具有分红和随机扰动相依模型的Gerber-Shiu惩罚函数

Gerber-Shiu Penalty Function with Dividend and Random Disturbance Dependent Model

摘要: 本文研究一类受布朗运动扰动的相依风险模型，并在模型中考虑分红策略，以Gerber-Shiu惩罚函数为研究对象，首先得到了其在有分红和无分红两种情况下的积分–微分方程，然后研究无红利支付与阈值分红策略下的Gerber-Shiu惩罚函数，最后基于矩阵商差理论和线性独立解构造，推导出有分红情形下Gerber-Shiu惩罚函数的显式表达式。研究结果为保险风险的精确度量与分红策略的优化设计提供了理论支撑，拓展了经典风险模型的应用场景。

Abstract: This paper investigates a class of dependent risk models perturbed by Brownian motion, incorporating dividend strategies within the model. Focusing on the Gerber-Shiu penalty function, we first derive its integro-differential equations under both dividend-paying and non-dividend-paying scenarios. Subsequently, we explore the Gerber-Shiu penalty function under non-dividend payment and threshold dividend strategies. Lastly, based on matrix quotient difference theory and linear independent solution construction, we deduce an explicit expression for the Gerber-Shiu penalty function in the dividend-paying case. The research findings provide theoretical support for the precise measurement of insurance risks and the optimal design of dividend strategies, expanding the application scenarios of classical risk models.

1. 引言

在保险精算文献中，用于描述保险业务盈余过程的经典Sparre-Andersen风险模型最早由Andersen提出，在保险风险理论及相关领域得到广泛应用，近年来，很多学者将相依结构引入风险模型以获得相依性质的研究(参见文献[1]-[5])。Vidmar研究了索赔时间与索赔额相依时对破产概率的影响，得出索赔时间与索赔额正依赖时，破产概率增大这一结论[6]，Adekambi将随机扰动与相依结构相结合，建立折现惩罚函数的微分–积分方程并求推导了折现惩罚函数的解析表达式[7]，Li和Sendova研究了受扰动的风险模型，其中索赔时间依赖于索赔大小，保费率取决于前一次索赔大小，且相依结构根据固定阈值水平来构建，得到了Gerber-Shiu惩罚函数满足的积分–微分方程以及指数索赔大小下的显示解[8] [9]，Sun和Zhang进一步研究了具有两类索赔和阈值分红策略的扰动风险模型，推导了Gerber-Shiu惩罚函数满足的矩阵形式的积分–微分方程，最后证明了Gerber-Shiu惩罚函数的闭合形式可由无红利支付情形下的Gerber-Shiu惩罚函数以及对应的齐次积分–微分方程的两个线性无关解构成的矩阵来表示[10]。

本文以文献[8] [9]中受布朗运动扰动的相依风险模型为基础，首先将该模型中的固定阈值拓展为随机阈值，然后参考文献[10]的研究过程，在模型中引入阈值分红策略(即盈余低于阈值时不分红、超过阈值时超出部分全部分红)，并假设保险费率为固定值，进一步增强模型的随机性和实用性，最后参考文献[10]的研究过程，对于改进后的风险模型，我们得到了Gerber-Shiu惩罚函数满足的积分–微分方程、广义伦德伯格方程的根以及该积分–微分方程的解析表达式。

2. 模型介绍

本文考虑以下盈余过程受布朗运动扰动的风险模型

$U\left(t\right)=u+ct-S\left(t\right)+\sigma W\left(t\right)\text{, }t\ge 0$ (2.1)

其中$u\ge 0$ 表示初始资本，$c>0$ 为保险费率，$\left\{W\left(t\right),t\ge 0\right\}$ 是标准布朗运动，$\sigma >0$ 是扩散率，$S\left(t\right)={\displaystyle \sum _{i=1}^{N(t)}{X}_i}$ 表示到达时间$t$ 的总索赔金额，$N\left(t\right)$ 表示索赔次数过程，它是一个更新过程，描述到时刻$t$ 的索赔次数，

索赔金额$X_1,X_2,\cdots$ 是独立同分布的随机变量，其累计分布函数为$B(\cdot )$ ，概率密度函数为$b(\cdot )$ ，均值为$\mu$ ，索赔间隔时间$\left\{W_i,i\ge 1\right\}$ 是独立同分布的随机变量序列，若索赔金额$X_i$ 大于某个随机阈值$Q$ ，则盈余过程被归为第1类，且到下一次索赔的时间服从速率为${\lambda }_1$ 的指数分布；若$X_i$ 小于$Q$ ，则盈余过程被归为第2类，且到下一次索赔的时间服从速率为${\lambda }_2$ 的指数分布，当盈余低于阈值b时不分红，超过b时，超出部分全部作为红利。

定义Gerber-Shiu函数为

$V_k\left(u\right)=E\left[{\text{e}}^{-\delta T_b}{w}_k\left(U_b\left(T_b{}^{-}\right),\left|U_b\left(T_b\right)\right|\right)I_{(T_b<\infty )}|U_b\left(0\right)=u\right]$

其中$\delta \ge 0$ 是贴现因子，$U_b\left(T_b^{-}\right)$ 是破产前盈余，$\left|U_b\left(T_b^{}\right)\right|$ 是破产后赤字，${\rm I}$ 是示性函数，$w_k$ 是关于破产前盈余和破产后赤字的二元函数，不失一般性，令$w_k\left(\text{0},\text{0}\right)=\text{1}$ 为确保破产不是必然事件，假设${\rm E}\left[cW-X\right]>0$ 恒成立。故当索赔金额$x>Q$ 时，下一次索赔的时间服从速率为${\lambda }_1$ 的指数分布，此时若盈余金额$0\le u<b$ ，则其Gerber-Shiu函数为$V_{11}\left(u\right)$ ，若盈余金额$u>b$ ，则其Gerber-Shiu函数为$V_{12}\left(u\right)$ ；当索赔金额$x<Q$ 时，下一次索赔的时间服从速率为${\lambda }_2$ 的指数分布，此时若盈余金额$0\le u<b$ ，则其Gerber-Shiu函数为$V_{21}\left(u\right)$ ，若盈余金额$u>b$ ，则其Gerber-Shiu函数为$V_{22}\left(u\right)$ 。在后续内容中，我们使用以下符号：

${\overrightarrow{V}}_1\left(u\right)=\{\begin{array}{l}{V}_{11}\left(u\right),0\le u<b\\ V_{12}\left(u\right),b\le u<\infty \end{array}$ ，${\overrightarrow{V}}_2\left(u\right)=\{\begin{array}{l}{V}_{21}\left(u\right),0\le u<b\\ V_{22}\left(u\right),b\le u<\infty \end{array}$

3. 积分–微分方程

在本节中，我们对Gerber-Shiu函数建立积分–微分方程。

命题2.1：函数${\overrightarrow{V}}_1\left(u\right)$ ，${\overrightarrow{V}}_2\left(u\right)$ 满足以下的积分–微分方程。

${{\overrightarrow{V}}^{″}}_1\left(u\right)-\frac{2}{{\sigma }^2}\left[c{\overrightarrow{V^{\prime }}}_1\left(u\right)+A{\overrightarrow{V}}_1\left(u\right)+{\displaystyle {\int }_0^{x}B}{\overrightarrow{V}}_1\left(u-x\right)\text{d}x+{\overrightarrow{\zeta }}_1\left(u\right)\right]$ (3.1)

${{\overrightarrow{V}}^{″}}_2\left(u\right)=-\frac{2}{{\sigma }^2}\left[\left(c-\alpha \right){{\overrightarrow{V}}^{\prime }}_2\left(u\right)+A{\overrightarrow{V}}_2\left(u\right)+{\displaystyle {\int }_0^{u-b}B{\overrightarrow{V}}_2\left(u-x\right)\text{d}x}+{\displaystyle {\int }_{u-b}^{u}B{\overrightarrow{V}}_1\left(u-x\right)+{\overrightarrow{\zeta }}_{\text{1}}\left(u\right)}\right]$ (3.2)

其中

$w_i\left(u\right)={\displaystyle {\int }_u^{\infty }{w}_i\left(u,x-u\right)p\left(x\right)\text{d}x}$

$\xi \left(y\right)=P\left(y>Q\right)b\left(y\right)$

$\chi \left(y\right)=P\left(y<Q\right)b\left(y\right)$

$A=\left(\begin{array}{cc}-\left({\lambda }_1+\delta \right)& 0\\ 0& -\left({\lambda }_2+\delta \right)\end{array}\right)$ ，$B=\left(\begin{array}{cc}{\lambda }_{\text{1}}\xi \left(x\right)& {\lambda }_{\text{1}}\chi \left(x\right)\\ {\lambda }_{\text{2}}\xi \left(x\right)& {\lambda }_{\text{2}}\chi \left(x\right)\end{array}\right)$

${\overrightarrow{V}}_1\left(u\right)=\left(\begin{array}{l}{V}_{11}\left(u\right)\\ V_{21}\left(u\right)\end{array}\right)$ ，${\overrightarrow{V}}_2\left(u\right)=\left(\begin{array}{l}{V}_{12}\left(u\right)\\ V_{22}\left(u\right)\end{array}\right)$ ，${\overrightarrow{\zeta }}_1\left(u\right)=\left(\begin{array}{l}{\lambda }_1{w}_1\left(u\right)\\ {\lambda }_2{w}_2\left(u\right)\end{array}\right)$

证明：对于$V_{i1}\left(u\right),i=1,2$ ，考虑一个长度为$dt$ 的小时间区间，并以在该区间内可能发生的首次索赔情况为条件，可以得到

$\begin{array}{c}{V}_{11}\left(u\right)={\text{e}}^{-\delta dt}\{\left(1-{\lambda }_{1}dt\right)E\left[V_{11}\left(u+cdt+\sigma W\left(dt\right)\right)\right]\\ +{\lambda }_{1}dt[{\displaystyle {\int }_0^{u+cdt+\sigma W\left(dt\right)}P\left(x>Q\right)}{V}_{11}\left(u+cdt+\sigma W\left(dt\right)-x\right)b\left(x\right)\text{d}x\\ +{\displaystyle {\int }_0^{u+cdt+\sigma W\left(dt\right)}P\left(x<Q\right)V_{21}\left(u+cdt+\sigma W\left(dt\right)-x\right)b\left(x\right)\text{d}x}]\end{array}$

$\begin{array}{c}{V}_{21}\left(u\right)={\text{e}}^{-\delta dt}\{\left(1-{\lambda }_{2}dt\right)E\left[V_{21}\left(u+cdt+\sigma W\left(dt\right)\right)\right]\\ +{\lambda }_{1}dt[{\displaystyle {\int }_0^{u+cdt+\sigma Wdt}P\left(x>Q\right)}{V}_{11}\left(u+cdt\right)+\sigma W\left(dt\right)-x)b\left(x\right)\text{d}x\\ +{\displaystyle {\int }_0^{u+cdt+\sigma Wdt}P\left(x<Q\right)V_{21}\left(u+cdt+\sigma W\left(dt\right)-x)b\left(x\right)\right)\text{d}x}]\end{array}$

由泰勒公式，以及同时除以$dt$ ，并令$dt$ 趋于0，得到

$\frac{{\sigma }^{\text{2}}}{\text{2}}{V^{″}}_{11}\left(u\right)+c{V^{\prime }}_{11}\left(u\right)-\left({\lambda }_1+\delta \right)V_{11}^{}\left(u\right)+{\lambda }_1{\displaystyle {\int }_0^u{V}_{11}^{}\left(u-x\right)\xi \left(x\right)\text{d}x}+{\lambda }_1{\displaystyle {\int }_0^u{V}_{21}^{}\left(u-x\right)}\chi \left(x\right)\text{d}x+{\lambda }_1{w}_1\left(u\right)=0$

$\frac{{\sigma }^{\text{2}}}{\text{2}}{V^{″}}_{21}\left(u\right)+c{V^{\prime }}_{21}\left(u\right)-\left({\lambda }_{\text{2}}+\delta \right)V_{\text{2}1}^{}\left(u\right)+{\lambda }_{\text{2}}{\displaystyle {\int }_0^u{V}_{11}^{}\left(u-x\right)\xi \left(x\right)\text{d}x}+{\lambda }_{\text{2}}{\displaystyle {\int }_0^u{V}_{21}^{}\left(u-x\right)}\chi \left(x\right)\text{d}x+{\lambda }_2{w}_2\left(u\right)=0$

同理，对于$V_{i2}\left(u\right),i=1,2$ ，考虑一个长度为$dt$ 的小时间区间，并以在该区间内可能发生的首次索赔情况为条件，我们可以得到

$\begin{array}{c}{V}_{12}\left(u\right)={\text{e}}^{-\delta dt}\{\alpha dt+\left(1-{\lambda }_{1}dt\right)E\left[V_{12}\left(u+\left(c-\alpha \right)dt+\sigma W\left(dt\right)\right)\right]\\ +{\lambda }_{1}dt[{\displaystyle {\int }_0^{u+\left(c-\alpha \right)dt+\sigma W\left(dt\right)-b}P\left(x>Q\right)V_{12}(u+\left(c-\alpha \right)dt}+\sigma W\left(dt\right)-x)b\left(x\right)\text{d}x\\ +{\displaystyle {\int }_0^{u+\left(c-\alpha \right)dt+\sigma W\left(dt\right)-b}P\left(x<Q\right)V_{22}}(u+\left(c-\alpha \right)dt+\sigma W\left(dt\right)-x)b\left(x\right)\text{d}y\\ +{\displaystyle {\int }_{u+\left(c-\alpha \right)dt+\sigma W\left(dt\right)-b}^{u+\left(c-\alpha \right)dt+\sigma W\left(dt\right)}P\left(x>Q\right)V_{11}\left(u+\left(c-\alpha \right)dt+\sigma W\left(dt\right)-x\right)b\left(x\right)\text{d}y}\\ +{\displaystyle {\int }_{u+\left(c-\alpha \right)dt+\sigma W\left(dt\right)-b}^{u+\left(c-\alpha \right)dt+\sigma W\left(dt\right)}P\left(x<Q\right)V_{21}\left(u+\left(c-\alpha \right)dt+\sigma W\left(dt\right)-x\right)b\left(x\right)}\text{d}y]\}\end{array}$

$\begin{array}{c}{V}_{22}\left(u\right)={\text{e}}^{-\delta dt}\{\alpha dt+\left(1-{\lambda }_{2}dt\right)E\left[V_{22}\left(u+\left(c-\alpha \right)dt+\sigma W\left(dt\right)\right)\right]\\ +{\lambda }_{2}dt[{\displaystyle {\int }_0^{u+\left(c-\alpha \right)dt+\sigma W\left(dt\right)-b}P\left(x>Q\right)V_{12}(u+\left(c-\alpha \right)dt}+\sigma W\left(dt\right)-x)b\left(x\right)\text{d}y\\ +{\displaystyle {\int }_0^{u+\left(c-\alpha \right)dt+\sigma W\left(dt\right)-b}P\left(x<Q\right)V_{22}}(u+\left(c-\alpha \right)dt+\sigma W\left(dt\right)-x)b\left(x\right)\text{d}y\\ +{\displaystyle {\int }_{u+\left(c-\alpha \right)dt+\sigma W\left(dt\right)-b}^{u+\left(c-\alpha \right)dt+\sigma W\left(dt\right)}P\left(x>Q\right)V_{11}\left(u+\left(c-\alpha \right)dt+\sigma W\left(dt\right)-x\right)b\left(x\right)\text{d}y}\\ +{\displaystyle {\int }_{u+\left(c-\alpha \right)dt+\sigma W\left(dt\right)-b}^{u+\left(c-\alpha \right)dt+\sigma W\left(dt\right)}P\left(x<Q\right)V_{21}\left(u+\left(c-\alpha \right)dt+\sigma W\left(dt\right)-x\right)b\left(x\right)}\text{d}y]\}\end{array}$

对上式先用泰勒公式展开，然后同时除以$dt$ ，并令$dt$ 趋于0，得到

$\begin{array}{c}\left({\lambda }_1+\delta \right)V_{12}\left(u\right)=\alpha +\left(c-\alpha \right){V^{\prime }}_{12}\left(u\right)+\frac{{\sigma }^2}{2}{V^{″}}_{12}\left(u\right)+{\lambda }_1[{\displaystyle {\int }_0^{u-b}P\left(x>Q\right)V_{12}\left(u-x\right)b\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_0^{u-b}P\left(x<Q\right)V_{22}\left(u-x\right)b\left(x\right)\text{d}x}+{\displaystyle {\int }_{u-b}^{u}P\left(x>Q\right)V_{11}\left(u-x\right)b\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_{u-b}^{u}P\left(x<Q\right)V_{21}\left(u-x\right)b\left(x\right)}\text{d}x]+{\lambda }_1{w}_1\left(u\right)\end{array}$

$\begin{array}{c}\left({\lambda }_2+\delta \right)V_{22}\left(u\right)=\alpha +\left(c-\alpha \right){V^{\prime }}_{22}\left(u\right)+\frac{{\sigma }^2}{2}{V^{″}}_{22}\left(u\right)+{\lambda }_2[{\displaystyle {\int }_0^{u-b}P\left(x>Q\right)V_{12}\left(u-x\right)b\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_0^{u-b}P\left(x<Q\right)V_{22}\left(u-x\right)b\left(x\right)\text{d}x}+{\displaystyle {\int }_{u-b}^{u}P\left(x>Q\right)V_{11}\left(u-x\right)b\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_{u-b}^{u}P\left(x<Q\right)V_{21}\left(u-x\right)b\left(x\right)}\text{d}x]+{\lambda }_2{w}_2\left(u\right)\end{array}$

由上式，我们可以得到

$\begin{array}{l}\frac{{\sigma }^{\text{2}}}{\text{2}}{V^{″}}_{12}\left(u\right)+\left(c-\alpha \right){V^{\prime }}_{12}\left(u\right)-\left({\lambda }_{\text{1}}+\delta \right)V_{\text{1}2}^{}\left(u\right)\\ +{\lambda }_{\text{1}}\{{\displaystyle {\int }_0^{u-b}\left[P\left(x>Q\right)V_{12}^{}\left(u-x\right)b\left(x\right)+P\left(x<Q\right)V_{22}\left(u-x\right)b\left(x\right)\right]\text{d}x}\\ +{\displaystyle {\int }_{u-b}^u\left[P\left(x>Q\right)V_{11}\left(u-x\right)b\left(x\right)+P\left(x<Q\right)V_{21}\left(u-x\right)b\left(x\right)\right]\text{d}x}\}+{\lambda }_1{w}_1\left(u\right)=\text{0}\end{array}$ $\begin{array}{l}\frac{{\sigma }^{\text{2}}}{\text{2}}{V^{″}}_{22}\left(u\right)+\left(c-\alpha \right){V^{\prime }}_{22}\left(u\right)-\left({\lambda }_{\text{2}}+\delta \right)V_{\text{2}2}^{}\left(u\right)\\ +{\lambda }_{\text{2}}\{{\displaystyle {\int }_0^{u-b}\left[P\left(x>Q\right)V_{12}^{}\left(u-x\right)b\left(x\right)+P\left(x<Q\right)V_{22}\left(u-x\right)b\left(x\right)\right]\text{d}x}\\ +{\displaystyle {\int }_{u-b}^u\left[P\left(x>Q\right)V_{11}\left(u-x\right)b\left(x\right)+P\left(x<Q\right)V_{21}\left(u-x\right)b\left(x\right)\right]\text{d}x}\}\text{+}{\lambda }_2{w}_2\left(u\right)=\text{0}\end{array}$

令${\overrightarrow{V}}_1\left(u\right)=\left(\begin{array}{l}{V}_{11}\left(u\right)\\ V_{21}\left(u\right)\end{array}\right),{\overrightarrow{V}}_2\left(u\right)=\left(\begin{array}{l}{V}_{12}\left(u\right)\\ V_{22}\left(u\right)\end{array}\right),{\overrightarrow{\zeta }}_1\left(u\right)=\left(\begin{array}{l}{\lambda }_1{w}_1\left(u\right)\\ {\lambda }_2{w}_2\left(u\right)\end{array}\right)$

从而

${V^{″}}_{11}\left(u\right)=\left(-\frac{2}{{\sigma }^2}\right)\left[c{V^{\prime }}_{11}\left(u\right)-\left({\lambda }_1+\delta \right)V_{11}\left(u\right)+{\lambda }_1{\displaystyle {\int }_0^u{V}_{11}\left(u\right)\xi \left(x\right)\text{d}x+{\lambda }_1}{\displaystyle {\int }_0^u{V}_{21}\left(u-x\right)\chi \left(x\right)\text{d}x}\right]\text{+}{\lambda }_1{w}_1\left(u\right)$

${V^{″}}_{21}\left(u\right)=\left(-\frac{2}{{\sigma }^2}\right)\left[c{V^{\prime }}_{21}\left(u\right)-\left({\lambda }_2+\delta \right)V_{21}\left(u\right)+{\lambda }_2{\displaystyle {\int }_0^u{V}_{11}\left(u\right)\xi \left(x\right)\text{d}x+{\lambda }_2{\displaystyle {\int }_0^u{V}_{21}\left(u-x\right)\chi \left(x\right)\text{d}x}}\right]\text{+}{\lambda }_2{w}_2\left(u\right)$

$\begin{array}{c}{V^{″}}_{12}\left(u\right)=\left(-\frac{2}{{\sigma }^2}\right)[\left(c-\alpha \right){V^{\prime }}_{12}\left(u\right)-\left({\lambda }_1+\delta \right)V_{12}\left(u\right)\\ +{\lambda }_1[{\displaystyle {\int }_0^{u-b}P\left(x>Q\right)V_{12}\left(u-x\right)b\left(x\right)\text{d}x}+{\displaystyle {\int }_0^{u-b}P\left(x<Q\right)V_{22}\left(u-x\right)b\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_{u-b}^{u}P\left(x>Q\right)V_{11}\left(u-x\right)b\left(x\right)\text{d}x}+{\displaystyle {\int }_{u-b}^{u}P\left(x<Q\right)V_{21}\left(u-x\right)b\left(x\right)\text{d}x}]+{\lambda }_1{w}_1\left(u\right)\end{array}$

$\begin{array}{c}{V^{″}}_{22}\left(u\right)=\left(-\frac{2}{{\sigma }^2}\right)[\left(c-\alpha \right){V^{\prime }}_{22}\left(u\right)-\left({\lambda }_2+\delta \right)V_{22}\left(u\right)\\ +{\lambda }_2[{\displaystyle {\int }_0^{u-b}P\left(x>Q\right)V_{12}\left(u-x\right)b\left(x\right)\text{d}x+}{\displaystyle {\int }_0^{u-b}P\left(x>Q\right)V_{22}\left(u-x\right)b\left(x\right)\text{d}x}\\ +{\displaystyle {\int }_{u-b}^{u}P\left(x>Q\right)V_{11}\left(u-x\right)b\left(x\right)\text{d}x}+{\displaystyle {\int }_{u-b}^{u}P\left(x<Q\right)V_{21}\left(u-x\right)b\left(x\right)\text{d}x}]\text{+}{\lambda }_2{w}_2\left(u\right)\end{array}$

整理可得(3.1)和(3.2)。

4. 广义伦德伯格方程的根

当$s\in C$ 且$R\left(s\right)\ge 0$ 时，对(3.1)式进行拉普拉斯变换得到

$s^2{\tilde{\overrightarrow{V}}}_1\left(s\right)-s{\tilde{\overrightarrow{V}}}_1\left(0\right)-{{\tilde{\overrightarrow{V}}}^{\prime }}_1\left(0\right)=\left(-\frac{2}{{\sigma }^2}\right)\left[c\left(s{\tilde{\overrightarrow{V}}}_1\left(s\right)-{\tilde{\overrightarrow{V}}}_1\left(0\right)\right)+A{\tilde{\overrightarrow{V}}}_1\left(s\right)+\overrightarrow{B}{\tilde{\overrightarrow{V}}}_1\left(s\right)+{\tilde{\overrightarrow{\zeta }}}_1\left(s\right)\right]$

然后将上式两边同时乘以$\frac{{\sigma }^2}{2}$ 并将包含${\tilde{\overrightarrow{V}}}_1\left(s\right)$ 的项移到等式左边，可得

$L_{\delta c}\left(s\right){\tilde{\overrightarrow{V}}}_1\left(s\right)=\frac{{\sigma }^2}{2}s{\tilde{\overrightarrow{V}}}_1\left(0\right)+\frac{{\sigma }^2}{2}{{\tilde{\overrightarrow{V}}}^{\prime }}_1\left(0\right)+c{\tilde{\overrightarrow{V}}}_1\left(0\right)-{\tilde{\overrightarrow{\zeta }}}_1\left(s\right),R\left(s\right)\ge 0$ (4.1)

其中

$L_{\delta c}\left(s\right)=\frac{{\sigma }^{\text{2}}}{\text{2}}{s}^{2}E+csE+A+\tilde{B}\left(s\right)$，${\tilde{\overrightarrow{\zeta }}}_1\left(s\right)={\displaystyle {\int }_{\text{0}}^{\infty }{\text{e}}^{-su}{\overrightarrow{\zeta }}_1\left(u\right)\text{d}u}$

定理3.1：对于$\delta >0$ ，广义伦德伯格方程$\det \left[L_{\delta c}\left(s\right)\right]=0$ 有两个实部为正的根${\rho }_{\text{1}}\left(\delta \right),{\rho }_{\text{2}}\left(\delta \right)$ 。

证明：$\det \left[L_{\delta c}\left(s\right)\right]=\left[\frac{{\sigma }^2}{2}{s}^2+cs-\left({\lambda }_1+\delta \right)+{\lambda }_1\tilde{\overrightarrow{\xi }}\left(s\right)\right]\left[\frac{{\sigma }^2}{2}{s}^2+cs-\left({\lambda }_2+\delta \right)+{\lambda }_1\tilde{\overrightarrow{\chi }}\left(s\right)\right]-{\lambda }_1{\lambda }_2\tilde{\overrightarrow{\xi }}\left(s\right)\tilde{\overrightarrow{\chi }}\left(s\right)$

令$\det \left[L_{\delta c}\left(s\right)\right]=\text{0}$ ，

则

$\left[\frac{{\sigma }^2}{2}{s}^2+cs-\left({\lambda }_1+\delta \right)+{\lambda }_1\tilde{\overrightarrow{\xi }}\left(s\right)\right]\left[\frac{{\sigma }^2}{2}{s}^2+cs-\left({\lambda }_2+\delta \right)+{\lambda }_1\tilde{\overrightarrow{\chi }}\left(s\right)\right]-{\lambda }_1{\lambda }_2\tilde{\overrightarrow{\xi }}\left(s\right)\tilde{\overrightarrow{\chi }}\left(s\right)=0$

令

${\gamma }_{\delta c}\left(s\right)=\frac{\left[\frac{{\sigma }^{\text{2}}}{\text{2}}{s}^2+cs-\left({\lambda }_1+\delta \right)+{\lambda }_1\tilde{\overrightarrow{\xi }}\left(s\right)\right]\left[\frac{{\sigma }^{\text{2}}}{\text{2}}{s}^2+cs-\left({\lambda }_2+\delta \right)+{\lambda }_2\tilde{\overrightarrow{\chi }}\left(s\right)\right]}{{\lambda }_1{\lambda }_2}=\tilde{\overrightarrow{\chi }}\left(s\right)\tilde{\overrightarrow{\xi }}\left(s\right)$ (4.2)

设轮廓$C_r$ 是以原点为圆心，半径为$r$ 的右半圆弧($r$ 足够大)，$C$ 是由$C_r$ 和虚轴共同构成的闭合轮廓，应用Rouché定理，令$f\left(s\right)={\gamma }_{\delta c}\left(s\right)$ ，$g\left(s\right)=-\tilde{\overrightarrow{\chi }}\left(s\right)\tilde{\overrightarrow{\xi }}\left(s\right)$ ，只需要证明在$C$ 上有

$\left|{\gamma }_{\delta c}\left(s\right)\right|>\left|\tilde{\overrightarrow{\chi }}\left(s\right)\tilde{\overrightarrow{\xi }}\left(s\right)\right|$ (4.3)

当$s\in C_r$ 时，$\left|s\right|=r\to \infty$ ，$\mathrm{Re}\left(s\right)\ge 0$ ，

$\left|{\gamma }_{\delta c}\left(s\right)\right|=\frac{\left[\frac{{\sigma }^{\text{2}}}{\text{2}}{r}^2+cr-\left({\lambda }_1+\delta \right)+{\lambda }_1\tilde{\overrightarrow{\xi }}\left(r\right)\right]\left[\frac{{\sigma }^{\text{2}}}{\text{2}}{r}^2+cr-\left({\lambda }_2+\delta \right)+{\lambda }_2\tilde{\overrightarrow{\chi }}\left(r\right)\right]}{{\lambda }_1{\lambda }_2}\to \infty \left(r\to \infty \right)$

由拉普拉斯变换的有界性得，$\left|\tilde{\overrightarrow{\xi }}\left(s\right)\right|\le 1$ ，$\left|\tilde{\overrightarrow{\chi }}\left(s\right)\right|\le 1$ ，故$\left|\tilde{\overrightarrow{\xi }}\left(s\right)\tilde{\overrightarrow{\chi }}\left(s\right)\right|\le 1$ ，显然当$r$ 足够大时，有

$\left|{\gamma }_{\delta c}\left(s\right)\right|>1>\left|\tilde{\overrightarrow{\xi }}\left(s\right)\tilde{\overrightarrow{\chi }}\left(s\right)\right|$ (4.4)

当$s=bi$ ，$\mathrm{Re}\left(s\right)=0$ 时，

$\left|\frac{{\sigma }^{\text{2}}}{\text{2}}{s}^2+cs-\left({\lambda }_1+\delta \right)\right|\ge \frac{{\sigma }^{\text{2}}}{\text{2}}{b}^2+\left({\lambda }_1+\delta \right)>{\lambda }_1\ge {\lambda }_1\left|\tilde{\overrightarrow{\xi }}\left(s\right)\right|$

$\left|\frac{{\sigma }^{\text{2}}}{\text{2}}{s}^2+cs-\left({\lambda }_{\text{2}}+\delta \right)\right|\ge \frac{{\sigma }^{\text{2}}}{\text{2}}{b}^2+\left({\lambda }_{\text{2}}+\delta \right)>{\lambda }_{\text{2}}\ge {\lambda }_{\text{2}}\left|\tilde{\overrightarrow{\chi }}\left(s\right)\right|$

则$\left|{\gamma }_{\delta c}\left(s\right)\right|>\left|\tilde{\overrightarrow{\chi }}\left(s\right)\tilde{\overrightarrow{\xi }}\left(s\right)\right|$ ，即(4.2)有两个根，下面证明这两个根都是实根。

先计算${\gamma }_{\delta c}\left(s\right)$ 的导数，得到下式

$\begin{array}{c}{{\gamma }^{\prime }}_{\delta c}\left(s\right)=\frac{\left[{\sigma }^{2}s+c+{\lambda }_1\widetilde{{\xi }^{\prime }}\left(s\right)\right]\left[\frac{{\sigma }^2}{2}{s}^2+cs-\left({\lambda }_2+\delta \right)+{\lambda }_2\tilde{\chi }\left(s\right)\right]}{{\lambda }_1{\lambda }_2}\\ +\frac{\left[\frac{{\sigma }^{\text{2}}}{\text{2}}+cs-\left({\lambda }_1+\delta \right)+{\lambda }_1\tilde{\xi }\left(s\right)\right]\left[{\sigma }^{2}s+c+{\lambda }_2\widetilde{{\chi }^{\prime }}\left(s\right)\right]}{{\lambda }_1{\lambda }_2}\end{array}$

再令${\gamma }_{\delta c}\left(s\right)=0$ 得到一个正实根$s_0$ ，并可验证${\gamma }_{\delta c}\left(s\right)$ 在$s\in \left[0,s_0\right]$ 上递减，在$s\in \left(s_0,\infty \right)$ 上递增，且在$s\ge 0$ 时有一个最小值${\gamma }_{\delta c}\left(s_0\right)\le {\gamma }_{\delta c}\left(0\right)$ ，再结合${\gamma }_{\delta c}\left(0\right)=\frac{\left({\lambda }_1+\delta \right)\left({\lambda }_1+\delta \right)}{{\lambda }_1{\lambda }_2}>1=\tilde{\chi }\left(0\right)\tilde{\xi }\left(0\right)$ ，则方程(4.2)有两个实根，证毕。

5. 无红利赔付的Gerber-Shiu惩罚函数

在本节中，记${\overrightarrow{v}}_1\left(u\right)=\left(\begin{array}{l}{v}_{11}\left(u\right)\\ v_{21}\left(u\right)\end{array}\right)$ 为无红利赔付$\left(b=\infty \right)$ 时的Gerber-Shiu惩罚函数，类似于上文命题2.1的证明过程，可以得到

${{\overrightarrow{v}}^{″}}_1\left(u\right)=-\frac{2}{{\sigma }^2}\left[c{{\overrightarrow{v}}^{\prime }}_1\left(u\right)+A{\overrightarrow{v}}_1\left(u\right)+{\displaystyle {\int }_0^{u}B}{\overrightarrow{v}}_1\left(u-x\right)\text{d}x+{\overrightarrow{\zeta }}_1\left(u\right)\right]$ (5.1)

5.1. ${\overrightarrow{v^{\prime }}}_1\left(0\right)$ 的表达式

矩阵$B\left(s\right)$ 对不同数的商差递归定义如下(见Lu和Li (2009))

$\begin{array}{l}B\left[r_1,s\right]=\frac{B\left(s\right)-B\left(r_1\right)}{s-r_1},\\ B\left[r_1,r_{\text{2}},s\right]=\frac{B\left[r_1,s\right]-B\left[r_1,r_2\right]}{s-r_2},\\ B\left[r_1,r_{\text{2}},r_{\text{3}},s\right]=\frac{B\left[r_1,r_{\text{2}},s\right]-B\left[r_1,r_2,r_{\text{3}}\right]}{s-r_{\text{3}}}，\end{array}$ (5.2)

定理4.1：${\overrightarrow{v^{\prime }}}_1\left(\text{0}\right)$的闭合形式由下式给出

${\overrightarrow{v^{\prime }}}_1\left(\text{0}\right)=\frac{2}{{\sigma }^2}\left[{\overrightarrow{\zeta }}_1\left({\rho }_2\right)+L_{\delta \text{c}}^{*}{\left[{\rho }_1,{\rho }_2\right]}^{-1}{L}_{\delta \text{c}}^{*}\left[{\rho }_1\right]{\overrightarrow{\zeta }}_1\left[{\rho }_1,{\rho }_2\right]\right]$ (5.3)

其中$A^{*}$ 为矩阵$A$ 的伴随矩阵。

证明：对于满足$R\left(s\right)\ge 0$ 的所有$s\in C$ ，${\tilde{\overrightarrow{v}}}_1\left(s\right)$ 都是有限的，对等式(4.1)两边同时乘以$L_{\delta \text{c}}^{*}\left(s\right)$ 并把$s={\rho }_i$ 代入得到：

$\left(-\frac{{\sigma }^{\text{2}}}{\text{2}}\right)L_{\delta c}^{*}\left({\rho }_i\right){\overrightarrow{v^{\prime }}}_1\left(0\right)=L_{\delta c}^{*}\left({\rho }_i\right)\left(\frac{{\sigma }^{\text{2}}}{\text{2}}{\rho }_i+c\right){\overrightarrow{v}}_1\left(0\right)-L_{\delta c}^{*}\left({\rho }_i\right){\tilde{\overrightarrow{\zeta }}}_1\left({\rho }_i\right),i=1,2$

运用(5.2)，结合边界条件${\overrightarrow{v}}_1\left(0\right)=\overrightarrow{0}$ ，得到

$\begin{array}{c}\left(-\frac{{\sigma }^{\text{2}}}{\text{2}}\right)L_{\delta c}^{*}\left[{\rho }_1,{\rho }_2\right]{\overrightarrow{v^{\prime }}}_1\left(0\right)=\left(-\frac{{\sigma }^2}{2}\right)\frac{L_{\delta c}^{*}\left({\rho }_2\right)-L_{\delta c}^{*}\left({\rho }_1\right)}{{\rho }_2-{\rho }_1}{\overrightarrow{v^{\prime }}}_1\left(0\right)\\ =\frac{L_{\delta c}^{*}\left({\rho }_2\right)\left(\frac{{\sigma }^2}{2}{\rho }_2+c\right)-L_{\delta c}^{*}\left({\rho }_1\right)\left(\frac{{\sigma }^2}{2}{\rho }_1+c\right)}{{\rho }_2-{\rho }_1}{\overrightarrow{v}}_1\left(0\right)\end{array}$

$\begin{array}{c}-\frac{L_{\delta c}^{*}\left({\rho }_2\right){\tilde{\overrightarrow{\zeta }}}_{\text{1}}\left({\rho }_2\right)-L_{\delta c}^{*}\left({\rho }_1\right){\tilde{\overrightarrow{\zeta }}}_{\text{1}}\left({\rho }_{\text{1}}\right)}{{\rho }_2-{\rho }_1}\\ =-\left[L_{\delta c}^{*}\left[{\rho }_1,{\rho }_2\right]{\tilde{\overrightarrow{\zeta }}}_{\text{1}}\left({\rho }_2\right)+L_{\delta c}^{*}\left({\rho }_1\right){\tilde{\overrightarrow{\zeta }}}_{\text{1}}\left[{\rho }_1,{\rho }_2\right]\right]\end{array}$$$

故

${\overrightarrow{v^{\prime }}}_1\left(0\right)=\frac{2}{{\sigma }^2}\left[{\overrightarrow{\zeta }}_1\left({\rho }_2\right)+L_{\delta \text{c}}^{*}{\left[{\rho }_1,{\rho }_2\right]}^{-1}{L}_{\delta \text{c}}^{*}\left[{\rho }_1\right]{\overrightarrow{\zeta }}_1\left[{\rho }_1,{\rho }_2\right]\right]$

5.2. ${\overrightarrow{v}}_1\left(u\right)$ 的显式表达式

方程(5.1)的常见齐次积分微分方程的形式为

${\overrightarrow{v^{″}}}_1\left(u\right)=\left(-\frac{2}{{\sigma }^2}\right)\left[c{\overrightarrow{v^{\prime }}}_1\left(u\right)+A{\overrightarrow{v}}_1\left(u\right)+{\displaystyle {\int }_0^{u}B}{\overrightarrow{v}}_1\left(u-x\right)\text{d}x\right]$ (5.4)

参考Bueton (2005)，得到以下定理。

定理4.2：令$Z\left(u\right)=\left({\overrightarrow{Z}}_1\left(u\right),{\overrightarrow{Z}}_2\left(u\right)\right)$ ，其中${\overrightarrow{Z}}_1\left(u\right),{\overrightarrow{Z}}_2\left(u\right)$ 是(5.4)的两个线性独立的解，使得$Z\left(\text{0}\right)=\text{0}$ ，并且$Z^{\prime }\left(\text{0}\right)=E$ ，有

${\overrightarrow{v}}_1\left(u\right)=Z\left(u\right){\overrightarrow{v^{\prime }}}_1\left(0\right)-\frac{2}{{\sigma }^2}{\displaystyle {\int }_0^{u}Z\left(u-x\right){\overrightarrow{\zeta }}_1\left(x\right)\text{d}x},0\le u<\infty$ (5.5)

证明：对以下等式两边进行拉普拉斯变换

$Z^{″}\left(u\right)=\left(-\frac{2}{{\sigma }^2}\right)\left[c{Z}^{\prime }\left(u\right)-AZ\left(u\right)+{\displaystyle {\int }_0^{u}BZ\left(u-x\right)\text{d}x}\right]$

其中$s\in C$ ，$R\left(s\right)\ge 0$ ，有

$\left[s^2\tilde{Z}\left(s\right)-sZ\left(0\right)-Z^{\prime }\left(\text{0}\right)\right]=\left(-\frac{\text{2}}{{\sigma }^{\text{2}}}\right)\left[c\left(s\tilde{Z}\left(s\right)-Z\left(0\right)\right)+A\tilde{Z}\left(s\right)+\tilde{B}\left(s\right)\tilde{Z}\left(s\right)\right]$

运用边界条件$Z\left(\text{0}\right)=\text{0}$ ，$Z^{\prime }\left(\text{0}\right)=E$ ，可得到

$s^2\tilde{Z}\left(s\right)\text{+}\frac{2}{{\sigma }^2}\left[csE+A+\tilde{B}\left(s\right)\right]\tilde{Z}\left(s\right)=E$

再将$csE+A+\tilde{B}\left(s\right)=L_{\delta c}\left(s\right)-\frac{{\sigma }^2}{2}{s}^{2}E$ 代入上式，可得

$\tilde{Z}\left(s\right)=\frac{{\sigma }^2}{2}{\left[L_{\delta c}\left(s\right)\right]}^{-1}$ (5.6)

又由式(4.1)和式(5.6)可得

${\tilde{\overrightarrow{v}}}_1\left(s\right)=s\tilde{Z}\left(s\right){\tilde{\overrightarrow{v}}}_1\left(0\right)+\tilde{Z}\left(s\right){\widetilde{\overrightarrow{v^{\prime }}}}_1\left(0\right)+\frac{2c}{{\sigma }^2}\tilde{Z}\left(s\right){\tilde{\overrightarrow{v}}}_1\left(0\right)-\frac{2}{{\sigma }^2}\tilde{Z}\left(s\right){\tilde{\overrightarrow{\zeta }}}_1\left(s\right)$

再结合边界条件${\overrightarrow{v}}_1\left(0\right)=\overrightarrow{0}$ ，并对上式进行拉普拉斯逆变换可得

${\overrightarrow{v}}_1\left(u\right)=Z\left(u\right){\overrightarrow{v^{\prime }}}_1\left(0\right)-\frac{2}{{\sigma }^2}{\displaystyle {\int }_0^{u}Z\left(u-x\right){\overrightarrow{\zeta }}_1\left(x\right)\text{d}x}$

故等式(5.5)得证。

6. 阈值分红策略下的Gerber-Shiu惩罚函数

通过定理4.2的类似推导，可得到

$\overrightarrow{V_1}\left(u\right)=Z\left(u\right)\overrightarrow{{V^{\prime }}_1}\left(0\right)-\frac{2}{{\sigma }^2}{\displaystyle {\int }_0^{u}Z\left(u-x\right){\overrightarrow{\varsigma }}_1\left(x\right)\text{d}x}\text{, }0\le u<b$ (6.1)

由方程(6.1)和(5.5)可计算得

$\begin{array}{c}{\overrightarrow{V}}_1\left(u\right)={\overrightarrow{v}}_1\left(u\right)+Z\left(u\right)\left[{\overrightarrow{V^{\prime }}}_1\left(0\right)-{\overrightarrow{v^{\prime }}}_1\left(0\right)\right]\\ ={\overrightarrow{v}}_1\left(u\right)+Z\left(u\right){\overrightarrow{K}}_1\left(b\right),0\le u<b\end{array}$ (6.2)

此处${\overrightarrow{K}}_1\left(b\right)$ 为未知向量，其表达式将在后文给出。参考Lu和Li (2009)的方法，可得到${\overrightarrow{V}}_2\left(u\right)$ 的表达式如下，对于$b\le u<\infty$ ，令$y=u-b$ ，$c_2=c-\alpha$ 且${\overrightarrow{\phi }}_1\left(y\right)={\overrightarrow{V}}_2\left(u\right)={\overrightarrow{V}}_2\left(y+b\right)\left(y>0\right)$ ，方程(3.2)可改写为

${\overrightarrow{{\phi }^{″}}}_1\left(y\right)=\left(-\frac{2}{{\sigma }^2}\right)\left[c_2{\overrightarrow{{\phi }^{\prime }}}_1\left(y\right)+A{\overrightarrow{\phi }}_1\left(y\right)+{\displaystyle {\int }_0^{y}B\left(x\right){\overrightarrow{\phi }}_1\left(y-x\right)\text{d}x}+{\eta }_1\left(y\right)\right],y>0$ (6.3)

其边界条件为

${\overrightarrow{\phi }}_1\left(0\right)={\overrightarrow{V}}_2\left(b+\right)={\overrightarrow{V}}_1\left(b-\right),{\overrightarrow{{\phi }^{\prime }}}_1\left(0\right)={\overrightarrow{V^{\prime }}}_2\left(b+\right)={\overrightarrow{V^{\prime }}}_1\left(b-\right)$

其中

${\overrightarrow{\eta }}_1\left(y\right)={\displaystyle {\int }_y^{y+b}B\left(x\right){\overrightarrow{V}}_1\left(y+b-x\right)\text{d}x}+{\overrightarrow{\zeta }}_1\left(y+b\right)$

参考定理4.2的证明方法，令$W\left(y\right)$ 为式(6.3)的解，则有

$\tilde{W}\left(s\right)=\frac{{\sigma }^2}{2}{\left[L_{\delta {\text{c}}_2}\left(s\right)\right]}^{-1}$

对式(6.3)两边进行拉普拉斯变换得到

$\frac{2}{{\sigma }^2}{L}_{\delta c_2}\left(s\right){\tilde{\overrightarrow{\phi }}}_1\left(s\right)=s{\overrightarrow{\phi }}_1\left(0\right)+{\overrightarrow{\phi }}_1{}^{\prime }\left(0\right)-\frac{2}{{\sigma }^2}{\tilde{\overrightarrow{\eta }}}_1\left(s\right)$

再令上式两边同乘$\tilde{W}\left(s\right)$ 得到

${\tilde{\overrightarrow{\phi }}}_1\left(s\right)=\tilde{W}\left(s\right)\left[s{\overrightarrow{\phi }}_1\left(0\right)+{\overrightarrow{\phi }}_1{}^{\prime }\left(0\right)\right]-\frac{2}{{\sigma }^2}\tilde{W}\left(s\right){\tilde{\overrightarrow{\eta }}}_1\left(s\right)$

最后对上式进行拉普拉斯逆变换得到

${\overrightarrow{\phi }}_1\left(y\right)=\left[W^{\prime }\left(y\right)+\frac{2c_2}{{\sigma }^2}W\left(y\right)\right]{\overrightarrow{\phi }}_1\left(0\right)+W\left(y\right){{\overrightarrow{\phi }}^{\prime }}_1\left(0\right)-\frac{2}{{\sigma }^2}{\displaystyle {\int }_0^{y}W\left(y-x\right){\overrightarrow{\eta }}_1\left(x\right)\text{d}x},y\ge 0$

即

${\overrightarrow{V}}_2\left(u\right)=\left[W^{\prime }\left(u-b\right)+\frac{2c_2}{{\sigma }^2}W\left(u-b\right)\right]{\overrightarrow{V}}_1\left(b\right)+W\left(u-b\right){\overrightarrow{V}}_1{}^{\prime }\left(b\right)-\frac{2}{{\sigma }^2}{\displaystyle {\int }_0^{u-b}W\left(u-b-x\right){\overrightarrow{\eta }}_1\left(x\right)\text{d}x},b\le u<\infty$ (6.4)

其中

$W\left(y\right)=W\left(u-b\right)=\frac{{\sigma }^2}{2}{ℒ}^{-1}\left\{{\left[L_{\delta c_2}\left(s\right)\right]}^{-1}\right\},y\ge 0$ (6.5)

接下来讨论${\overrightarrow{K}}_1\left(b\right)$ 的表达式，回顾Lu和Li (2009)中定义的算子$T_{\text{r}}$ ，其作用于矩阵函数$B\left(y\right)$ 上为

$T_{\text{r}}B\left(y\right)={\displaystyle {\int }_y^{\infty }{\text{e}}^{-r\left(x-y\right)}B\left(x\right)\text{d}x,r\in C,y\ge 0}$

则对于不同的${\text{r}}_1,r_2\in C$ 和$y\ge 0$ ，有

$T_{r_1}{T}_{r_2}B\left(y\right)=T_{r_2}{T}_{r_1}B\left(y\right)=\frac{T_{r_1}B\left(y\right)-T_{r_2}B\left(y\right)}{r_2-r_1}$ (6.6)

将方程(3.2)两边乘以${\text{e}}^{-\text{s}\left(u-b\right)}$ ，并对$u$ 从$b$ 到$\infty$ 积分，可得对于满足$\Re \left(\text{s}\right)\ge 0$ 的$\text{s}\in ℂ$ ，有

$L_{\delta c_2}\left(s\right)T_s{\overrightarrow{V}}_2\left(b\right)=\left(\frac{{\sigma }^{2}s}{2}+c_2\right){\overrightarrow{V}}_2\left(b\right)+\frac{{\sigma }^2}{2}{{\overrightarrow{V}}^{\prime }}_2\left(b\right)-{\displaystyle {\int }_0^b{T}_{s}B\left(b-y\right){\overrightarrow{V}}_1\left(y\right)\text{d}y}-T_s{\overrightarrow{\varsigma }}_1\left(b\right)$

等式两边同时乘以$L_{\delta c_2}^{*}\left(s\right)$ 得到

$T_s{\overrightarrow{V}}_2\left(b\right)\text{=}\frac{L_{\delta c_2}^{*}\left(s\right)\left\{\left[\left(\frac{{\sigma }^{2}s}{2}+c_2\right){\overrightarrow{V}}_2\left(b\right)+\frac{{\sigma }^2}{2}{{\overrightarrow{V}}^{\prime }}_2\left(b\right)\right]-{\overrightarrow{\beta }}_1\left(s\right)\right\}}{\det \left[L_{\delta c_2}\left(s\right)\right]}$ (6.7)

其中

${\overrightarrow{\beta }}_1\left(s\right)={\displaystyle {\int }_0^b{T}_{s}B\left(b-y\right){\overrightarrow{V}}_1\left(y\right)\text{d}y}+T_s{\overrightarrow{\varsigma }}_1\left(b\right)$

将${\rho }_i,i=1,2$ 代入式(6.7)，则式(6.7)的分母为0，此时分子也为0，得到

$L_{\delta c_2}^{*}\left(s\right)\left\{\left[\left(\frac{{\sigma }^{2}s}{2}+c_2\right){\overrightarrow{V}}_2\left(b\right)+\frac{{\sigma }^2}{2}{{\overrightarrow{V}}^{\prime }}_2\left(b\right)\right]-{\overline{\beta }}_1\left(s\right)\right\}=0$

即

$L_{\delta c_2}\left({\rho }_i\right){\overrightarrow{V^{\prime }}}_2\left(b\right)=\left(-\frac{2}{{\sigma }^2}\right)\left[\left(\frac{{\sigma }^2{\rho }_i}{2}+c_2\right)L_{\delta c_2}^{*}\left({\rho }_i\right){\overrightarrow{V}}_2\left(b\right)-L_{\delta c_2}^{*}\left({\rho }_i\right){\overrightarrow{\beta }}_1\left({\rho }_i\right)\right],i=1,2$

再由式(5.2)，可推导出

$\begin{array}{l}{L}_{\delta c_2}^{*}\left[{\rho }_1,{\rho }_2\right]\overrightarrow{{V^{\prime }}_2}\left(b\right)\\ =\left(-\frac{2}{{\sigma }^2}\right)[\frac{\left(\frac{{\sigma }^2{\rho }_2}{2}+c_2\right)L_{\delta c_2}^{*}\left({\rho }_2\right)-\left(\frac{{\sigma }^2{\rho }_1}{2}+c_2\right)L_{\delta c_2}^{*}\left({\rho }_1\right)}{{\rho }_2-{\rho }_1}{\overrightarrow{V}}_2\left(b\right)-\frac{L_{\delta c_2}^{*}\left({\rho }_2\right){\overrightarrow{\beta }}_1\left({\rho }_2\right)-L_{\delta c_2}^{*}\left({\rho }_1\right){\overrightarrow{\beta }}_1\left({\rho }_1\right)}{{\rho }_2-{\rho }_1}]\\ \text{=}\left(-\frac{2}{{\sigma }^2}\right)\{\left[\left(\frac{{\sigma }^2{\rho }_2}{2}+c_2\right)L_{\delta c_2}^{*}\left[{\rho }_1,{\rho }_2\right]\text{+}\frac{{\sigma }^2}{2}{L}_{\delta c_2}^{*}\left({\rho }_1\right)\right]{\overrightarrow{V}}_2\left(b\right)-\left[L_{\delta c_2}^{*}\left[{\rho }_1,{\rho }_2\right]{\overrightarrow{\beta }}_1\left({\rho }_2\right)\text{+}{L}_{\delta c_2}^{*}\left({\rho }_1\right){\overrightarrow{\beta }}_1\left[{\rho }_1,{\rho }_2\right]\right]\}\end{array}$ (6.8)

利用在$b$ 时的边界条件，结合式(6.2)和(6.6)，可将式(6.8)重写为

$\begin{array}{c}{L}_{\delta c_2}^{*}\left[{\rho }_1,{\rho }_2\right]{\overrightarrow{v^{\prime }}}_1\left(b\right)+L_{\delta c_2}^{*}\left[{\rho }_1,{\rho }_2\right]Z^{\prime }\left(b\right){\overrightarrow{K}}_1\left(b\right)\\ =[\left(-{\rho }_1-\frac{2c_2}{{\sigma }^2}\right)L_{\delta c_2}^{*}\left[{\rho }_1,{\rho }_2\right]-L_{\delta c_2}^{*}\left({\rho }_1\right)]{\overrightarrow{v}}_1\left(b\right)\\ +[\left(-{\rho }_2-\frac{2c_2}{{\sigma }^2}\right)L_{\delta c_2}^{*}\left[{\rho }_1,{\rho }_2\right]-L_{\delta c_2}^{*}\left({\rho }_1\right)]Z\left(b\right){\overrightarrow{K}}_1\left(b\right)\\ +\frac{2}{{\sigma }^2}{L}_{\delta c_2}^{*}\left[{\rho }_1,{\rho }_2\right]{\displaystyle {\int }_0^b{T}_{{\rho }_2}B\left(b-y\right){\overrightarrow{v}}_1\left(y\right)\text{d}y}\\ +\frac{2}{{\sigma }^2}{L}_{\delta c_2}^{*}\left[{\rho }_1,{\rho }_2\right]{\displaystyle {\int }_0^b{T}_{{\rho }_2}B\left(b-y\right)Z\left(y\right)\text{d}y{\overrightarrow{K}}_1\left(b\right)}\\ +\frac{2}{{\sigma }^2}{L}_{\delta c_2}^{*}\left[{\rho }_1,{\rho }_2\right]T_{{\rho }_2}{\overrightarrow{\zeta }}_1\left(b\right)\\ -\frac{2}{{\sigma }^2}{L}_{\delta c_2}^{*}\left({\rho }_1\right){\displaystyle {\int }_0^b{T}_{{\rho }_1}{T}_{{\rho }_2}B\left(b-y\right){\overrightarrow{v}}_1\left(y\right)\text{d}y}\\ -\frac{2}{{\sigma }^2}{L}_{\delta c_2}^{*}\left({\rho }_1\right){\displaystyle {\int }_0^b{T}_{{\rho }_1}{T}_{{\rho }_2}B\left(b-y\right)Z\left(y\right)\text{d}y{\overrightarrow{K}}_1\left(b\right)}\\ -\frac{2}{{\sigma }^2}{L}_{\delta c_2}^{*}\left({\rho }_1\right)T_{{\rho }_1}{T}_{{\rho }_2}{\overrightarrow{\zeta }}_1\left(b\right)\end{array}$

整理包含${\overrightarrow{K}}_1\left(b\right)$ 的项构成${\rm H}{\overrightarrow{K}}_1\left(b\right)$ ，其余常数项为${\overrightarrow{R}}_1$ ，由此可得

${\overrightarrow{K}}_1\left(b\right)={{\rm H}}^{-1}{\overrightarrow{R}}_1$ (6.9)

其中

$\begin{array}{c}{\rm H}=L_{\delta c_2}^{*}\left[{\rho }_1,{\rho }_2\right]Z^{\prime }\left(b\right)+[\left({\rho }_2+\frac{2c_2}{{\sigma }^2}\right)L_{\delta c_2}^{*}\left[{\rho }_1,{\rho }_2\right]+L_{\delta c_2}^{*}\left({\rho }_1\right)]Z\left(b\right)\\ -\frac{2}{{\sigma }^2}{L}_{\delta c_2}^{*}\left[{\rho }_1,{\rho }_2\right]{\displaystyle {\int }_0^b{T}_{{\rho }_2}B\left(b-y\right)Z\left(y\right)\text{d}y}+\frac{2}{{\sigma }^2}{L}_{\delta c_2}^{*}\left({\rho }_1\right){\displaystyle {\int }_0^b{T}_{{\rho }_1}{T}_{{\rho }_2}B\left(b-y\right)Z\left(y\right)\text{d}y}\end{array}$

$\begin{array}{c}{\overrightarrow{R}}_1=-L_{\delta c_2}^{*}\left[{\rho }_1,{\rho }_2\right]{\overrightarrow{v^{\prime }}}_1\left(b\right)-[\left({\rho }_2+\frac{2c_2}{{\sigma }^2}\right)L_{\delta c_2}^{*}\left[{\rho }_1,{\rho }_2\right]+L_{\delta c_2}^{*}\left({\rho }_1\right)]{\overrightarrow{v}}_1\left(b\right)\\ +\frac{2}{{\sigma }^2}{L}_{\delta c_2}^{*}\left[{\rho }_1,{\rho }_2\right]{\displaystyle {\int }_0^b{T}_{{\rho }_2}B\left(b-y\right){\overrightarrow{v}}_1\left(y\right)\text{d}y}-\frac{2}{{\sigma }^2}{L}_{\delta c_2}^{*}\left({\rho }_1\right){\displaystyle {\int }_0^b{T}_{{\rho }_1}{T}_{{\rho }_2}B\left(b-y\right){\overrightarrow{v}}_1\left(y\right)\text{d}y}\\ +\frac{2}{{\sigma }^2}{L}_{\delta c_2}^{*}\left[{\rho }_1,{\rho }_2\right]T_{{\rho }_2}{\overrightarrow{\zeta }}_1\left(b\right)-\frac{2}{{\sigma }^2}{L}_{\delta c_2}^{*}\left({\rho }_1\right)T_{{\rho }_1}{T}_{{\rho }_2}{\overrightarrow{\zeta }}_1\left(b\right)\end{array}$

最后，将主要结果总结为以下定理。

定理5.1：函数${\overrightarrow{V}}_1\left(u\right),{\overrightarrow{V}}_2\left(u\right)$ 的解析表达式如下

$\{\begin{array}{l}{\overrightarrow{V}}_1\left(u\right)={\overrightarrow{v}}_1\left(u\right)+Z\left(u\right){\overrightarrow{K}}_1\left(b\right),0\le u<b,\hfill \\ \begin{array}{c}{\overrightarrow{V}}_2\left(u\right)=\left[W^{\prime }\left(u-b\right)+\frac{2c_2}{{\sigma }^2}W\left(u-b\right)\right]{\overrightarrow{V}}_1\left(b\right)+W\left(u-b\right){\overrightarrow{V}}_1{}^{\prime }\left(u\right)\\ -\frac{2}{{\sigma }^2}{\displaystyle {\int }_0^{u-b}W\left(u-b-x\right){\overrightarrow{\eta }}_1\left(x\right)\text{d}x,b\le u<\infty }\end{array}\hfill \end{array}$

其中，${\overrightarrow{V}}_1\left(u\right)$ 由定理4.2给出，而$Z\left(u\right)$ ，$W\left(u\right)$ ，${\overrightarrow{K}}_1\left(b\right)$ 分别由(5.6)，(6.5)，(6.9)定义。

7. 数值分析

若对任意$x,y>0$ ，都有$w\left(x,y\right)=1$ ，则此时Gerber-Shiu函数$V_k\left(u\right)$ 为破产时间的拉普拉斯变换，记为${\tilde{V}}_k\left(u\right)$ ，若进一步满足$\delta =0$ 且对任意$x,y\in R^{\text{+}}$ ，都有$w\left(x,y\right)=1$ ，则此时Gerber-Shiu函数$V_k\left(u\right)$ 为破产概率，下面我们考虑一个数值示例来阐明破产概率的变化。

Figure 1. Bankruptcy probability chart

图1. 破产概率图示

在本节中，假设$\delta =0$ ，分红阈值$b=5$ ，保费率$c=2$ ，布朗运动扩散率$\sigma =0.5$ ，索赔额服从指数分布$\text{X~Exp}\left(\alpha =1\right)$ ，并以阈值$Q=1$ 划分两类风险状态，当$X>1$ 时索赔间隔时间${\text{W}}_1\text{~Exp}\left({\lambda }_1=0.8\right)$ ，当$X<1$ 时索赔间隔时间${\text{W}}_2\text{~Exp}\left({\lambda }_2=1.\text{2}\right)$ ，且经过计算安全负载条件$\text{E}\left[cW-X\right]\approx 0.97>0$ ，满足破产非必然条件，则伦德伯格方程的根为${\rho }_1=0,{\rho }_2=0.26,R_1=-2.22,R_2=-1.98$ ，由本节之前的计算结果可以得到，当$0<u<5$ 时，破产概率为

$\begin{array}{l}{V}_{11}\left(u\right)=0.73{\text{e}}^{-0.26u}-0.02{\text{e}}^{-2.22u}\begin{array}{cc}& \end{array}{V}_{12}\left(u\right)=0.65{\text{e}}^{-0.26u}-0.03{\text{e}}^{-2.22u}\\ V_{21}\left(u\right)=0.58{\text{e}}^{-0.26u}-0.04{\text{e}}^{-1.98u}\begin{array}{cc}& \end{array}{V}_{22}\left(u\right)=0.50{\text{e}}^{-0.26u}-0.05{\text{e}}^{-1.98u}\end{array}$

当μ ≥ 5时，破产概率为

$\begin{array}{l}{V}_{11}\left(u\right)=0.70{\text{e}}^{-0.26u}-0.015{\text{e}}^{-2.22u}\begin{array}{cc}& \end{array}{V}_{12}\left(u\right)=0.62{\text{e}}^{-0.26u}-0.025{\text{e}}^{-2.22u}\\ V_{21}\left(u\right)=0.55{\text{e}}^{-0.26u}-0.035{\text{e}}^{-1.98u}\begin{array}{cc}& \end{array}{V}_{22}\left(u\right)=0.47{\text{e}}^{-0.26u}-0.045{\text{e}}^{-1.98u}\end{array}$

图1表明，随着初始盈余u的不断增大，不同风险状态下的破产概率逐渐降低，初始盈余越高，风险抵御能力越强。