数学物理学报  2015, Vol. 35 Issue (5): 1004-1017   PDF (395 KB)    
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杨龙
多层分红策略下以FGM Copula为相依结构的风险模型
杨龙    
广西省桂林市广西师范大学数学与统计学院 桂林 541004
收稿日期: 2014-03-14; 修订日期: 2015-02-24
作者简介: 杨龙,yanglong1773@126.com
摘要: 该文考虑了多层分红策略下相依的风险模型,用Farlie-Gumbel-Morgenstern(FGM) copula定义了索赔间隔时间和索赔额之间的相依结构,研究了Gerber-Shiu期望折扣罚金函数,导出了其所满足的积分微分方程和瑕疵更新方程,并给出了它们的解析解.最后,以索赔额分布服从指数分布为例,给出了破产概率所满足的具体解.
关键词: FGM copula     相依结构     Gerber-Shiu函数     多段分红策略     瑕疵更新方程    
The Risk Process with Dependence Based on FGM Copula under A Multi-Layer Dividend Strategy
Yang Long    
School of Mathematics and Statistics, Guangxi Normal University, Guilin 541004
Received: 2014-03-14; Revised: 2015-02-24
Abstract: In this paper, we consider an extension to the classical compound Poisson risk model under a multi-layer dividend strategy, in which the claim size and inter-claim time are dependent. We assume that the dependence structure between the claim size and the inter-claim time is based on a Farlie-Gumbel-Morgenstern copula. Some piecewise integro-differential equations with certain boundary conditions for the Gerber-Shiu function are derived. Then, applying these results, some defective renewal equations for the Gerber-Shiu function are obtained and explicit expressions are solved. Finally, to illustrate the solution procedure, explicit expressions for the ruin probability are given for exponential claim size.
Key words: Dependence structure     Multi-layer dividend strategy     Farlie-Gumbel-Morgenstern copula     Gerber-Shiu function     Defective renewal equation    
1 引言

在精算数学中,经典风险模型作为核心研究对象已长达一个多世纪. 该模型假定保险公司在时刻$t$的盈余过程为

\begin{equation} U(t)=u+ct-\sum\limits_{i=1}^{N(t)}{X_i}=u+ct-S(t),\quad \quad t\geq 0, \end{equation} (1.1)

其中,$u$表示保险公司的初始准备金,$c>0$表示保费收入率. 计数过程$\{N(t);t\geq0\}$ 是参数为$\lambda$的泊松过程, 它表示到时刻$t$的索赔总次数. 索赔间隔时间序列$\{V_i;i\geq1\}$是指数分布列, 具有共同的概率密度函数$k(t)=\lambda {\rm e}^{-\lambda t}$和分布函数$K(t)=1-{\rm e}^{-\lambda t}$. 非负随机变量序列$\{X_i;i\geq1\}$为独立同分布的个体索赔额序列, 其共同的分布函数为$F$,密度函数为$f$. 在经典风险模型中,通常假设索赔间隔时间变量$V_i$和个体索赔额变量$X_i$是相互独立的. 尽管这一假设确实使得破产问题的研究得到简化,但和一些现实情形不相吻合,如灾难保险. 为了克服这一缺陷,一些考虑索随机变量$V_i$和$X_i$具有相依性的风险模型被提了出来, 参见文献[1, 2, 3]及其参考文献.

考虑到分红在现实生活中的重要应用,同时也为了更实际地反映出保险投资组合的盈余现金流, De finetti[4]首次在风险模型中引入了分红策略. 从那时起,带分红策略的风险模型受到许多学者的广泛关注和研究. 他们通常采用的分红策略有barrier分红策略和threshold分红策略, 而主要研究工作集中于探讨相应模型下Gerber-Shiu函数的计算方法问题.

多层分红策略作为barrier分红策略和threshold分红策略的一个重要推广也得到了一些精算学者的关注, 可参见文献[5, 6, 7]. 受到Zhang和Yang[7]的启发,该文考虑了多层分红策略下以FGM copula方程定义了索赔间隔时间和索赔额之间的相依结构的风险模型. 类似于 Albrecher和Hartinger[5],我们考虑下面的多层分红策略. 令$N$为一正整数,定义分红层$0=b_0< b_1<\cdots< b_{N-1}< b_N=\infty.$ 在时刻$t$,若盈余处于第$i$个分红层,即$b_{i-1}\leq U(t)< b_{i},$ 则公司收取的保费率为$c_i$, 直到盈余水平上升到第$i+1$个分层或由于理赔而降到分红层$i$以下. 为了表述方便, 我们将该多层分红策略下的盈余过程仍然记为$U(t),$ 则(1.1)式变为

\begin{equation} {\rm d}U(t)=c_i{\rm d}t-{\rm d}S(t),\;\;\;\;\;\;\;\;b_{i-1}\leq U(t)< b_i, \end{equation} (1.2)

对于$i=1,2,\cdots,N.$

该文主要探讨了多层分红策略下相依风险模型(1.2)的Gerber-Shiu 期望折扣罚金函数的计算方法并给出了相应的计算结果.

2 相依结构和风险测度

本文中用FGM copula建立了索赔间隔时间和个体索赔额之间的相依关系. FGM copula的定义式为

\begin{equation} C(u,v)=uv+\theta uv(1-u)(1-v),\;\;\;\;0\leq u,v\leq1, \end{equation} (2.1)

其中$-1\leq\theta\leq1.$ FGM copula包含了正相关和负相关,同时也暗含了独立性假设即$\theta=0.$ 关于FGM copula的更多性质,参见文献[8].

在文中接下来的部分,我们假定二维随机变量序列$\{(X_i,V_i),i \in {\mathbb N}^+\}$ 是独立同分布的并与通常表示变量$(X,V)$具有相同的分布. 令$F_{X,V}(x,t)$ 和 $f_{X,V}(x,t)$分别表示 $(X,V)$的分布函数和概率密度函数.

由个体索赔额的边际分布函数$F$,索赔间隔时间的边际分布函数$K$以及(2.1)式可得$(X,V)$ 的分布函数和概率密度函数分别为

\begin{eqnarray*} F_{X,V}(x,t)&=&C(F(x),K(t))\\ &=&F(x)K(t)+\theta F(x)K(t)(1-F(x))(1-K(t)),\;\;(x,t)\in {\mathbb R}^+\times{\mathbb R}^+, \\ f_{X,V}(x,t)&=&c(F(x),K(t))f(x)k(t)\\ &=&\lambda {\rm e}^{-\lambda t}f(x)+\theta(2\lambda {\rm e}^{-2\lambda t}-\lambda {\rm e}^{-\lambda t})h(x),\;\;(x,t)\in {\mathbb R}^+\times{\mathbb R}^+, \end{eqnarray*}

其中$c(u,v)=\frac{\partial^2}{\partial u \partial v}C(u,v)$且$h(x)=(1-2F(x))f(x).$ 特别地,个体索赔额的条件概率密度函数为

$$f_{X|V=t}(x)=f(x)+\theta(2{\rm e}^{-\lambda t}-1)h(x).$$

另外,我们可以得出多层分红策略下经典的风险模型是本文中的模型在 $\theta=0$时的特殊情形.

风险模型(1.2)的破产时刻定义为$T=\inf\{t>0:U(t)<0\},$ 其中若对任意$t\geq0$都有$U(t)\geq0,$则记$T=\infty.$ 相应地,给定初始盈余$U(0)=u,$ 破产概率可定义为

$$\psi(u)=P(T<\infty|U(0)=u).$$

为了研究更多的与破产相关的风险测度,我们引入Gerber-Shiu期望折扣罚金函数,其定义为

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

其中$\delta>0$是折现因子,$w$为一非负的惩罚函数, $U(T-)$和$|U(T)|$分别表示破产前的瞬时盈余和破产时的赤字,$I(A)$为事件$A$的示性函数. 特别地,当$w(x_1,x_2)=1$时,Gerber-Shiu函数变为破产时刻的拉普拉斯变换$\phi_T(u)$; 当$\delta=0,w(x_1,x_2)=1$时,Gerber-Shiu函数变为破产概率$\psi(u)$. 另外, 我们还可以得到其它与破产相关的函数,这里不再一一介绍.

3 积分微分方程

本节主要给出了Gerber-Shiu期望折扣罚金函数所满足的积分微分方程. 首先当$i=1,2,\cdots,N,$令

$$A_i({\cal D})=\Big(\frac{2\lambda+\delta}{c_i}{\cal I}-{\cal D}\Big) \Big(\frac{\lambda+\delta}{c_i}{\cal I}-{\cal D}\Big),$$

其中${\cal I}$和${\cal D}$分别表示恒等算子和微分算子.

当$b_{i-1}\leq u< b_i$,令$c(u)=c_i$为一非负的保费收入率函数并对第一次索赔时间和索赔额取条件, 同时利用$f_{X,V}(x,t)$的定义有

\begin{eqnarray} \Phi(u)&=&\displaystyle\int_0^{\infty}\int_0^{U(t)} {\rm e}^{-\delta t}\Phi(U(t)-x)f_{X,V}(x,t){\rm d}x{\rm d}t \nonumber\\ &&+\displaystyle\int_0^{\infty}\int_{U(t)}^{\infty}{\rm e}^{-\delta t} w(U(t),x-U(t))f_{X,V}(x,t){\rm d}x{\rm d}t \nonumber\\ &=&\displaystyle\lambda\int_0^{\infty}\int_0^{U(t)} {\rm e}^{-\delta t}\Phi(U(t)-x)f(x){\rm e}^{-\lambda t}{\rm d}x{\rm d}t \nonumber\\ &&+\displaystyle\lambda\int_0^{\infty}\int_{U(t)}^{\infty}{\rm e}^{-\delta t}w(U(t),x-U(t))f(x){\rm e}^{-\lambda t}{\rm d}x{\rm d}t \nonumber\\ &&+\displaystyle\lambda\theta\int_0^{\infty}\int_{0}^{U(t)}{\rm e}^{-\delta t}\Phi(U(t)-x)h(x)(2{\rm e}^{-2\lambda t}-{\rm e}^{-\lambda t}){\rm d}x{\rm d}t \nonumber\\ &&+\displaystyle\lambda\theta\int_0^{\infty}\int_{U(t)}^{\infty}{\rm e}^{-\delta t}w(U(t),x-U(t))h(x)(2{\rm e}^{-2\lambda t}-{\rm e}^{-\lambda t}){\rm d}x{\rm d}t, \end{eqnarray} (3.1)

其中$U(t)$为第一次索赔到达时刻$t$时的瞬时盈余并满足$U(0)=u$和${\rm d}U(t)=c(U(t)){\rm d}t.$ 令

$$\xi_1(u)=\int_u^\infty w(u,x-u)f(x){\rm d}x,\;\;\;\;\;\;\;\;\gamma_1(u)=\int_0^u\Phi(u-x)f(x){\rm d}x+\xi_1(u),$$

$$\xi_2(u)=\int_u^\infty w(u,x-u)h(x){\rm d}x,\;\;\;\;\;\;\;\;\gamma_2(u)=\int_0^u \Phi(u-x)h(x){\rm d}x+\xi_2(u).$$

则(3.1)式变为

\begin{equation}\begin{array}{l} \Phi(u)=\displaystyle\int_0^\infty\lambda {\rm e}^{-(\lambda+\delta)t}\big[\gamma_1(U(t))-\theta\gamma_2(U(t))\big]{\rm d}t +\displaystyle\int_0^\infty2\lambda\theta {\rm e}^{-(2\lambda+\delta)t}\gamma_2(U(t)){\rm d}t. \end{array}\end{equation} (3.2)

对(3.2)式作变量替换$s=U(t),$ 并注意到在$U(0)=u$的条件下, ${\rm d}s=c(s){\rm d}t$和$t=\int_u^s[c(y)]^{-1}{\rm d}y$,可得

\begin{eqnarray} \Phi(u)&=&\displaystyle\int_u^\infty\lambda {\rm e}^{-(\lambda+\delta)\int_u^s[c(y)]^{-1}{\rm d}y}[\gamma_1(s)-\theta\gamma_2(s)][c(s)]^{-1}{\rm d}s \nonumber\\ &&+\displaystyle\int_u^\infty2\lambda\theta {\rm e}^{-(2\lambda+\delta)\int_u^s[c(y)]^{-1}{\rm d}y}\gamma_2(s)[c(s)]^{-1}{\rm d}s. \end{eqnarray} (3.3)

对(3.3)式关于$u$求导并由(3.3)式和$c(u)=c_i$给出

\begin{equation} \Phi'(u)=\displaystyle\frac{\lambda+\delta}{c_i}\Phi(u)+\displaystyle\frac{\lambda}{c_i}\displaystyle\int_u^\infty2\lambda\theta {\rm e}^{-(2\lambda+\delta)\int_u^s[c(y)]^{-1}{\rm d}y}\gamma_2(s)[c(s)]^{-1}{\rm d}s -\displaystyle\frac{\lambda}{c_i}(\gamma_1(u)+\theta\gamma_2(u)). \end{equation} (3.4)

再对(3.4)式关于$u$求导并利用(3.4)式有

\begin{eqnarray} A_i({\cal D})\Phi(u)&=&\displaystyle\frac{\lambda}{c_i}\Big(\displaystyle\frac{2\lambda+\delta}{c_i}{\cal I}-{\cal D}\Big)\Big(\displaystyle\int_0^u\Phi(u-x)f(x){\rm d}x+\xi_1(u)\Big) \nonumber\\ &&+\displaystyle\frac{\lambda\theta}{c_i}\Big(\displaystyle\frac{\delta}{c_i}{\cal I}-{\cal D}\Big)\Big(\displaystyle\int_0^u\Phi(u-x)h(x){\rm d}x+\xi_2(u)\Big), \end{eqnarray} (3.5)

其中$b_{i-1}\leq u< b_i$且$i=1,2,\cdots,N.$

利用(3.3)式很容易得到Gerber-Shiu函数在$u=b_i\;(i=1,2,\cdots,N-1)$处是连续的,即

\begin{equation} \Phi(b_i-)=\Phi(b_i+). \end{equation} (3.6)

再者由(3.4)式和边界条件(3.6)式可得Gerber-Shiu函数在$u=b_i\;(i=1,2,\cdots,N-1)$处满足

\begin{equation} c_i\Phi'(b_i-)=c_{i+1}\Phi'(b_i+), \end{equation} (3.7)

边界条件(3.6)和(3.7)式表明Gerber-Shiu函数在$u=b_i\;(i=1,2,\cdots,N-1)$处是连续但不可导的, 因此(3.5)式中$\Phi(u)$的导数是指$\Phi(u)$的右导数.

4 分析积分微分方程

该节的主要内容是对积分微分方程(3.5)进行分析和讨论. 在后面的内容中,我们将在一个字母上加一个``帽子"来表示相应函数的拉普拉斯变换.

由文献[2,推论4.1]可知,对于任意的$c_i\;(i=1,2,\cdots,N)$下面的广义Lundberg 基本方程

\begin{equation} \Big(\frac{2\lambda+\delta}{c_i}-s\Big)\Big(\frac{\lambda+\delta}{c_i}-s\Big) -\Big(\frac{\lambda}{c_i}\Big(\frac{2\lambda+\delta}{c_i}-s\Big)\hat{f}(s) +\frac{\lambda\theta}{c_i}\Big(\frac{\delta}{c_i}-s\Big)\hat{h}(s)\Big)=0 \end{equation} (4.1)

在右半复平面内恰好有两个不同的根,记为$\rho_{i1}(\delta)$和$\rho_{i2}(\delta)$. 不失一般性,我们将这两个根中实部最小的记为$\rho_{i1}(\delta)$, 则有$\lim\limits_{\delta\rightarrow 0}{\rho_{i1}(\delta)=0}.$ 为了表达的方便, 文中后面的部分将$\rho_{i1}(\delta)$和$\rho_{i2}(\delta)$简记为$\rho_{i1}$和$\rho_{i2}.$

接下来介绍一下Dickson-Hipp算子及其部分性质. 对于一个定义在${\mathbb R}^+$ 的可积函数$h$,Dickson-Hipp算子的定义如下

$$T_sh(x)=\int_x^\infty {\rm e}^{-s(y-x)}h(y){\rm d}y\;\;\;x>0,$$

其中$s$要使得上面的积分绝对可积. 其具有很多优良性质,该文将主要应用下面两个性质:

1. $T_sh(0)=\hat{h}(s);$

2. $T_s T_rh(x)=T_r T_sh(x)= \frac{T_rh(x)-T_sh(x)}{s-r},r\neq s.$

关于Dickson-Hipp算子的更多性质,可参见文献[9].

类似于文献[10], 我们将方程(3.5)中的约束条件$b_{i-1}\leq u< b_i$放松为$b_{i-1}\leq u,$ 并记$\Phi_i(u)$为下面方程的特解

\begin{eqnarray} &&A_i({\cal D})\Phi_i(u) \nonumber\\ &=&\frac{\lambda}{c_i}\Big(\displaystyle\frac{2\lambda+\delta}{c_i}{\cal I}-{\cal D}\Big) \Big(\displaystyle\int_0^{u-b_{i-1}}\Phi_i(u-x)f(x){\rm d}x+\displaystyle\int_{u-b_{i-1}}^u\Phi(u-x)f(x){\rm d}x+\xi_1(u)\Big) \nonumber\\ &&+\displaystyle\frac{\lambda\theta}{c_i}\Big(\displaystyle\frac{\delta}{c_i}{\cal I}-{\cal D}\Big) \Big(\displaystyle\int_0^{u-b_{i-1}}\Phi_i(u-x)h(x){\rm d}x+\displaystyle\int_{u-b_{i-1}}^u\Phi(u-x)h(x){\rm d}x+\xi_2(u)\Big), \end{eqnarray} (4.2)

其中$|T_s\Phi_i(b_{i-1})|<\infty.$ $\Phi_1(u)$正是不考虑分红策略时的Gerber-Shiu函数, 其已被Cossette 等[2]研究.

由积分微分方程理论知方程(3.5)的通解可表示为

\begin{equation} \Phi(u)=\Phi_i(u)+r_{i1}v_{i1}(u)+r_{i2}v_{i2}(u),\;\;\;\;b_{i-1}\leq u< b_i, \end{equation} (4.3)

其中$r_{i1}$和$r_{i2}$是常系数,$v_{i1}(u)$和$v_{i2}(u)$是下面齐次积分微分方程的两个线性无关的解

\begin{eqnarray} A_i({\cal D})v_i(u)&=&\displaystyle\frac{\lambda}{c_i}\Big(\displaystyle\frac{2\lambda+\delta}{c_i}{\cal I} -{\cal D}\Big)\displaystyle\int_0^{u-b_{i-1}}v_i(u-x)f(x){\rm d}x \nonumber\\ &&+\displaystyle\frac{\lambda\theta}{c_i}\Big(\displaystyle\frac{\delta}{c_i}{\cal I} -{\cal D}\Big)\displaystyle\int_0^{u-b_{i-1}}v_i(u-x)h(x){\rm d}x,\;\;u\geq b_{i-1}, \end{eqnarray} (4.4)

且当$n=N$时有$r_{N1}=r_{N2}=0.$

4.1 $\Phi_i(u)$满足的瑕疵更新方程

该小节给出了$\Phi_i(u)$满足的瑕疵更新方程,其在计算Gerber-Shiu函数$\Phi(u)$的显示表达式起着至关重要的作用.

定理 4.1 当$i=1,2,\cdots,N$和Re$(s)\geq 0$时,给定条件$\hat{\Phi}_i(s)<\infty,$ 则方程 (4.2)存在唯一解,且该解满足下面的积分方程

\begin{equation} \Phi_i(u)=k_i\bigg(\displaystyle\int_0^{u-b_{i-1}}\Phi_i(u-x)g_i(x){\rm d}x+\int_{u-b_{i-1}}^u\Phi(u-x)g_i(x){\rm d}x\bigg)+\eta_i(u), \end{equation} (4.5)

其中

\begin{eqnarray*} k_i&=&\displaystyle\frac{\lambda}{c_i}\bigg(\Big(\frac{2\lambda+\delta}{c_i}-\rho_{i1}\Big)T_0T_{\rho_{i1}}T_{\rho_{i2}}f(0) +\theta\Big(\frac{\delta}{c_i}-\rho_{i1}\Big)T_0T_{\rho_{i1}}T_{\rho_{i2}}h(0)\\ &&+T_0T_{\rho_{i2}}f(0)+\theta T_0T_{\rho_{i2}}h(0)\bigg), \end{eqnarray*}

\begin{eqnarray*} g_i(x)&=&\displaystyle\frac{\lambda}{c_ik_i}\bigg(\Big(\frac{2\lambda+\delta}{c_i}-\rho_{i1}\Big)T_{\rho_{i1}}T_{\rho_{i2}}f(x) +\theta\Big(\frac{\delta}{c_i}-\rho_{i1}\Big)T_{\rho_{i1}}T_{\rho_{i2}}h(x)\\ &&+T_{\rho_{i2}}f(x)+\theta T_{\rho_{i2}}h(x)\bigg), \end{eqnarray*}

\begin{eqnarray*} \eta_i(x)&=&\displaystyle\frac{\lambda}{c_i}\bigg(\Big(\frac{2\lambda+\delta}{c_i}-\rho_{i1}\Big)T_{\rho_{i1}}T_{\rho_{i2}}\xi_1(x) +\theta\Big(\frac{\delta}{c_i}-\rho_{i1}\Big)T_{\rho_{i1}}T_{\rho_{i2}}\xi_2(x)\\ &&+T_{\rho_{i2}}\xi_1(x)+\theta T_{\rho_{i2}}\xi_2(x)\bigg). \end{eqnarray*}

证 证明环境 作变量代换$y=u-b_{i-1}$并令$Q_i(y)=Q_i(u-b_{i-1})=\Phi_i(u)$,则(4.2)式变为

\begin{equation} A_i({\cal D})Q_i(y)=\displaystyle\frac{\lambda}{c_i}\Big(\displaystyle\frac{2\lambda+\delta}{c_i}{\cal I}-{\cal D}\Big)H_i(y)+\displaystyle\frac{\lambda\theta}{c_i} \Big(\displaystyle\frac{\delta}{c_i}{\cal I}-{\cal D}\Big)K_i(y), \end{equation} (4.6)

其中

\begin{equation} H_i(y)=\displaystyle\int_0^yQ_i(y-x)f(x){\rm d}x+M_i(y),\;M_i(y)=\displaystyle\int_0^{b_{i-1}}\Phi(x)f(y+b_{i-1}-x){\rm d}x+\xi_1(y+b_{i-1}), \end{equation} (4.7)
\begin{equation} K_i(y)=\displaystyle\int_0^yQ_i(y-x)h(x){\rm d}x+N_i(y),\;N_i(y)=\displaystyle\int_0^{b_{i-1}}\Phi(x)h(y+b_{i-1}-x){\rm d}x+\xi_2(y+b_{i-1}). \end{equation} (4.8)

对(4.6)式两边同取拉普拉斯变换并由拉普拉斯变换的性质可得

\begin{eqnarray} &&A_i(s)\hat{Q}_i(s)-Q_i(0)s-\Big(Q'_i(0)-\frac{2\delta+3\lambda}{c_i}Q_i(0)\Big) \nonumber\\ &=&\displaystyle\frac{\lambda}{c_i}\Big(\displaystyle\frac{2\lambda+\delta}{c_i}-s\Big)\hat{H}_i(s) +\displaystyle\frac{\lambda\theta}{c_i}\Big(\displaystyle\frac{\delta}{c_i}-s\Big)\hat{K}_i(s)+\frac{\lambda}{c_i}H_i(0)+\frac{\lambda\theta}{c_i}K_i(0). \end{eqnarray} (4.9)

由(4.7)和(4.8)式可得

\begin{equation} \hat{H}_i(s)=\hat{Q}_i(s)\hat{f}(s)+\hat{M}_i(s),\;\;\;\;\;\;\hat{K}_i(s)=\hat{Q}_i(s)\hat{h}(s)+\hat{N}_i(s), \end{equation} (1.10)

将(4.10)式代入(4.9)式并经过计算得

\begin{equation} \hat{Q}_i(s)=\displaystyle\frac{\alpha_i(s)+\beta_i(s)}{\big(\frac{2\lambda+\delta}{c_i}-s\big)\big(\frac{\lambda+\delta}{c_i}-s\big) -\big(\frac{\lambda}{c_i}(\frac{2\lambda+\delta}{c_i}-s)\hat{f}(s)+\frac{\lambda\theta}{c_i}(\frac{\delta}{c_i}-s)\hat{h}(s)\big)}, \end{equation} (4.11)

其中

\begin{equation} \alpha_i(s)=Q_i(0)s+Q'_i(0)+\frac{\lambda}{c_i}H_i(0)+\frac{\lambda\theta}{c_i}K_i(0)-\frac{2\delta+3\lambda}{c_i}Q_i(0), \end{equation} (4.12)
\begin{equation} \beta_i(s)=\displaystyle\frac{\lambda}{c_i}\Big(\frac{2\lambda+\delta}{c_i}-s\Big) \hat{M}_i(s)+\displaystyle\frac{\lambda\theta}{c_i}\Big(\frac{\delta}{c_i}-s\Big)\hat{N}_i(s). \end{equation} (4.13)

由文献[2,推论7.1]知(4.11)式的分母可变为$(s-\rho_{i1})(s-\rho_{i2})(1-T_sT_{\rho_{i1}}T_{\rho_{i2}}p_i(0)),$ 其中$p_i$满足

$$T_sp_i(0)=\displaystyle\frac{\lambda}{c_i}\Big(\frac{2\lambda+\delta}{c_i}-s\Big) \hat{f}(s)+\displaystyle\frac{\lambda\theta}{c_i}\Big(\frac{\delta}{c_i}-s\Big)\hat{h}(s).$$

由条件Re$(s)\geq 0$时$\hat{Q}_i(s)$是解析函数知,(4.11)式中分母的根同样也是分子的根. 于是有$\alpha_i(\rho_{ij})=-\beta_i(\rho_{ij})\;(j=1,2).$ 从(4.12)式可知$\alpha_i(s)$是$s$的一阶多项式. 使用拉格朗日插值定理,(4.11)式的分子可重新写为

\begin{eqnarray} \alpha_i(s)+\beta_i(s)&=&\displaystyle\frac{\rho_{i2}-s}{\rho_{i1}-\rho_{i2}}\beta_i(\rho_{i1})+\displaystyle\frac{\rho_{i1}-s}{\rho_{i2}-\rho_{i1}}\beta_i(\rho_{i2})+\beta_i(s) \nonumber\\ &=&\displaystyle\frac{(s-\rho_{i1})(s-\rho_{i2})}{\rho_{i1}-\rho_{i2}}\bigg(\displaystyle\frac{\beta_i(s)-\beta_i(\rho_{i2})}{\rho_{i2}-s} -\displaystyle\frac{\beta_i(s)-\beta_i(\rho_{i1})}{\rho_{i1}-s}\bigg). \end{eqnarray} (4.14)

从(4.13)式很容易得,当$j=1,2$时

\begin{eqnarray*} \displaystyle\frac{\beta_i(s)-\beta_i(\rho_{ij})}{\rho_{ij}-s} &=&\displaystyle\frac{\lambda}{c_i}\Big(\displaystyle\frac{2\lambda+\delta}{c_i} -\rho_{ij}\Big)\frac{\hat{M}_i(s)-\hat{M}_i(\rho_{ij})}{\rho_{ij}-s} +\displaystyle\frac{\lambda}{c_i}\hat{M}_i(s)\\ &&+\displaystyle\frac{\lambda\theta}{c_i}\Big(\frac{\delta}{c_i}-\rho_{ij}\Big)\frac{\hat{N}_i(s)-\hat{N}_i(\rho_{ij})}{\rho_{ij}-s}+\displaystyle\frac{\lambda\theta}{c_i}\hat{N}_i(s). \end{eqnarray*}

将上式代入(4.14)式右边并经仔细计算得

\begin{eqnarray} \alpha_i(s)+\beta_i(s)&=&\displaystyle\frac{(s-\rho_{i1}) (s-\rho_{i2})}{\rho_{i1}-\rho_{i2}}\bigg[\displaystyle\frac{\lambda}{c_i} \Big(\displaystyle\frac{2\lambda+\delta}{c_i}-\rho_{i1}\Big) \Big(\frac{\hat{M}_i(s)-\hat{M}_i(\rho_{i1})}{s-\rho_{i1}}-\frac{\hat{M}_i(s)- \hat{M}_i(\rho_{i2})}{s-\rho_{i2}}\Big) \nonumber\\ &&+\displaystyle\frac{\lambda\theta}{c_i}\Big(\frac{\delta}{c_i}-\rho_{i1}\Big) \Big(\frac{\hat{N}_i(s)-\hat{N}_i(\rho_{i1})}{s-\rho_{i1}}-\frac{\hat{N}_i(s) -\hat{N}_i(\rho_{i2})}{s-\rho_{i2}}\Big) \nonumber\\ &&-\displaystyle\frac{\lambda(\rho_{i1}-\rho_{i2})}{c_i}\frac{\hat{M}_i(s)-\hat{M}_i(\rho_{i2})}{s-\rho_{i2}}-\displaystyle\frac{\lambda\theta(\rho_{i1}-\rho_{i2})}{c_i} \frac{\hat{N}_i(s)-\hat{N}_i(\rho_{i2})}{s-\rho_{i2}}\bigg]. \end{eqnarray} (4.15)

现在我们回顾一个函数$M(s)$对于不同的常数$r_1,r_2,r_3,\cdots,$的差商,其递归定义如下

$$M[r_1,s]=\displaystyle\frac{M(s)-M(r_1)}{s-r_1},\;\;\;\;M[r_1,r_2,s]=\displaystyle\frac{M[r_1,s]-M[r_2,s]}{r_1-r_2},$$

等等. 那么,应用函数$\hat{M}_i(s)$和$\hat{N}_i(s)$对于$\rho_{i1}$和$\rho_{i2}$的差商,(4.15)式可变为

\begin{eqnarray} \alpha_i(s)+\beta_i(s)&=&(s-\rho_{i1})(s-\rho_{i2}) \bigg(\displaystyle\frac{\lambda}{c_i}\Big(\displaystyle\frac{2\lambda+\delta}{c_i}-\rho_{i1}\Big) \hat{M}_i[\rho_{i1},\rho_{i2},s] -\displaystyle\frac{\lambda}{c_i}\hat{M}_i[\rho_{i2},s] \nonumber\\ &&+\displaystyle\frac{\lambda\theta}{c_i}\Big(\frac{\delta}{c_i}-\rho_{i1}\Big) \hat{N}_i[\rho_{i1},\rho_{i2},s]-\displaystyle\frac{\lambda\theta}{c_i} \hat{N}_i[\rho_{i2},s]\bigg). \end{eqnarray} (4.16)

将(4.16)式代入(4.11)式并经简单计算得

\begin{eqnarray} \hat{Q}_i(s)&=&\hat{Q}_i(s)\cdot T_sT_{\rho_{i1}}T_{\rho_{i2}}p_i(0) +\displaystyle\frac{\lambda}{c_i}\Big(\displaystyle\frac{2\lambda+\delta}{c_i} -\rho_{i1}\Big)\hat{M}_i[\rho_{i1},\rho_{i2},s] -\displaystyle\frac{\lambda}{c_i}\hat{M}_i[\rho_{i2},s] \nonumber\\ &&+\displaystyle\frac{\lambda\theta}{c_i}\Big(\displaystyle\frac{\delta}{c_i} -\rho_{i1}\Big)\hat{N}_i[\rho_{i1},\rho_{i2},s] -\displaystyle\frac{\lambda\theta}{c_i}\hat{N}_i[\rho_{i2},s]. \end{eqnarray} (4.17)

令$f(x)$为一实值可积函数,Gerber和Shiu[11] 证明了$\hat{f}(r_1,r_2,\cdots,r_n,s)$的拉普拉斯逆变换满足

\begin{equation} {\cal L}^{-1}\big(\hat{f}(r_1,r_2,\cdots,r_n,s)\big)=(-1)^n\bigg(\prod\limits_{i=1}^{n}T_{r_{i}}f(x)\bigg). \end{equation} (4.18)

因此,根据(4.18)式,对(4.17)式应用拉普拉斯逆变换得

\begin{eqnarray} Q_i(y)&=&\displaystyle\int_0^yQ_i(y-x)T_{\rho_{i1}}T_{\rho_{i2}}p_i(x){\rm d}x +\displaystyle\frac{\lambda}{c_i}\Big(\displaystyle\frac{2\lambda+\delta}{c_i}-\rho_{i1}\Big)T_{\rho_{i1}}T_{\rho_{i2}}M_i(y) +\displaystyle\frac{\lambda}{c_i}T_{\rho_{i2}}M_i(y) \nonumber\\ &&+\displaystyle\frac{\lambda\theta}{c_i}\Big(\displaystyle\frac{\delta}{c_i}-\rho_{i1}\Big)T_{\rho_{i1}}T_{\rho_{i2}}N_i(y) +\displaystyle\frac{\lambda\theta}{c_i}T_{\rho_{i2}}N_i(y). \end{eqnarray} (4.19)

最后,通过(4.19)式可得定理4.1. 由文献[2]可知$k_i<1$且$g_i(x)$是一个适定概率密度函数. 因此,(4.5)式是一个瑕疵更新方程.

4.2 齐次积分微分方程

该小节讨论了齐次积分微分方程(4.4)的解$v_{i1}(u)$和$v_{i2}(u)$. 我们将利用拉普拉斯变换的方法给出齐次积分微分方程(4.4)的两个线性无关的解. 首先,假定初始条件$v_{ij}^{(k)}(b_{i-1})={I}(k=j-1),\;k=0,1$成立. 容易验证,在该初始条件下,$v_{i1}(u)$和$v_{i2}(u)$是线性无关的.作变量替换$y=u-b_{i-1}$,$\chi_i(y)=v_i(y+b_{i-1})=v_i(u)$,则(4.4)式变为

\begin{eqnarray} A_i({\cal D})\chi_i(y)&=&\displaystyle\frac{\lambda}{c_i}\Big(\displaystyle\frac{2\lambda+\delta}{c_i}{\cal I}-{\cal D}\Big)\displaystyle\int_0^y\chi_i(u-x)f(x){\rm d}x \nonumber\\ &&+\displaystyle\frac{\lambda\theta}{c_i}\Big(\displaystyle\frac{\delta}{c_i}{\cal I}-{\cal D}\Big)\displaystyle\int_0^y\chi_i(u-x)h(x){\rm d}x. \end{eqnarray} (4.20)

对(4.20)式两边同取拉普拉斯变换并令$C_i(s)=(\frac{2\lambda+\delta}{c_i}-s)(\frac{\lambda+\delta}{c_i}-s)=\sum\limits_{j=0}^{2}C_{ij}s^j.$ 当$\rho_{i1}$和$\rho_{i2}$不同时, Li和Garrido[12]证明了

$$\chi_i(y)=\displaystyle\int_0^y\chi_i(u-x)g_i(x){\rm d}x+\sum\limits_{j=1}^{2}k_{ij}{\rm e}^{\rho_{ij}y},$$

其中

$$k_{ij}=-\displaystyle\frac{\sum\limits_{m=0}^{1}\chi^{(m)}_i(0)\sum\limits_{k=m+1}^{2}C_{ik}\rho_{ij}^{k-m-1}}{\prod\limits_{k=1,k\neq j}^{2}(\rho_{ik}-\rho_{ij})}.$$

因此,有

\begin{equation} v_i(u)=\displaystyle\int_0^{u-b_{i-1}}v_i(u-x)g_i(x){\rm d}x+\sum\limits_{j=1}^{2}k_{ij}{\rm e}^{\rho_{ij}(u-b_{i-1})}. \end{equation} (4.21)

现在考虑拉普拉斯变换$\hat{f}$和$\hat{h}$属于有理族的情形,即$\hat{f}(s)=\frac{q_{k-1}(s)}{p_{k}(s)}$,$\hat{h}(s)=\frac{q_{l-1}(s)}{p_{l}(s)}$,其中,$k,l\in{\mathbb N^+},$ $p_{k}(s)$和$p_{l}(s)$是最高次项系数为1且最高次分别为$k$和$l$的两个多项式,它们在右半复平面内没有零点. $q_{k-1}(s)$和$q_{l-1}(s)$分别表示最高次为$k-1$和$l-1$的多项式,且满足 $q_{k-1}(0)=p_k(0),$ $q_{l-1}(0)=p_l(0).$

由(4.1)式易知$\rho_{i1}$和$\rho_{i2}$是方程

\begin{equation} C_i(s)p_{k}(s)p_{l}(s) -\bigg(\frac{\lambda}{c_i}\Big(\frac{2\lambda+\delta}{c_i}-s\Big)q_{k-1}(s)p_{l}(s) +\frac{\lambda\theta}{c_i}\Big(\frac{\delta}{c_i}-s\Big)q_{l-1}(s)p_{k}(s)\bigg)=0 \end{equation} (4.22)

的两个根. 由于上式左边是一个最高次项系数为1的$k+l+2$次的多项式, 所以方程(4.22)在左半复平面内必有$k+l$个根,记为$-R_{i1},\cdots,-R_{i k+l}.$ 由文献[12]知

\begin{equation} \hat{\chi}_i(s)=\displaystyle\frac{d_i(s)p_{k}(s)p_{l}(s)} {\Big[\prod\limits_{j=1}^{2}(s-\rho_{ij})\Big]\Big[\prod\limits_{j=1}^{k+l}(s+R_{ij})\Big]}, \end{equation} (4.23)

其中,$d_i(s)=\sum\limits_{j=0}^{1}s^j\sum\limits_{k=j+1}^{2}C_{ik}\chi_i^{(k-j-1)}(0)$是一个1次多项式. 进一步,如果$\rho_{i1},\rho_{i2},R_{i1},\cdots,R_{i k+l}$ 互不相同时,由部分分式展开得

\begin{equation} \hat{\chi}_i(s)=\sum\limits_{j=1}^{2}\displaystyle\frac{a_{ij}}{s-\rho_{ij}}+\sum\limits_{j=1}^{k+l}\displaystyle\frac{b_{ij}}{s+R_{ij}}, \end{equation} (4.24)

其中

$$a_{ij}=-\displaystyle\frac{d_i(\rho_{ij})p_k(\rho_{ij})p_l(\rho_{ij})} {\Big[\prod\limits_{m=1}^{k+l}(R_{im}+\rho_{ij})\Big] \Big[\prod\limits_{n=1,n\neq j}^{2}(\rho_{in}-\rho_{ij})\Big]},$$

$$b_{ij}=\displaystyle\frac{d_i(-R_{ij})p_k(-R_{ij})p_l(-R_{ij})} {\Big[\prod\limits_{m=1}^{2}(R_{ij}+\rho_{im})\Big] \Big[\prod\limits_{n=1,n\neq j}^{k+l}(R_{in}-R_{ij})\Big]}.$$

对(4.23)式进行拉普拉斯逆变换得

$$\chi_i(y)=\sum\limits_{j=1}^{2}a_{ij}{\rm e}^{\rho_{ij}y}+\sum\limits_{j=1}^{k+l}b_{ij}{\rm e}^{-R_{ij}y},\;\;\;\;y\geq 0,$$

因此,有

\begin{equation} v_i(u)=\sum\limits_{j=1}^{2}a_{ij}{\rm e}^{\rho_{ij}(u-b_{i-1})}+\sum\limits_{j=1}^{k+l}b_{ij}{\rm e}^{-R_{ij}(u-b_{i-1})},\;\;\;\;u\geq b_{i-1}. \end{equation} (4.25)

(4.25)式给出了齐次积分微分方程(4.4)的解的一般表达式. 尽管根$\rho_{i1}$, $\rho_{i2}$,$-R_{i1}$,$\cdots$, $-R_{i k+l}$可能会出现复数,但若出现,则它们必然以成对的共轭复数形式出现, 所以(4.25)式中的$v_i(u)$必是一个实值函数. 由该表达式再结合初始条件$v_{ij}^{(k)}(b_{i-1})={I}(k=j-1),\;k=0,1,$ 我们可得齐次积分微分方程(4.4)的两个线性无关的解$v_{i1}(u)$和$v_{i2}(u)$.

5 Gerber-Shiu函数的计算方法

该小节将给出一个计算Gerber-Shiu函数的方法. 利用上一节的结果和文献[13] 中关于瑕疵更新方程的结果,我们推导Gerber-Shiu函数的显式表达式.

首先,当$i=1,2,\cdots,N$时,定义下面的复合几何分布

\begin{equation} K_i(x)=1-\bar{K}_i(x)=1-\sum\limits_{n=1}^{\infty}{(1-k_i)k_i^n\overline{G^{*n}}_i(x)},\;\;x\geq 0, \end{equation} (5.1)

其中$\overline{G^{*n}}_i(x)$是$g_i(x)$的自身$n$-重卷积所对应的尾分布函数.

下面考虑瑕疵更新方程(4.5),作变换$y=u-b_{i-1},$ $Q_i(y)=\Phi_i(y+b_{i-1})=\Phi_i(u),$ 则(4.5)式可写为

\begin{equation} Q_i(y)=k_i\displaystyle\int_0^yQ_i(y-x){\rm d}G_i(x)+k_iP_i(y+b_{i-1}), \end{equation} (5.2)

其中

\begin{equation} P_i(y+b_{i-1})=P_i(u)=\displaystyle\int_{u-b_{i-1}}^u\Phi(u-x){\rm d}G_i(x)+\eta_i(u)/k_i. \end{equation} (5.3)

由文献[13,定理2.1]知瑕疵更新方程(5.2)的解可表示为

$$Q_i(y)=\frac{k_i}{1-k_i}\displaystyle\int_0^yP_i(y+b_{i-1}-x){\rm d}K_i(x)+k_iP_i(y+b_{i-1}). $$

因此,有

\begin{equation} \Phi_i(u)=\frac{k_i}{1-k_i}\displaystyle\int_0^{u-b_{i-1}}P_i(u-x){\rm d}K_i(x)+k_iP_i(u). \end{equation} (5.4)

将(5.3)式代入(5.4)式得

\begin{equation} \Phi_i(u)=\displaystyle\int_0^{b_{i-1}}\Phi(y)\varphi_i(u,y){\rm d}y+\zeta_i(u), \end{equation} (5.5)

其中

$$\varphi_i(u,y)=\displaystyle\frac{k_i}{1-k_i}\displaystyle\int_0^{u-b_{i-1}}g_i(u-x-y){\rm d}K_i(x)+k_ig_i(u-y),$$

$$\zeta_i(u)=\displaystyle\frac{1}{1-k_i}\displaystyle\int_0^{u-b_{i-1}}\eta_i(u-x){\rm d}K_i(x)+\eta_i(u).$$

公式(5.5)给出了一个递推计算函数$\Phi_1(u),$ $\Phi_2(u),$ $\cdots,$ $\Phi_N(u)$的方法,其中初始值$\Phi_1(u)=\zeta_1(u).$ 然而,未知系数$r_{i1},$ $r_{i2}$在(5.5)式中并没有显式地表示出来,这样不方便我们确定这些未知系数. 为此,下面给出了$\Phi_1(u),$ $\Phi_2(u),$ $\cdots,$ $\Phi_N(u)$的另外一种表示方法.

定理 5.1 当$i=1,2,\cdots,N$和Re$(s)\geq 0$时,给定条件$\hat{\Phi}_i(s)<\infty,$ 则当$n=1,2,\cdots,N$时,$\Phi_n(u)$的解可表示为

\begin{equation} \Phi_n(u)=L_n(u)+\sum\limits_{i=1}^{n-1}\sum\limits_{j=1}^{2}r_{ij}L_{n,i,j}(u), \end{equation} (5.6)

其中

\begin{equation} L_n(u)=\zeta_n(u)+\sum\limits_{i=1}^{n-1}\displaystyle\int_{b_{i-1}}^{b_i}L_i(y)\varphi_n(u,y){\rm d}y, \end{equation} (5.7)
\begin{equation} L_{n,i,j}(u)=\sum\limits_{l=i+1}^{n-1}\displaystyle\int_{b_{l-1}}^{b_l}L_{l,i,j}(y)\varphi_n(u,y){\rm d}y+\int_{b_{i-1}}^{b_i}v_{ij}(y)\varphi_n(u,y){\rm d}y. \end{equation} (5.8)

证 参见文献[6,定理3].

接下来,我们来确定未知系数$r_{ij},$ $i=1,2,\cdots,N,$ $j=1,2.$ 注意到$r_{N1}=r_{N2}=0.$ 考虑在$u=b_n\;(n=1,2,\cdots,N-1)$处的边界条件(3.6)和(3.7)以及初始条件$v_{ij}^{(k)}(b_{i-1})={I}(k=j-1),\;k=0,1,$ 有下面的递推关系

\begin{eqnarray} r_{n+1,1}&=&L_n(b_n)-L_{n+1}(b_n)+\sum\limits_{i=1}^{n-1}\sum\limits_{j=1}^{2}r_{ij}(L_{n,i,j}(b_n)-L_{n+1,i,j}(b_n)) \nonumber\\ &&+\sum\limits_{j=1}^{2}r_{nj}(v_{nj}(b_n)-L_{n+1,n,j}(b_n)), &&-L'_{n+1}(b_n)-\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{2}r_{ij}L'_{n+1,i,j}(b_n). \end{eqnarray} (5.9)
\begin{eqnarray} r_{n+1,2}&=&\displaystyle\frac{c_n}{c_{n+1}}\bigg[L'_n(b_n)+\sum\limits_{i=1}^{n-1}\sum\limits_{j=1}^{2}r_{ij}L'_{n,i,j}(b_n) +\sum\limits_{j=1}^{2}r_{nj}v'_{nj}(b_n)\bigg] \nonumber\\ &&-L'_{n+1}(b_n)-\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{2}r_{ij}L'_{n+1,i,j}(b_n). \end{eqnarray} (5.10)

结合(4.3)式和定理5.1,我们可得下面的定理

定理 5.2 当$i=1,2,\cdots,N$和Re$(s)\geq 0$时,给定条件$\hat{\Phi}_i(s)<\infty,$ 则当$n=1,2,\cdots,N$时,期望折扣罚金函数$\Phi(u)$的解可表示为

\begin{equation} \Phi(u)=L_n(u)+\sum\limits_{i=1}^{n-1}\sum\limits_{j=1}^{2}r_{ij}L_{n,i,j}(u)+\sum\limits_{j=1}^{2}r_{nj}v_{nj}(u),\;\;\;b_{n-1}\leq u< b_n, \end{equation} (5.11)

其中系数$r_{ij}$可由(5.9)和(5.10)式确定.

注 5.1 公式(5.11)非常依赖于复合几何分布函数$K_i(x)$. 由文献[13]知, 当$g_i(x)$是混合指数函数或混合Erlang函数时, 我们可以得到$K_i(x)$显式表达式.

6 计算实例

该节给了一个例子来演示上节介绍的计算Gerber-Shiu函数的方法. 假定个别理赔额服从指数分布, 即$f(x)=\alpha {\rm e}^{-\alpha x}$且其拉普拉斯变换为 $\hat{f}(s)=\frac{\alpha}{s+\alpha}.$ 由此可知$h(x)=2\alpha {\rm e}^{-2\alpha x}-\alpha {\rm e}^{-\alpha x},\;\;x\geq0$且$\hat{h}(s)=\frac{2\alpha}{s+2\alpha}-\frac{\alpha}{s+\alpha}.$ 同时,假定有3个分红层,即$N=3$,$0=b_0< b_1< b_2< b_3=\infty.$

在上面的假定下,广义Lundberg基本方程(4.1)变为

\begin{eqnarray} &&\Big(\frac{2\lambda+\delta}{c_i}-s\Big)\Big(\frac{\lambda+\delta}{c_i}-s\Big) -\frac{\lambda}{c_i}\bigg(\Big(\frac{2\lambda+\delta}{c_i}-s\Big)\frac{\alpha}{s+\alpha} \nonumber\\ && +\theta\Big(\frac{\delta}{c_i}-s\Big)\Big(\frac{2\alpha}{s+2\alpha} -\frac{\alpha}{s+\alpha}\Big)\bigg)=0,~~i=1,2,3. \end{eqnarray} (6.1)

该方程有5个根,

$$\rho_{i1}>0,~~ \rho_{i2}>0,~~-R_{i1}<0,~~-R_{i2}<0,~~-R_{i3}<0. $$

对于$i=1,2,$ 令$v_{i1}(u)$和$v_{i2}(u)$ 是满足下面初值条件的齐次积分微分方程(4.4)的解

$$v_{i1}(b_{i-1})=1, ~~v'_{b_{i-1}}=0,~~v_{i2}(b_{i-1})=0,~~v'_{i2}(b_{i-1})=1.$$

由公式(4.25)可得,当$u\geq b_{i-1}$时

\begin{eqnarray} v_{i1}(u)&=&\displaystyle\frac{[c_i\rho_{i1}-(3\lambda+2\delta)](\rho_{i1}+\alpha)^2(\rho_{i1}+2\alpha)}{c_i(R_{i1}+\rho_{i1})(R_{i2}+\rho_{i1})(R_{i3}+\rho_{i1})(\rho_{i1}-\rho_{i2})}{\rm e}^{\rho_{i1}(u-b_{i-1})} \nonumber\\ &&+\displaystyle\frac{[c_i\rho_{i2}-(3\lambda+2\delta)](\rho_{i2}+\alpha)^2(\rho_{i2}+2\alpha)}{c_i(R_{i1}+\rho_{i2})(R_{i2}+\rho_{i2})(R_{i3}+\rho_{i2})(\rho_{i2}-\rho_{i1})}{\rm e}^{\rho_{i2}(u-b_{i-1})} \nonumber\\ &&-\displaystyle\frac{(c_iR_{i1}+3\lambda+2\delta)(\alpha-R_{i1})^2(2\alpha-R_{i1})}{c_i(R_{i1}+\rho_{i1})(R_{i1}+\rho_{i2})(R_{i2}-R_{i1})(R_{i3}-R_{i1})}{\rm e}^{-R_{i1}(u-b_{i-1})} \nonumber\\ &&-\displaystyle\frac{(c_iR_{i2}+3\lambda+2\delta)(\alpha-R_{i2})^2(2\alpha-R_{i2})}{c_i(R_{i2}+\rho_{i1})(R_{i2}+\rho_{i2})(R_{i1}-R_{i2})(R_{i3}-R_{i2})}{\rm e}^{-R_{i2}(u-b_{i-1})} \nonumber\\ &&-\displaystyle\frac{(c_iR_{i3}+3\lambda+2\delta)(\alpha-R_{i3})^2(2\alpha-R_{i3})}{c_i(R_{i3}+\rho_{i1})(R_{i3}+\rho_{i2})(R_{i1}-R_{i3})(R_{i2}-R_{i3})}{\rm e}^{-R_{i3}(u-b_{i-1})}, \end{eqnarray} (6.2)
\begin{eqnarray} v_{i2}(u)&=&\displaystyle\frac{(\rho_{i1}+\alpha)^2(\rho_{i1}+2\alpha)}{(R_{i1}+\rho_{i1})(R_{i2}+\rho_{i1})(R_{i3}+\rho_{i1})(\rho_{i1}-\rho_{i2})}{\rm e}^{\rho_{i1}(u-b_{i-1})} \nonumber\\ &&+\displaystyle\frac{(\rho_{i2}+\alpha)^2(\rho_{i2}+2\alpha)}{(R_{i1}+\rho_{i2})(R_{i2}+\rho_{i2})(R_{i3}+\rho_{i2})(\rho_{i2}-\rho_{i1})}{\rm e}^{\rho_{i2}(u-b_{i-1})} \nonumber\\ &&+\displaystyle\frac{(\alpha-R_{i1})^2(2\alpha-R_{i1})}{(R_{i1}+\rho_{i1})(R_{i1}+\rho_{i2})(R_{i2}-R_{i1})(R_{i3}-R_{i1})}{\rm e}^{-R_{i1}(u-b_{i-1})} \nonumber\\ &&+\displaystyle\frac{(\alpha-R_{i2})^2(2\alpha-R_{i2})}{(R_{i2}+\rho_{i1})(R_{i2}+\rho_{i2})(R_{i1}-R_{i2})(R_{i3}-R_{i2})}{\rm e}^{-R_{i2}(u-b_{i-1})} \nonumber\\ &&+\displaystyle\frac{(\alpha-R_{i3})^2(2\alpha-R_{i3})}{(R_{i3}+\rho_{i1})(R_{i3}+\rho_{i2})(R_{i1}-R_{i3})(R_{i2}-R_{i3})}{\rm e}^{-R_{i3}(u-b_{i-1})}. \end{eqnarray} (6.3)

通过细心计算有

$$k_i=\displaystyle\frac{\frac{\lambda}{c_i}\big(\frac{2\lambda}{c_i}+(1-\theta)(\frac{\delta}{c_i}+\alpha)\big)}{(\rho_{i1}+\alpha)(\rho_{i2}+\alpha)} +\displaystyle\frac{\frac{\lambda\theta}{c_i}\big(\frac{\delta}{c_i}+2\alpha\big)}{(\rho_{i1}+2\alpha)(\rho_{i2}+2\alpha)}, $$

$$g_i(x)=\displaystyle\frac{\frac{\lambda}{c_i}\big(\frac{2\lambda}{c_i}+(1-\theta)(\frac{\delta}{c_i}+\alpha)\big)}{k_i(\rho_{i1}+\alpha)(\rho_{i2}+\alpha)}\alpha {\rm e}^{-\alpha x} +\displaystyle\frac{\frac{\lambda\theta}{c_i}\big(\frac{\delta}{c_i}+2\alpha\big)}{k_i(\rho_{i1}+2\alpha)(\rho_{i2}+2\alpha)}2\alpha {\rm e}^{-2\alpha x},$$

$$\eta_i(x)=\displaystyle\frac{\frac{\lambda}{c_i}\big(\frac{2\lambda}{c_i}+(1-\theta)(\frac{\delta}{c_i}+\alpha)\big)}{(\rho_{i1}+\alpha)(\rho_{i2}+\alpha)} {\rm e}^{-\alpha x} +\displaystyle\frac{\frac{\lambda\theta}{c_i}\big(\frac{\delta}{c_i}+2\alpha\big)}{(\rho_{i1}+2\alpha)(\rho_{i2}+2\alpha)}{\rm e}^{-2\alpha x}. $$

由文献[13,(4.1)式]知$\bar{K}_i(x)$的表达式为

$$\bar{K}_i(x)=A_{i1}{\rm e}^{-S_{i1}x}+A_{i2}{\rm e}^{-S_{i2}x},\;\;\;\;x\geq0,$$

其中$S_{i1}$ 和 $S_{i2}$ 是下面等式的根

$$\displaystyle\frac{\frac{\lambda}{c_i}\big(\frac{2\lambda}{c_i}+(1-\theta)(\frac{\delta}{c_i}+\alpha)\big)}{(\rho_{i1}+\alpha)(\rho_{i2}+\alpha)}\frac{\alpha}{\alpha-S} +\displaystyle\frac{\frac{\lambda\theta}{c_i}\big(\frac{\delta}{c_i}+2\alpha\big)}{(\rho_{i1}+2\alpha)(\rho_{i2}+2\alpha)}\frac{2\alpha}{2\alpha-S}=1,$$

且

$$A_{i1}=\displaystyle\frac{\displaystyle\frac{\frac{\lambda}{c_i}\big(\frac{2\lambda}{c_i}+(1-\theta)(\frac{\delta}{c_i}+\alpha)\big)}{(\rho_{i1}+\alpha)(\rho_{i2}+\alpha)}\frac{1}{\alpha-S_{i1}} +\displaystyle\frac{\frac{\lambda\theta}{c_i}\big(\frac{\delta}{c_i}+2\alpha\big)}{(\rho_{i1}+2\alpha)(\rho_{i2}+2\alpha)}\frac{1}{2\alpha-S_{i1}}} {\displaystyle\frac{\frac{\lambda}{c_i}\big(\frac{2\lambda}{c_i}+(1-\theta)(\frac{\delta}{c_i}+\alpha)\big)}{(\rho_{i1}+\alpha)(\rho_{i2}+\alpha)}\frac{\alpha}{(\alpha-S_{i1})^2} +\displaystyle\frac{\frac{\lambda\theta}{c_i}\big(\frac{\delta}{c_i}+2\alpha\big)}{(\rho_{i1}+2\alpha)(\rho_{i2}+2\alpha)}\frac{2\alpha}{(2\alpha-S_{i1})^2}}, $$

$$A_{i2}=\displaystyle\frac{\displaystyle\frac{\frac{\lambda}{c_i}\big(\frac{2\lambda}{c_i}+(1-\theta)(\frac{\delta}{c_i}+\alpha)\big)}{(\rho_{i1}+\alpha)(\rho_{i2}+\alpha)}\frac{1}{\alpha-S_{i2}} +\displaystyle\frac{\frac{\lambda\theta}{c_i}\big(\frac{\delta}{c_i}+2\alpha\big)}{(\rho_{i1}+2\alpha)(\rho_{i2}+2\alpha)}\frac{1}{2\alpha-S_{i2}}} {\displaystyle\frac{\frac{\lambda}{c_i}\big(\frac{2\lambda}{c_i}+(1-\theta)(\frac{\delta}{c_i}+\alpha)\big)}{(\rho_{i1}+\alpha)(\rho_{i2}+\alpha)}\frac{\alpha}{(\alpha-S_{i2})^2} +\displaystyle\frac{\frac{\lambda\theta}{c_i}\big(\frac{\delta}{c_i}+2\alpha\big)}{(\rho_{i1}+2\alpha)(\rho_{i2}+2\alpha)}\frac{2\alpha}{(2\alpha-S_{i2})^2}}.$$

例 1 (破产概率) 本例中,令$\delta=0$和$w(x,y)\equiv1$, 则Gerber-Shiu函数$\Phi(u)$变为破产概率$\psi(u)$. 令

$$\lambda=1,~~\alpha=1.01,~~c_1=1.6,~~c_2=1.4,~~c_3=1.2,~~ b_1=5,~~b_2=10. $$

在该假设下,应用上节的计算过程(使用Matlab), 我们得到了$\psi(u)$的具体值,下面给出了$\psi(u) $ 对不同的相依参数$\theta=1,0.5,-0.5,-1$下的具体值.

$\bullet $ 当$\theta=1$时,有

$$ \psi(u)=\left\{ \begin{array}{l} 9.3652\cdot10^{-17}{\rm e}^{-1.0100u}+1.0880\cdot10^{-16}{\rm e}^{-2.0200u}-4.1067\cdot10^{-6}{\rm e}^{1.0995u}\\ ~~~+0.0254{\rm e}^{0.0158u}0.5194{\rm e}^{-0.5129u}+0.0405{\rm e}^{-1.7509u}\\ ~~~-0.0128{\rm e}^{-0.5208u}-0.0011{\rm e}^{-1.7495u},\;\;\;\;\;\;\;0\leq u<5,\\ 8.9050\cdot10^{-15}{\rm e}^{-1.0100u}-7.6156\cdot10^{-13}{\rm e}^{-2.0200u}-5.3591\cdot10^{-10}{\rm e}^{1.2495u}\\ ~~~+0.0053{\rm e}^{0.0228u}+0.4796{\rm e}^{-0.4049u}-0.2536{\rm e}^{-1.7471u}\\ ~~~-0.0282{\rm e}^{-0.4139u}-1.1449{\rm e}^{-1.7456u},\;\;\;\;\;\;5\leq u<10,\\ -1.3040\cdot10^{-9}{\rm e}^{-2.0200u}-1.5571\cdot10^{-13}{\rm e}^{-1.0100u}\\ ~~~+0.2045{\rm e}^{-0.2651u}-1.3153\cdot10^{3} {\rm e}^{-1.7462u},\;\;\;u\geq10. \end{array}\right. $$

$\bullet $ 当$\theta=0.5$时,有

$$\psi(u)=\left\{ \begin{array}{l} 1.0635\cdot10^{-16}{\rm e}^{-1.0100u}-1.7575\cdot10^{-16}{\rm e}^{-2.0200u}-1.4292\cdot10^{-6}{\rm e}^{1.1753u}\\ ~~~+0.0354{\rm e}^{0.0157u}-0.0196{\rm e}^{-0.4523u}-6.1385\cdot10^{-4}{\rm e}^{-1.8937u}\\ ~~~+0.5696{\rm e}^{-0.4455u}+0.0168{\rm e}^{-1.8943u},\;\;\;\;\;\;\;0\leq u<5,\\ 0.0089{\rm e}^{0.0226u}-3.8038\cdot10^{-15}{\rm e}^{-1.0100u}-1.3440\cdot10^{-10}{\rm e}^{1.3403u}\\ ~~~-3.9251\cdot10^{-13}{\rm e}^{-2.0200u}-0.0382{\rm e}^{-0.3592u}-1.7013{\rm e}^{-1.8909u}\\ ~~~+0.5429{\rm e}^{-0.3515u}+0.3516{\rm e}^{-1.8915u},\;\;\;\;\;\;5\leq u<10,\\ 7.4345\cdot10^{-9}{\rm e}^{-2.0200u}-4.6601\cdot10^{-13}{\rm e}^{-1.0100u}\\ ~~~+0.2665{\rm e}^{-0.2322u}-3.4104\cdot10^{3}{\rm e}^{-1.8898u},\;\;\;u\geq10. \end{array}\right. $$

$\bullet $ 当$\theta=-0.5$时,有

$$\psi(u)=\left\{ \begin{array}{l} 1.2927\cdot10^{-16}{\rm e}^{-2.0200u}-1.4948\cdot10^{-16}{\rm e}^{-1.0100u}+5.5384\cdot10^{-7}{\rm e}^{1.3233u}\\ ~~~+0.0559{\rm e}^{0.0155u}-0.0350{\rm e}^{-0.3599u}+7.2575\cdot10^{-4}{\rm e}^{-2.1340u}\\ ~~~+0.6419{\rm e}^{-0.3543u}-0.0125{\rm e}^{-2.1336u},\;\;\;\;\;\;\;\;0\leq u<5,\\ 2.1525\cdot10^{-12}{\rm e}^{-2.0200u}-8.3096\cdot10^{-15}{\rm e}^{-1.0100u}+3.0010\cdot10^{-11}{\rm e}^{1.5142u}\\ ~~~+0.0174{\rm e}^{0.0223u}-0.0565{\rm e}^{-0.2860u}+8.9402{\rm e}^{-2.1376u}\\ ~~~+0.6404{\rm e}^{-0.2797u}-4.9875{\rm e}^{-2.1372u},\;\;\;\;\;\;\;5\leq u<10,\\ 5.1135\cdot10^{4}{\rm e}^{-2.1402u}-2.1233\cdot10^{-12}{\rm e}^{-1.0100u}\\ ~~~+0.3779{\rm e}^{-0.1880u}+1.6776\cdot10^{-8}{\rm e}^{-2.0200u},\;\;\;u\geq10. \end{array}\right. $$

$\bullet $ 当$\theta=-1$时,有

$$\psi(u)=\left\{ \begin{array}{l} 7.2619\cdot10^{-7}{\rm e}^{1.3952u}-4.7929\cdot10^{-17}{\rm e}^{-1.0100u}-1.1166\cdot10^{-16}{\rm e}^{-2.0200u}\\ ~~~+0.0657{\rm e}^{0.0155u}-0.0429{\rm e}^{-0.3268u}+0.0015{\rm e}^{-2.2388u}\\ ~~~+0.6691{\rm e}^{-0.3218u}-0.0221{\rm e}^{-2.2382u}\;\;\;\;\;\;\;\;\;\;0\leq u<5,\\ 1.8042\cdot10^{-12}{\rm e}^{-2.0200u}-6.0132\cdot10^{-15}{\rm e}^{-1.0100u} +2.6466\cdot10^{-11}{\rm e}^{1.5973u}\\ ~~~+0.0220{\rm e}^{0.0220u}-0.0644{\rm e}^{-0.2601u}+34.9355{\rm e}^{-2.2464u}\\ ~~~+0.6780{\rm e}^{-0.2543u}-22.4101{\rm e}^{-2.24570u},\;\;\;\;\;\;5\leq u<10,\\ 1.5468\cdot10^{-12}{\rm e}^{-1.0100u}+3.2705\cdot10^{5}{\rm e}^{-2.2526u}\\ ~~~-3.2148\cdot10^{-8}{\rm e}^{-2.0200u}+0.4263{\rm e}^{-0.1723u},\;\;\;u\geq10. \end{array}\right. $$

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