基于广义Erlang(2)分布和随机收入风险模型的破产概率

Ruin Probability Based on a Generalized Erlang(2) Distribution and Stochastic Income Risk Model

摘要: 在本研究中，我们构建了一个包含随机收入的风险模型。在此模型中，索赔间隔时间假定遵循广义Erlang(2)分布，同时保费收入假定服从指数分布。通过数学推导，我们获得了破产概率的拉普拉斯变换以及瑕疵更新方程。此外，在索赔额服从指数分布的条件下，我们推导出了破产概率的显示表达式。

Abstract: In this study, we construct a risk model incorporating stochastic income. In this model, the claim inter-arrival times are assumed to follow a generalized Erlang(2) distribution, while the premium income is assumed to follow an exponential distribution. Through mathematical derivation, we obtain the Laplace transform of the ruin probability as well as a defective renewal equation. Furthermore, under the condition that claim sizes follow an exponential distribution, we derive an explicit expression for the ruin probability.

1. 引言

在保险精算研究领域，众多学者和专家致力于探讨和分析保险公司的破产概率分布问题，例如，在文献[1]中，研究者深入研究了一种具有随机收入特征的风险模型，该模型考虑了索赔规模与索赔区间之间存在的依赖关系，通过研究，他们成功推导出了期望折现惩罚函数及其拉普拉斯变换，并基于此，进一步得到了破产概率的渐进公式。此外，Erlang分布作为排队理论中最常使用的分布之一，它在风险理论中也扮演着重要的角色，研究者们发现，Erlang分布与风险理论之间存在着紧密的联系。例如，在文献[2]中，研究者研究了两个独立的索赔计数过程，分别是具有广义Erlang(2)索赔到达时间的泊松分布和Sparre Andersen过程。同时，他的论文也引用许多关于这方面研究的文献。研究发现许多论文开始探讨如何将经典风险模型的方法和结果与基于Erlang或广义Erlang分布的索赔到达时间的Sparre Andersen风险模型的方法和结果进行有效的结合，在这方面，Li在[3]的研究工作也是一个典型例子，也就是他的研究为这一领域提供了重要的参考，同时，他的研究中也引用了其他相关文献，为后续的研究者提供了丰富的研究资料和参考方向。

在本研究中，我们对一个特定的风险过程模型进行了深入的探讨，该模型基于广义Erlang(2)分布和随机收入风险模型。在研究的第二部分，我们对这一风险模型进行了详尽的阐述和分析。随后，在第三部分，我们特别关注了保费收入呈指数分布的特定情形，并成功推导出了破产概率函数的微分方程。在第四部分，通过严密的数学推导，我们得到了破产概率函数的缺陷更新方程。最终，在第五部分，针对索赔金额服从指数分布的情况，我们进一步推导出了破产概率的显式表达式。这些成果为理解风险过程中的破产风险提供了坚实的理论基础。

2. 模型结构

假设索赔额${\left(Y_i\right)}_{i\ge 1}$ 为一个非负独立同分布(i.i.d.)的随机变量序列，其分布函数为$Q\left(y\right)$ ，概率密度函数为$q\left(y\right)$ ，均值为${\mu }_{\text{Y}}$ ，索赔次数过程$N_2\left(t\right)$ ，索赔间隔时间${\left(V_i\right)}_{i\ge 1}$ 为广义的Erlang(2)分布，即两个独立随机变量和$V_i=L_{i1}+L_{i2}$ ，其中$L_{i1}$ 是独立同分布的参数为${\lambda }_1$ 的指数分布，其中$L_{i2}$ 是独立同分布的参数为${\lambda }_2$ 的指数分布。

保费收入${\left(X_i\right)}_{i\ge 1}$ 为非负i.i.d.的随机变量序列，其分布函数为$P\left(y\right)$ ，概率密度函数为$p\left(y\right)$ ，均值为${\mu }_X$ ，保费收入到达次数$N_1\left(t\right)$ 是强度参数为$\lambda ＞0$ 的泊松过程并且与${\left(X_i\right)}_{i\ge 1}$ 相互独立，则保费收入间隔时间${\left(W_i\right)}_{i\ge 1}$ 为i.i.d.的参数为$\lambda$ 的指数分布。此外，假设在$\left[0,t\right]$ 上的总索赔额${\displaystyle \sum _{i=1}^{N_2\left(t\right)}{Y}_i}$ 和总保费${\displaystyle \sum _{i=1}^{N_1\left(t\right)}{X}_i}$ 相互独立，如果$N_1\left(t\right)=0$ 那么${\displaystyle \sum _{i=1}^0{X}_i}=0$ ，对于${\displaystyle \sum _{i=1}^{N_2\left(t\right)}{Y}_i}$ 有类似的约定，用$U\left(t\right)$ 表示保险公司的非负初始盈余，则保险公司的盈余过程$U\left(t\right)$ 遵循以下方程

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

定义破产时间为$T=\inf \left\{T\ge 0:U\left(T\right)<0\right\}$ ，(如果集合为空则$T=\infty$ )，最终破产概率 $\psi \left(u\right)=P\left(T<\infty |U\left(0\right)=u\right)$ ，$u\ge 0$ 。$\psi \left(u\right)$ 表示索赔在状态1时的破产概率，即没有发生真正索赔时的破产概率，${\psi }_1\left(u\right)$ 表示索赔在状态2时的破产概率，即发生了真正索赔的破产概率。

为确保证破产不是必然事件，假设安全负载条件$\lambda {\mu }_X>\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1+{\lambda }_2}{\mu }_Y$ 成立。在下文中，我们约定函数$\psi \left(x\right)$ 的拉普拉斯变换记为$\tilde{\psi }\left(s\right)$ ，为便于全文统一，本文采用符号“~”来标识函数的拉普拉斯变换。

3. 破产概率的积分–微分方程

因为服从广义Erlang(2)分布的随机变量可以分解为2个相互独立的服从参数为${\lambda }_1$ ，${\lambda }_2$ 的指数随机变量之和。因此，在风险模型(1)中索赔过程可视为：在第一个子索赔中索赔量为0，仅仅在第二次子索赔产生了实际的索赔，索赔量分布函数为$Q$ 。

在充分小的时间区间$\left(0,\epsilon \right)$ 内，第一类索赔可以早于第一次保费，也可以在第一次索赔之后，设$M=W_1\wedge L_{11}$ (即$M$ 取$W_1,L_{11}$ 的最小)，对于$u\ge 0$ ，以第一次事件(保费或索赔)的时间为条件，我们可以得到破产概率方程：

$\psi \left(u\right)={\displaystyle {\int }_0^{\infty }P\left(M=dt,M=L_{11}\right){\psi }_1\left(u\right)}+{\displaystyle {\int }_0^{\infty }P\left(M=dt,M=W_1\right)\psi \left(u+x\right)p\left(x\right)\text{d}x}$ , (2)

其中，

$p\left(M=dt,M=L_{11}\right)={\lambda }_1{\text{e}}^{-\left(\lambda +{\lambda }_1\right)t}$ , $p\left(M=dt,M=W_1\right)=\lambda {\text{e}}^{-\left(\lambda +{\lambda }_1\right)t}$ .

使用这些概率，(2)可以改写为

$\psi \left(u\right)=\frac{{\lambda }_1}{\lambda +{\lambda }_1}{\psi }_1\left(u\right)+\frac{\lambda }{\lambda +{\lambda }_1}{\displaystyle {\int }_0^{\infty }\psi \left(u+x\right)p\left(x\right)\text{d}x}$ . (3)

令$A\left(u\right)={\displaystyle {\int }_0^{\infty }\psi \left(u+x\right)p\left(x\right)\text{d}x}$ ，

对(3)进行拉普拉斯变换得到

$\tilde{\psi }\left(s\right)={\displaystyle {\int }_0^{\infty }{\text{e}}^{-su}\psi \left(u\right)\text{d}u}={\displaystyle {\int }_0^{\infty }{\text{e}}^{-su}\left(\frac{{\lambda }_1}{\lambda +{\lambda }_1}{\psi }_1\left(u\right)+\frac{\lambda }{\lambda +{\lambda }_1}A\left(u\right)\right)\text{d}u}=\frac{{\lambda }_1}{\lambda +{\lambda }_1}{\tilde{\psi }}_1\left(s\right)+\frac{\lambda }{\lambda +{\lambda }_1}\tilde{A}\left(s\right)$. (4)

$Z=W_1\wedge L_{12}$ ，通过相似论证，我们得到

$\begin{array}{c}{\psi }_1\left(u\right)={\displaystyle {\int }_0^{\infty }P\left(Z=dt,Z=L_{12}\right)\left({\displaystyle {\int }_0^u\psi \left(u-y\right)q\left(y\right)\text{d}y}+{\displaystyle {\int }_u^{\infty }q\left(y\right)\text{d}y}\right)}\\ +{\displaystyle {\int }_0^{\infty }P\left(Z=dt,Z=W_1\right)}{\displaystyle {\int }_0^{\infty }{\psi }_1\left(u+x\right)p\left(x\right)\text{d}x}\end{array}$ (5)

其中，

$p\left(Z=dt,Z=L_{12}\right)={\lambda }_2{\text{e}}^{-\left(\lambda +{\lambda }_2\right)t}$ , $p\left(Z=dt,Z=W_1\right)=\lambda {\text{e}}^{-\left(\lambda +{\lambda }_2\right)t}$ .

同理，(5)可以改写为

${\psi }_1\left(u\right)=\frac{{\lambda }_2}{\lambda +{\lambda }_2}\left({\displaystyle {\int }_0^u\psi \left(u-y\right)q\left(y\right)\text{d}y}+{\displaystyle {\int }_u^{\infty }q\left(y\right)\text{d}y}\right)+\frac{\lambda }{\lambda +{\lambda }_2}{\displaystyle {\int }_0^{\infty }{\psi }_1}\left(u+x\right)p\left(x\right)\text{d}x$ (6)

令$A_1\left(u\right)={\displaystyle {\int }_0^{\infty }{\psi }_1\left(u+x\right)p\left(x\right)\text{d}x}$ ，$W\left(u\right)={\displaystyle {\int }_u^{\infty }q\left(y\right)\text{d}y}$ 。

对(6)进行拉普拉斯变换得到

$\begin{array}{c}{\tilde{\psi }}_1\left(s\right)={\displaystyle {\int }_0^{\infty }{\text{e}}^{-su}{\psi }_1\left(u\right)\text{d}u}\\ ={\displaystyle {\int }_0^{\infty }{\text{e}}^{-su}\left(\frac{{\lambda }_2}{\lambda +{\lambda }_2}\left({\displaystyle {\int }_0^u\psi \left(u-y\right)q\left(y\right)\text{d}y}+W\left(u\right)\right)+\frac{\lambda }{\lambda +{\lambda }_1}{A}_1\left(u\right)\right)\text{d}u}\\ =\frac{{\lambda }_2}{\lambda +{\lambda }_2}\left(\tilde{\psi }\left(s\right)\tilde{q}\left(s\right)+\tilde{W}\left(s\right)\right)+\frac{\lambda }{\lambda +{\lambda }_2}{\tilde{A}}_1\left(s\right).\end{array}$ (7)

在本节其余部分，我们假设保费服从指数分布$P\left(x\right)=1-{\text{e}}^{-\frac{x}{{\mu }_x}}$ ，${\mu }_{\text{x}}>0$ 。

与[4]中一样，我们定义一个关于复数$r$ 的实值函数$f$ 的算子$T_{r}f$ 为

$T_{r}f\left(x\right)={\displaystyle {\int }_x^{\infty }{\text{e}}^{-r\left(y-x\right)}f\left(y\right)\text{d}y}$ , $x\ge 0$ .

显然，$f$ 的拉普拉斯变换$\tilde{f}\left(s\right)$ 可以表示为$T_{s}f\left(0\right)$ ，对于不同的$r_1$ 和$r_2$ ，

$T_{r_1}{T}_{r_2}f\left(x\right)=T_{r_2}{T}_{r_1}f\left(x\right)=\frac{T_{r_1}f\left(x\right)-T_{r_2}f\left(x\right)}{r_2-r_1}$ , $x\ge 0$ .

若$r_1=r_2=r$ ，

$T_{r_1}{T}_{r_2}f\left(x\right)={\displaystyle {\int }_x^{\infty }\left(y-x\right){\text{e}}^{-r\left(y-x\right)}f\left(y\right)\text{d}y}$ , $x\ge 0$ .

这个运算符的属性可以在[5] [6]中找到。

对于$\mathrm{Re}s\ge \frac{1}{{\mu }_x}$ ，通过改变积分顺序

$\begin{array}{c}\tilde{A}\left(s\right)={\displaystyle {\int }_0^{\infty }{\text{e}}^{-su}{\displaystyle {\int }_x^{\infty }\psi \left(u+x\right)\frac{{\text{e}}^{-\frac{x}{{\mu }_x}}}{{\mu }_x}\text{d}x\text{d}u}}\\ ={\displaystyle {\int }_0^{\infty }{\displaystyle {\int }_x^{\infty }{\text{e}}^{-su}\psi \left(u+x\right)\text{d}u}\frac{{\text{e}}^{-\frac{x}{{\mu }_x}}}{{\mu }_x}\text{d}x\text{d}u}\\ =\frac{1}{{\mu }_x}{\displaystyle {\int }_0^{\infty }{\displaystyle {\int }_x^{\infty }{\text{e}}^{-s\left(u-x\right)}\psi \left(u\right)\text{d}u}{\text{e}}^{-\frac{x}{{\mu }_x}}}\text{d}x,\end{array}$

从算子$T_r$ 的定义和性质来看，上面的方程可以简化为

$\tilde{A}\left(s\right)=\frac{1}{{\mu }_x}{\displaystyle {\int }_0^{\infty }{T}_s\psi \left(x\right){\text{e}}^{-\frac{x}{{\mu }_x}}\text{d}x}=\frac{1}{{\mu }_x}{T}_1{T}_s\psi \left(0\right)=\frac{1}{{\mu }_x}\frac{T_s\psi \left(0\right)-T_{\frac{1}{{\mu }_x}}\psi \left(0\right)}{\frac{1}{{\mu }_x}-s}=\frac{\tilde{\psi }\left(s\right)-\tilde{\psi }\left(\frac{1}{{\mu }_x}\right)}{1-s{\mu }_x},$

相似地我们可以得到${\tilde{A}}_1\left(s\right)=\frac{{\tilde{\psi }}_1\left(s\right)-{\tilde{\psi }}_1\left(\frac{1}{{\mu }_x}\right)}{1-s{\mu }_x}$ 。

下文研究焦点将集中于破产概率函数$\psi \left(u\right)$ 和${\psi }_1\left(u\right)$ 上。他们的拉普拉斯变换推导如下。将上述$\tilde{A}\left(s\right)$ 和${\tilde{A}}_1\left(s\right)$ 简化结果带入(4)，(7)，我们可以得到方程

$\tilde{\psi }\left(s\right)=\frac{\frac{{\lambda }_1}{\lambda +{\lambda }_1}{B}_2\left(s\right)+C_2\left(s\right)B_1\left(s\right)}{\frac{{\lambda }_1}{\lambda +{\lambda }_1}\frac{{\lambda }_2}{\lambda +{\lambda }_2}\tilde{q}\left(s\right)-C_1\left(s\right)C_2\left(s\right)}$ , (8)

${\tilde{\psi }}_1\left(s\right)=\frac{\frac{{\lambda }_2}{\lambda +{\lambda }_2}\tilde{q}\left(s\right)B_1+C_1{B}_2}{\frac{{\lambda }_2}{\lambda +{\lambda }_2}\frac{{\lambda }_1}{\lambda +{\lambda }_1}\tilde{q}\left(s\right)-C_1{C}_2}$ , (9)

其中$C_1\left(s\right)=1-\frac{\lambda }{\lambda +{\lambda }_1}\frac{1}{1-s{\mu }_x}$ ，$C_2\left(s\right)=1-\frac{\lambda }{\lambda +{\lambda }_2}\frac{1}{1-s{\mu }_x}$ ，$B_1\left(s\right)=\frac{\lambda }{\lambda +{\lambda }_1}\frac{1}{1-s{\mu }_x}\tilde{\psi }\left(\frac{1}{{\mu }_x}\right)$ ， $B_2\left(s\right)=\frac{\lambda }{\lambda +{\lambda }_2}\frac{1}{1-s{\mu }_x}{\tilde{\psi }}_1\left(\frac{1}{{\mu }_x}\right)-\frac{{\lambda }_2}{\lambda +{\lambda }_2}\tilde{W}\left(s\right)$ 。

为了得到$\tilde{\psi }\left(s\right)$ 和${\tilde{\psi }}_1\left(s\right)$ ，为了进一步推导破产概率$\psi \left(u\right)$ 和${\psi }_1\left(u\right)$ ，我们只需要得到$\tilde{\psi }\left(\frac{1}{{\mu }_x}\right)$ 和${\tilde{\psi }}_1\left(\frac{1}{{\mu }_x}\right)$ 。

注意，(8)和(9)右边分母相同。现在我们讨论方程分母的根

$\frac{{\lambda }_1}{\lambda +{\lambda }_1}\frac{{\lambda }_2}{\lambda +{\lambda }_2}\tilde{q}\left(s\right)-C_1\left(s\right)C_2\left(s\right)=0$ . (10)

引理1：$\frac{{\lambda }_1}{\lambda +{\lambda }_1}\frac{{\lambda }_2}{\lambda +{\lambda }_2}\tilde{q}\left(s\right)-C_1\left(s\right)C_2\left(s\right)=0$ 在右半复平面上恰好有两个根，记为${\rho }_1$ 和${\rho }_2$ 且$\mathrm{Re}{\rho }_1>0$ ，${\rho }_2=0$ 。

Klimenok定理：设函数$f\left(z\right)$ 和$\varphi \left(z\right)$ 在开盘上是解析的$\left|z\right|<1$ ，并且以下关系成立：

${\left|f\left(z\right)\right|}_{\left|z\right|=1,z\ne 1}>{\left|\varphi \left(z\right)\right|}_{\left|z\right|=1,z\ne 1}$ , $f\left(1\right)=-\varphi \left(1\right)\ne 0$ .

设函数$f\left(z\right)$ 和$\varphi \left(z\right)$ 在$\left|z\right|=1$ 处有导数，且下列不等式成立：$\frac{f^{\prime }\left(1\right)+{\varphi }^{\prime }\left(z\right)}{f\left(1\right)}>0$ 。

则$f\left(z\right)+\varphi \left(z\right)$ 和$f\left(z\right)$ 的零点个数$N_{f+\varphi }$ 和$N_f$ 的关系如下：$N_{f+\varphi }=N_f-1$ 。

我们参照参考文献[7]的推论2进行证明，应用Rouché’s定理的扩展定理Klimenok定理[8]来证明Lundberg广义方程(10)的根的个数。Klimenok定理是Rouché’s定理的一个扩展，Rouché’s定理要求函数$F\left(z\right)=f\left(z\right)+g\left(z\right)$ 中$F\left(z\right)$ 在$\left|z\right|\le 1$ 是解析的，而Klimenok (2001)定理只要求在$\left|z\right|<1$ 上是解析的，$\left|z\right|=1$ 边界上连续即可。本文中我们令$F\left(z\right)=\frac{{\lambda }_1}{\lambda +{\lambda }_1}\frac{{\lambda }_2}{\lambda +{\lambda }_2}\tilde{q}\left(z\right)-C_1\left(z\right)C_2\left(z\right)$ ，其中$z=\frac{k-s}{k}$ ，当$\left|z\right|=1$ 时$F\left(z\right)=0$ ，不满足Rouché’s定理的$\left|z\right|\le 1$ 是解析的条件，所以我们应用Klimenok定理进行证明，证明过程如下：

证明：$\frac{{\lambda }_1}{\lambda +{\lambda }_1}\frac{{\lambda }_2}{\lambda +{\lambda }_2}\tilde{q}\left(s\right)-\left(1-\frac{\lambda }{\lambda +{\lambda }_1}\frac{1}{1-s{\mu }_x}\right)\left(1-\frac{\lambda }{\lambda +{\lambda }_2}\frac{1}{1-s{\mu }_x}\right)=0$ ，

这就等价于

$\left(1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_1}\right)\left(1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_2}\right)-\frac{{\lambda }_1}{\lambda +{\lambda }_1}\frac{{\lambda }_2}{\lambda +{\lambda }_2}{\left(1-s{\mu }_x\right)}^2\tilde{q}\left(s\right)=0$ .

我们定义一个轮廓$D_k=\left\{s:\left|z\right|=1\right\}$ ，其中$z=\frac{k-s}{k}$ 。轮廓$D_k$ 是以$k$ 为半径，$k$ 为圆心的圆，用$s$ 表示。当$k\to \infty$ 时，$z$ 是以原点为圆心，以1为半径的圆。

令$g\left(s\right)=\left(1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_1}\right)\left(1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_2}\right)$ ，$f\left(s\right)=-\frac{{\lambda }_1}{\lambda +{\lambda }_1}\frac{{\lambda }_2}{\lambda +{\lambda }_2}{\left(1-s{\mu }_x\right)}^2\tilde{q}\left(s\right)$ ，则$g\left(z\right)=\left(1-\left(k-kz\right){\mu }_x-\frac{\lambda }{\lambda +{\lambda }_1}\right)\left(1-\left(k-kz\right){\mu }_x-\frac{\lambda }{\lambda +{\lambda }_2}\right)$ ，$f\left(z\right)=-\frac{{\lambda }_1}{\lambda +{\lambda }_1}\frac{{\lambda }_2}{\lambda +{\lambda }_2}{\left(1-\left(k-kz\right){\mu }_x\right)}^2\tilde{q}\left(k-kz\right)$ 。显然，$g\left(z\right)$ ，$f\left(z\right)$ 在$\left|z\right|<1$ 上解析，且在$\left|z\right|=1$ 上连续，下证$g\left(z\right)$ ，$f\left(z\right)$ 满足Klimenok定理。

① 证${\left|g\left(z\right)\right|}_{\left|z\right|=1,z\ne 1}>{\left|f\left(z\right)\right|}_{\left|z\right|=1,z\ne 1}$ 等价于证明$\left|\frac{f\left(s\right)}{g\left(s\right)}\right|<1$ 。

证明：

$\begin{array}{c}{\left(1-s{\mu }_x\right)}^2=\left(1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_1}+\frac{\lambda }{\lambda +{\lambda }_1}\right)\left(1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_2}+\frac{\lambda }{\lambda +{\lambda }_2}\right)\\ =\left(1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_1}\right)\left(1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_2}\right)+\frac{\lambda }{\lambda +{\lambda }_2}\left(1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_1}\right)\\ +\frac{\lambda }{\lambda +{\lambda }_1}\left(1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_2}\right)+\frac{\lambda }{\lambda +{\lambda }_1}\frac{\lambda }{\lambda +{\lambda }_2}\\ =g\left(s\right)+\frac{\lambda }{\lambda +{\lambda }_2}\left(1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_1}\right)+\frac{\lambda }{\lambda +{\lambda }_1}\left(1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_2}\right)+\frac{\lambda }{\lambda +{\lambda }_1}\frac{\lambda }{\lambda +{\lambda }_2},\end{array}$

所以，

$\begin{array}{c}\left|\frac{f\left(s\right)}{g\left(s\right)}\right|=\frac{{\lambda }_1}{\lambda +{\lambda }_1}\frac{{\lambda }_2}{\lambda +{\lambda }_2}|\tilde{q}\left(s\right)(1+\frac{\lambda }{\lambda +{\lambda }_2}\frac{1}{1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_2}}+\frac{\lambda }{\lambda +{\lambda }_1}\frac{1}{1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_1}}\\ +\frac{\lambda }{\lambda +{\lambda }_1}\frac{\lambda }{\lambda +{\lambda }_2}\frac{1}{1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_1}}\frac{1}{1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_2}})|\\ \le \frac{{\lambda }_1}{\lambda +{\lambda }_1}\frac{{\lambda }_2}{\lambda +{\lambda }_2}|1+\frac{\lambda }{\lambda +{\lambda }_2}\frac{1}{1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_2}}+\frac{\lambda }{\lambda +{\lambda }_1}\frac{1}{1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_1}}\\ +\frac{\lambda }{\lambda +{\lambda }_1}\frac{\lambda }{\lambda +{\lambda }_2}\frac{1}{1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_1}}\frac{1}{1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_2}}|.\end{array}$

当$s\to \infty$ 时，

$\frac{1}{1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_1}}\to 0$ , $\frac{1}{1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_2}}\to 0$ , $\frac{1}{1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_1}}\frac{1}{1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_2}}\to 0$ ,

所以$s\to \infty$ 时，$\left|\frac{f\left(s\right)}{g\left(s\right)}\right|\le \frac{{\lambda }_1}{\lambda +{\lambda }_1}\frac{{\lambda }_2}{\lambda +{\lambda }_2}<1$ ，即证得${\left|g\left(z\right)\right|}_{\left|z\right|=1,z\ne 1}>{\left|f\left(z\right)\right|}_{\left|z\right|=1,z\ne 1}$ 。

② 证$z=1,g\left(z\right)=f\left(z\right)$ ，即证$s=0$ 时，$g\left(s\right)=-f\left(s\right)$ 。

证：$g\left(s\right)=\left(1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_1}\right)\left(1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_2}\right)$ ，则$g\left(0\right)=\frac{{\lambda }_1}{\lambda +{\lambda }_1}\frac{{\lambda }_2}{\lambda +{\lambda }_2}$ ， $f\left(s\right)=-\frac{{\lambda }_1}{\lambda +{\lambda }_1}\frac{{\lambda }_2}{\lambda +{\lambda }_2}{\left(1-s{\mu }_x\right)}^2\tilde{q}\left(s\right)$ 则$f\left(0\right)=-\frac{{\lambda }_1}{\lambda +{\lambda }_1}\frac{{\lambda }_2}{\lambda +{\lambda }_2}$ 。

③ 证$z=1,\frac{g^{\prime }\left(z\right)+f^{\prime }\left(z\right)}{g\left(z\right)}>0$ ，

$\begin{array}{l}\frac{\text{d}}{\text{d}z}(\frac{{\lambda }_1}{\lambda +{\lambda }_1}\frac{{\lambda }_2}{\lambda +{\lambda }_2}\tilde{q}\left(s\right)(1+\frac{\lambda }{\lambda +{\lambda }_2}\frac{1}{1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_2}}+\frac{\lambda }{\lambda +{\lambda }_1}\frac{1}{1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_1}}\\ {+\frac{\lambda }{\lambda +{\lambda }_1}\frac{\lambda }{\lambda +{\lambda }_2}\frac{1}{1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_1}}\frac{1}{1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_2}}))|}_{z=1}\\ =k\frac{{\lambda }_1+{\lambda }_2}{{\lambda }_1{\lambda }_2}\left(\lambda {\mu }_x-\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1+{\lambda }_2}{\mu }_y\right)>0\end{array}$

即证$z=1,\frac{g^{\prime }\left(z\right)+f^{\prime }\left(z\right)}{g\left(z\right)}>0$ ，由Klimenok定理知(10)有两个根，记为${\rho }_1$ ，${\rho }_2$ 并且${\rho }_2=0$ 。

4. 破产概率函数的瑕疵更新方程

在本节中，我们通过破产概率的拉普拉斯变换推导，得到了破产概率函数的瑕疵更新方程。

通过计算，式(8)与式(9)可被重新表述为：

$\tilde{\psi }\left(s\right)=\frac{{\tilde{f}}_{1,1}\left(s\right)+{\tilde{f}}_{1,2}\left(s\right)}{{\tilde{h}}_1\left(s\right)-{\tilde{h}}_2\left(s\right)}$ , ${\tilde{\psi }}_1\left(s\right)=\frac{{\tilde{f}}_{2,1}\left(s\right)+{\tilde{f}}_{2,2}\left(s\right)}{{\tilde{h}}_1\left(s\right)-{\tilde{h}}_2\left(s\right)}$ . (11)

其中，

${\tilde{h}}_1\left(s\right)=\frac{{\lambda }_1{\lambda }_2}{\left(\lambda +{\lambda }_1\right)\left(\lambda +{\lambda }_2\right)}{\left(1-s{\mu }_x\right)}^2\tilde{q}\left(s\right)$ ,

${\tilde{h}}_2\left(s\right)={\left(1-s{\mu }_x\right)}^2-\left(1-s{\mu }_x\right)\left(\frac{\lambda }{\lambda +{\lambda }_1}+\frac{\lambda }{\lambda +{\lambda }_2}\right)+\frac{{\lambda }^2}{\left(\lambda +{\lambda }_1\right)\left(\lambda +{\lambda }_2\right)}$ ,

${\tilde{f}}_{1,1}\left(s\right)=\left(1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_2}\right)\frac{\lambda }{\lambda +{\lambda }_1}\tilde{\psi }\left(\frac{1}{{\mu }_x}\right)$ ,

${\tilde{f}}_{1,2}\left(s\right)=\frac{{\lambda }_1\left(1-s{\mu }_x\right)\left(\lambda {\tilde{\psi }}_1\left(\frac{1}{{\mu }_x}\right)-{\lambda }_2\left(1-s{\mu }_x\right)\tilde{W}\left(s\right)\right)}{\left(\lambda +{\lambda }_1\right)\left(\lambda +{\lambda }_2\right)}$ ,

${\tilde{f}}_{2,1}\left(s\right)=\left(1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_1}\right)\frac{\lambda {\tilde{\psi }}_1\left(\frac{1}{{\mu }_x}\right)}{\lambda +{\lambda }_2}$ ,

${\tilde{f}}_{2,2}\left(s\right)=\frac{\lambda }{\lambda +{\lambda }_1}\frac{{\lambda }_2}{\lambda +{\lambda }_2}\tilde{q}\left(s\right)\left(1-s{\mu }_x\right)\tilde{\psi }\left(\frac{1}{{\mu }_x}\right)-\frac{{\lambda }_2}{\lambda +{\lambda }_2}\left(1-s{\mu }_x-\frac{\lambda }{\lambda +{\lambda }_1}\right)\left(1-s{\mu }_x\right)\tilde{W}\left(s\right)$ .

我们利用拉格朗日插值定理对(11)式进行改写，最终得到破产概率函数的瑕疵更新方程。

命题1：破产概率的拉普拉斯变换$\tilde{\psi }\left(s\right)$ ，${\tilde{\psi }}_1\left(s\right)$ 满足

$\tilde{\psi }\left(s\right)=\frac{T_s{T}_{\rho 2}{T}_{\rho 1}{h}_1\left(0\right)}{{\mu }_x^2}\tilde{\psi }\left(s\right)-\frac{T_s{T}_{\rho 2}{T}_{\rho 1}{f}_{1,2}\left(0\right)}{{\mu }_x^2}$ ; ${\tilde{\psi }}_1\left(s\right)=\frac{T_s{T}_{\rho 2}{T}_{\rho 1}{h}_1\left(0\right)}{{\mu }_x^2}{\tilde{\psi }}_1\left(s\right)-\frac{T_s{T}_{\rho 2}{T}_{\rho 1}{f}_{2,2}\left(0\right)}{{\mu }_x^2}$ (12)

证明：对于$\mathrm{Re}s\ge 0$ ，$\psi \left(s\right)$ ，${\psi }_1\left(s\right)$ 解析式，如果$s={\rho }_1$ 和$s={\rho }_2$ ，则(8)和(9)的分子为零。因此，可以得出，对于$i=1,2$ ，$j=1,2$ ，$f_{i,1}\left({\rho }_j\right)=-f_{i,2}\left({\rho }_j\right)$ ，容易看出$f_{i,1}\left(s\right)$ 是$s$ 中的1次多项式，使用拉格朗日插值定理，推导出

${\tilde{f}}_{i,1}\left(s\right)={\tilde{f}}_{i,1}\left({\rho }_1\right)\left(\frac{s-{\rho }_2}{{\rho }_1-{\rho }_2}\right)+{\tilde{f}}_{i,1}\left({\rho }_2\right)\left(\frac{s-{\rho }_1}{{\rho }_2-{\rho }_1}\right)=-\frac{{\tilde{f}}_{i,1}\left({\rho }_2\right)\left(s-{\rho }_2\right)-{\tilde{f}}_{i,2}\left({\rho }_2\right)\left(s-{\rho }_1\right)}{{\rho }_1-{\rho }_2}.$

这意味着

$\begin{array}{c}{\tilde{f}}_{i,1}\left(s\right)+{\tilde{f}}_{i,2}\left(s\right)=\frac{\left(s-{\rho }_2\right)\left({\tilde{f}}_{i,2}\left(s\right)-{\tilde{f}}_{i,2}\left({\rho }_1\right)\right)-\left(s-{\rho }_1\right)\left({\tilde{f}}_{i,2}\left(s\right)-{\tilde{f}}_{i,2}\left({\rho }_2\right)\right)}{{\rho }_1-{\rho }_2}\\ =\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)\frac{T_s{T}_{\rho 2}{f}_{i,2}\left(0\right)-T_s{T}_{\rho 1}{f}_{i,2}\left(0\right)}{{\rho }_1-{\rho }_2}\\ =\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)T_s{T}_{\rho 2}{T}_{\rho 1}{f}_{i,2}\left(0\right).\end{array}$ (13)

我们利用类似的过程来寻找$\psi \left(s\right)$ ，${\psi }_1\left(s\right)$ 的分母$h_1\left(s\right)-h_2\left(s\right)$ 的代替表达式，从引理1我们可以知道对于$i=1,2$ ，$h_1\left({\rho }_i\right)=h_2\left({\rho }_i\right)$ ，同样，易看出$h_2\left(s\right)$ 是$s$ 中的2次多项式，使用拉格朗日插值定理，我们得到

$\begin{array}{c}{\tilde{h}}_2\left(s\right)={\tilde{h}}_2\left(0\right)\frac{\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)}{{\rho }_1{\rho }_2}+s\left(\frac{{\tilde{h}}_2\left({\rho }_1\right)}{{\rho }_1}\frac{s-{\rho }_2}{{\rho }_1-{\rho }_2}+\frac{{\tilde{h}}_2\left({\rho }_2\right)}{{\rho }_2}\frac{s-{\rho }_1}{{\rho }_2-{\rho }_1}\right)\\ ={\tilde{h}}_2\left(0\right)\frac{\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)}{{\rho }_1{\rho }_2}+s\left(\frac{{\tilde{h}}_1\left({\rho }_1\right)}{{\rho }_1}\frac{s-{\rho }_2}{{\rho }_1-{\rho }_2}+\frac{{\tilde{h}}_1\left({\rho }_2\right)}{{\rho }_2}\frac{s-{\rho }_1}{{\rho }_2-{\rho }_1}\right)\\ ={\tilde{h}}_2\left(0\right)\frac{\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)}{{\rho }_1{\rho }_2}+\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)\times \left(\frac{{\tilde{h}}_1\left({\rho }_1\right)}{{\rho }_1}\frac{1}{{\rho }_1-{\rho }_2}+\frac{{\tilde{h}}_1\left({\rho }_1\right)}{{\rho }_2}\frac{1}{{\rho }_2-{\rho }_1}\right)\\ +{\tilde{h}}_1\left({\rho }_1\right)\frac{s-{\rho }_2}{{\rho }_1-{\rho }_2}+{\tilde{h}}_1\left({\rho }_2\right)\frac{s-{\rho }_1}{{\rho }_2-{\rho }_1}.\end{array}$

因此，利用Dickson-Hipp算子的性质6，$h_1\left(s\right)-h_2\left(s\right)$ 变成

$\begin{array}{c}{\tilde{h}}_1\left(s\right)-{\tilde{h}}_2\left(s\right)=-{\tilde{h}}_2\left(0\right)\frac{\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)}{{\rho }_1{\rho }_2}-\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)\times \left(\frac{{\tilde{h}}_1\left({\rho }_1\right)}{{\rho }_1\left({\rho }_1-{\rho }_2\right)}+\frac{{\tilde{h}}_1\left({\rho }_2\right)}{{\rho }_2\left({\rho }_2-{\rho }_1\right)}\right)\\ +\left({\tilde{h}}_1\left(s\right)-{\tilde{h}}_1\left({\rho }_1\right)\frac{s-{\rho }_2}{{\rho }_1-{\rho }_2}-{\tilde{h}}_1\left({\rho }_2\right)\frac{s-{\rho }_1}{{\rho }_2-{\rho }_1}\right)\\ =\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)\left(-T_0{T}_{\rho 2}{T}_{\rho 1}{h}_2\left(0\right)+\left(\frac{{\tilde{h}}_1\left(s\right)}{\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)}-\frac{{\tilde{h}}_1\left({\rho }_1\right)}{\left(s-{\rho }_1\right)\left({\rho }_2-{\rho }_1\right)}-\frac{{\tilde{h}}_1\left({\rho }_2\right)}{\left(s-{\rho }_2\right)\left({\rho }_2-{\rho }_1\right)}\right)\right)\\ =\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)\left(T_0{T}_{\rho 2}{T}_{\rho 1}{h}_1\left(0\right)-T_s{T}_{\rho 2}{T}_{\rho 1}{h}_2\left(0\right)\right).\end{array}$ (14)

容易证明，

$\begin{array}{c}{T}_s{T}_{\rho 2}{T}_{\rho 1}{h}_2\left(0\right)=\frac{T_s{T}_{\rho 1}{h}_2\left(0\right)-T_s{T}_{\rho 2}{h}_2\left(0\right)}{{\rho }_2-{\rho }_1}=\frac{\frac{{\tilde{h}}_2\left(s\right)-{\tilde{h}}_2\left({\rho }_2\right)}{s-{\rho }_2}-\frac{{\tilde{h}}_2\left(s\right)-{\tilde{h}}_2\left({\rho }_1\right)}{s-{\rho }_1}}{{\rho }_2-{\rho }_1}\\ =\frac{\frac{{\left(1-s{\mu }_x\right)}^2-{\left(1-{\rho }_2{\mu }_x\right)}^2}{s-{\rho }_2}-\frac{{\left(1-s{\mu }_x\right)}^2-{\left(1-{\rho }_1{\mu }_x\right)}^2}{s-{\rho }_1}}{{\rho }_2-{\rho }_1}\\ =\frac{\left(-2{\mu }_x+\left(s+{\rho }_2\right){\mu }_x^2\right)-\left(-2{\mu }_x+\left(s+{\rho }_1\right){\mu }_x^2\right)}{{\rho }_2-{\rho }_1}={\mu }_x^2.\end{array}$ (15)

这意味着(14)变成

${\tilde{h}}_1\left(s\right)-{\tilde{h}}_2\left(s\right)=\left(s-{\rho }_1\right)\left(s-{\rho }_2\right)\left(T_s{T}_{\rho 2}{T}_{\rho 1}{h}_1\left(0\right)-{\mu }_x^2\right).$ (16)

将(13)和(16)带入(11)中，得到

$\tilde{\psi }\left(s\right)=\frac{T_s{T}_{\rho 2}{T}_{\rho 1}{f}_{1,2}\left(0\right)}{T_s{T}_{\rho 2}{T}_{\rho 1}{h}_1\left(0\right)-{\mu }_x^2}$ , ${\tilde{\psi }}_1\left(s\right)=\frac{T_s{T}_{\rho 2}{T}_{\rho 1}{f}_{2,2}\left(0\right)}{T_s{T}_{\rho 2}{T}_{\rho 1}{h}_1\left(0\right)-{\mu }_x^2}$ .

综上(12)式得证。

利用命题1，我们可以得到$\psi \left(u\right),{\psi }_1\left(u\right)$ 的瑕疵更新方程。

命题2：$\psi \left(u\right),{\psi }_1\left(u\right)$ 分别满足下列瑕疵更新方程

$\psi \left(u\right)=k_{\delta }{\displaystyle {\int }_0^u\psi \left(u-y\right)\varsigma \left(y\right)\text{d}y}+{\epsilon }_1\left(u\right)$ , ${\psi }_1\left(u\right)=k_{\delta }{\displaystyle {\int }_0^u{\psi }_1\left(u-y\right)\varsigma \left(y\right)\text{d}y}+{\epsilon }_2\left(u\right)$ . (17)

其中，

$\begin{array}{c}{k}_{\delta }=\frac{{\lambda }_1{\lambda }_2}{\left(\lambda +{\lambda }_1\right)\left(\lambda +{\lambda }_2\right){\mu }_x^2}{T}_0{T}_{\rho 2}{T}_{\rho 1}q\left(0\right)-2\frac{{\lambda }_1{\lambda }_2}{\left(\lambda +{\lambda }_1\right)\left(\lambda +{\lambda }_2\right){\mu }_x}\left({\rho }_1{T}_0{T}_{\rho 2}{T}_{\rho 1}q\left(0\right)-T_0{T}_{\rho 2}q\left(0\right)\right)\\ +\frac{{\lambda }_1{\lambda }_2}{\left(\lambda +{\lambda }_1\right)\left(\lambda +{\lambda }_2\right)}\left(1-\left({\rho }_2+{\rho }_1\right)T_0{T}_{{\rho }_2}q\left(0\right)+{\rho }_1^2{T}_0{T}_{\rho 2}{T}_{\rho 1}q\left(0\right)\right),\end{array}$

$\begin{array}{c}\varsigma (y)=\frac{1}{k_{\delta }}(\frac{{\lambda }_1{\lambda }_2}{\left(\lambda +{\lambda }_1\right)\left(\lambda +{\lambda }_2\right){\mu }_x^2}{T}_{\rho 2}{T}_{\rho 1}q\left(y\right)-2\frac{{\lambda }_1{\lambda }_2}{\left(\lambda +{\lambda }_1\right)\left(\lambda +{\lambda }_2\right){\mu }_x}\left({\rho }_1{T}_{\rho 2}{T}_{\rho 1}q\left(y\right)-T_{\rho 2}q\left(y\right)\right)\\ +\frac{{\lambda }_1{\lambda }_2}{\left(\lambda +{\lambda }_1\right)\left(\lambda +{\lambda }_2\right)}\left(\tilde{q}\left(y\right)-\left({\rho }_2+{\rho }_1\right)T_{\rho 2}q\left(y\right)+{\rho }_1^2{T}_{\rho 2}{T}_{\rho 1}q\left(y\right)\right)),\end{array}$

$\begin{array}{c}{\xi }_1(u)=\frac{{\lambda }_1{\lambda }_2}{\left(\lambda +{\lambda }_1\right)\left(\lambda +{\lambda }_2\right){\mu }_x^2}{T}_{\rho 2}{T}_{\rho 1}W\left(u\right)-\frac{2u_x{\lambda }_1}{\left(\lambda +{\lambda }_1\right)\left(\lambda +{\lambda }_2\right)}\left({\rho }_1{T}_{\rho 2}{T}_{\rho 1}W\left(u\right)-T_{\rho 2}W\left(u\right)\right)\\ +\frac{{\lambda }_1}{\left(\lambda +{\lambda }_1\right)\left(\lambda +{\lambda }_2\right)}\left(\tilde{W}\left(u\right)-\left({\rho }_2+{\rho }_1\right)T_{\rho 2}W\left(u\right)+{\rho }_1^2{T}_{\rho 2}{T}_{\rho 1}W\left(u\right)\right),\end{array}$

$\begin{array}{c}{\xi }_2\left(u\right)=-\frac{1}{u_x^2}{T}_{\rho 2}{T}_{\rho 1}q\left(u\right)+\frac{{\lambda }_2\lambda }{\left(\lambda +{\lambda }_2\right)\left(\lambda +{\lambda }_1\right)u_x}\tilde{\psi }\left(\frac{1}{u_x}\right)\left({\rho }_1{T}_{\rho 1}q\left(u\right)-T_{\rho 2}q\left(u\right)\right)\\ +\frac{{\lambda }_1{\lambda }_2}{\left(\lambda +{\lambda }_2\right)\left(\lambda +{\lambda }_1\right)u_x^2}{T}_{\rho 2}{T}_{\rho 1}W\left(u\right)+\frac{{\lambda }_2\lambda +2{\lambda }_1{\lambda }_2}{\left(\lambda +{\lambda }_2\right)\left(\lambda +{\lambda }_1\right)u_x}\left({\rho }_1{T}_{\rho 2}{T}_{\rho 1}W\left(u\right)-T_{\rho 2}W\left(u\right)\right)\\ +\frac{{\lambda }_2}{\lambda +{\lambda }_2}\left(\tilde{W}\left(u\right)-\left({\rho }_1+{\rho }_2\right)T_{\rho 2}W\left(u\right)+{\rho }_2^2{T}_{\rho 2}{T}_{\rho 1}W\left(u\right)\right).\end{array}$

证明：回顾Dickson-Hipp运算符的性质，我们有

$\frac{\frac{\tilde{f}\left(s\right)-\tilde{f}\left({\rho }_2\right)}{s-{\rho }_2}-\frac{\tilde{f}\left(s\right)-\tilde{f}\left({\rho }_1\right)}{s-{\rho }_1}}{{\rho }_2-{\rho }_1}=\frac{T_s{T}_{{\rho }_1}f\left(0\right)-T_s{T}_{\rho 2}f\left(0\right)}{{\rho }_2-{\rho }_1}=T_s{T}_{\rho 2}{T}_{{\rho }_1}f\left(0\right),$ (T-1)

$\begin{array}{c}\frac{\frac{s\tilde{f}\left(s\right)-{\rho }_2\tilde{f}\left({\rho }_2\right)}{s-{\rho }_2}-\frac{s\tilde{f}\left(s\right)-{\rho }_1\tilde{f}\left({\rho }_1\right)}{s-{\rho }_1}}{{\rho }_2-{\rho }_1}=\frac{{\rho }_1{T}_s{T}_{{\rho }_1}f\left(0\right)-{\rho }_2{T}_s{T}_{\rho 2}f\left(0\right)}{{\rho }_2-{\rho }_1}\\ ={\rho }_1{T}_s{T}_{\rho 2}{T}_{{\rho }_1}f\left(0\right)-T_s{T}_{\rho 2}f\left(0\right),\end{array}$ (T-2)

$\begin{array}{c}\frac{\frac{s^2\tilde{f}\left(s\right)-{\rho }_2^2\tilde{f}\left({\rho }_2\right)}{s-{\rho }_2}-\frac{s^2\tilde{f}\left(s\right)-{\rho }_1^2\tilde{f}\left({\rho }_1\right)}{s-{\rho }_1}}{{\rho }_2-{\rho }_1}=\frac{{\rho }_2\tilde{f}\left(s\right)-{\rho }_1\tilde{f}\left(s\right)}{{\rho }_2-{\rho }_1}-\frac{{\rho }_2^2{T}_s{T}_{\rho 2}f\left(0\right)-{\rho }_1^2{T}_s{T}_{\rho 1}f\left(0\right)}{{\rho }_2-{\rho }_1}\\ =\tilde{f}\left(s\right)-\left({\rho }_2+{\rho }_1\right)T_s{T}_{\rho 2}f\left(0\right)-{\rho }_1^2{T}_s{T}_{\rho 2}{T}_{\rho 1}f\left(0\right).\end{array}$ (T-3)

从Dickson-Hipp算子$T_r$ 和(T-1)~(T-3)的定义可以推导出

$\begin{array}{c}{T}_s{T}_{\rho 2}{T}_{\rho 1}{h}_1\left(0\right)=\frac{{\lambda }_1{\lambda }_2}{\left(\lambda +{\lambda }_1\right)\left(\lambda +{\lambda }_2\right)}(T_s{T}_{\rho 2}{T}_{\rho 1}q\left(0\right)-2u_x\left({\rho }_1{T}_s{T}_{\rho 2}{T}_{\rho 1}q\left(0\right)-T_s{T}_{\rho 2}q\left(0\right)\right)\\ +u_x^2\left(\tilde{q}\left(s\right)-\left({\rho }_1+{\rho }_2\right)T_s{T}_{\rho 2}q\left(0\right)+{\rho }_1^2{T}_s{T}_{\rho 2}{T}_{\rho 1}q\left(0\right)\right)),\end{array}$ (18)

$\begin{array}{c}{T}_s{T}_{\rho 2}{T}_{\rho 1}{f}_{1,2}\left(0\right)=\frac{{\lambda }_1{\lambda }_2}{\left(\lambda +{\lambda }_1\right)\left(\lambda +{\lambda }_2\right)}{T}_s{T}_{\rho 2}{T}_{\rho 1}W\left(0\right)+\frac{2u_x{\lambda }_1}{\left(\lambda +{\lambda }_1\right)\left(\lambda +{\lambda }_2\right)}\left({\rho }_1{T}_s{T}_{\rho 2}{T}_{\rho 1}W\left(0\right)-T_s{T}_{\rho 2}W\left(0\right)\right)\\ -\frac{u_x{}^2{\lambda }_1{\lambda }_2}{\left(\lambda +{\lambda }_1\right)\left(\lambda +{\lambda }_2\right)}\left(\tilde{W}\left(s\right)-\left({\rho }_1+{\rho }_2\right)T_s{T}_{\rho 2}W\left(0\right)+{\rho }_1^2{T}_s{T}_{\rho 2}{T}_{\rho 1}W\left(0\right)\right)\\ =-u_x^2{T}_s{\xi }_1\left(0\right),\end{array}$ (19)

$\begin{array}{c}{T}_s{T}_{\rho 2}{T}_{\rho 1}{f}_{2,2}\left(0\right)=\frac{{\lambda }_2\lambda }{\left(\lambda +{\lambda }_2\right)\left(\lambda +{\lambda }_1\right)}{T}_s{T}_{\rho 2}{T}_{\rho 1}q\left(0\right)-\frac{{\lambda }_2\lambda }{\left(\lambda +{\lambda }_2\right)\left(\lambda +{\lambda }_1\right)}{u}_x\tilde{\psi }\left(\frac{1}{u_x}\right)\left({\rho }_1{T}_s{T}_{\rho 1}q\left(0\right)-T_s{T}_{\rho 2}q\left(0\right)\right)\\ -\frac{{\lambda }_1{\lambda }_2}{\left(\lambda +{\lambda }_2\right)\left(\lambda +{\lambda }_1\right)}{T}_s{T}_{\rho 2}{T}_{\rho 1}W\left(0\right)+\frac{{\lambda }_2\lambda +2{\lambda }_1{\lambda }_2}{\left(\lambda +{\lambda }_2\right)\left(\lambda +{\lambda }_1\right)}{u}_x\left({\rho }_1{T}_s{T}_{\rho 2}{T}_{\rho 1}W\left(0\right)-T_s{T}_{\rho 2}W\left(0\right)\right)\\ -\frac{{\lambda }_2}{\lambda +{\lambda }_2}{u}_x^2\left(\tilde{W}\left(s\right)-\left({\rho }_2+{\rho }_1\right)T_s{T}_{\rho 2}W\left(0\right)+{\rho }_2^2{T}_s{T}_{\rho 2}{T}_{\rho 1}W\left(0\right)\right)\\ =-u_x^2{T}_s{\xi }_2\left(0\right).\end{array}$ (20)

将(18)~(20)带入(12)，可以推导出

$\tilde{\psi }\left(s\right)=\frac{T_s{T}_{\rho 2}{T}_{\rho 1}{h}_1\left(0\right)}{u_x^2}\tilde{\psi }\left(s\right)+T_s{\xi }_1\left(0\right)$ ; ${\tilde{\psi }}_1\left(s\right)=\frac{T_s{T}_{\rho 2}{T}_{\rho 1}{h}_1\left(0\right)}{u_x^2}{\tilde{\psi }}_1\left(s\right)+T_s{\xi }_2\left(0\right)$ . (21)

求(21)的拉普拉斯逆变换得到

$\psi \left(u\right)=\frac{T_0{T}_{\rho 2}{T}_{\rho 1}{h}_1\left(0\right)}{u_x^2}{\displaystyle {\int }_0^u\psi }\left(u-y\right)\frac{T_{\rho 2}{T}_{\rho 1}{h}_1\left(y\right)}{T_0{T}_{\rho 2}{T}_{\rho 1}{h}_1\left(0\right)}\text{d}y+{\xi }_1\left(u\right)$ ,

${\psi }_1\left(u\right)=\frac{T_0{T}_{\rho 2}{T}_{\rho 1}{h}_1\left(0\right)}{u_x^2}{\displaystyle {\int }_0^u{\psi }_1}\left(u-y\right)\frac{T_{\rho 2}{T}_{\rho 1}{h}_1\left(y\right)}{T_0{T}_{\rho 2}{T}_{\rho 1}{h}_1\left(0\right)}\text{d}y+{\xi }_2\left(u\right)$ .

对应于(17)式。

下证瑕疵更新方程(17)式的$k_{\delta }<1$ ，这里参考[1]中$\delta =0$ 的情况证明。

$s={\rho }_1\left(\delta \right)$ 是(10)式的根，

即$\frac{{\lambda }_1{\lambda }_2}{\left(\lambda +{\lambda }_1+\delta \right)\left(\lambda +{\lambda }_2+\delta \right)}\tilde{q}\left({\rho }_1\left(\delta \right)\right)=\left(1-\frac{\lambda }{\lambda +{\lambda }_2+\delta }\frac{1}{1-{\rho }_1\left(\delta \right)u_x}\right)\left(1-\frac{\lambda }{\lambda +{\lambda }_1+\delta }\frac{1}{1-{\rho }_1\left(\delta \right)u_x}\right)$ 。当$\delta =0$ 时，${\rho }_1\left(\delta \right)={\rho }_1\left(0\right)=0$ 。

对$\delta$ 求导，我么们可以得到${{\rho }^{\prime }}_1\left(0\right)=\frac{{\lambda }_1+{\lambda }_2}{\lambda \left({\lambda }_1+{\lambda }_2\right)u_x-{\lambda }_1{\lambda }_2{u}_y}$ 。

由安全负载条件知：${{\rho }^{\prime }}_1\left(0\right)>0$ ，令$k_{\delta }$ 中的极限$\delta \to 0^{+}$ ，应用洛必达求导法则得：

$\begin{array}{c}{k}_{\delta }=\frac{T_0{T}_0{T}_{\rho 2\left(0\right)}{h}_1\left(0\right)}{u_x^2}=1+\frac{{\tilde{h}}_1\left(0\right)-{\tilde{h}}_2\left(0\right)}{u_x^2{\rho }_1{\rho }_2}\\ =1-\frac{{\lambda }^2+\lambda \left({\lambda }_2+{\lambda }_1\right)}{u_x^2{\rho }_2\left(\lambda +{\lambda }_1\right)\left(\lambda +{\lambda }_2\right)}\times \underset{\delta \to 0^{+}}{\lim }\frac{\delta \left(2\lambda +{\lambda }_1+{\lambda }_2\right)}{{\rho }_1\left(\delta \right)}\\ =1-\frac{\left({\lambda }^2+\lambda \left({\lambda }_2+{\lambda }_1\right)\right)\left(2\lambda +{\lambda }_1+{\lambda }_2\right)}{u_x^2{\rho }_2\left(\lambda +{\lambda }_1\right)\left(\lambda +{\lambda }_2\right){{\rho }^{\prime }}_1\left(0\right)}<1.\end{array}$

由此可知(17)为瑕疵更新方程，证明完毕。

5. 数值图解

在本研究部分，我们通过提供具体的数值案例，深入分析了不同参数对破产概率的影响。采用控制变量法，我们系统地研究了各个参数对破产概率的作用机制。

参数$\lambda$ 对破产概率的影响

当保费和索赔分别服从均值为${\mu }_x=1$ ，${\mu }_y=0.5$ 指数分布时，设${\lambda }_1=0.5$ ，${\lambda }_2=2$ ，在上述设置下，我们讨论$\lambda =1$ ，$\lambda =1.5$ ，$\lambda =2$ 时的破产概率(我们记$\lambda =i$ 时，相对应的破产概率记为${\psi }_{*i}\left(s\right)$ )，很容易查证安全负载条件是满足的。

$\lambda =1$ 时，参数带入(10)式解等式的根：

解得：0，0.6156，−1.5084。

将${\rho }_1$ ，${\rho }_2$ 带入(18)，(19)中的分母，得$T_s{T}_{\rho 2}{T}_{\rho 1}{f}_{1,2}\left(0\right)$ ，$T_s{T}_{\rho 2}{T}_{\rho 1}{h}_1\left(0\right)$ ，带入(12)式得$\tilde{\psi }\left(s\right)$ 。

$\lambda =1.5$ ，$\lambda =2$ 时，同理

最后，(12)进行拉普拉斯逆变换得到

${\psi }_{*1}\left(u\right)=0.2057{\text{e}}^{-1.2607u}$ ,

${\psi }_{*1.5}\left(u\right)=0.1365{\text{e}}^{-1.5084u}$ ,

${\psi }_{*2}\left(u\right)=0.0981{\text{e}}^{-1.6466u}$ .

下图显示了上述例子中$u\in \left[0,5\right]$ 值的破产概率${\psi }_{*1}\left(u\right)$ 。${\psi }_{*1.5}\left(u\right)$ ${\psi }_{*2}\left(u\right)$ ，由图1可得随着参数$\lambda$ 的增大，破产概率也增大。

Figure 1. Ruin probabilities $\lambda$ for different parameters

图1. 不同参数下的破产概率$\lambda$

当保费和索赔分别服从均值为${\mu }_x=1$ ，${\mu }_y=0.5$ 指数分布时，设$\lambda =0.5$ ，${\lambda }_2=2$ ，在上述设置下，我们讨论${\lambda }_1=0.5$ ，${\lambda }_1=1$ ，${\lambda }_1=1.5$ 时的破产概率(我们记${\lambda }_1=i$ 时，相对应的破产概率记为${\psi }_{*i}\left(s\right)$ )，很容易查证安全负载条件是满足的。

相似的步骤求得$\psi \left(s\right)$ 得：

${\psi }_{*0.5}\left(u\right)=0.2057{\text{e}}^{-1.2607u}$ ,

${\psi }_{*1}\left(u\right)=0.3044{\text{e}}^{-1.1882u}$ ,

${\psi }_{*1.5}\left(u\right)=0.3617{\text{e}}^{-1.1470u}$ .

下图显示了上述例子中$u\in \left[0,5\right]$ 值的破产概率${\psi }_{*0.5}\left(u\right)$ 。${\psi }_{*1}\left(u\right)$ ${\psi }_{*1.5}\left(u\right)$ 。由图2可得随着参数${\lambda }_1$ 的增大，破产概率也增大。

Figure 2. Ruin probabilities ${\lambda }_1$ for different parameters

图2. 不同参数下的破产概率${\lambda }_1$

当保费和索赔分别服从均值为${\mu }_x=1$ ，${\mu }_y=0.5$ 指数分布时，设$\lambda =1$ ，${\lambda }_1=0.5$ ，在上述设置下，我们讨论${\lambda }_2=2$ ，${\lambda }_2=3$ ，${\lambda }_2=4$ 时的破产概率(我们记${\lambda }_2=i$ 时，相对应的破产概率记为${\psi }_{*i}\left(s\right)$ )，很容易查证安全负载条件是满足的。

相似的步骤求得$\psi \left(s\right)$ 得：

${\psi }_{*2}\left(u\right)=0.2057{\text{e}}^{-1.2607u}$ ,

${\psi }_{*3}\left(u\right)=0.1653{\text{e}}^{-1.1882u}$ ,

${\psi }_{*4}\left(u\right)=0.1422{\text{e}}^{-1.1470u}$ .

下图显示了上述例子中$u\in \left[0,5\right]$ 值的破产概率${\psi }_{*2}\left(u\right)$ ，${\psi }_{*3}\left(u\right)$ ，${\psi }_{*4}\left(u\right)$ 。由图3可得随着参数${\lambda }_2$ 的增大，破产概率反而减小。

Figure 3. Ruin probabilities ${\lambda }_2$ for different parameters

图3. 不同参数下的破产概率${\lambda }_2$

破产概率受上面参数$\lambda$ ，${\lambda }_1$ ，${\lambda }_2$ 的影响，但总体上随着初值$u$ 的增大而减小，因为在复合泊松模型中，破产概率$\psi \left(u\right)$ 是初始资本$u$ 的函数，满足$\psi \left(u\right)\approx C{\text{e}}^{-Ru}$ ，其中$R$ 是调节系数(与保费和索赔分布有关)。