具有两类索赔干扰风险模型的Gerber-Shiu函数

The Gerber-Shiu Function with Two Types of Claim Interference Risk Model

摘要: 考虑一个受布朗运动扰动的两类索赔风险模型，两个索赔过程分别是独立的泊松过程和广义Erlang(2)过程。以Gerber-Shiu函数为研究对象，首先得到了由索赔导致破产和扰动导致破产的两种情况下惩罚函数满足的积分–微分方程及拉普拉斯变换的表达式。当两类的索赔额均呈指数分布时，给出了一个具体的数值例子。

Abstract: Consider a two-class claim risk model perturbed by Brownian motion, where the two claim processes are independent Poisson process and generalized Erlang(2) process respectively. Taking the Gerber-Shiu function as the research object, the integral-differential equations and the Laplace transform expressions satisfied by the penalty functions in the cases of ruin caused by claims and perturbation are obtained first. When the claim amounts of both classes follow an exponential distribution, a specific numerical example is given.

1. 引言

传统风险理论的研究焦点长期集中于破产概率分析领域。随着精算学科的发展，研究者逐步拓展了对多维破产测度的探索，包括但不限于破产时刻的联合密度函数、边际密度函数、破产赤字分布以及破产前盈余特征等关键指标(参见文献[1]-[4])。具有里程碑意义的是，精算学权威Hans Gerber与Elias Shiu于1998年开创性地引入期望折现惩罚函数(expected discounted penalty function) [5]，将其应用于复合泊松风险模型的破产问题研究。这一突破性方法不仅显著提升了风险理论的研究维度，更使得该函数体系发展成为现代破产理论分析的核心工具。在经典的风险模型中，通常假设索赔额和索赔间隔时间之间相互独立，这在一定程度上简化了数学计算。

1957年，Sparre Andersen在经典风险模型的基础上进行推广，首次提出更新风险模型。这一模型不再限定索赔次数必须服从泊松过程，而是改用更普适的更新计数过程，同时将索赔间隔时间的分布从指数分布扩展到了任意一般分布。此后大量学者开始关注除了复合泊松风险过程以外的其它更新风险模型，Erlang分布是其中最常用的分布之一，Dickson和Hipp就考虑了Erlang(2)风险模型中的破产概率问题。之后，Tsai和Sun研究了在Erlang(2)风险模型下的期望折现惩罚函数。对于两类索赔到达风险模型，文献[6]讨论了两类过程分别为Poission过程和广义Erlang(n)过程的情形，文献[7]研究了两类索赔到达过程分别为Poission过程和广义Erlang(2)过程的情形，本文在文献[7]的基础上研究干扰情形下的期望折现惩罚函数。对于此类风险模型，我们得到了Gerber-Shiu惩罚函数满足的积分–微分方程、拉普拉斯变换以及瑕疵更新方程。

2. 模型的建立

本文考虑索赔间隔时间服从泊松分布和Erlang(2)分布的一类干扰风险模型$\left\{U\left(t\right);t\ge 0\right\}$ ，

$U\left(t\right)=u+ct-{\displaystyle \sum _{i=1}^{N_1\left(t\right)}{Y}_i}-{\displaystyle \sum _{i=1}^{N_2\left(t\right)}{Z}_i}+\sigma B\left(t\right)$

式中$u\ge \text{0}$ 为初始盈余，$c$ 为保费率，设第一类索赔额${\left\{Y_i\right\}}_{i\ge 1}$ 是一列严格正的随机变量序列，其分布函数为$F\left(x\right)$ ，密度函数为$f\left(x\right)$ ；二类索赔额${\left\{Z_i\right\}}_{i\ge 1}$ 是一列严格正的随机变量序列，其分布函数为$V\left(x\right)$ ，密度函数为$v\left(x\right)$ 。用${\mu }_Y$ 和${\mu }_Z$ 表示$Y$ 和$Z$ 的均值。$\left\{N_1\left(t\right):t\ge 0\right\}$ 是参数为$\lambda$ 的泊松分布，表示到时刻$t$ 时第一类索赔的索赔次数。$\left\{N_2\left(t\right):t\ge 0\right\}$ 表示到时刻$t$ 时第二类索赔的索赔次数，$N_2\left(t\right)=\sup \left\{n:T_1+T_2+\cdots +T_n\le t\right\}$ ，其中$\left\{T_1,T_2,\cdots \right\}$ 是独立同分布的随机变量序列，表示第2类索赔时间间距，服从Erlang(2)分布，$T_i=L_{i1}+L_{i2}$ ，其中${\left\{L_{i1}\right\}}_{i\ge 1}$ 为$i$ 个参数为${\lambda }_1$ 的指数分布随机变量，${\left\{L_{i2}\right\}}_{i\ge 1}$ 为$i$ 个参数为${\lambda }_2$ 的指数分布随机变量。此外，$U\left(t\right)$ 包含一个扰动项，$\left\{B\left(t\right),t\ge 0\right\}$ 是标准的布朗运动，$\sigma$ 为扩散参数。

最后我们假设${\left\{Y_i\right\}}_{i\ge 1}$ 和${\left\{Z_i\right\}}_{i\ge 1}$ 是相互独立的，也独立于$N_1\left(t\right)$ 和$N_2\left(t\right)$ ，为保证破产不是必然事件，安全负载条件为$c>\lambda {\mu }_Y+\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1+{\lambda }_2}{\mu }_Z$ 。定义破产时间为$T=\inf \left\{t\ge 0,U\left(t\right)<0\right\},T=\infty$ ，Gerber-Shiu函数为：

$m\left(u\right)=E\left[{\text{e}}^{-\delta T}w\left(U\left(T-\right),\left|U\left(T\right)\right|\right)I\left(T<\infty ,U\left(T\right)<0\right)|U\left(0\right)=u\right]$ ,

其中$\delta \ge \text{0}$ 是贴现因子，$U\left(T-\right)$ 是破产前盈余，$\left|U\left(T\right)\right|$ 是破产后赤字，$I$ 是示性函数，$w\left(\cdot ,\cdot \right)$ 是关于破产前盈余和破产后赤字的二元函数。特别地，令$w\left(\text{0},\text{0}\right)=1$ 。易知，Gerber-Shiu函数$m\left(u\right)$ 可分解为

$m\left(u\right)=\varphi \left(u\right)+\psi \left(u\right)$ , (1)

其中

$\varphi \left(u\right)=E\left[{\text{e}}^{-\delta T}w\left(U\left(T-\right),\left|U\left(T\right)\right|\right)I\left(T<\infty ,U\left(T\right)<0\right)|U\left(0\right)=u\right]$ ,

$\psi \left(u\right)=\varphi \left(u\right)=E\left[{\text{e}}^{-\delta T}I\left(T<\infty ,U\left(T\right)=0\right)|U\left(0\right)=u\right]$ ,

表示由索赔导致的破产以及由布朗运动导致的破产。

假设${\varphi }_{i,j}\left(u\right),j=1,2$ ，表示破产由第$i,i=1,2$ 类索赔引起，且第二类索赔在状态$j$ 时破产的Gerber-Shiu函数。则函数(1)可表示为

$m_{i,j}\left(u\right)={\varphi }_{i,j}\left(u\right)+{\psi }_j\left(u\right)$ 。 (2)

3. 林德伯格方程

对于第2节中建立的风险模型，设$T_0=0$ ，$T_k={\displaystyle \sum _{j=1}^k{V}_j}$ 为来自第二类的第$k$ 次索赔的到达时间。当$U_0=0$ 时，对于$k=1,2,\cdots$ ，

$U_k=u+c{T}_k-{\displaystyle \sum _{j=1}^k{Z}_j}-{\displaystyle \sum _{i=1}^{N_1\left(T_k\right)}{Y}_j}+\sigma B\left(T_k\right)=u+{\displaystyle \sum _{j=1}^k\left[c{V}_j-Z_j-{\displaystyle \sum _{i=1}^{N_1\left(V_j\right)}{Y}_{ij}}+\sigma B\left(V_j\right)\right]}$ ,

为第二类第$k$ 次索赔后的盈余。我们要找到一个$s\in C$ 使得$E{\left\{{\text{e}}^{-\delta T_k+s{U}_k}\right\}}_{k\ge 0}$ 形成鞅当且仅当

$E\left[{\text{e}}^{\left(cs-\delta \right)V_1-s{Z}_1-s{\displaystyle \sum _{i=1}^{N_1\left(V_1\right)}{Y}_i}+\sigma sB\left(V_1\right)}\right]=E\left[{\text{e}}^{\left(cs-\delta \right)V_1-s{\displaystyle \sum _{i=1}^{N_1\left(V_1\right)}{Y}_i}+\sigma sB\left(V_1\right)}\right]E\left[{\text{e}}^{-s{Z}_1}\right]=1$ . (3)

由于

$\begin{array}{c}E\left[{\text{e}}^{\left(cs-\delta \right)V_1-s{Z}_1-s{\displaystyle \sum _{i=1}^{N_1\left(V_1\right)}{Y}_i}+\sigma sB\left(V_1\right)}\right]=E\left[E\left[{\text{e}}^{\left(cs-\delta \right)V_1-s{\displaystyle \sum _{i=1}^{N_1\left(V_1\right)}{Y}_i}+\sigma sB\left(V_1\right)}|V_1\right]\right]\\ ={\displaystyle {\int }_0^{\infty }E\left[{\text{e}}^{\left(cs-\delta \right)V_1-s{\displaystyle \sum _{i=1}^{N_1\left(V_1\right)}{Y}_i}+\sigma sB\left(V_1\right)}|V_1=t\right]f_{V_1}\left(t\right)\text{d}t}\\ ={\displaystyle {\int }_0^{\infty }{\text{e}}^{\left(cs-\delta \right)t}E\left\{{\text{e}}^{-s{\displaystyle \sum _{i=1}^{N_1\left(t\right)}{Y}_i}}\right\}E\left\{{\text{e}}^{\sigma sB\left(t\right)}\right\}{f}_{V_1}\left(t\right)\text{d}t}\\ =\frac{{\lambda }_{\text{1}}{\lambda }_{\text{2}}}{{\lambda }_{\text{2}}-{\lambda }_{\text{1}}}\left[{\displaystyle {\int }_{\text{0}}^{\infty }{\text{e}}^{\left[\frac{{\sigma }^{\text{2}}}{2}{s}^{\text{2}}+cs-\left(\delta +\lambda +{\lambda }_1\right)+\lambda \tilde{f}\left(s\right)\right]t}\text{d}t}-{\displaystyle {\int }_{\text{0}}^{\infty }{\text{e}}^{\left[\frac{{\sigma }^{\text{2}}}{2}{s}^{\text{2}}+cs-\left(\delta +\lambda +{\lambda }_2\right)+\lambda \tilde{f}\left(s\right)\right]t}\text{d}t}\right],\end{array}$

(3)式可简化为

$\frac{{\lambda }_{\text{1}}{\lambda }_{\text{2}}}{{\lambda }_{\text{2}}-{\lambda }_{\text{1}}}\left[{\displaystyle {\int }_{\text{0}}^{\infty }{\text{e}}^{\left[\frac{{\sigma }^{\text{2}}}{2}{s}^{\text{2}}+cs-\left(\delta +\lambda +{\lambda }_1\right)+\lambda \tilde{f}\left(s\right)\right]t}\text{d}t}-{\displaystyle {\int }_{\text{0}}^{\infty }{\text{e}}^{\left[\frac{{\sigma }^{\text{2}}}{2}{s}^{\text{2}}+cs-\left(\delta +\lambda +{\lambda }_2\right)+\lambda \tilde{f}\left(s\right)\right]t}\text{d}t}\right]\tilde{v}\left(s\right)=1$,

整理得：

$\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_1+\delta \right)+\lambda \tilde{f}\left(s\right)\right]\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_{\text{2}}+\delta \right)+\lambda \tilde{f}\left(s\right)\right]-{\lambda }_{\text{1}}{\lambda }_{\text{2}}\tilde{v}\left(s\right)=\text{0}$, (4)

记(4)式为林德伯格方程。

引理3.1 当$\delta >0$ 时，林德伯格方程有两个实部为正的根，分别记为${\rho }_{\text{1}}$ ，${\rho }_{\text{2}}$ 。

证明：由(2)式可得

$\text{1}=\frac{{\lambda }_{\text{1}}{\lambda }_{\text{2}}\tilde{v}\left(s\right)-{\left(\lambda \tilde{f}\left(s\right)\right)}^2-\left\{\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_1+\delta \right)\right]\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_{\text{2}}+\delta \right)\right]\right\}\lambda \tilde{f}\left(s\right)}{\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_1+\delta \right)\right]\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_{\text{2}}+\delta \right)\right]}$,

记

$f_1\left(s\right)={\lambda }_{\text{1}}{\lambda }_{\text{2}}\tilde{v}\left(s\right)-{\left(\lambda \tilde{f}\left(s\right)\right)}^2-\left\{\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_1+\delta \right)\right]\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_{\text{2}}+\delta \right)\right]\right\}\lambda \tilde{f}\left(s\right)$,

$f_2\left(s\right)=\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_1+\delta \right)\right]\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_{\text{2}}+\delta \right)\right]$ .

设一集合$C_r=\left\{s:\left|s\right|=r,\mathrm{Re}\left(s\right)\ge 0,r>0\right\}$ 。令$r\to \infty$ ，在$C_r$ 上应用Rouch é定理，当$s\to \infty$ 时，

$\begin{array}{c}\left|\frac{f_1\left(s\right)}{f_2\left(s\right)}\right|=\left|\frac{{\lambda }_{\text{1}}{\lambda }_{\text{2}}\tilde{v}\left(s\right)-{\left(\lambda \tilde{f}\left(s\right)\right)}^2-\left\{\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_1+\delta \right)\right]+\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_{\text{2}}+\delta \right)\right]\right\}\lambda \tilde{f}\left(s\right)}{\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_1+\delta \right)\right]\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_{\text{2}}+\delta \right)\right]}\right|\\ <\frac{{\lambda }_{\text{1}}{\lambda }_{\text{2}}\left|\tilde{v}\left(s\right)\right|}{\left|\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_1+\delta \right)\right]\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_{\text{2}}+\delta \right)\right]\right|}+\lambda \frac{1}{\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_{\text{2}}+\delta \right)\right]}\left|\tilde{f}\left(s\right)\right|\\ +{\lambda }^2\frac{{\left|\tilde{f}\left(s\right)\right|}^2}{\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_1+\delta \right)\right]\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_{\text{2}}+\delta \right)\right]}+\lambda \frac{1}{\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_1+\delta \right)\right]}\left|\tilde{f}\left(s\right)\right|\\ <{\lambda }^2\frac{1}{\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_1+\delta \right)\right]\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_{\text{2}}+\delta \right)\right]}+\lambda \frac{1}{\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_{\text{2}}+\delta \right)\right]}\\ +{\lambda }^2\frac{1}{\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_1+\delta \right)\right]\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_{\text{2}}+\delta \right)\right]}+\lambda \frac{1}{\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_1+\delta \right)\right]}\\ \to \text{0,}\end{array}$

即$\left|f_1\left(s\right)\right|<\left|f_2\left(s\right)\right|$ 。 □

引理3.2 当$\delta =0$ 时，林德伯格方程有一个实部为正的根，记为${\rho }_{\text{1}}$ ，另一个根为0。

证明：定义$C=\left\{s:\left|z\right|=1\right\}$ ，令$z=\frac{k-s}{k}$ 。当$z=\text{1}$ 时，$s=0$ 。

当$k\to \infty$ 时，则有

$\begin{array}{l}\frac{\text{d}}{\text{d}z}{\left\{1-\frac{\begin{array}{l}{\lambda }_{\text{1}}{\lambda }_{\text{2}}\tilde{v}\left(k-kz\right)-{\left(\lambda \tilde{f}\left(k-kz\right)\right)}^2\\ -\left\{\left[\frac{1}{2}{\sigma }^2{\left(k-kz\right)}^2+c\left(k-kz\right)-\left(\lambda +{\lambda }_1\right)\right]\left[\frac{1}{2}{\sigma }^2{\left(k-kz\right)}^2+c\left(k-kz\right)-\left(\lambda +{\lambda }_{\text{2}}\right)\right]\right\}\lambda \tilde{f}\left(k-kz\right)\\ +\left[\frac{1}{2}{\sigma }^2{\left(k-kz\right)}^2+c\left(k-kz\right)-\left(\lambda +{\lambda }_1\right)\right]\left[\frac{1}{2}{\sigma }^2{\left(k-kz\right)}^2+c\left(k-kz\right)-\left(\lambda +{\lambda }_{\text{2}}\right)\right]\end{array}}{\left[\frac{1}{2}{\sigma }^2\left(k-kz\right)+c\left(k-kz\right)-\left(\lambda +{\lambda }_1\right)\right]\left[\frac{1}{2}{\sigma }^2\left(k-kz\right)+c\left(k-kz\right)-\left(\lambda +{\lambda }_{\text{2}}\right)\right]}\right\}}_{z=1}\\ =\frac{\text{d}}{\text{d}z}{\left\{1-E\left[{\text{e}}^{\left[c\left(k-kz\right)\right]V_1-\left(k-kz\right)Z_1-\left(k-kz\right){\displaystyle \sum _{i=1}^{N_1\left(V_1\right)}{Y}_i}+\sigma \left(k-kz\right)B\left(V_1\right)}\right]\right\}}_{z=1}=\frac{\text{d}}{\text{d}z}{\left(1-E\left[{\text{e}}^{k\left(1-z\right)\left({\displaystyle \sum _{i=1}^{N_1\left(V_1\right)}{Y}_i}+Z-c{V}_1-\sigma B\left(V_1\right)\right)}\right]\right)}_{z=1}\\ =kE\left[{\displaystyle \sum _{i=1}^{N_1\left(V\right)}{Y}_i}+Z-cV-\sigma B\left(V\right)\right]>0.\end{array}$

根据Klimenok可知，林德伯格方程有一个实部为正的根，记为${\rho }_{\text{1}}$ ，另一个根为0。 □

4. Gerber-Shiu函数

定理3.1 拉普拉斯变换${\tilde{\varphi }}_{i,j}\left(s\right),i,j=1,2,s\ge 0$ 满足以下方程

${\tilde{\varphi }}_{1,1}\left(s\right)=\frac{\left[\frac{1}{2}{\sigma }^2{{\varphi }^{\prime }}_{1,1}\left(0\right)-\lambda {\tilde{\zeta }}_1\left(s\right)\right]A_2\left(s\right)-{\lambda }_1\left[\frac{1}{2}{\sigma }^2{{\varphi }^{\prime }}_{1,2}-\lambda {\tilde{\zeta }}_1\left(s\right)\right]}{A_1\left(s\right)A_2\left(s\right)-{\lambda }_1{\lambda }_2\tilde{v}\left(s\right)}$ , (5)

${\tilde{\varphi }}_{1,2}\left(s\right)=\frac{\left[\frac{1}{2}{\sigma }^2{{\varphi }^{\prime }}_{1,2}\left(0\right)-\lambda {\tilde{\zeta }}_1\left(s\right)\right]A_1\left(s\right)-{\lambda }_2\tilde{v}\left(s\right)\left[\frac{1}{2}{\sigma }^2{{\varphi }^{\prime }}_{1,1}\left(0\right)-\lambda {\tilde{\zeta }}_1\left(s\right)\right]}{A_1\left(s\right)A_2\left(s\right)-{\lambda }_1{\lambda }_2\tilde{v}\left(s\right)}$ , (6)

${\tilde{\varphi }}_{2,1}\left(s\right)=\frac{\frac{1}{2}{\sigma }^2{{\varphi }^{\prime }}_{2,1}\left(0\right)A_2\left(s\right)-{\lambda }_1\left[\frac{1}{2}{\sigma }^2{{\varphi }^{\prime }}_{2,2}\left(0\right)-{\lambda }_2{\tilde{\zeta }}_2\left(s\right)\right]}{A_1\left(s\right)A_2\left(s\right)-{\lambda }_1{\lambda }_2\tilde{v}\left(s\right)}$ , (7)

${\tilde{\varphi }}_{2,2}\left(s\right)=\frac{\left[\frac{1}{2}{\sigma }^2{{\varphi }^{\prime }}_{2,2}\left(0\right)-{\lambda }_2{\tilde{\zeta }}_2\left(s\right)\right]A_1\left(s\right)-\frac{1}{2}{\sigma }^2{{\varphi }^{\prime }}_{2,1}\left(0\right){\lambda }_2\tilde{v}\left(s\right)}{A_1\left(s\right)A_2\left(s\right)-{\lambda }_1{\lambda }_2\tilde{v}\left(s\right)}$ , (8)

其中

$A_1\left(s\right)=\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_{\text{1}}+\delta \right)$ , $A_2\left(s\right)=\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_{\text{2}}+\delta \right)+\lambda \tilde{f}\left(s\right)$,

${\zeta }_{\text{1}}\left(u\right)={\displaystyle {\int }_u^{\infty }w\left(u,y-u\right)f\left(y\right)\text{d}y}$ , ${\zeta }_{\text{2}}\left(u\right)={\displaystyle {\int }_u^{\infty }w\left(u,z-u\right)v\left(z\right)\text{d}z}$ .

证明：以第一次索赔发生时间和索赔金额为条件应用全概率公式，我们考虑模型满足的Gerber-Shiu函数满足下列方程

$\begin{array}{c}{\varphi }_{1,1}\left(u\right)=\left(1-\lambda dt\right)\left(1-{\lambda }_{1}dt\right){\text{e}}^{-\delta dt}E\left[{\varphi }_{1,1}\left(u+cdt+\sigma W\left(dt\right)\right)\right]\\ +\lambda dt\left(1-{\lambda }_{1}dt\right){\text{e}}^{-\delta dt}[{\displaystyle {\int }_0^{u+cdt+\sigma W\left(dt\right)}{\varphi }_{1,2}\left(u+cdt+\sigma W\left(dt\right)-y\right)f\left(y\right)\text{d}y}\\ +{\displaystyle {\int }_{u+cdt+\sigma W\left(dt\right)}^{\infty }w\left(u+cdt+\sigma W\left(dt\right),y-u-cdt-\sigma W\left(dt\right)\right)f\left(y\right)\text{d}y}]\\ +{\lambda }_{1}dt\left(1-\lambda dt\right){\text{e}}^{-\delta dt}{\varphi }_{1,2}\left(u+cdt+\sigma W\left(dt\right)\right)+o\left(dt\right),\end{array}$ (9)

$\begin{array}{c}{\varphi }_{1,2}\left(u\right)=\left(1-\lambda dt\right)\left(1-{\lambda }_{2}dt\right){\text{e}}^{-\delta dt}E\left[{\varphi }_{1,2}\left(u+cdt+\sigma W\left(dt\right)\right)\right]\\ +\lambda dt\left(1-{\lambda }_{2}dt\right){\text{e}}^{-\delta dt}[{\displaystyle {\int }_0^{u+cdt+\sigma W\left(dt\right)}{\varphi }_{1,2}\left(u+cdt+\sigma W\left(dt\right)-y\right)f\left(y\right)\text{d}y}\\ +{\displaystyle {\int }_{u+cdt+\sigma W\left(dt\right)}^{\infty }w\left(u+cdt+\sigma W\left(dt\right),y-u-cdt-\sigma W\left(dt\right)\right)f\left(y\right)\text{d}y}]\\ +{\lambda }_{2}dt\left(1-\lambda dt\right){\text{e}}^{-\delta dt}{\displaystyle {\int }_0^{u+cdt+\sigma W\left(dt\right)}{\varphi }_{1,1}\left(u+cdt+\sigma W\left(dt\right)-z\right)v\left(z\right)\text{d}z}+o\left(dt\right),\end{array}$ (10)

$\begin{array}{c}{\varphi }_{2,1}\left(u\right)=\left(1-\lambda dt\right)\left(1-{\lambda }_{1}dt\right){\text{e}}^{-\delta dt}E\left[{\varphi }_{2,1}\left(u+cdt+\sigma W\left(dt\right)\right)\right]\\ +\lambda dt\left(1-{\lambda }_{1}dt\right){\text{e}}^{-\delta dt}{\displaystyle {\int }_0^{u+cdt+\sigma W\left(dt\right)}{\varphi }_{2,1}\left(u+cdt+\sigma W\left(dt\right)-y\right)f\left(y\right)\text{d}y}\\ +{\lambda }_{1}dt\left(1-\lambda dt\right){\text{e}}^{-\delta dt}{\varphi }_{2,2}\left(u+cdt+\sigma W\left(dt\right)\right)+o\left(dt\right),\end{array}$ (11)

$\begin{array}{c}{\varphi }_{2,2}\left(u\right)=\left(1-\lambda dt\right)\left(1-{\lambda }_{2}dt\right){\text{e}}^{-\delta dt}E\left[{\varphi }_{2,2}\left(u+cdt+\sigma W\left(dt\right)\right)\right]\\ +\lambda dt\left(1-{\lambda }_{2}dt\right){\text{e}}^{-\delta dt}[{\displaystyle {\int }_0^{u+cdt+\sigma W\left(dt\right)}{\varphi }_{2,2}\left(u+cdt+\sigma W\left(dt\right)-y\right)f\left(y\right)\text{d}y}\\ +{\displaystyle {\int }_{u+cdt+\sigma W\left(dt\right)}^{\infty }w\left(u+cdt+\sigma W\left(dt\right),y-u-cdt-\sigma W\left(dt\right)\right)v\left(z\right)\text{d}z}+o\left(dt\right).\end{array}$ (12)

利用泰勒展开公式

$E\left[{\varphi }_{i,j}\left(u+cdt+\sigma W\left(dt\right)\right)\right]={\varphi }_{i,j}\left(u\right)+c{{\varphi }^{\prime }}_{i,j}\left(u\right)dt+\frac{1}{2}{\sigma }^2{{\varphi }^{″}}_{i,j}\left(u\right)dt+o\left(dt\right)$ ,

对(9)、(10)式整理后等式两边同时除以$dt$ ，并令$dt\to \text{0}$ 得

$\left(\delta +\lambda +{\lambda }_1\right){\varphi }_{1,1}\left(u\right)=c{{\varphi }^{\prime }}_{1,1}\left(u\right)+\frac{1}{2}{\sigma }^2{{\varphi }^{″}}_{1,1}\left(u\right)+\lambda \left[{\displaystyle {\int }_0^u{\varphi }_{1,1}\left(u-y\right)f\left(y\right)\text{d}y}+{\zeta }_1\left(u\right)\right]+{\lambda }_1{\varphi }_{1,2}\left(u\right)$ ,

$\begin{array}{c}\left(\delta +\lambda +{\lambda }_2\right){\varphi }_{1,2}\left(u\right)=c{{\varphi }^{\prime }}_{1,2}\left(u\right)\text{d}t+\frac{1}{2}{\sigma }^2{{\varphi }^{″}}_{1,2}\left(u\right)dt+\lambda \left[{\displaystyle {\int }_0^u{\varphi }_{1,2}\left(u-y\right)f\left(y\right)\text{d}y}+\zeta \left(u\right)\right]\\ +{\lambda }_2{\displaystyle {\int }_0^u{\varphi }_{1,1}\left(u-z\right)v\left(z\right)\text{d}z},\end{array}$

对上式取拉普拉斯变换有

$\begin{array}{c}\left(\delta +\lambda +{\lambda }_1\right){\tilde{\varphi }}_{1,1}\left(s\right)=c\left[s{\tilde{\varphi }}_{1,1}\left(s\right)-{\varphi }_{1,1}\left(0\right)\right]+\frac{1}{2}{\sigma }^2\left[s^2{\tilde{\varphi }}_{1,1}\left(s\right)-s{\varphi }_{1,1}\left(0\right)-{{\varphi }^{\prime }}_{1,1}\left(0\right)\right]\\ +\lambda {\tilde{\varphi }}_{1,1}\left(s\right)\tilde{f}\left(s\right)+\lambda {\tilde{\zeta }}_1\left(s\right)+{\lambda }_1{\tilde{\varphi }}_{1,2}\left(s\right),\end{array}$

$\begin{array}{c}\left(\delta +\lambda +{\lambda }_2\right){\tilde{\varphi }}_{1,2}\left(s\right)=c\left[s{\tilde{\varphi }}_{1,2}\left(s\right)-{\varphi }_{1,2}\left(0\right)\right]+\frac{1}{2}{\sigma }^2\left[s^2{\tilde{\varphi }}_{1,2}(s)-s{\varphi }_{1,2}\left(0\right)-{{\varphi }^{\prime }}_{1,2}\left(0\right)\right]\\ +\lambda {\varphi }_{1,2}\left(s\right)\tilde{f}\left(s\right)+\lambda \tilde{\zeta }\left(s\right)+{\lambda }_2\tilde{v}\left(s\right){\tilde{\varphi }}_{1,1}\left(s\right).\end{array}$

由于${\varphi }_{1,1}\left(0\right)={\varphi }_{1,2}\left(0\right)=0$ ，整理得

$\{\begin{array}{l}\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_1+\delta \right)+\lambda \tilde{f}\left(s\right)\right]{\tilde{\varphi }}_{1,1}\left(s\right)=\frac{1}{2}{\sigma }^2{{\varphi }^{\prime }}_{1,1}\left(0\right)-\lambda {\tilde{\zeta }}_1\left(s\right)-{\lambda }_1{\tilde{\varphi }}_{1,2}\left(s\right)\\ \left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_2+\delta \right)+\lambda \tilde{f}\left(s\right)\right]{\tilde{\varphi }}_{1,2}\left(s\right)=\frac{1}{2}{\sigma }^2{{\varphi }^{\prime }}_{1,2}\left(0\right)-\lambda {\tilde{\zeta }}_1\left(s\right)-{\lambda }_2\tilde{v}\left(s\right){\tilde{\varphi }}_{1,1}\left(s\right)\end{array}$

由此推导得到(5)、(6)。对(11)、(12)利用同样的方法即可推导(7)、(8)。 □

将Gerber-Shiu惩罚函数的拉普拉斯变换重新整理为

${\tilde{\varphi }}_{1,1}\left(s\right)=\frac{{\tilde{\vartheta }}_{1,\text{1}}\left(s\right)+{\tilde{\beta }}_{1,\text{1}}\left(s\right)}{{\tilde{h}}_{1,\delta }\left(s\right)-{\tilde{h}}_{2,\delta }\left(s\right)}$, ${\tilde{\varphi }}_{1,2}\left(s\right)=\frac{{\tilde{\vartheta }}_{1,2}\left(s\right)+{\tilde{\beta }}_{1,2}\left(s\right)}{{\tilde{h}}_{1,\delta }\left(s\right)-{\tilde{h}}_{2,\delta }\left(s\right)}$

${\tilde{\varphi }}_{2,1}\left(s\right)=\frac{{\tilde{\vartheta }}_{2,1}\left(s\right)+{\tilde{\beta }}_{2,1}\left(s\right)}{{\tilde{h}}_{1,\delta }\left(s\right)-{\tilde{h}}_{2,\delta }\left(s\right)}$ , ${\tilde{\varphi }}_{2,2}\left(s\right)=\frac{{\tilde{\vartheta }}_{2,2}\left(s\right)+{\tilde{\beta }}_{2,2}\left(s\right)}{{\tilde{h}}_{1,\delta }\left(s\right)-{\tilde{h}}_{2,\delta }\left(s\right)}$

其中

${\tilde{h}}_{1,\delta }\left(s\right)=\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_1+\delta \right)\right]\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_{\text{2}}+\delta \right)\right]$,

${\tilde{h}}_{\text{2},\delta }(s)={\lambda }_{\text{1}}{\lambda }_{\text{2}}\tilde{v}\left(s\right)-\lambda \left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_1+\delta \right)\right]\tilde{f}\left(s\right)-\lambda \left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_{\text{2}}+\delta \right)\right]\tilde{f}\left(s\right)-{\lambda }^2{\left[\tilde{f}\left(s\right)\right]}^2$,

${\tilde{\vartheta }}_{1,1}\left(s\right)=\left[\frac{1}{2}{\sigma }^2{{\varphi }^{\prime }}_{1,1}\left(0\right)-\lambda {\tilde{\zeta }}_1\left(s\right)\right]\left[\frac{1}{2}{\sigma }^2{s}^2+cs-\left(\lambda +{\lambda }_{\text{2}}+\delta \right)+\lambda \tilde{f}\left(s\right)\right]$, ${\tilde{\beta }}_{\text{1,1}}\left(s\right)={\lambda }_1\left[\lambda {\tilde{\zeta }}_{\text{1}}\left(s\right)-\frac{1}{2}{\sigma }^2{{\varphi }^{\prime }}_{\text{1,2}}\left(0\right)\right]$,

${\tilde{\vartheta }}_{1,\text{2}}\left(s\right)=\left[\frac{1}{2}{\sigma }^2{{\varphi }^{\prime }}_{\text{1,2}}\left(0\right)-\lambda {\tilde{\zeta }}_{\text{1}}\left(s\right)\right]A_1\left(s\right)$, ${\tilde{\beta }}_{\text{1},2}\left(s\right)={\lambda }_{\text{2}}\tilde{v}\left(s\right)\left[\lambda {\tilde{\zeta }}_{\text{2}}\left(s\right)-\frac{1}{2}{\sigma }^2{{\varphi }^{\prime }}_{\text{1,1}}\left(0\right)\right]$,

${\tilde{\vartheta }}_{\text{2,1}}\left(s\right)=\frac{1}{2}{\sigma }^2{{\varphi }^{\prime }}_{2,1}\left(0\right)A_2\left(s\right)$, ${\tilde{\beta }}_{2,\text{1}}\left(s\right)={\lambda }_1\left[{\lambda }_2{\tilde{\zeta }}_2\left(s\right)-\frac{1}{2}{\sigma }^2{{\varphi }^{\prime }}_{\text{2,2}}\left(0\right)\right]$,

${\tilde{\vartheta }}_{\text{2},2}\left(s\right)=\left[\frac{1}{2}{\sigma }^2{{\varphi }^{\prime }}_{\text{2,2}}\left(0\right)-{\lambda }_2{\tilde{\zeta }}_2\left(s\right)\right]A_1\left(s\right)$, ${\tilde{\beta }}_{2,\text{2}}\left(s\right)=-\frac{1}{2}{\sigma }^2{{\varphi }^{\prime }}_{\text{2,1}}\left(0\right){\lambda }_{\text{2}}\tilde{v}\left(s\right)$.

定理3.2 拉普拉斯变换${\tilde{\psi }}_i\left(s\right),i=1,2,s\ge 0$ 满足以下方程

${\tilde{\psi }}_{\text{1}}\left(s\right)=\frac{\left[\frac{1}{2}{\sigma }^{2}s+c+\frac{1}{2}{\sigma }^2{{\psi }^{\prime }}_{\text{1}}\left(0\right)\right]A_2\left(s\right)-{\lambda }_1\left[\frac{1}{2}{\sigma }^{2}s+c+\frac{1}{2}{\sigma }^2{{\psi }^{\prime }}_{\text{2}}\left(0\right)\right]}{A_1\left(s\right)A_2\left(s\right)-{\lambda }_{\text{1}}{\lambda }_{\text{2}}\tilde{v}\left(s\right)}$, (13)

${\tilde{\psi }}_{\text{2}}\left(s\right)=\frac{\left[\frac{1}{2}{\sigma }^{2}s+c+\frac{1}{2}{\sigma }^2{{\psi }^{\prime }}_{\text{2}}\left(0\right)\right]A_1\left(s\right)-{\lambda }_{\text{2}}\tilde{v}\left(s\right)\left[\frac{1}{2}{\sigma }^{2}s+c+\frac{1}{2}{\sigma }^2{{\psi }^{\prime }}_{\text{1}}\left(0\right)\right]}{A_1\left(s\right)A_2\left(s\right)-{\lambda }_{\text{1}}{\lambda }_{\text{2}}\tilde{v}\left(s\right)}$. (14)

证：以第一次索赔时间发生之前是否发生破产为条件应用全概率公式，我们考虑模型满足的Gerber-Shiu函数满足下列方程

$\begin{array}{c}{\psi }_{\text{1}}\left(u\right)=\left(1-\lambda dt\right)\left(1-{\lambda }_{1}dt\right){\text{e}}^{-\delta dt}{\varphi }_{\text{1,1}}\left(u+cdt+\sigma W\left(dt\right)\right)\\ +\lambda dt\left(1-{\lambda }_{1}dt\right){\text{e}}^{-\delta dt}{\displaystyle {\int }_0^{u+cdt+\sigma W\left(dt\right)}{\psi }_{\text{1}}\left(u+cdt+\sigma W\left(dt\right)-y\right)f\left(y\right)\text{d}y}\\ +{\lambda }_{1}dt\left(1-\lambda dt\right){\text{e}}^{-\delta dt}{\psi }_{\text{2}}\left(u+cdt+\sigma W\left(dt\right)\right)+o\left(dt\right),\end{array}$ (15)

$\begin{array}{c}{\psi }_{\text{2}}\left(u\right)=\left(1-\lambda dt\right)\left(1-{\lambda }_{\text{2}}dt\right){\text{e}}^{-\delta dt}{\psi }_{\text{2}}\left(u+cdt+\sigma W\left(dt\right)\right)\\ +\lambda dt\left(1-{\lambda }_{\text{2}}dt\right){\text{e}}^{-\delta dt}{\displaystyle {\int }_0^{u+cdt+\sigma W\left(dt\right)}{\psi }_{\text{2}}\left(u+cdt+\sigma W\left(dt\right)-y\right)f\left(y\right)\text{d}y}\\ +{\lambda }_{\text{2}}dt\left(1-\lambda dt\right){\text{e}}^{-\delta dt}{\displaystyle {\int }_{\text{0}}^{u+cdt+\sigma W\left(dt\right)}{\phi }_{\text{1}}\left(u+cdt+\sigma W\left(dt\right)-z\right)v\left(z\right)\text{d}z}+o\left(dt\right).\end{array}$ (16)

由于${\psi }_1\left(0\right)={\psi }_2\left(0\right)=1$ ，利用相同的推导过程，我们可得到(15)、(16)。 □

将Gerber-Shiu惩罚函数的拉普拉斯变换重新整理为

${\tilde{\psi }}_{\text{1}}\left(s\right)=\frac{{\tilde{\eta }}_{\text{1}}\left(s\right)+{\tilde{\alpha }}_{\text{1}}\left(s\right)}{{\tilde{h}}_{1,\delta }\left(s\right)-{\tilde{h}}_{2,\delta }\left(s\right)}$, ${\tilde{\psi }}_{\text{2}}\left(s\right)=\frac{{\tilde{\eta }}_2\left(s\right)+{\tilde{\alpha }}_2\left(s\right)}{{\tilde{h}}_{1,\delta }\left(s\right)-{\tilde{h}}_{2,\delta }\left(s\right)}$,

其中

${\tilde{\eta }}_{\text{1}}\left(s\right)=\left[\frac{1}{2}{\sigma }^{2}s+c+\frac{1}{2}{\sigma }^2{{\psi }^{\prime }}_{\text{1}}\left(0\right)\right]A_2\left(s\right)$, ${\tilde{\alpha }}_{\text{1}}\left(s\right)=-{\lambda }_1\left[\frac{1}{2}{\sigma }^{2}s+c+\frac{1}{2}{\sigma }^2{{\psi }^{\prime }}_{\text{2}}\left(0\right)\right]$,

${\tilde{\eta }}_{\text{2}}\left(s\right)=\left[\frac{1}{2}{\sigma }^{2}s+c+\frac{1}{2}{\sigma }^2{{\psi }^{\prime }}_{\text{2}}\left(0\right)\right]A_1\left(s\right)$, ${\tilde{\alpha }}_2\left(s\right)=-{\lambda }_1\left[\frac{1}{2}{\sigma }^{2}s+c+\frac{1}{2}{\sigma }^2{{\psi }^{\prime }}_{\text{1}}\left(0\right)\right]$.

5. 瑕疵更新方程

这一节研究索赔或扩散导致破产发生的Gerber-Shiu惩罚函数满足的瑕疵更新方程。

引理1 Gerber-Shiu惩罚函数${\tilde{\varphi }}_{i,j}\left(s\right)$ 和${\tilde{\psi }}_i\left(s\right)$ 的拉普拉斯变换可以写成如下形式：

${\tilde{\varphi }}_{1，1}\left(s\right)=\frac{T_s{T}_{{\rho }_2}{T}_{{\rho }_1}{\vartheta }_{1,1}\left(0\right)}{1-T_s{T}_{{\rho }_2}{T}_{{\rho }_1}{h}_{2,\delta }\left(0\right)}$ , (17)

${\tilde{\varphi }}_{1,2}\left(s\right)=\frac{T_s{T}_{{\rho }_2}{T}_{{\rho }_1}{\vartheta }_{1,2}\left(0\right)}{1-T_s{T}_{{\rho }_2}{T}_{{\rho }_1}{h}_{2,\delta }\left(0\right)}$ , (18)

${\tilde{\varphi }}_{\text{2,1}}\left(s\right)=\frac{T_s{T}_{{\rho }_2}{T}_{{\rho }_1}{\vartheta }_{2,1}\left(0\right)}{1-T_s{T}_{{\rho }_2}{T}_{{\rho }_1}{h}_{2,\delta }\left(0\right)}$, (19)

${\tilde{\varphi }}_{\text{2,2}}\left(s\right)=\frac{T_s{T}_{{\rho }_2}{T}_{{\rho }_1}{\vartheta }_{2,\text{2}}\left(0\right)}{1-T_s{T}_{{\rho }_2}{T}_{{\rho }_1}{h}_{2,\delta }\left(0\right)}$, (20)

${\tilde{\psi }}_{\text{1}}\left(s\right)=\frac{T_s{T}_{{\rho }_2}{T}_{{\rho }_1}{\eta }_1\left(0\right)}{1-T_s{T}_{{\rho }_2}{T}_{{\rho }_1}{h}_{2,\delta }\left(0\right)}$, (21)

${\tilde{\psi }}_{\text{2}}\left(s\right)=\frac{T_s{T}_{{\rho }_2}{T}_{{\rho }_1}{\eta }_{\text{2}}\left(0\right)}{1-T_s{T}_{{\rho }_2}{T}_{{\rho }_1}{h}_{2,\delta }\left(0\right)}$, (22)

证明：林德伯格方程的两个非负根分别为${\rho }_{\text{1}}$ 和${\rho }_{\text{2}}$ ，这表明${\tilde{h}}_{1,\delta }\left({\rho }_{\text{1}}\right)={\tilde{h}}_{\text{2},\delta }\left({\rho }_{\text{1}}\right)$、${\tilde{h}}_{1,\delta }\left({\rho }_{\text{2}}\right)={\tilde{h}}_{\text{2,}\delta }\left({\rho }_{\text{2}}\right)$。则

$\begin{array}{c}{\tilde{h}}_{1,\delta }\left(s\right)={\tilde{h}}_{1,\delta }\left(0\right)\frac{\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)}{{\rho }_1{\rho }_2}+s\left(\frac{{\tilde{h}}_{1,\delta }\left({\rho }_1\right)}{{\rho }_1}\frac{s-{\rho }_2}{{\rho }_1-{\rho }_2}+\frac{{\tilde{h}}_{1,\delta }\left({\rho }_2\right)}{{\rho }_2}\frac{s-{\rho }_1}{{\rho }_2-{\rho }_1}\right)\\ ={\tilde{h}}_{1,\delta }\left(0\right)\frac{\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)}{{\rho }_1{\rho }_2}+s\left(\frac{{\tilde{h}}_{2,\delta }\left({\rho }_1\right)}{{\rho }_1}\frac{s-{\rho }_2}{{\rho }_1-{\rho }_2}+\frac{{\tilde{h}}_{2,\delta }\left({\rho }_2\right)}{{\rho }_2}\frac{s-{\rho }_1}{{\rho }_2-{\rho }_1}\right)\\ ={\tilde{h}}_{1,\delta }\left(0\right)\frac{\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)}{{\rho }_1{\rho }_2}+\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)\left(\frac{{\tilde{h}}_{2,\delta }\left({\rho }_1\right)}{{\rho }_1}\frac{1}{{\rho }_1-{\rho }_2}+\frac{{\tilde{h}}_{2,\delta }\left({\rho }_2\right)}{{\rho }_2}\frac{1}{{\rho }_2-{\rho }_1}\right)\\ +{\tilde{h}}_{2,\delta }\left(s_1\right)\frac{s-{\rho }_2}{{\rho }_1-{\rho }_2}+{\tilde{h}}_{2,\delta }\left(s_2\right)\frac{s-{\rho }_1}{{\rho }_2-{\rho }_1},\end{array}$

根据拉格朗日插值定理与文献[8]中定义的算子$T_r$ 的性质可知，

$\begin{array}{c}{\tilde{h}}_{1,\delta }\left(s\right)-{\tilde{h}}_{\text{2},\delta }\left(s\right)={\tilde{h}}_{1,\delta }\left(0\right)\frac{\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)}{{\rho }_1{\rho }_2}+\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)\left(\frac{{\tilde{h}}_{2,\delta }\left({\rho }_1\right)}{{\rho }_1}\frac{1}{{\rho }_1-{\rho }_2}+\frac{{\tilde{h}}_{2,\delta }\left({\rho }_2\right)}{{\rho }_2}\frac{1}{{\rho }_2-{\rho }_1}\right)\\ -\left({\tilde{h}}_{2,\delta }\left(s\right)-{\tilde{h}}_{2,\delta }\left(s_1\right)\frac{s-{\rho }_2}{{\rho }_1-{\rho }_2}-{\tilde{h}}_{2,\delta }\left(s_2\right)\frac{s-{\rho }_1}{{\rho }_2-{\rho }_1}\right)\\ =\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)T_0{T}_{s{}_2}{T}_{s_1}{h}_{1,\delta }(0)\\ -\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)\left(\frac{{\tilde{h}}_{\text{2},\delta }\left(s\right)}{\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)}-\frac{{\tilde{h}}_{2,\delta }\left({\rho }_1\right)}{s-{\rho }_1}\frac{1}{{\rho }_1-{\rho }_2}-\frac{{\tilde{h}}_{2,\delta }\left({\rho }_2\right)}{s-{\rho }_2}\frac{1}{{\rho }_2-{\rho }_1}\right)\\ =\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)\left(T_0{T}_{{\rho }_2}{T}_{{\rho }_1}{h}_{1,\delta }\left(0\right)-T_s{T}_{{\rho }_2}{T}_{{\rho }_1}{h}_{2,\delta }\left(0\right)\right).\end{array}$

明显得到$T_0{T}_{{\rho }_2}{T}_{{\rho }_1}{h}_{1,\delta }\left(0\right)=1$ ，那么上式可写成如下形式

${\tilde{h}}_{1,\delta }\left(s\right)-{\tilde{h}}_{\text{2},\delta }\left(s\right)=\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)\left(1-T_s{T}_{{\rho }_2}{T}_{{\rho }_{\text{1}}}{h}_{2,\delta }\left(0\right)\right)$.

同理，对于$Res\ge 0$ ，分母的根也是分子的根。

因此，对于$i=1,2$ 有${\tilde{\vartheta }}_{\text{1},1}\left({\rho }_i\right)=-{\tilde{\beta }}_{\text{1},1}\left({\rho }_i\right)$，则

${\tilde{\beta }}_{\text{1},1}\left(s\right)={\tilde{\beta }}_{\text{1},1}\left({\rho }_{\text{1}}\right)\left(\frac{s-{\rho }_2}{{\rho }_1-{\rho }_2}\right)+{\tilde{\beta }}_{\text{1},1}\left({\rho }_{\text{2}}\right)\left(\frac{s-{\rho }_1}{{\rho }_2-{\rho }_1}\right)=-\frac{{\tilde{\vartheta }}_{\text{1},1}\left({\rho }_{\text{1}}\right)\left(s-{\rho }_2\right)-{\tilde{\vartheta }}_{\text{1},1}\left({\rho }_{\text{2}}\right)\left(s-{\rho }_1\right)}{{\rho }_1-{\rho }_2}$,

那么有

$\begin{array}{c}{\tilde{\vartheta }}_{\text{1},1}\left(s\right)+{\tilde{\beta }}_{\text{1},1}\left(s\right)=\frac{\left(s-{\rho }_2\right)-\left(s-{\rho }_1\right)}{{\rho }_1-{\rho }_2}{\tilde{\vartheta }}_{\text{1},1}\left(s\right)-\frac{{\tilde{\vartheta }}_{\text{1},1}\left({\rho }_1\right)\left(s-{\rho }_2\right)-{\tilde{\beta }}_{1,\delta }\left({\rho }_2\right)\left(s-{\rho }_1\right)}{{\rho }_1-{\rho }_2}\\ =\frac{\left(s-{\rho }_2\right)\left({\tilde{\vartheta }}_{\text{1},1}\left(s\right)-{\tilde{\vartheta }}_{\text{1},1}\left({\rho }_1\right)\right)-\left(s-{\rho }_1\right)\left({\tilde{\vartheta }}_{\text{1},1}\left(s\right)-{\tilde{\vartheta }}_{\text{1},1}\left({\rho }_2\right)\right)}{{\rho }_1-{\rho }_2}\\ =\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)\frac{T_s{T}_{{\rho }_2}{\vartheta }_{\text{1},1}\left(0\right)-T_s{T}_{{\rho }_1}{\vartheta }_{\text{1},1}\left(0\right)}{{\rho }_1-{\rho }_2}\\ =\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)T_s{T}_{{\rho }_2}{T}_{{\rho }_1}{\vartheta }_{\text{1},1}\left(0\right),\end{array}$

综上我们得到了(17)式。

同理可以得到(18)~(22)式。 □

定理4.1 当$u\ge 0$ 时，Gerber-Shiu函数满足的瑕疵更新方程表示如下：

$m_{\text{1,1}}\left(u\right)=k_{\delta }{\displaystyle {\int }_0^u{\varphi }_{1,1}}\left(u-y\right)\eta \left(y\right)\text{d}y+{\sigma }_1\left(u\right)$ , $m_{\text{1}，\text{2}}\left(u\right)=k_{\delta }{\displaystyle {\int }_0^u{\varphi }_{\text{1}，\text{2}}}\left(u-y\right)\eta \left(y\right)\text{d}y+{\sigma }_{\text{2}}\left(u\right)$ ,

$m_{\text{2,1}}\left(u\right)=k_{\delta }{\displaystyle {\int }_0^u{\varphi }_{\text{2,1}}}\left(u-y\right)\eta \left(y\right)\text{d}y+{\sigma }_{\text{3}}\left(u\right)$ , $m_{2,\text{2}}\left(u\right)=k_{\delta }{\displaystyle {\int }_0^u{\varphi }_{\text{2}，\text{2}}}\left(u-y\right)\eta \left(y\right)\text{d}y+{\sigma }_{\text{4}}\left(u\right)$ ,

其中

$k_{\delta }={\displaystyle {\int }_0^{\infty }{T}_{{\rho }_2}{T}_{{\rho }_1}{h}_{2,\delta }\left(y\right)\text{d}y}$ , $\eta \left(y\right)=\frac{T_{{\rho }_2}{T}_{{\rho }_1}{h}_{2,\delta }\left(y\right)}{T_0{T}_{{\rho }_2}{T}_{{\rho }_1}{h}_{2,\delta }\left(0\right)}$ ,

${\sigma }_1\left(u\right)=T_{{\rho }_2}{T}_{{\rho }_1}{\vartheta }_{\text{1},1}\left(u\right)+T_{{\rho }_2}{T}_{{\rho }_1}{\eta }_{\text{1}}\left(u\right)$ , ${\sigma }_2\left(u\right)=T_{{\rho }_2}{T}_{{\rho }_1}{\vartheta }_{\text{1},\text{2}}\left(u\right)+T_{{\rho }_2}{T}_{{\rho }_1}{\eta }_{\text{2}}\left(u\right)$ ,

${\sigma }_3\left(u\right)=T_{{\rho }_2}{T}_{{\rho }_1}{\vartheta }_{\text{2},1}\left(u\right)+T_{{\rho }_2}{T}_{{\rho }_1}{\eta }_{\text{1}}\left(u\right)$ , ${\sigma }_{\text{4}}\left(u\right)=T_{{\rho }_2}{T}_{{\rho }_1}{\vartheta }_{\text{2},\text{2}}\left(u\right)+T_{{\rho }_2}{T}_{{\rho }_1}{\eta }_{\text{2}}\left(u\right)$ .

证明：结合(5)，(6)，(7)，(8)，(13)，(14)，以及(2)，我们可以得到

${\tilde{m}}_{i,j}\left(s\right)=\frac{T_s{T}_{{\rho }_2}{T}_{{\rho }_1}{\vartheta }_{i,j}\left(0\right)+T_s{T}_{{\rho }_2}{T}_{{\rho }_1}{\eta }_j\left(0\right)}{1-T_s{T}_{{\rho }_2}{T}_{{\rho }_1}{h}_{2,\delta }\left(0\right)},i,j=1,2$.

根据拉普拉斯变换的反演公式，我们可以得到Gerber-Shiu函数满足的更新方程形式，这只需证明更新方程是瑕疵的，往证$0<k_{\delta }<1$ 。

当$\delta >0$ 时，${\rho }_{\text{1}},{\rho }_{\text{2}}>0$ ，有

$\begin{array}{c}{k}_{\delta }=T_0{T}_{{\rho }_2}{T}_{{\rho }_1}{h}_{2,\delta }\left(y\right)=1-\frac{{\tilde{h}}_{\text{1},\delta }\left(0\right)-{\tilde{h}}_{\text{2},\delta }\left(0\right)}{{\rho }_2{\rho }_1}\\ =1-\frac{\left(\lambda +{\lambda }_1+\delta \right)\left(\lambda +{\lambda }_2+\delta \right)-\lambda \left(\lambda +{\lambda }_1+\delta \right)\tilde{f}\left(0\right)-\lambda \left(\lambda +{\lambda }_2+\delta \right)\tilde{f}\left(0\right)+{\lambda }^2-{\lambda }_1{\lambda }_2\tilde{v}\left(0\right)}{{\rho }_2{\rho }_1}\\ =1-\frac{\left({\lambda }_1+{\lambda }_2+\delta \right)\delta }{{\rho }_2{\rho }_1}，\end{array}$

由于$\delta ,{\lambda }_1,{\lambda }_2,{\rho }_2,{\rho }_1>0$ ，因此$k_{\delta }<1$ 。

当$\delta =0$ 时，${\rho }_{\text{1}}>0$ ，往证$k_{\text{0}}<1$ 。我们令(1)式中的$s={\rho }_1$ ，有$E\left[{\text{e}}^{\left(c{\rho }_{\text{1}}-\delta \right)V-{\rho }_{\text{1}}Z-{\rho }_{\text{1}}{\displaystyle \sum _{i=1}^{N_1\left(V\right)}{Y}_i}+\sigma {\rho }_{\text{1}}B\left(V\right)}\right]=1$ 。

对于上式方程关于$\delta$ 求导，且令$\delta =\text{0}$ ，我们有${{\rho }^{\prime }}_1=\frac{EV}{E\left(cV-Z-Y\right)}>0$ 。

令$\delta \to \text{0}$ ，我们可以得到

$k_0=T_0{T}_0{T}_{{\rho }_1}{h}_{2,\text{0}}\left(0\right)=1-\underset{\delta \to 0}{\lim }\frac{\left({\lambda }_1+{\lambda }_2+\delta \right)\delta }{{\rho }_1{\rho }_2}=1-\frac{\left({\lambda }_1+{\lambda }_2\right)E\left(cV-Z-Y\right)}{EV}<1.$

6. 数值分析

众所周知，Gerber-Shiu惩罚函数是统一各种破产测度的统一工具。例如当$w\left(x,y\right)=1,\delta =0$ 时我们得到破产概率，$\delta >0$ 时得到破产时间的拉普拉斯变换；当$w\left(x,y\right)={\rm I}\left\{x\le z\right\},\delta =0$ 时我们获得破产前盈余的分布；当$w\left(x,y\right)={\rm I}\left\{y\le z\right\},\delta =0$ 时我们得到破产赤字的分布；当$w\left(x,y\right)={\rm I}\left\{x+y\le z\right\},\delta =0$ 时我们得到导致破产的索赔的分布。由定理4.1，我们可以获得上述破产精算量的更新方程。为计算方便，本节仅以破产概率为例，给出一个数值例子。

在本节中，我们假设$\delta =0$ ，第一类的索赔金额密度为$f\left(x\right)=\alpha {\text{e}}^{-\alpha x},\alpha >0,x>0$ ，第二类的索赔金额密度为$v\left(x\right)=\beta {\text{e}}^{-\beta x},\beta >0,x>0$ 。例如：我们取$c=\text{4},\lambda =1,\lambda =\text{1},{\lambda }_{\text{2}}=2,\sigma =\text{4},\alpha =1,\beta =\text{2}$ ，它们满足安全负载条件。则林德伯格方程有7个根分别为

${\rho }_1=0,{\rho }_2=0.4280,R_1=-0.2797,R_2=-0.6925,R_3=-1.1673,R_4=-1.2983,R_5=-1.9902$ ,

利用前一节的结果，可以得到当$u\ge 0$ 时，最终破产概率为

${\Phi }_1\left(u\right)=\text{0}{\text{.5393e}}^{-\text{0}\text{.6925}u}+0.0873{\text{e}}^{-\text{0}.2797u}-0.0067{\text{e}}^{-1.9902u}+0.0146{\text{e}}^{-\text{1}\text{.1673}u}+0.0733{\text{e}}^{-1.2983u}$ ,

${\Phi }_{\text{2}}\left(u\right)=-0.0479{\text{e}}^{-\text{0}\text{.6925}u}+\text{1}{\text{.0033e}}^{-\text{0}.2797u}-0.0038{\text{e}}^{-1.9902u}+0.4012{\text{e}}^{-\text{1}\text{.1673}u}-0.3529{\text{e}}^{-1.2983u}$ ,

${\Phi }_{\text{3}}\left(u\right)=\text{0}{\text{.6698e}}^{-\text{0}\text{.6925}u}-0.061{\text{e}}^{-\text{0}.2797u}-0.0094{\text{e}}^{-1.9902u}+0.2797{\text{e}}^{-\text{1}\text{.1673}u}+0.109{\text{e}}^{-1.2983u}$ ,

${\Phi }_{\text{4}}\left(u\right)=-0.{\text{0183e}}^{-\text{0}\text{.6925}u}+\text{0}{\text{.8723e}}^{-\text{0}.2797u}-0.0\text{243}8{\text{e}}^{-1.9902u}+0.{\text{2225e}}^{-\text{1}\text{.1673}u}-0.{\text{0526e}}^{-1.2983u}$ .

Figure 1. Bankruptcy probability chart

图1. 破产概率图示

图1表明了不同值$u\in \left[0,10\right]$ 的破产概率。从这张图中我们可以看到，正如预期的那样，这些破产概率随着初始盈余$u$ 的增加而减少。