dual risk model, the bankruptcy time of the surplus process $\{U_b\left(t\right),t\ge 0\}$ is

$$\begin{array}{c}\hfill T_{U_b}=\inf \{\mathrm{t}>0:{\mathrm{U}}_{\mathrm{b}}\left(\mathrm{t}\right)\le 0\}.\end{array}$$

Then, the Laplace transform of the bankruptcy time $T_{U_b}$ in a risk model with a Parisian implementation delay in dividends is denoted as:

$$\begin{array}{c}\hfill {\mathbb{E}}_u\left[e^{-q{T}_{U_b}}\right]=\varphi (u;b)=\left\{\begin{array}{cc}{\varphi }_L(u;b);\hfill & 0\le u\le b,\hfill \\ {\varphi }_A(u;b);\hfill & u>b,\hfill \end{array}\right.\end{array}$$

(3)

where $q\ge 0$ is the Laplace transform operator, and ${\mathbb{E}}_u$ represents the expected value when $U_b\left(0\right)=U\left(0\right)=u$.

In addition, let

$$U_k\left(t\right)=\left\{\begin{array}{cc}U\left(t\right);\hfill & t\ge 0,k=1,\hfill \\ b+U_b\left(t\right)-U_b\left({\eta }_{k-1}\right);\hfill & t\ge {\eta }_{k-1},k=2,3,\cdots .\hfill \end{array}\right.$$

And, for $k=1,2,\cdots$, let

$$\begin{array}{ccc}\hfill {\nu }_k& =& \inf \{t\ge {\eta }_{k-1}:U_k\left(t\right)>b\},\hfill \\ \hfill {\xi }_k& =& \inf \{t\ge {\nu }_k:U_k\left(t\right)=b\},\hfill \\ \hfill {\eta }_k& =& ({\nu }_k+r)\wedge {\xi }_k,{\eta }_0=0.\hfill \end{array}$$

So, we have

$$\begin{array}{c}\hfill U_b\left(t\right)=U_k\left(t\right),{\eta }_{k-1}\le t\le {\eta }_k,k=1,2,\cdots ,\end{array}$$

(4)

and

$$\begin{array}{c}\hfill U_b\left(t\right)=U\left(t\right),0\le t<{\nu }_1.\end{array}$$

(5)

For the convenience of research, let

$$\begin{array}{c}\hfill U\left(t\right)=u-X\left(t\right),t\ge 0.\end{array}$$

(6)

Then,

$$\begin{array}{c}\hfill U_b\left(t\right)=u-X\left(t\right)-L_t,t\ge 0.\end{array}$$

(7)

The risk model $X\left(t\right)$ is defined here as:

$$\begin{array}{c}\hfill X\left(t\right)=ct-\sum _{i=1}^{N\left(t\right)}{Z}_i,t\ge 0,\end{array}$$

where c (>0) represents the income of the insurance company per unit time, and $Z_i$ represents the claim amount paid by the insurance company for the ith time. $N\left(t\right)$ represents the number of claims made by the insurance company within the time period of $[0,t]$, and $\{N_t,t\ge 0\}$ is a Poisson process with an intensity of $\lambda$. We assume that $\left\{N\right(t),t\ge 0\}$ and ${\left\{Z_i\right\}}_{i=1,2,\dots }$ are independent of each other.

Definition 2.

We assume that the initial time of process ${\left\{X\left(t\right)\right\}}_{t\ge 0}$ is ${\tau }_c^{\pm }$, that is, for any real number c, we have

$$\begin{array}{ccc}\hfill {\tau }_c^{+}& =& \inf \{\mathrm{t}\ge 0,{\mathrm{X}}_{\mathrm{t}}\ge \mathrm{c}\};\hfill \\ \hfill {\tau }_c^{-}& =& \inf \{\mathrm{t}\ge 0,{\mathrm{X}}_{\mathrm{t}}<\mathrm{c}\}.\hfill \end{array}$$

Because $Z_i$ represents the claim amount paid by the insurance company for the ith time, much of the literature assumes that $Z_i$ follows an exponential distribution, which is more in line with reality. In this article, we assume that $Z_i$ follows a more general situation, and we investigate the case where ${\left\{Z_i\right\}}_{i=1,2,\dots }$ follows a mixed exponential distribution. We assume that the random income ${\left\{Z_i\right\}}_{i=1,2,\dots }$ of process $U_b\left(t\right)$ is a column of independent and identically distributed positive random variables, and we assume its probability density function is

$$f\left(z\right)=\sum _{i=1}^n{a}_i{\alpha }_i{e}^{-{\alpha }_{i}z},z>0,$$

where $n\in Z^{+},0<{\alpha }_1<{\alpha }_2<\cdots <{\alpha }_n,{\displaystyle \sum _{i=1}^n}{a}_i=1$, and $a_i>0$ for any $i=1,2,\cdots n$. It is not difficult to find that the Laplace index of $X\left(t\right)$ is

$$\psi \left(\theta \right)=log\mathbb{E}\left[e^{\theta X_1}\right]=c\theta -\lambda +\lambda \sum _{i=1}^n\frac{a_i{\alpha }_i}{\theta +{\alpha }_i},\theta >-\alpha .$$

According to [23], we can easily know the scaling functions $W^q\left(y\right)$ and $Z^q\left(y\right)$ of $X\left(t\right)$; that is, for $q>0$ or $q=0$ and ${\psi }^{\prime }\left(\theta \right)\ne 0$,

$$\begin{array}{ccc}\hfill W^{\left(q\right)}\left(y\right)& =& \sum _{i=1}^N\frac{e^{{\rho }_{i,q}y}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)},y\ge 0.\hfill \end{array}$$

$$Z^{\left(q\right)}\left(y\right)=1+q\sum _{i=1}^N\frac{e^{{\rho }_{i,q}y}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}}$$

$$=\left\{\begin{array}{cc}q{\displaystyle \sum _{i=1}^N}\frac{e^{{\rho }_{i,q}y}}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_i^{\left(q\right)}};\hfill & \mathrm{if}q>0,y\ge 0,\hfill \\ 1;\hfill & \mathrm{if}q=0,y\ge 0.\hfill \end{array}\right.$$

$$\begin{array}{ccc}\hfill {\mathcal{W}}_b^{(p,q)}\left(y\right)& =& \sum _{i=1}^N\frac{e^{{\rho }_{i,q+p}y}}{{\psi }^{\prime }\left({\rho }_{i,q+p}\right)}-q{\int }_0^b\sum _{i=1}^N\frac{e^{{\rho }_{i,q+p}(y-z)}}{{\psi }^{\prime }\left({\rho }_{i,q+p}\right)}\sum _{i=1}^N\frac{e^{{\rho }_{i,p}z}}{{\psi }^{\prime }\left({\rho }_{i,p}\right)}dz\hfill \\ & =& \sum _{i=1}^N\frac{e^{{\rho }_{i,q+p}y}}{{\psi }^{\prime }\left({\rho }_{i,q+p}\right)}\left[1-q\sum _{j=1}^N\frac{e^{({\rho }_{j,p}-{\rho }_{i,q+p})b}-1}{({\rho }_{j,p}-{\rho }_{i,q+p}){\psi }^{\prime }\left({\rho }_{j,p}\right)}\right],\hfill \end{array}$$

(8)

where ${\rho }_{1,q}>{\rho }_{2,q}>\cdots >{\rho }_{N,q}$ is the root of $\psi \left(\theta \right)=q$, $N=n+1$.

According to [15],

$$\begin{array}{c}\hfill \mathbb{P}(X_r\in dz)=e^{-\lambda r}\sum _{i=1}^{\infty }\frac{{(-\lambda r)}^i}{i!}{\int }_0^{\infty }{F}^{\ast i}\left(dy\right)\mathcal{N}\left((z+y-cr)\sigma \sqrt{r}\right)dz,\end{array}$$

where $\mathcal{N}$ is the distribution function of a standard normal random variable, and $F^{\ast i}$ is the i-fold integral of F.

Lemma 1.

For any $0\le u\le b$, $q\ge 0$, we have

$$\begin{array}{ccc}\hfill {\mathbb{E}}_u(e^{q{\tau }_b^{+}};{\tau }_b^{+}<{\tau }_0^{-})& =& \frac{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}u}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}b}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}\hfill \end{array}$$

(9)

$$\begin{array}{ccc}\hfill {\mathbb{E}}_u(e^{q{\tau }_0^{-}};{\tau }_0^{-}<{\tau }_b^{+})& =& \left(1+q\sum _{i=1}^N\frac{e^{{\rho }_{i,q}u}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}}\right)\hfill \\ & & -\left(1+q\sum _{i=1}^N\frac{e^{{\rho }_{i,q}b}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}}\right)\frac{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}u}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}b}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}.\hfill \end{array}$$

(10)

Lemma 2.

For any $q\ge 0$, $r>0$, we have

$$\begin{array}{ccc}\hfill \left(i\right){\int }_0^{\infty }\sum _{i=1}^N\frac{e^{{\rho }_{i,0}z}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\frac{z}{r}·\mathbb{P}(X_r\in dz)& =& 1,\hfill \end{array}$$

(11)

$$\begin{array}{ccc}\hfill \left(ii\right){\int }_0^{\infty }\sum _{i=1}^N\frac{e^{{\rho }_{i,q}z}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}\frac{z}{r}·\mathbb{P}(X_r\in dz)& =& e^{qr}.\hfill \end{array}$$

(12)

Lemma 3.

For any q, $p\ge 0$, $a<x<c$, we have

$$\begin{array}{c}{\displaystyle {\mathbb{E}}_x\left(e^{-p{\tau }_a^{-}}\sum _{i=1}^N\frac{e^{{\rho }_{i,q}X\left({\tau }_a^{-}\right)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}{I}_{\{{\tau }_a^{-}<{\tau }_c^{+}\}}\right)}\hfill \\ =\sum _{i=1}^N\frac{e^{{\rho }_{i,q}x}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}-(q-p){\int }_0^{x-a}\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(x-z)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}\sum _{i=1}^N\frac{e^{{\rho }_{i,p}z}}{{\psi }^{\prime }\left({\rho }_{i,p}\right)}dz\hfill \\ -\frac{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,p}(x-a)}}{{\psi }^{\prime }\left({\rho }_{i,p}\right)}}{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,p}(c-a)}}{{\psi }^{\prime }\left({\rho }_{i,p}\right)}}\left(\sum _{i=1}^N\frac{e^{{\rho }_{i,q}c}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}-(q-p){\int }_0^{c-a}\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(c-z)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}\sum _{i=1}^N\frac{e^{{\rho }_{i,p}z}}{{\psi }^{\prime }\left({\rho }_{i,p}\right)}dz\right).\hfill \end{array}$$

(13)

Lemma 4.

For any $y<0$, $r>0$, $q\ge 0$, we have

$$\begin{array}{ccc}\hfill {\mathbb{E}}_y\left(e^{-q{\tau }_0^{+}};{\tau }_0^{+}\le r\right)& =& {\int }_0^{\infty }{e}^{-qr}\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(y+z)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}\frac{z}{r}·\mathbb{P}(X_r\in dz)\hfill \end{array}$$

(14)

$$\begin{array}{ccc}\hfill {\mathbb{P}}_y({\tau }_0^{+}\le r)& =& {\int }_0^{\infty }\sum _{i=1}^N\frac{e^{{\rho }_{i,0}(y+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\frac{z}{r}·\mathbb{P}(X_r\in dz).\hfill \end{array}$$

(15)

3. Main Result

In this section, we consider the case that ${\left\{Z_i\right\}}_{i=1,2,\dots }$ follows a mixed exponential distribution, and we give the Laplace transformation of the ruin time for a risk model with a Parisian implementation delay. Finally, we give display expressions using the scale function that can be further used to calculate dividends.

Theorem 1.

Assume that $U_b\left(t\right)$ is a risk model with a mixed exponential distribution of random income with a Parisian implementation delay in dividends. For any $b\ge 0$, $r>0$, we have

$\left(i\right)$ when $q\ge 0$, $0\le u\le b$,

$$\begin{array}{c}\hfill {\varphi }_L(b;b)={\left({\int }_0^{\infty }{e}^{-qr}\mathcal{M}(b,z)\frac{z}{r}\mathbb{P}(X_r\in dz)+e^{-qr}\left(1+q\sum _{i=1}^N\frac{e^{{\rho }_{i,q}b}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}}\right)\right)}^{-1};\end{array}$$

(16)

$\left(ii\right)$ when $q\ge 0$, $0\le u\le b$,

$$\begin{array}{ccc}\hfill {\varphi }_L(u;b)& =& \frac{{\int }_0^{\infty }{e}^{-qr}\mathcal{M}(b-u,z)\frac{z}{r}\mathbb{P}(X_r\in dz)+e^{-qr}(1+q{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}})}{{\int }_0^{\infty }{e}^{-qr}\mathcal{M}(b,z)\frac{z}{r}\mathbb{P}(X_r\in dz)+e^{-qr}(1+q{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}b}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}})};\hfill \end{array}$$

(17)

$\left(iii\right)$ when $q\ge 0$,$u>b$,

$$\begin{array}{ccc}\hfill {\varphi }_A(u;b)& =& \left(e^{-qr}-{\int }_0^{\infty }{e}^{-qr}\left(\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b-u+z)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}-\sum _{i=1}^N\frac{e^{{\rho }_{i,0}(b-u+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\right)\frac{z}{r}\mathbb{P}(X_r\in dz)\right){\varphi }_L(b;b),\hfill \end{array}$$

(18)

where

$$\begin{array}{c}\hfill \mathcal{M}(b,z)=\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b+z)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}-\sum _{i=1}^N\frac{e^{{\rho }_{i,0}(b+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\left[1+q\sum _{j=1}^N\frac{e^{({\rho }_{j,q}-{\rho }_{i,0})b}-1}{({\rho }_{j,q}-{\rho }_{i,0}){\psi }^{\prime }\left({\rho }_{j,q}\right)}\right].\end{array}$$

According to the reflection relationship between the dual risk model and the classical risk model, it can be concluded that

$$\begin{array}{ccc}\hfill {\mathbb{E}}_u\left(e^{-q{T}_{U_b}}{I}_{\{T_{U_b}<{\nu }_1\}}\right)& =& {\mathbb{E}}_{b-u}\left(e^{-q{\tau }_b^{+}}{I}_{\{{\tau }_b^{+}<{\tau }_0^{-}\}}\right),\hfill \end{array}$$

(19)

$$\begin{array}{ccc}\hfill {\mathbb{E}}_u\left(e^{-q{\nu }_1}{I}_{\{T_{U_b}>{\nu }_1\}}\right)& =& {\mathbb{E}}_{b-u}\left(e^{-q{\tau }_0^{-}}{I}_{\{{\tau }_b^{+}>{\tau }_0^{-}\}}\right).\hfill \end{array}$$

(20)

Proof of Theorem 1.

We know that the path of $U_b\left(t\right)$ is bounded variation; then, $W^{\left(q\right)}\left(0\right)=\frac{1}{c}$. For $0\le u\le b$, due to the strong Markov property of ${\eta }_1$ at rest, Equation (19), and $U_b\left({\eta }_1\right)=b$, we can export

$$\begin{array}{ccc}\hfill {\varphi }_L(u;b)& =& {\mathbb{E}}_u\left[e^{-q{T}_{U_b}}\right]\hfill \\ & =& {\mathbb{E}}_u\left[e^{-q{T}_{U_b}}{I}_{\{T_{U_b}<{\nu }_1\}}\right]+{\mathbb{E}}_u\left[e^{-q{T}_{U_b}}{I}_{\{T_{U_b}>{\nu }_1\}}\right]\hfill \\ & =& {\mathbb{E}}_u\left[e^{-q{T}_{U_b}}{I}_{\{T_{U_b}<{\nu }_1\}}\right]+{\mathbb{E}}_u\left[e^{-q{\eta }_1}{I}_{\{T_{U_b}>{\nu }_1\}}\right]{\varphi }_L(b;b)\hfill \\ & =& {\mathbb{E}}_{b-u}\left[e^{-q{\tau }_b^{+}}{I}_{\{{\tau }_b^{+}<{\tau }_0^{-}\}}\right]\hfill \\ & +& {\mathbb{E}}_u\left[e^{-q{\nu }_1}{I}_{\{T_{U_b}>{\nu }_1\}}{\mathbb{E}}_{U_b\left({\nu }_1\right)}\left(e^{-q{\eta }_1}\right)\right]{\varphi }_L(b;b).\hfill \end{array}$$

(21)

This is recorded as $\Delta \left(u\right)={\mathbb{E}}_u\left[e^{-q{\nu }_1}{I}_{\{T_{U_b}>{\nu }_1\}}{\mathbb{E}}_{U_b\left({\nu }_1\right)}\left(e^{-q{\eta }_1}\right)\right]$; then,

$$\begin{array}{ccc}\hfill {\varphi }_L(u;b)& =& {\mathbb{E}}_{b-u}\left[e^{-q{\tau }_b^{+}}{I}_{\{{\tau }_b^{+}<{\tau }_0^{-}\}}\right]+\Delta \left(u\right){\varphi }_L(b;b).\hfill \end{array}$$

(22)

From the spatial homogeneity of the model, the relationship between the classical risk model X and the dual risk model U, and the calculation of $\Delta \left(u\right)$ in Equation (20), we can obtain

$$\begin{array}{ccc}\hfill \Delta \left(u\right)& =& {\mathbb{E}}_u\left[e^{-q{\nu }_1}{I}_{\{T_{U_b}>{\nu }_1\}}{\mathbb{E}}_{U_b\left({\nu }_1\right)}\left(e^{-q{\eta }_1}\right)\right]\hfill \\ & =& {\mathbb{E}}_u\left[e^{-q{\nu }_1}{I}_{\{T_{U_b}>{\nu }_1\}}{\mathbb{E}}_{U_b\left({\nu }_1\right)-b}\left(e^{-q(T_{U_b}\wedge r)}\right)\right]\hfill \\ & =& {\mathbb{E}}_{b-u}\left[e^{-q{\tau }_0^{-}}{I}_{\{{\tau }_b^{+}>{\tau }_0^{-}\}}{\mathbb{E}}_{U_b\left({\nu }_1\right)-b}\left(e^{-q(T_{U_b}\wedge r)}\right)\right]\hfill \\ & =& {\mathbb{E}}_{b-u}\left[e^{-q{\tau }_0^{-}}{I}_{\{{\tau }_b^{+}>{\tau }_0^{-}\}}{\mathbb{E}}_{X\left({\tau }_0^{-}\right)}\left(e^{-q{\tau }_0^{+}}{I}_{\{{\tau }_0^{+}<r\}}\right)\right]\hfill \\ & +& {\mathbb{E}}_{b-u}\left[e^{-q{\tau }_0^{-}}{I}_{\{{\tau }_b^{+}>{\tau }_0^{-}\}}{e}^{-qr}{\mathbb{P}}_{X\left({\tau }_0^{-}\right)}({\tau }_0^{+}>r)\right]\hfill \\ & =& {\Delta }_1\left(u\right)+{\Delta }_2\left(u\right).\hfill \end{array}$$

(23)

Using the Tonelli theorem and Equation (14), as well as the spatial homogeneity of X,

$$\begin{array}{ccc}\hfill {\Delta }_1\left(u\right)& =& {\mathbb{E}}_{b-u}\left[e^{-q{\tau }_0^{-}}{I}_{\{{\tau }_b^{+}>{\tau }_0^{-}\}}{\mathbb{E}}_{X\left({\tau }_0^{-}\right)}\left(e^{-q{\tau }_0^{+}}{I}_{\{{\tau }_0^{+}<r\}}\right)\right]\hfill \\ & =& {\mathbb{E}}_{b-u}\left[e^{-q{\tau }_0^{-}}{I}_{\{{\tau }_b^{+}>{\tau }_0^{-}\}}{\int }_0^{\infty }{e}^{-qr}\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(X\left({\tau }_0^{-}\right)+z)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}\frac{z}{r}\mathbb{P}(X_r\in dz)\right]\hfill \\ & =& {\int }_0^{\infty }{e}^{-qr}{\mathbb{E}}_{b-u}\left(e^{-q{\tau }_0^{-}}\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(X\left({\tau }_0^{-}\right)+z)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}{I}_{\{{\tau }_b^{+}>{\tau }_0^{-}\}}\right)\frac{z}{r}\mathbb{P}(X_r\in dz)\hfill \\ & =& {\int }_0^{\infty }{e}^{-qr}{\mathbb{E}}_{b-u+z}\left(e^{-q{\tau }_z^{-}}\sum _{i=1}^N\frac{e^{{\rho }_{i,q}\left(X\left({\tau }_z^{-}\right)\right)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}{I}_{\{{\tau }_{b+z}^{+}>{\tau }_z^{-}\}}\right)\frac{z}{r}\mathbb{P}(X_r\in dz).\hfill \end{array}$$

(24)

Let $q=p$ in Equation (13); then, for $z<b-u+z<b+z$, we have

$$\begin{array}{c}{\displaystyle {\mathbb{E}}_{b-u+z}\left[e^{-q{\tau }_z^{-}}\sum _{i=1}^N\frac{e^{{\rho }_{i,q}X\left({\tau }_z^{-}\right)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}{I}_{\{{\tau }_z^{-}<{\tau }_{b+z}^{+}\}}\right]}\hfill \\ =\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b-u+z)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}-\frac{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}b}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b+z)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}.\hfill \end{array}$$

(25)

Substituting this into Equation (24) yields

$$\begin{array}{ccc}\hfill {\Delta }_1\left(u\right)& =& {\int }_0^{\infty }{e}^{-qr}\left(\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b-u+z)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}-\frac{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}b}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b+z)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}\right)\frac{z}{r}\mathbb{P}(X_r\in dz).\hfill \end{array}$$

(26)

Using the Tonelli theorem and Equation (15), as well as the spatial homogeneity of X,

$$\begin{array}{ccc}\hfill {\Delta }_2\left(u\right)& =& {\mathbb{E}}_{b-u}\left[e^{-q{\tau }_0^{-}}{I}_{\{{\tau }_b^{+}>{\tau }_0^{-}\}}{e}^{-qr}{\mathbb{P}}_{X\left({\tau }_0^{-}\right)}({\tau }_0^{+}>r)\right]\hfill \\ & =& {\mathbb{E}}_{b-u}\left[e^{-q{\tau }_0^{-}}{e}^{-qr}(1-{\mathbb{P}}_{X\left({\tau }_0^{-}\right)}({\tau }_0^{+}\le r))I_{\{{\tau }_0^{-}<{\tau }_b^{+}\}}\right]\hfill \\ & =& e^{-qr}{\mathbb{E}}_{b-u}\left[e^{-q{\tau }_0^{-}}{I}_{\{{\tau }_0^{-}<{\tau }_b^{+}\}}\right]-e^{-qr}{\mathbb{E}}_{b-u}\left[e^{-q{\tau }_0^{-}}{\mathbb{P}}_{X\left({\tau }_0^{-}\right)}({\tau }_0^{+}\le r)I_{\{{\tau }_0^{-}<{\tau }_b^{+}\}}\right]\hfill \\ & =& e^{-qr}{\mathbb{E}}_{b-u}\left[e^{-q{\tau }_0^{-}}{I}_{\{{\tau }_0^{-}<{\tau }_b^{+}\}}\right]\hfill \\ & -& e^{-qr}{\mathbb{E}}_{b-u}\left[e^{-q{\tau }_0^{-}}{\int }_0^{\infty }\sum _{i=1}^N\frac{e^{{\rho }_{i,0}(X\left({\tau }_0^{-}\right)+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\frac{z}{r}·\mathbb{P}(X_r\in dz)I_{\{{\tau }_0^{-}<{\tau }_b^{+}\}}\right]\hfill \\ & =& e^{-qr}{\mathbb{E}}_{b-u}\left[e^{-q{\tau }_0^{-}}{I}_{\{{\tau }_0^{-}<{\tau }_b^{+}\}}\right]\hfill \\ & -& {\int }_0^{\infty }{e}^{-qr}{\mathbb{E}}_{b-u+z}\left[e^{-q{\tau }_z^{-}}\sum _{i=1}^N\frac{e^{{\rho }_{i,0}X\left({\tau }_z^{-}\right)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}{I}_{\{{\tau }_z^{-}<{\tau }_{b+z}^{+}\}}\right]\frac{z}{r}\mathbb{P}(X_r\in dz).\hfill \end{array}$$

(27)

Let $q=p,q=0$ in Equation (13) and $p=q,q=-q$ in Equation (8); then,

$$\begin{array}{c}{\displaystyle {\mathbb{E}}_{b-u+z}\left[e^{-q{\tau }_z^{-}}\sum _{i=1}^N\frac{e^{{\rho }_{i,0}X\left({\tau }_z^{-}\right)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}{I}_{\{{\tau }_z^{-}<{\tau }_{b+z}^{+}\}}\right]}\hfill \\ =\sum _{i=1}^N\frac{e^{{\rho }_{i,0}(b-u+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}+q{\int }_0^{b-u}\sum _{i=1}^N\frac{e^{{\rho }_{i,0}(b-u+z-y)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\sum _{i=1}^N\frac{e^{{\rho }_{i,q}y}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}dy\hfill \\ -\frac{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}\left(b\right)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}\left(\sum _{i=1}^N\frac{e^{{\rho }_{i,0}(b+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}+q{\int }_0^b\sum _{i=1}^N\frac{e^{{\rho }_{i,0}(b+z-y)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\sum _{i=1}^N\frac{e^{{\rho }_{i,q}y}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}dy\right)\hfill \\ =\sum _{i=1}^N\frac{e^{{\rho }_{i,0}(b-u+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\left[1+q\sum _{j=1}^N\frac{e^{({\rho }_{j,q}-{\rho }_{i,0})(b-u)}-1}{({\rho }_{j,q}-{\rho }_{i,0}){\psi }^{\prime }\left({\rho }_{j,q}\right)}\right]\hfill \\ -\frac{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}b}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}\sum _{i=1}^N\frac{e^{{\rho }_{i,0}(b+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\left[1+q\sum _{j=1}^N\frac{e^{({\rho }_{j,q}-{\rho }_{i,0})b}-1}{({\rho }_{j,q}-{\rho }_{i,0}){\psi }^{\prime }\left({\rho }_{j,q}\right)}\right].\hfill \end{array}$$

(28)

According to Equation (10), it can be seen that

$$\begin{array}{ccc}\hfill {\mathbb{E}}_{b-u}\left(e^{-q{\tau }_0^{-}}{I}_{\{{\tau }_0^{-}<{\tau }_b^{+}\}}\right)& =& \left(1+q\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}}\right)\hfill \\ & & -\left(1+q\sum _{i=1}^N\frac{e^{{\rho }_{i,q}b}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}}\right)\frac{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}b}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}.\hfill \end{array}$$

(29)

Substituting the above simplified results into Equation (27), we can obtain

$$\begin{array}{ccc}\hfill {\Delta }_2\left(u\right)& =& e^{-qr}\left(\left(1+q\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}}\right)-\left(1+q\sum _{i=1}^N\frac{e^{{\rho }_{i,q}b}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}}\right)\frac{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}b}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}\right)\hfill \\ & -& {\int }_0^{\infty }{e}^{-qr}(\sum _{i=1}^N\frac{e^{{\rho }_{i,0}(b-u+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\left[1+q\sum _{j=1}^N\frac{e^{({\rho }_{j,q}-{\rho }_{i,0})(b-u)}-1}{({\rho }_{j,q}-{\rho }_{i,0}){\psi }^{\prime }\left({\rho }_{j,q}\right)}\right]\hfill \\ & -& \frac{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}b}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}\sum _{i=1}^N\frac{e^{{\rho }_{i,0}(b+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\left[1+q\sum _{j=1}^N\frac{e^{({\rho }_{j,q}-{\rho }_{i,0})b}-1}{({\rho }_{j,q}-{\rho }_{i,0}){\psi }^{\prime }\left({\rho }_{j,q}\right)}\right])\frac{z}{r}\mathbb{P}(X_r\in dz)\hfill \\ & =& e^{-qr}(1+q\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}})\hfill \\ & -& {\int }_0^{\infty }{e}^{-qr}\left(\sum _{i=1}^N\frac{e^{{\rho }_{i,0}(b-u+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\left[1+q\sum _{j=1}^N\frac{e^{({\rho }_{j,q}-{\rho }_{i,0})(b-u)}-1}{({\rho }_{j,q}-{\rho }_{i,0}){\psi }^{\prime }\left({\rho }_{j,q}\right)}\right]\right)\frac{z}{r}\mathbb{P}(X_r\in dz)\hfill \\ & -& \frac{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}b}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}(e^{-qr}\left(1+q\sum _{i=1}^N\frac{e^{{\rho }_{i,q}b}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}}\right)\hfill \\ & -& {\int }_0^{\infty }{e}^{-qr}\sum _{i=1}^N\frac{e^{{\rho }_{i,0}(b+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\left[1+q\sum _{j=1}^N\frac{e^{({\rho }_{j,q}-{\rho }_{i,0})b}-1}{({\rho }_{j,q}-{\rho }_{i,0}){\psi }^{\prime }\left({\rho }_{j,q}\right)}\right]\frac{z}{r}\mathbb{P}(X_r\in dz)).\hfill \end{array}$$

(30)

Combining Equations (26) and (30), we can obtain

$$\begin{array}{ccc}\hfill \Delta \left(u\right)& =& {\Delta }_1\left(u\right)+{\Delta }_2\left(u\right)\hfill \\ & =& {\int }_0^{\infty }{e}^{-qr}\left(\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b-u+z)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}-\frac{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}b}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b+z)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}\right)\frac{z}{r}\mathbb{P}(X_r\in dz)\hfill \\ & +& e^{-qr}\left(1+q\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}}\right)\hfill \\ & -& {\int }_0^{\infty }{e}^{-qr}\left(\sum _{i=1}^N\frac{e^{{\rho }_{i,0}(b-u+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\left[1+q\sum _{j=1}^N\frac{e^{({\rho }_{j,q}-{\rho }_{i,0})(b-u)}-1}{({\rho }_{j,q}-{\rho }_{i,0}){\psi }^{\prime }\left({\rho }_{j,q}\right)}\right]\right)\frac{z}{r}\mathbb{P}(X_r\in dz)\hfill \\ & -& \frac{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}b}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}(e^{-qr}\left(1+q\sum _{i=1}^N\frac{e^{{\rho }_{i,q}b}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}}\right)\hfill \\ & -& {\int }_0^{\infty }{e}^{-qr}\sum _{i=1}^N\frac{e^{{\rho }_{i,0}(b+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\left[1+q\sum _{j=1}^N\frac{e^{({\rho }_{j,q}-{\rho }_{i,0})b}-1}{({\rho }_{j,q}-{\rho }_{i,0}){\psi }^{\prime }\left({\rho }_{j,q}\right)}\right]\frac{z}{r}\mathbb{P}(X_r\in dz))\hfill \\ & =& e^{-qr}\left(\left(1+q\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}}\right)-\left(1+q\sum _{i=1}^N\frac{e^{{\rho }_{i,q}b}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}}\right)\frac{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}b}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}\right)\hfill \\ & +& {\int }_0^{\infty }{e}^{-qr}(\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b-u+z)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}\hfill \\ & -& \sum _{i=1}^N\frac{e^{{\rho }_{i,0}(b-u+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\left[1+q\sum _{j=1}^N\frac{e^{({\rho }_{j,q}-{\rho }_{i,0})(b-u)}-1}{({\rho }_{j,q}-{\rho }_{i,0}){\psi }^{\prime }\left({\rho }_{j,q}\right)}\right])\frac{z}{r}\mathbb{P}(X_r\in dz)\hfill \\ & -& \frac{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}b}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}{\int }_0^{\infty }{e}^{-qr}(\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b+z)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)})\hfill \\ & -& \sum _{i=1}^N\frac{e^{{\rho }_{i,0}(b+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\left[1+q\sum _{j=1}^N\frac{e^{({\rho }_{j,q}-{\rho }_{i,0})b}-1}{({\rho }_{j,q}-{\rho }_{i,0}){\psi }^{\prime }\left({\rho }_{j,q}\right)}\right]\frac{z}{r}\mathbb{P}(X_r\in dz).\hfill \end{array}$$

(31)

Let

$$\begin{array}{c}\hfill \mathcal{M}(b,z)=\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b+z)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}-\sum _{i=1}^N\frac{e^{{\rho }_{i,0}(b+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\left[1+q\sum _{j=1}^N\frac{e^{({\rho }_{j,q}-{\rho }_{i,0})b}-1}{({\rho }_{j,q}-{\rho }_{i,0}){\psi }^{\prime }\left({\rho }_{j,q}\right)}\right].\end{array}$$

(32)

Substituting Equation (32) into Equation (31) gives

$$\begin{array}{ccc}\hfill \Delta \left(u\right)& =& {\int }_0^{\infty }{e}^{-qr}\mathcal{M}(b-u,z)\frac{z}{r}\mathbb{P}(X_r\in dz)\hfill \\ & +& e^{-qr}(1+q\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}})\hfill \\ & -& \frac{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}b}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}({\int }_0^{\infty }{e}^{-qr}\mathcal{M}(b,z)\frac{z}{r}\mathbb{P}(X_r\in dz)\hfill \\ & +& e^{-qr}(1+q\sum _{i=1}^N\frac{e^{{\rho }_{i,q}b}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}})).\hfill \end{array}$$

(33)

Let $u=b$ in the above equation; then,

$$\begin{array}{ccc}\hfill 1-\Delta \left(b\right)& =& \frac{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}0}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}b}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}({\int }_0^{\infty }{e}^{-qr}\mathcal{M}(b,z)\frac{z}{r}\mathbb{P}(X_r\in dz)\hfill \\ & & +e^{-qr}(1+q\sum _{i=1}^N\frac{e^{{\rho }_{i,q}b}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}})).\hfill \end{array}$$

(34)

Let $u=b$ in Equation (22); then,

$$\begin{array}{c}\hfill {\varphi }_L(b;b)=\mathbb{E}\left(e^{-q{\tau }_b^{+}}{I}_{\{{\tau }_b^{+}<{\tau }_0^{-}\}}\right)+\Delta \left(b\right){\varphi }_L(b;b).\end{array}$$

According to Equation (9),

$$\begin{array}{c}\hfill (1-\Delta \left(b\right)){\varphi }_L(b;b)=\mathbb{E}\left(e^{-q{\tau }_b^{+}}{I}_{\{{\tau }_b^{+}<{\tau }_0^{-}\}}\right)=\frac{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}0}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}{{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}b}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}}.\end{array}$$

(35)

Then,

$$\begin{array}{c}\hfill \left({\int }_0^{\infty }{e}^{-qr}\mathcal{M}(b,z)\frac{z}{r}\mathbb{P}(X_r\in dz)+e^{-qr}(1+q\sum _{i=1}^N\frac{e^{{\rho }_{i,q}b}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}})\right){\varphi }_L(b;b)=1;\end{array}$$

(36)

that is,

$$\begin{array}{ccc}\hfill {\varphi }_L(b;b)& =& {\left({\int }_0^{\infty }{e}^{-qr}\mathcal{M}(b,z)\frac{z}{r}\mathbb{P}(X_r\in dz)+e^{-qr}(1+q\sum _{i=1}^N\frac{e^{{\rho }_{i,q}b}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}})\right)}^{-1},\hfill \end{array}$$

(37)

where

$$\begin{array}{c}\hfill \mathcal{M}(b,z)=\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b+z)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}-\sum _{i=1}^N\frac{e^{{\rho }_{i,0}(b+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\left[1+q\sum _{j=1}^N\frac{e^{({\rho }_{j,q}-{\rho }_{i,0})b}-1}{({\rho }_{j,q}-{\rho }_{i,0}){\psi }^{\prime }\left({\rho }_{j,q}\right)}\right].\end{array}$$

(38)

Theorem 1(i) is proven.

Substitute $\Delta \left(u\right)$ and ${\varphi }_L(b;b)$ into Equation (22) and simplify it to

$$\begin{array}{ccc}\hfill {\varphi }_L(u;b)& =& {\mathbb{E}}_{b-u}\left(e^{-q{\tau }_b^{+}}{I}_{\{{\tau }_b^{+}<{\tau }_0^{-}\}}\right)+\Delta \left(u\right){\varphi }_L(b;b)\hfill \\ & =& \frac{{\int }_0^{\infty }{e}^{-qr}\mathcal{M}(b-u,z)\frac{z}{r}\mathbb{P}(X_r\in dz)+e^{-qr}(1+q{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}(b-u)}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}})}{{\int }_0^{\infty }{e}^{-qr}\mathcal{M}(b,z)\frac{z}{r}\mathbb{P}(X_r\in dz)+e^{-qr}(1+q{\sum }_{i=1}^N\frac{e^{{\rho }_{i,q}b}-1}{{\psi }^{\prime }\left({\rho }_{i,q}\right){\rho }_{i,q}})}.\hfill \end{array}$$

(39)

Theorem 1(ii) is proven.

When $u>b$, it is easy to know that ${\nu }_1=0$; then, the first Parisian implementation delay starts from time 0. Utilizing the strong Markov property of process X at stop time ${\nu }_1$, it can be calculated that

$${\varphi }_A(u;b)={\mathbb{E}}_u\left[e^{-q{T}_{U_b}}\right]={\mathbb{E}}_u\left[e^{-q{\eta }_1}\right]{\varphi }_L(b;b).$$

Using spatial homogeneity, using Equation (19), and substituting Equations (14) and (15), it can be inferred that if $u>b$, then

$$\begin{array}{ccc}\hfill {\mathbb{E}}_u\left[e^{-q{\eta }_1}\right]& =& {\mathbb{E}}_{u-b}\left(e^{-q(T_U\wedge r)}\right)\hfill \\ & =& {\mathbb{E}}_b\left(e^{-q{\tau }_u^{+}}{I}_{\{{\tau }_u^{+}<r\}}\right)+{\mathbb{E}}_b\left(e^{-qr}{I}_{\{{\tau }_u^{+}>r\}}\right)\hfill \\ & =& {\mathbb{E}}_{b-u}\left(e^{-q{\tau }_0^{+}}{I}_{\{{\tau }_0^{+}<r\}}\right)+e^{-qr}{\mathbb{P}}_{b-u}({\tau }_0^{+}>r)\hfill \\ & =& {\int }_0^{\infty }{e}^{-qr}\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b-u+z)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}\frac{z}{r}·\mathbb{P}(X_r\in dz)\hfill \\ & +& e^{-qr}\left(1-{\int }_0^{\infty }\sum _{i=1}^N\frac{e^{{\rho }_{i,0}(b-u+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\frac{z}{r}·\mathbb{P}(X_r\in dz)\right)\hfill \\ & =& e^{-qr}-{\int }_0^{\infty }{e}^{-qr}\left(\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b-u+z)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}-\sum _{i=1}^N\frac{e^{{\rho }_{i,0}(b-u+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\right)\frac{z}{r}\mathbb{P}(X_r\in dz).\hfill \end{array}$$

(40)

Next,

$$\begin{array}{ccc}\hfill {\varphi }_A(u;b)& =& \left(e^{-qr}-{\int }_0^{\infty }{e}^{-qr}\left(\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b-u+z)}}{{\psi }^{\prime }\left({\rho }_{i,q}\right)}-\sum _{i=1}^N\frac{e^{{\rho }_{i,q}(b-u+z)}}{{\psi }^{\prime }\left({\rho }_{i,0}\right)}\right)\frac{z}{r}\mathbb{P}(X_r\in dz)\right){\varphi }_L(b;b).\hfill \end{array}$$

(41)

This completes the proof. □

4. Discussion and Conclusions

This paper applies the concept of the Parisian implementation delay to dividend payments, gives the Laplace transformation of the ruin time for a risk model with a Parisian implementation delay, and gives the display expressions using the scale function. However, this article only considers the situation where the company’s i-th claim amount follows a mixed exponential process, which is relatively narrow.

Next, we will apply the concept of the Parisian implementation delay to a wider range of situations and provide companies with more reasonable models. And, we will discuss the Laplace transform, where the claim amount follows other processes, for the direct use of statisticians and practical applications.