辽宁师范大学数学学院，辽宁 大连

收稿日期：2023年6月25日；录用日期：2023年7月19日；发布日期：2023年7月28日

摘要

在本文中我们考虑了经典复合泊松模型的一个推广，该模型的累计索赔额过程的增量是独立的。其中索赔间隔时间的分布是两个独立的指数随机变量之和，在本文中我们考虑了索赔时间与后续索赔规模之间的一种特殊的依赖结构，导出了Gerber-Shiu惩罚函数的积分微分方程，利用Rouché定理研究了林德伯格等式的根，导出了惩罚函数的拉普拉斯变换及其满足的瑕疵更新方程。最后在索赔时间间隔服从两个独立的指数随机变量之和分布时给出破产概率的解析解，对理论结果做数值分析，对不同相依参数下的破产概率进行对比。

关键词

两个独立的指数随机变量之和分布，积分微分方程，林德伯格等式，Gerber-Shiu期望贴现惩罚函数，拉普拉斯变换，瑕疵更新方程

Research on the Ruin Problem of Dependent Risk Model Based on the Sum of Two Independent Exponential Random Variables

Jingxuan Wang

School of Mathematics, Liaoning Normal University, Dalian Liaoning

Received: Jun. 25^th, 2023; accepted: Jul. 19^th, 2023; published: Jul. 28^th, 2023

ABSTRACT

In this paper, we consider an extension of the classical compound Poisson model where the incremental cumulative claims process is independent. The distribution of claim interval follows the sum of two independent random variables. In this paper, we consider a special dependence structure between claim time and subsequent claim scale, derive the Gerber-Shiu penalty function integral differential equation, and use Rouché’s theorem to study the roots of Lundberg equation. The Laplace transform of penalty function and its defect renewal equation are derived. Finally, the analytic solution of ruin probability is given when the claim interval follows the sum distribution of two independent exponential random variables. The theoretical results are analyzed numerically, the ruin probabilities under different dependent parameters are compared.

Keywords:Sum Distribution of Two Independent Exponential Random Variables, Integral Differential Equation, Lundberg Equation, Gerber-Shiu Expected Discounted Penalty Function, The Laplace Transform, Defective Renewal Equation

Copyright © 2023 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

在精算文献中有两个著名的风险模型。即经典的复合泊松风险模型和基于连续时间更新过程的风险模型(后者也称为Sparre Andersen [1] 风险模型)。已被广泛地分析了破产概率和许多破产相关的量，如破产时间的边际分布和联合缺陷分布。在这两种风险模型中，破产时的赤字，破产前的盈余和导致破产的索赔规模已经得到了广泛的研究(Dickson (1992, 1993) [2] [3] ，Gerber and Shiu (1997) [4] ，Dickson and Hipp (1998) [5] ，Rolski et al. (1999) [6] 和其中的参考文献)，Gerber和Shiu (1998) [7] 在复合Poisson风险模型的框架内提出了一个统一的折现罚函数方法，并对Sparre Andersen风险模型的一些类进行了研究(见Dickson and Hipp (2001年) [8] ，Gerber和Shiu (2005年) [9] 和Li and Garrido (2004年，2005年) [10] )。

请注意，对于这两个风险模型，明确假设两个连续索赔之间的到达间限时间和索赔金额是独立的。然而，有许多现实世界的情况下，这样的假设是不恰当的。例如，对于一个承保地震造成的损失的业务，索赔之间的时间越长，损失越大，Albrecher和Boxma (2004) [11] 对经典复合泊松风险模型提出了一个扩展，其中两次索赔之间的时间分布取决于前一次索赔的大小。在本文中，我们考虑对经典复合泊松风险模型的一个扩展，它考虑反向依赖结构，意味着下一次索赔的分布大小取决于上次索赔后的时间。我们将看到，后者的风险模型为破产理论分析提供了一个很好的框架，因为盈余过程增量之间的独立性假设得到了保留，我们提到Albrecher和Teuoels (2006) [12] 考虑了区间时间内的任意相关结构和通过Copula表示的后续索赔额，他们得到了有限时间和无限时间破产机率的渐近结果。Zhang和Liu [13] 在索赔相依情况下讨论了分红策略问题。

本文在构造了基于两个独立的指数随机变量之和分布的相依索赔破产概率模型。第一节介绍了风险模型，然后提出了本文研究的是相依索赔模型，间隔时间服从两个独立的指数随机变量之和分布，相依结构由两个指数分布函数复合而成。第二节建立了期望惩罚函数的微分积分方程。第三节讨论了林德伯格方程的根。第四、五节给出了惩罚函数的拉普拉斯变换及其显示表达以及瑕疵更新方程第六节基于索赔相依结构的联合密度函数和间隔时间服从两个独立的指数随机变量之和分布等假设下，得到了破产概率的显式表达式，并与索赔相依指数模型进行了对比。

2. 模型构建

在时刻t的盈余被定义为 $U\left(t\right)=u+ct-{\displaystyle \underset{k=1}{\overset{N\left(t\right)}{\sum }}{X}_k},t\ge 0$ 。其中 $U\left(0\right)=u$ 表示初始盈余，c是保费率即单位时间收取的保险费金额； $\left\{N\left(t\right),t\ge 0\right\}$ 是一个更新过程为到时刻t为止发生的索赔次数。

在时刻t的盈余被定义为 $U\left(t\right)=u+ct-{\displaystyle \underset{k=1}{\overset{N\left(t\right)}{\sum }}{X}_k},t\ge 0$ 。其中 $U\left(0\right)=u$ 表示初始盈余，c是保费率即单位时间收取的保险费金额； $\left\{N\left(t\right),t\ge 0\right\}$ 是一个更新过程为到时刻t为止发生的索赔次数， $\left\{X_i,i=1,2,\cdots \right\}$ 表示第i次的索赔额，第 $i-1$ 到第i次的索赔间隔时间用 $V_i$ 表示， $V_1$ 表示第一次索赔的时间。索赔额 $\left\{X_i,i=1,2,\cdots \right\}$ 是恒正的独立同分布的随机变量序列且与 $\left\{N\left(t\right),t\ge 0\right\}$ 独立，记 $\left\{X_i,i=1,2,\cdots \right\}$ 的密度函数为 $f_X$ ，分布函数为 $F_X$ 。索赔间隔时间 $\left\{W_i,i\in {\mathbb{N}}^{+}\right\}$ 形成了独立同分布的随机变量序列， $W=W_1+W_2$ ， $W_j$ 是两个独立的参数为 ${\lambda }_j,j=1,2$ 的指数分布，W的密度函数和分布函数分别为 $f_W,F_W$ 当 ${\lambda }_1\ne {\lambda }_2$ 时

$f_W\left(t\right)=\frac{{\lambda }_1{\lambda }_2}{{\lambda }_2-{\lambda }_1}\left({\text{e}}^{-{\lambda }_{1}t}-{\text{e}}^{-{\lambda }_{2}t}\right),F_W\left(t\right)=1-\frac{1}{{\lambda }_2-{\lambda }_1}\left({\lambda }_2{\text{e}}^{-{\lambda }_{1}t}-{\lambda }_1{\text{e}}^{-{\lambda }_{2}t}\right),$ (1)

令 $\frac{{\lambda }_1{\lambda }_2}{{\lambda }_2-{\lambda }_1}=\lambda$ ，

$f_W\left(t\right)=\lambda \left({\text{e}}^{-{\lambda }_{1}t}-{\text{e}}^{-{\lambda }_{2}t}\right)$ (2)

我们定义 $X_k|W_k$ 的密度函数是由两个边际密度函数 $f_1$ 和 $f_2$ (期望分别为 ${\eta }_1$ 和 ${\eta }_2$ )复合而成的，其表达式为：

$f_{X_k|W_k}\left(x\right)={\text{e}}^{-\beta W_k}{f}_1\left(x\right)+\left(1-{\text{e}}^{-\beta W_k}\right)f_2\left(x\right),x\ge 0,k=1,2,\cdots ,$

其中， $\beta$ 是正的常数。 $X_k$ 的边际分布函数为

$f_{X_K}\left(x\right)=M_W\left(-\beta \right)f_1\left(x\right)+\left[1-M_W\left(-\beta \right)\right]f_2(\; x\; )$

计算得 $M_W\left(-\beta \right)={\displaystyle {\int }_0^{+\infty }{\text{e}}^{-\beta t}}\lambda \left({\text{e}}^{-{\lambda }_{1}t}-{\text{e}}^{-{\lambda }_{2}t}\right)\text{d}t=\lambda \left(\frac{1}{\beta +{\lambda }_1}-\frac{1}{\beta +{\lambda }_2}\right)=\frac{{\lambda }_1{\lambda }_2}{\left(\beta +{\lambda }_1\right)\left(\beta +{\lambda }_2\right)}$ ，即

$f_{X_K}\left(x\right)=\frac{{\lambda }_1{\lambda }_2}{\left(\beta +{\lambda }_1\right)\left(\beta +{\lambda }_2\right)}{f}_1\left(x\right)+\frac{\beta \left(\beta +{\lambda }_1+{\lambda }_2\right)}{\left(\beta +{\lambda }_1\right)\left(\beta +{\lambda }_2\right)}{f}_2\left(x\right)$ (3)

为了确保破产不发生，保费率c应满足安全负载条件：

$E\left[c{W}_j-X_j\right]>0,j=1,2,\cdots ,$

$E\left[c{W}_j-X_j\right]=\frac{c\left({\lambda }_1+{\lambda }_2\right)}{{\lambda }_1{\lambda }_2}-\frac{{\lambda }_1{\lambda }_2{\eta }_1+\beta \left(\beta +{\lambda }_1+{\lambda }_2\right){\eta }_2}{\left(\beta +{\lambda }_1\right)\left(\beta +{\lambda }_2\right)}>0$

在经典风险理论中，风险模型的盈余为负时的破产，是指理论上的破产。而在实际情况中，保险公司会采取一些补救的措施来缓解财务的紧张。但它仍可以作为预测保险公司财务状况的重要指标之一，破产概率代表着破产的可能性。将T定义为破产时间

$T={\inf }_{t\ge 0}\left\{t,U\left(t\right)<0\right\},$

为保险公司的破产时间，即盈余小于0的时刻。当 $U_t\ge 0$ 时 $\tau =\infty$ ，即理论上的永不破产。破产赤字定义为 $\left|U_{\tau }\right|$ ，破产前盈余为 $U_{\tau -}$ 。令

$\psi \left(u,t\right)=P\left\{\tau <t|U\left(0\right)=u\right\}$

为保险公司有限时间破产概率。令

$\psi \left(u,t\right)=P\left\{\tau <\infty |U\left(0\right)=u\right\}$

为无限时间破产概率，又称为最终破产概率。

3. 微分–积分方程

在这一节中我们推导了惩罚函数 $m_{\delta }\left(u\right)$ 的积分微分方程，通过第一次索赔的时间和金额条件，我们有

$\begin{array}{l}{m}_{\delta }\left(u\right)={\displaystyle {\int }_0^{\infty }\lambda \left({\text{e}}^{-{\lambda }_{1}t}-{\text{e}}^{-{\lambda }_{2}t}\right){\text{e}}^{-\delta t}}{\displaystyle {\int }_0^{u+ct}{m}_{\delta }\left(u+ct-y\right)}\left({\text{e}}^{-\beta t}{f}_1\left(x\right)+\left(1-{\text{e}}^{-\beta t}\right)f_2\left(x\right)\right)\text{d}y\text{d}t\\ +{\displaystyle {\int }_0^{\infty }\lambda \left({\text{e}}^{-{\lambda }_{1}t}-{\text{e}}^{-{\lambda }_{2}t}\right){\text{e}}^{-\delta t}{\displaystyle {\int }_{u+ct}^{\infty }w\left(u+ct,y-\left(u+ct\right)\right)}}\left({\text{e}}^{-\beta t}{f}_1\left(x\right)+\left(1-{\text{e}}^{-\beta t}\right)f_2\left(x\right)\right)\text{d}y\text{d}t\end{array}$ (4)

定理1 在第二节介绍的风险模型中惩罚函数 $m_{\delta }\left(u\right)$ 满足下列方程

$\alpha \left(D\right)\beta \left(D\right)m_{\delta }\left(u\right)=\alpha \left(D\right)\left({\sigma }_1\left(u\right)-{\sigma }_2\left(u\right)\right)+\beta \left(D\right){\sigma }_2\left(u\right)$ (5)

其中

$\begin{array}{l}\alpha \left(D\right)=\left(\frac{{\lambda }_1+\delta }{{\lambda }_1}I-\frac{c}{{\lambda }_1}D\right)\left(\frac{{\lambda }_2+\delta }{{\lambda }_2}I-\frac{c}{{\lambda }_2}D\right)\\ \beta \left(D\right)=\left(\frac{{\lambda }_1+\delta +\beta }{{\lambda }_1}I-\frac{c}{{\lambda }_1}D\right)\left(\frac{{\lambda }_2+\delta +\beta }{{\lambda }_2}I-\frac{c}{{\lambda }_2}D\right)\end{array}$

其中I为恒等因子，D为微分因子

证明：令 $u+ct=t^{\prime }$

$\begin{array}{c}{m}_{\delta }\left(u\right)={\displaystyle {\int }_u^{\infty }\frac{\lambda }{c}}\left({\text{e}}^{-{\lambda }_1\frac{t-u}{c}}-{\text{e}}^{-{\lambda }_2\frac{t-u}{c}}\right){\text{e}}^{-\delta \frac{t-u}{c}}{\displaystyle {\int }_0^t{m}_{\delta }\left(t-y\right)\left({\text{e}}^{-\beta \frac{t-u}{c}}{f}_1\left(x\right)+\left(1-{\text{e}}^{-\beta \frac{t-u}{c}}\right)f_2\left(x\right)\right)\text{d}y\text{d}t}\\ +{\displaystyle {\int }_u^{\infty }\frac{\lambda }{c}}\left({\text{e}}^{-{\lambda }_1\frac{t-u}{c}}-{\text{e}}^{-{\lambda }_2\frac{t-u}{c}}\right){\text{e}}^{-\delta \frac{t-u}{c}}{\displaystyle {\int }_t^{\infty }w\left(t,y-t\right)\left({\text{e}}^{-\beta \frac{t-u}{c}}{f}_1\left(x\right)+\left(1-{\text{e}}^{-\beta \frac{t-u}{c}}\right)f_2\left(x\right)\right)\text{d}y\text{d}t}\\ =\frac{\lambda }{c}\left[{\displaystyle {\int }_u^{\infty }\left({\text{e}}^{-\frac{{\lambda }_1+\delta +\beta }{c}\left(t-u\right)}-{\text{e}}^{-\frac{{\lambda }_2+\delta +\beta }{c}\left(t-u\right)}\right)}\left({\sigma }_{1,\delta }\left(t\right)-{\sigma }_{2,\delta }\left(t\right)\right)\text{d}t+{\displaystyle {\int }_u^{\infty }\left({\text{e}}^{-\frac{{\lambda }_1+\delta }{c}\left(t-u\right)}-{\text{e}}^{-\frac{{\lambda }_2+\delta }{c}\left(t-u\right)}\right){\sigma }_{2,\delta }\left(t\right)\text{d}t}\right]\\ =\frac{\lambda }{c}\xi \left(u\right)+\frac{\lambda }{c}\eta (\; u\; )\end{array}$

其中

${\gamma }_i\left(t\right)={\displaystyle {\int }_t^{\infty }w\left(t,y-t\right)f_i\left(y\right)\text{d}y},{\sigma }_{i,\delta }\left(t\right)={\displaystyle {\int }_0^t{m}_{\delta }\left(t-y\right)f_i\left(y\right)\text{d}y}+{\gamma }_i\left(t\right)$ (6)

$\xi \left(u\right)={\displaystyle {\int }_u^{\infty }\left({\text{e}}^{-\frac{{\lambda }_1+\delta +\beta }{c}\left(t-u\right)}-{\text{e}}^{-\frac{{\lambda }_2+\delta +\beta }{c}\left(t-u\right)}\right)}\left({\sigma }_{1,\delta }\left(t\right)-{\sigma }_{2,\delta }\left(t\right)\right)\text{d}t$ (7)

$\eta \left(u\right)={\displaystyle {\int }_u^{\infty }\left({\text{e}}^{-\frac{{\lambda }_1+\delta }{c}\left(t-u\right)}-{\text{e}}^{-\frac{{\lambda }_2+\delta }{c}\left(t-u\right)}\right){\sigma }_{2,\delta }\left(t\right)\text{d}t}$ (8)

$D\xi \left(u\right)={\displaystyle {\int }_u^{\infty }\left(\frac{{\lambda }_1+\delta +\beta }{c}{\text{e}}^{-\frac{{\lambda }_1+\delta +\beta }{c}\left(t-u\right)}-\frac{{\lambda }_2+\delta +\beta }{c}{\text{e}}^{-\frac{{\lambda }_2+\delta +\beta }{c}\left(t-u\right)}\right)}\left({\sigma }_{1,\delta }\left(t\right)-{\sigma }_{2,\delta }\left(t\right)\right)\text{d}t$

$\begin{array}{c}{D}^2\xi \left(u\right)={\displaystyle {\int }_u^{\infty }\left({\left(\frac{{\lambda }_1+\delta +\beta }{c}\right)}^2{\text{e}}^{-\frac{{\lambda }_1+\delta +\beta }{c}\left(t-u\right)}-{\left(\frac{{\lambda }_2+\delta +\beta }{c}\right)}^2{\text{e}}^{-\frac{{\lambda }_2+\delta +\beta }{c}\left(t-u\right)}\right)}\left({\sigma }_{1,\delta }\left(t\right)-{\sigma }_{2,\delta }\left(t\right)\right)\text{d}t\\ -\frac{{\lambda }_1-{\lambda }_2}{c}({\sigma }_{1,\delta }\left(t\right)-{\sigma }_{2,\delta }(\; t\; )\; )\end{array}$

可以得到 $\beta \left(D\right)\xi \left(u\right)=\frac{c}{\lambda }({\sigma }_{1,\delta }\left(t\right)-{\sigma }_{2,\delta }(\; t\; )\; )$

$D\eta \left(u\right)={\displaystyle {\int }_u^{\infty }\frac{{\lambda }_1+\delta }{c}\left({\text{e}}^{-\frac{{\lambda }_1+\delta }{c}\left(t-u\right)}-\frac{{\lambda }_2+\delta }{c}{\text{e}}^{-\frac{{\lambda }_2+\delta }{c}\left(t-u\right)}\right){\sigma }_{2,\delta }\left(t\right)\text{d}t}$

$D^2\eta \left(u\right)={\displaystyle {\int }_u^{\infty }{\left(\frac{{\lambda }_1+\delta }{c}\right)}^2\left({\text{e}}^{-\frac{{\lambda }_1+\delta }{c}\left(t-u\right)}-{\left(\frac{{\lambda }_2+\delta }{c}\right)}^2{\text{e}}^{-\frac{{\lambda }_2+\delta }{c}\left(t-u\right)}\right){\sigma }_{2,\delta }\left(t\right)\text{d}t}-\frac{{\lambda }_1-{\lambda }_2}{c}{\sigma }_{2,\delta }(\; u\; )$

$\begin{array}{c}{D}^3\eta \left(u\right)={\displaystyle {\int }_u^{\infty }{\left(\frac{{\lambda }_1+\delta }{c}\right)}^3\left({\text{e}}^{-\frac{{\lambda }_1+\delta }{c}\left(t-u\right)}-{\left(\frac{{\lambda }_2+\delta }{c}\right)}^3{\text{e}}^{-\frac{{\lambda }_2+\delta }{c}\left(t-u\right)}\right){\sigma }_{2,\delta }\left(t\right)\text{d}t}\\ -\left[{\left(\frac{{\lambda }_1+\delta }{c}\right)}^2-{\left(\frac{{\lambda }_2+\delta }{c}\right)}^2\right]{\sigma }_{2,\delta }\left(u\right)-\frac{{\lambda }_1-{\lambda }_2}{c}D{\sigma }_{2,\delta }(\; u\; )\end{array}$

由上式可以得到 $\alpha \left(D\right)\eta \left(u\right)=\frac{c}{\lambda }\gamma \left(D\right){\sigma }_2(\; u\; )$

$\alpha \left(D\right)\beta \left(D\right)m_{\delta }\left(u\right)=\alpha \left(D\right)\beta \left(D\right)\left[\frac{\lambda }{c}\xi \left(u\right)+\frac{\lambda }{c}\eta \left(u\right)\right]=\alpha \left(D\right)\left({\sigma }_1\left(u\right)-{\sigma }_2\left(u\right)\right)+\beta \left(D\right){\sigma }_2(\; u\; )$

证毕。

4. 林德伯格等式

令 $U_0=u$ ，在k次索赔之后的盈余则为：

$U_k=u+{\displaystyle \underset{i=1}{\overset{k}{\sum }}\left(c{W}_i-X_i\right),k=1,2,\cdots ,}$

当 $s>0$ ，过程 $\left\{{\text{e}}^{-\delta {\displaystyle {\sum }_{i=1}^k{W}_i+s{U}_k}},k=0,1,2,\cdots \right\}$ 是一个鞅，当且仅当

$E\left[{\text{e}}^{-\delta W}{\text{e}}^{s\left(cW-X\right)}\right]=1,$

这个等式被称为林德伯格等式，计算整理可得

$\begin{array}{c}E\left[{\text{e}}^{-\delta W}{\text{e}}^{-S\left(X-CW\right)}\right]=E\left[{\text{e}}^{-\left(\delta -cs\right)W}E\left[{\text{e}}^{-sX}|W\right]\right]\\ =E\left[{\text{e}}^{-\left(\delta -cs\right)W}\left({\text{e}}^{-\beta W}{\hat{f}}_1\left(x\right)+\left(1-{\text{e}}^{-\beta W}\right){\hat{f}}_2\left(x\right)\right)\right]\\ =E\left[{\text{e}}^{-\left(\beta +\delta -cs\right)W}\right]{\hat{f}}_1\left(x\right)+E\left[{\text{e}}^{-\left(\delta -cs\right)W}\right]{\hat{f}}_2\left(x\right)-E\left[{\text{e}}^{-\left(\beta +\delta -cs\right)W}\right]{\hat{f}}_2\left(x\right)\\ =\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)}{\hat{f}}_1\left(x\right)\\ +\left(\frac{{\lambda }_1{\lambda }_2}{\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)}-\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)}\right){\hat{f}}_2\left(x\right)\\ ={\lambda }_1{\lambda }_2\frac{\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right){\hat{f}}_1\left(x\right)+\beta \left(\beta +2\delta +{\lambda }_1+{\lambda }_2-2cs\right){\hat{f}}_2\left(x\right)}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)}\end{array}$ (9)

即

$\frac{{\lambda }_1{\lambda }_2}{c^2}\frac{\left(\frac{\delta +{\lambda }_1}{c}-s\right)\left(\frac{\delta +{\lambda }_2}{c}-s\right){\hat{f}}_1\left(x\right)+\left[\left(\frac{\beta +\delta +{\lambda }_1}{c}-s\right)\left(\frac{\beta +\delta +{\lambda }_2}{c}-s\right)-\left(\frac{\delta +{\lambda }_1}{c}-s\right)\left(\frac{\delta +{\lambda }_2}{c}-s\right)\right]{\hat{f}}_2\left(x\right)}{\left(\frac{\beta +\delta +{\lambda }_1}{c}-s\right)\left(\frac{\beta +\delta +{\lambda }_2}{c}-s\right)\left(\frac{\delta +{\lambda }_1}{c}-s\right)\left(\frac{\delta +{\lambda }_2}{c}-s\right)}=1$

整理可得：

$\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)}={\left(\frac{\beta +\delta +{\lambda }_1}{{\lambda }_1}-\frac{c}{{\lambda }_1}s\right)}^{-1}{\left(\frac{\beta +\delta +{\lambda }_2}{{\lambda }_2}-\frac{c}{{\lambda }_2}s\right)}^{-1}=\frac{1}{\beta (\; s\; )}$

$\begin{array}{l}\frac{{\lambda }_1{\lambda }_2\beta \left(\beta +2\delta +{\lambda }_1+{\lambda }_2-2cs\right)}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)}\\ =\frac{{\lambda }_1{\lambda }_2}{\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)}-\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)}=\frac{1}{\alpha \left(s\right)}-\frac{1}{\beta (\; s\; )}\end{array}$

上式整理后得 ${\hat{f}}_1\left(s\right)\frac{1}{\beta \left(s\right)}+{\hat{f}}_2\left(s\right)\left(\frac{1}{\alpha \left(s\right)}-\frac{1}{\beta \left(s\right)}\right)=1$ 即

$\alpha \left(s\right)\beta \left(s\right)-\alpha \left(s\right){\hat{f}}_1\left(s\right)-\left(\beta \left(s\right)-\alpha \left(s\right)\right){\hat{f}}_2\left(s\right)=0$ (10)

下面我们将应用Rouché’s定理来求解林德伯格等式根。

定理2 对于 $\delta >0$ ，我们要证明林德伯格等式有四个根，这四个根分别记为 ${\rho }_1\left(\delta \right)$ ， ${\rho }_2\left(\delta \right)$ ， ${\rho }_3\left(\delta \right)$ ， ${\rho }_4\left(\delta \right)$ ， $Re\left({\rho }_i\left(\delta \right)\right)>0,i=1,2,3,4$ 。

证明：林德伯格等式可以写成

$\begin{array}{l}\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)\\ ={\lambda }_1{\lambda }_2\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right){\hat{f}}_1\left(s\right)+\left[\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)-\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)\right]{\hat{f}}_2(\; s\; )\end{array}$

可以看出上面的等式有四个实部大于零的根，令 $r>0$ ， $C_r$ 表示Contour集，包含−ir到ir的虚轴和一个半径为r的半圈，顺时针方向从−ir到ir，即 $C_r=\left\{s\in C:\left|s\right|=r,Re\left(s\right)\ge 0,r>0\right\}$ 。令 $r\to \infty$ ，用C表

示有限Contour集。我们在封闭Contour集C上应用Rouché’s定理证明结果。

1) 当 $Re\left(s\right)>0$ ，即当s在半圆上时，当 $r\to \infty$ ，我们有 $\left|\beta +\delta +{\lambda }_1-cs\right|\to \infty$ ， $\left|\beta +\delta +{\lambda }_2-cs\right|\to \infty$ ， $\left|\delta +{\lambda }_1-cs\right|\to \infty$ ， $\left|\delta +{\lambda }_2-cs\right|\to \infty$ ，因此在C上有

$\begin{array}{l}|\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)}{\hat{f}}_1\left(s\right)+(\frac{{\lambda }_1{\lambda }_2}{\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)}\\ -\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)}){\hat{f}}_2\left(s\right)|\\ \le \left|{\hat{f}}_1\left(s\right)\right|\left|\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)}\right|\\ +\left|{\hat{f}}_2\left(s\right)\right|\left|\frac{{\lambda }_1{\lambda }_2}{\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)}-\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)}\right|\to 0\end{array}$ (11)

因此 $\begin{array}{l}|\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)}{\hat{f}}_1\left(s\right)+(\frac{{\lambda }_1{\lambda }_2}{\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)}\\ -\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)}){\hat{f}}_2\left(s\right)|<1\end{array}$ 在C上成立

2) $Re\left(s\right)=0$ ，即s在虚轴上时，对 $$ ，令

${\hat{d}}_{\delta }\left(s\right)=\frac{{\lambda }_1{\lambda }_2}{\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)}-\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)}$

$\begin{array}{c}\left|{\hat{d}}_{\delta }\left(s\right)\right|=\left|\frac{{\lambda }_1{\lambda }_2}{\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)}-\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)}\right|\\ ={\lambda }_1{\lambda }_2\left|\frac{\beta \left(\beta +2\delta +{\lambda }_1+{\lambda }_2-2cs\right)}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)}\right|\\ \le {\lambda }_1{\lambda }_2\beta \left|\frac{{\left(\beta +2\delta +{\lambda }_1+{\lambda }_2\right)}^2+{\left(2cs\right)}^2}{\left(\beta +\delta +{\lambda }_1\right)\left(\beta +\delta +{\lambda }_2\right)\left(\delta +{\lambda }_1\right)\left(\delta +{\lambda }_2\right)\left(\beta +2\delta +{\lambda }_1+{\lambda }_2\right)}\right|\\ \le {\lambda }_1{\lambda }_2\beta \left|\frac{{\left(\beta +2\delta +{\lambda }_1+{\lambda }_2\right)}^2}{\left(\beta +\delta +{\lambda }_1\right)\left(\beta +\delta +{\lambda }_2\right)\left(\delta +{\lambda }_1\right)\left(\delta +{\lambda }_2\right)\left(\beta +2\delta +{\lambda }_1+{\lambda }_2\right)}\right|={\hat{d}}_{\delta }\left(0\right)\end{array}$ (12)

此时

$\begin{array}{l}|\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)}{\hat{f}}_1\left(s\right)+(\frac{{\lambda }_1{\lambda }_2}{\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)}\\ -\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)}){\hat{f}}_2\left(s\right)|\\ =\left|\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)}{\hat{f}}_1\left(s\right)+{\hat{f}}_2\left(s\right){\hat{d}}_{\delta }\left(s\right)\right|\\ \le \left|\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)}\right|+\left|{\hat{d}}_{\delta }\left(s\right)\right|\\ \le \left|\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1\right)\left(\beta +\delta +{\lambda }_2\right)}\right|+\left|{\hat{d}}_{\delta }\left(0\right)\right|\end{array}$

因为 $\delta >0,{\hat{d}}_{\delta }\left(0\right)>0$ ，所以 $\frac{{\lambda }_1{\lambda }_2}{\left(\delta +{\lambda }_1\right)\left(\delta +{\lambda }_2\right)}-\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1\right)\left(\beta +\delta +{\lambda }_2\right)}>0$ ，因此，当s在虚轴上时， $\delta >0$ 等式变为

$\begin{array}{l}|\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)}{\hat{f}}_1\left(s\right)+(\frac{{\lambda }_1{\lambda }_2}{\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)}\\ -\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)}){\hat{f}}_2\left(s\right)|\\ \le \left|\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1\right)\left(\beta +\delta +{\lambda }_2\right)}\right|+\left|{\hat{d}}_{\delta }\left(0\right)\right|<1\end{array}$

综上所述，我们证明出在这两种情况下不等式

$\begin{array}{l}\left|{\lambda }_1{\lambda }_2\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right){\hat{f}}_1\left(s\right)+\left[\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)-\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)\right]{\hat{f}}_2\left(s\right)\right|\\ <\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)\end{array}$

应用Rouché’s定理，可知等式与 $\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)=0$ 在 $C_r$ 内有相同个数的根，因此林德伯格等式在 $C_r$ 内也有四个实部为正数的根，证毕。

当 $\delta =0$ ，由于

$\begin{array}{l}|\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)}{\hat{f}}_1\left(s\right)+(\frac{{\lambda }_1{\lambda }_2}{\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)}\\ -\frac{{\lambda }_1{\lambda }_2}{\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)}){\hat{f}}_2\left(s\right)|\\ =\left|\frac{{\lambda }_1{\lambda }_2}{\left(\beta +{\lambda }_1\right)\left(\beta +{\lambda }_2\right)}+1-\frac{{\lambda }_1{\lambda }_2}{\left(\beta +{\lambda }_1\right)\left(\beta +{\lambda }_2\right)}\right|=1\end{array}$ (13)

Rouché’s定理不再满足，可以利用Klimenok (2007)中的方法来求解林德伯格方程中根的个数。

定理3 当 $\delta =0$ 时，林德伯格等式有三个正实部的根，记为 ${\rho }_1\left(0\right),{\rho }_2\left(0\right),{\rho }_3\left(0\right)$ ，还有一个根为0。

证明：定义Contour集 $D_k=s:\left|z\right|=1$ ，令 $z=1-\frac{s}{k}$ 。根据s，Contour集 $D_k$ 是一个圆心为k，半径为k的圆。与定理2类似，此时令 $k\to \infty$ ，记D是有限Contour集，函数

${\lambda }_1{\lambda }_2\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right){\hat{f}}_1\left(s\right)+\left[\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)-\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)\right]{\hat{f}}_2\left(s\right)$ 和 $\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)$ 在D上都连续。根据Klimenok (2001)中的定理1，

需证：

$\begin{array}{l}\frac{\text{d}}{\text{d}z}\{1-\frac{{\lambda }_1{\lambda }_2}{\left(\beta +{\lambda }_1-ck\left(1-z\right)\right)\left(\beta +{\lambda }_2-ck\left(1-z\right)\right)}{\hat{f}}_1\left(k-kz\right)[\frac{{\lambda }_1{\lambda }_2}{\left({\lambda }_1-ck\left(1-z\right)\right)\left({\lambda }_2-ck\left(1-z\right)\right)}\\ -{\frac{{\lambda }_1{\lambda }_2}{\left(\beta +{\lambda }_1-ck\left(1-z\right)\right)\left(\beta +{\lambda }_2-ck\left(1-z\right)\right)}]{\hat{f}}_2\left(k-kz\right)\}|}_{z=1}>0\\ ={\frac{\text{d}}{\text{d}z}\left\{1-E\left[{\text{e}}^{\left(k-kz\right)\left(cW-X\right)}\right]\right\}|}_{z=1}=kE\left(cW-X\right)>0\end{array}$

由于 $E\left(cW-X\right)>0$ 成立，因此上述证明成立。由参考文献Klimenok (2001)可推出在D中等式根有3个，也就是

$\left(\beta +\delta +{\lambda }_1-cs\right)\left(\beta +\delta +{\lambda }_2-cs\right)\left(\delta +{\lambda }_1-cs\right)\left(\delta +{\lambda }_2-cs\right)$ 根的个数减少一个，且知有一根为0。

5. 惩罚函数的拉普拉斯变换

设惩罚函数的拉普拉斯变换为 ${\hat{m}}_{\delta }\left(s\right)$ ，即 ${\hat{m}}_{\delta }\left(s\right)={\displaystyle {\int }_0^{\infty }{\text{e}}^{-su}{m}_{\delta }\left(u\right)\text{d}u}$ 对于 $u\ge 0$

$c{m}_{\delta }\left(u\right)=\lambda \left[{\displaystyle {\int }_u^{\infty }\left({\text{e}}^{-\frac{{\lambda }_1+\delta +\beta }{c}\left(t-u\right)}-{\text{e}}^{-\frac{{\lambda }_2+\delta +\beta }{c}\left(t-u\right)}\right)}\left({\sigma }_{1,\delta }\left(t\right)-{\sigma }_{2,\delta }\left(t\right)\right)\text{d}t+{\displaystyle {\int }_u^{\infty }\left({\text{e}}^{-\frac{{\lambda }_1+\delta }{c}\left(t-u\right)}-{\text{e}}^{-\frac{{\lambda }_2+\delta }{c}\left(t-u\right)}\right){\sigma }_{2,\delta }\left(t\right)\text{d}t}\right]$

其拉普拉斯变换为

$\begin{array}{c}c{\hat{m}}_{\delta }\left(s\right)=\lambda {\displaystyle {\int }_0^{\infty }{\text{e}}^{-\frac{{\lambda }_1+\delta +\beta }{c}t}}\left({\sigma }_{1,\delta }\left(t\right)-{\sigma }_{2,\delta }\left(t\right)\right){\displaystyle {\int }_0^t{\text{e}}^{-\left(s-\frac{{\lambda }_1+\delta +\beta }{c}\right)u}\text{d}u\text{d}t}\\ -\lambda {\displaystyle {\int }_0^{\infty }{\text{e}}^{-\frac{{\lambda }_2+\delta +\beta }{c}t}}\left({\sigma }_{1,\delta }\left(t\right)-{\sigma }_{2,\delta }\left(t\right)\right){\displaystyle {\int }_0^t{\text{e}}^{-\left(s-\frac{{\lambda }_2+\delta +\beta }{c}\right)u}\text{d}u\text{d}t}\\ +\lambda {\displaystyle {\int }_0^{\infty }{\text{e}}^{-\frac{{\lambda }_1+\delta }{c}t}}{\sigma }_{2,\delta }\left(t\right){\displaystyle {\int }_0^t{\text{e}}^{-\left(s-\frac{{\lambda }_1+\delta }{c}\right)u}\text{d}u\text{d}t}-\lambda {\displaystyle {\int }_0^{\infty }{\text{e}}^{-\frac{{\lambda }_2+\delta }{c}t}}{\sigma }_{2,\delta }\left(t\right){\displaystyle {\int }_0^t{\text{e}}^{-\left(s-\frac{{\lambda }_2+\delta }{c}\right)u}\text{d}u\text{d}t}\end{array}$ (14)

$\begin{array}{c}c{\hat{m}}_{\delta }\left(s\right)=\lambda \left[\frac{1}{-\left(s-\frac{{\lambda }_1+\delta +\beta }{c}\right)}-\frac{1}{-\left(s-\frac{{\lambda }_2+\delta +\beta }{c}\right)}\right]\left({\hat{\sigma }}_{1,\delta }\left(s\right)-{\hat{\sigma }}_{2,\delta }\left(s\right)\right)\\ +\lambda \frac{1}{s-\frac{{\lambda }_1+\delta +\beta }{c}}{\displaystyle {\int }_0^{\infty }{\text{e}}^{-\frac{{\lambda }_1+\delta +\beta }{c}t}}\left({\sigma }_{1,\delta }\left(t\right)-{\sigma }_{2,\delta }\left(t\right)\right)\text{d}t\\ -\lambda \frac{1}{s-\frac{{\lambda }_2+\delta +\beta }{c}}{\displaystyle {\int }_0^{\infty }{\text{e}}^{-\frac{{\lambda }_2+\delta +\beta }{c}t}}\left({\sigma }_{1,\delta }\left(t\right)-{\sigma }_{2,\delta }\left(t\right)\right)\text{d}t\\ +\lambda \left[\frac{1}{-\left(s-\frac{{\lambda }_1+\delta }{c}\right)}-\frac{1}{-\left(s-\frac{{\lambda }_2+\delta }{c}\right)}\right]{\hat{\sigma }}_{2,\delta }\left(s\right)\\ +\lambda \frac{1}{s-\frac{{\lambda }_1+\delta }{c}}{\displaystyle {\int }_0^{\infty }{\text{e}}^{-\frac{{\lambda }_1+\delta }{c}t}}{\sigma }_{2,\delta }\left(t\right)\text{d}t-\lambda \frac{1}{s-\frac{{\lambda }_2+\delta }{c}}{\displaystyle {\int }_0^{\infty }{\text{e}}^{-\frac{{\lambda }_2+\delta }{c}t}}{\sigma }_{2,\delta }\left(t\right)\text{d}t\end{array}$ (15)