#### 期刊菜单

The Gerber-Shiu Discounted Penalty Function for the Risk Model with Dependence Structure and a Constant Dividend Barrier
DOI: 10.12677/pm.2024.145161, PDF, HTML, XML, 下载: 59  浏览: 120

Abstract: We consider a compound Poisson risk model with dependence and a constant dividend barrier, where the claim amount and the interclaim time follow some bivariate distribution. An integrodifferential equation for the Gerber-Shiu discounted penalty function is derived. We also solve the integrodifferential equation and show that the solution is a linear combination of the Gerber-Shiu function with no barrier and the solution of an associated homogeneous integrodifferential equation.

1. 引言

2. 模型建构

$U\left(t\right)=u+ct-\underset{i=1}{\overset{N\left(t\right)}{\sum }}{X}_{i},\text{\hspace{0.17em}}t\ge 0,$

${f}_{W}\left(t\right)=\lambda {\text{e}}^{-\lambda t},\text{\hspace{0.17em}}t\ge 0.$

${f}_{X,W}\left(x,t\right)=\underset{i=1}{\overset{n}{\sum }}{\lambda }_{i}{\text{e}}^{-{\lambda }_{i}t}{f}_{i}\left(x\right),$ (1)

$\left\{\begin{array}{ll}\text{d}{U}_{b}\left(t\right)=c\text{d}t-\text{d}S\left(t\right),\hfill & \text{if}\text{\hspace{0.17em}}{U}_{b}\left(t\right)

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

$E\left[c{W}_{i}-{X}_{i}\right]>0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,2,\cdots$

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

3. Gerber-Shiu贴现惩罚函数

$\underset{i=1}{\overset{n}{\prod }}\left(\frac{{\lambda }_{i}+\delta }{c}I-D\right){m}_{b,\delta }\left(u\right)=\underset{i=1}{\overset{n}{\sum }}\underset{j=1,j\ne i}{\overset{n}{\prod }}\frac{{\lambda }_{i}}{c}\left(\frac{{\lambda }_{j}+\delta }{c}I-D\right){\sigma }_{i}\left(u\right)$ (2)

$0\le u\le b<\infty$ 边界条件为

${{m}^{\prime }}_{b,\delta }\left(b\right)=0,$ (3)

${m}_{b,\delta }^{\left(n\right)}\left(b\right)=-\underset{i=1}{\overset{n}{\sum }}\frac{{\lambda }_{i}}{c}{\sigma }_{i}^{\left(n-1\right)}\left(b\right),$ (4)

${\sigma }_{i}\left(u\right)={\int }_{0}^{u}{m}_{b,\delta }\left(u-x\right){f}_{i}\left(x\right)\text{d}x+{\omega }_{i}\left(u\right),$ (5)

${\omega }_{i}\left(u\right)={\int }_{u}^{\infty }\omega \left(u,x-u\right){f}_{i}\left(x\right)\text{d}x.$ (6)

$\left\{{X}_{1}\le u+c{W}_{1}\le b\right\},\left\{{X}_{1}>u+c{W}_{1}\right\}\cap \left\{u+c{W}_{1}\le b\right\},$

$\left\{{X}_{1}{}_{1}\le bb\right\}\cap \left\{u+c{W}_{1}>b\right\}$

$\begin{array}{l}{m}_{b,\delta }\left(u\right)={\int }_{0}^{\left(b-u\right)/c}{\int }_{0}^{u+ct}{\text{e}}^{-\delta t}{m}_{b,\delta }\left(u+ct-x\right){f}_{X,W}\left(x,t\right)\text{d}x\text{d}t\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\int }_{0}^{\left(b-u\right)/c}{\int }_{u+ct}^{\infty }{\text{e}}^{-\delta t}\omega \left(u+ct,x-u-ct\right){f}_{X,W}\left(x,t\right)\text{d}x\text{d}t\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\int }_{\left(b-u\right)/c}^{\infty }{\int }_{0}^{b}{\text{e}}^{-\delta t}{m}_{b,\delta }\left(b-x\right){f}_{X,W}\left(x,t\right)\text{d}x\text{d}t\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\int }_{\left(b-u\right)/c}^{\infty }{\int }_{b}^{\infty }{\text{e}}^{-\delta t}\omega \left(b,x-b\right){f}_{X,W}\left(x,t\right)\text{d}x\text{d}t.\end{array}$ (7)

${m}_{b,\delta }\left(u\right)=\underset{i=1}{\overset{n}{\sum }}{\lambda }_{i}{\int }_{0}^{\left(b-u\right)/c}{\text{e}}^{-\left({\lambda }_{i}+\delta \right)}{\sigma }_{i}\left(u+ct\right)\text{d}t+\underset{i=1}{\overset{n}{\sum }}{\lambda }_{i}{\int }_{\left(b-u\right)/c}^{\infty }{\text{e}}^{-\left({\lambda }_{i}+\delta \right)}{\sigma }_{i}\left(b\right)\text{d}t,$ (8)

${m}_{b,\delta }\left(u\right)=\underset{i=1}{\overset{n}{\sum }}\frac{{\lambda }_{i}}{c}{\int }_{u}^{\infty }{\text{e}}^{-\left(\delta +{\lambda }_{i}\right)\left(\left(t-u\right)/c\right)}{\sigma }_{i}\left(t\wedge b\right)\text{d}t$

${g}_{i}\left(u\right)={\int }_{u}^{\infty }{\text{e}}^{-\left(\delta +{\lambda }_{i}\right)\left(\left(t-u\right)/c\right)}{\sigma }_{i}\left(t\wedge b\right)\text{d}t,$

${{g}^{\prime }}_{i}\left(u\right)=\frac{{\lambda }_{i}\text{+}\delta }{c}{g}_{i}\left(u\right)-{\sigma }_{i}\left(u\right),$

$n=2$

$\begin{array}{l}{m}_{b,\delta }\left(u\right)=\frac{{\lambda }_{1}}{c}{\int }_{u}^{\infty }{\text{e}}^{-\left(\delta +{\lambda }_{1}\right)\left(\left(t-u\right)/c\right)}{\sigma }_{1}\left(t\wedge b\right)\text{d}t+\frac{{\lambda }_{2}}{c}{\int }_{u}^{\infty }{\text{e}}^{-\left(\delta +{\lambda }_{2}\right)\left(\left(t-u\right)/c\right)}{\sigma }_{2}\left(t\wedge b\right)\text{d}t\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }=\frac{{\lambda }_{1}}{c}{g}_{1}\left(u\right)\text{+}\frac{{\lambda }_{\text{2}}}{c}{g}_{\text{2}}\left(u\right),\end{array}$ (9)

$\begin{array}{l}{{m}^{\prime }}_{b,\delta }\left(u\right)=\frac{{\lambda }_{1}+\delta }{c}\frac{{\lambda }_{1}}{c}{g}_{1}\left(u\right)-\frac{{\lambda }_{2}+\delta }{c}\frac{{\lambda }_{2}}{c}{g}_{\text{2}}\left(u\right)-\frac{{\lambda }_{1}}{c}{\sigma }_{1}\left(u\right)-\frac{{\lambda }_{2}}{c}{\sigma }_{2}\left(u\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{{\lambda }_{1}+\delta }{c}\left({m}_{b,\delta }\left(u\right)-\frac{{\lambda }_{2}}{c}{g}_{\text{2}}\left(u\right)\right)+\frac{{\lambda }_{2}+\delta }{c}\left({m}_{b,\delta }\left(u\right)-\frac{{\lambda }_{1}}{c}{g}_{1}\left(u\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\frac{{\lambda }_{1}}{c}{\sigma }_{1}\left(u\right)-\frac{{\lambda }_{2}}{c}{\sigma }_{2}\left(u\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(\frac{{\lambda }_{1}+\delta }{c}+\frac{{\lambda }_{2}+\delta }{c}\right){m}_{b,\delta }\left(u\right)-\frac{{\lambda }_{1}+\delta }{c}\frac{{\lambda }_{2}}{c}{g}_{\text{2}}\left(u\right)-\frac{{\lambda }_{2}+\delta }{c}\frac{{\lambda }_{1}}{c}{g}_{1}\left(u\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\frac{{\lambda }_{1}}{c}{\sigma }_{1}\left(u\right)-\frac{{\lambda }_{2}}{c}{\sigma }_{2}\left(u\right),\end{array}$ (10)

(10)式继续对u求导

$\begin{array}{l}{{m}^{″}}_{b,\delta }\left(u\right)=\left(\frac{{\lambda }_{1}+\delta }{c}+\frac{{\lambda }_{2}+\delta }{c}\right){{m}^{\prime }}_{b,\delta }\left(u\right)-\frac{{\lambda }_{2}}{c}\frac{{\lambda }_{1}+\delta }{c}\frac{{\lambda }_{2}+\delta }{c}{g}_{\text{2}}\left(u\right)-\frac{{\lambda }_{1}}{c}\frac{{\lambda }_{1}+\delta }{c}\frac{{\lambda }_{2}+\delta }{c}{g}_{1}\left(u\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{{\lambda }_{1}}{c}\frac{{\lambda }_{2}+\delta }{c}{\sigma }_{1}\left(u\right)+\frac{{\lambda }_{2}}{c}\frac{{\lambda }_{1}+\delta }{c}{\sigma }_{2}\left(u\right)-\frac{{\lambda }_{1}}{c}{{\sigma }^{\prime }}_{1}\left(u\right)-\frac{{\lambda }_{2}}{c}{{\sigma }^{\prime }}_{2}\left(u\right),\end{array}$

$\begin{array}{l}{{m}^{″}}_{b,\delta }\left(u\right)=\left(\frac{{\lambda }_{1}+\delta }{c}+\frac{{\lambda }_{2}+\delta }{c}\right){{m}^{\prime }}_{b,\delta }\left(u\right)+\frac{{\lambda }_{1}+\delta }{c}\frac{{\lambda }_{2}+\delta }{c}{m}_{b,\delta }\left(u\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{{\lambda }_{1}}{c}\frac{{\lambda }_{2}+\delta }{c}{\sigma }_{1}\left(u\right)+\frac{{\lambda }_{2}}{c}\frac{{\lambda }_{1}+\delta }{c}{\sigma }_{2}\left(u\right)-\frac{{\lambda }_{1}}{c}{{\sigma }^{\prime }}_{1}\left(u\right)-\frac{{\lambda }_{2}}{c}{{\sigma }^{\prime }}_{2}\left(u\right),\end{array}$

$\begin{array}{l}\left(\frac{{\lambda }_{1}+\delta }{c}I-D\right)\left(\frac{{\lambda }_{2}+\delta }{c}I-D\right){m}_{b,\delta }\left(u\right)\\ =\frac{{\lambda }_{1}}{c}\left(\frac{{\lambda }_{2}+\delta }{c}I-D\right){\sigma }_{1}\left(u\right)+\frac{{\lambda }_{2}}{c}\left(\frac{{\lambda }_{1}+\delta }{c}I-D\right){\sigma }_{2}\left(u\right).\end{array}$

$\underset{i=1}{\overset{k}{\prod }}\left(\frac{{\lambda }_{i}+\delta }{c}I-D\right){m}_{b,\delta }\left(u\right)=\underset{i=1}{\overset{k}{\sum }}\underset{j=1,j\ne i}{\overset{k}{\prod }}\frac{{\lambda }_{i}}{c}\left(\frac{{\lambda }_{j}+\delta }{c}I-D\right){\sigma }_{i}\left( u \right)$

$\begin{array}{l}\underset{i=1}{\overset{k+1}{\prod }}\left(\frac{{\lambda }_{i}+\delta }{c}I-D\right){m}_{b,\delta }\left(u\right)\\ =\underset{i=1}{\overset{k}{\prod }}\left(\frac{{\lambda }_{i}+\delta }{c}I-D\right)\left(\frac{{\lambda }_{k+1}+\delta }{c}I-D\right){m}_{b,\delta }\left(u\right)\\ =\underset{i=1}{\overset{k}{\prod }}\left(\frac{{\lambda }_{i}+\delta }{c}I-D\right)\left(\frac{{\lambda }_{k+1}+\delta }{c}I-D\right)\underset{l}{\overset{k}{\sum }}\frac{{\lambda }_{l}}{c}{g}_{l}\left(u\right)\text{\hspace{0.17em}}+\underset{i=1}{\overset{k+1}{\prod }}\left(\frac{{\lambda }_{i}+\delta }{c}I-D\right)\frac{{\lambda }_{k+1}}{c}{g}_{{}_{k+1}}\left(u\right)\\ =\underset{i=1}{\overset{k}{\sum }}\underset{j=1,j\ne i}{\overset{k}{\prod }}\frac{{\lambda }_{i}}{c}\left(\frac{{\lambda }_{j}+\delta }{c}I-D\right){\sigma }_{i}\left(u\right)+\underset{j=1,j\ne i}{\overset{k+1}{\prod }}\frac{{\lambda }_{i}}{c}\left(\frac{{\lambda }_{j}+\delta }{c}I-D\right){\sigma }_{i}\left(u\right)\\ =\underset{i=1}{\overset{k+1}{\sum }}\underset{j=1,j\ne i}{\overset{k+1}{\prod }}\frac{{\lambda }_{i}}{c}\left(\frac{{\lambda }_{j}+\delta }{c}I-D\right){\sigma }_{i}\left(u\right).\end{array}$

${{m}^{\prime }}_{b,\delta }\left(u\right)$$u=b$ 时可以得出边界条件(3)，同时，(2)式在 $u=b$ 时可得到 式。

$\underset{k=1}{\overset{n}{\prod }}\left(\frac{{\lambda }_{k}+\delta }{c}I-D\right){m}_{\infty ,\delta }\left(u\right)=\underset{k=1}{\overset{n}{\sum }}\underset{j=1,j\ne k}{\overset{n}{\prod }}\frac{{\lambda }_{k}}{c}\left(\frac{{\lambda }_{j}+\delta }{c}I-D\right){\sigma }_{k}\left(u\right);$

${\sigma }_{k}\left(u\right)={\int }_{0}^{u}{m}_{\infty ,\delta }\left(u-x\right){f}_{i}\left(x\right)\text{d}x+{\omega }_{i}\left(u\right).$

4. 贴现惩罚函数的一个表示方法

${T}_{k}f\left(x\right)={\int }_{x}^{\infty }{\text{e}}^{-r\left(u-x\right)}f\left(u\right)\text{d}u,\text{\hspace{0.17em}}r\in C.$

$\underset{i=1}{\overset{n}{\prod }}\left(\frac{{\lambda }_{i}+\delta }{c}I-D\right)y\left(u\right)=\underset{i=1}{\overset{n}{\sum }}\underset{j=1,j\ne i}{\overset{n}{\prod }}\frac{{\lambda }_{i}}{c}\left(\frac{{\lambda }_{j}+\delta }{c}I-D\right){\int }_{0}^{u}y\left(u-x\right){f}_{i}\left(x\right)\text{d}x.$ (11)

$\stackrel{^}{y}\left(s\right)={\int }_{0}^{s}{\text{e}}^{-sx}y\left(x\right)\text{d}x$ ，对(11)式微积分方程两边进行拉普拉斯变换，我们能得到

$\stackrel{^}{y}\left(s\right)=\underset{k=1}{\overset{n}{\sum }}{\rho }_{k}\left(s\right){y}^{\left(k-1\right)}\left(0\right){\left[\underset{i=1}{\overset{n}{\prod }}\left(\frac{{\lambda }_{i}+\delta }{c}-s\right)-\underset{i=1}{\overset{n}{\sum }}\frac{{\lambda }_{i}}{c}\underset{j=1,j\ne i}{\overset{n}{\prod }}\left(\frac{{\lambda }_{j}-\delta }{c}-s\right){\stackrel{^}{f}}_{i}\left(s\right)\right]}^{-1}$ (12)

${\rho }_{1}\left(s\right)=-\frac{1}{s}\left[\underset{i=1}{\overset{n}{\prod }}\left(\frac{{\lambda }_{i}\text{+}\delta }{c}\text{-}s\right)-{\tau }_{\lambda ,0}\right],$

${\rho }_{2}\left(s\right)=\frac{{\left(-1\right)}^{2}}{s}\left[{\rho }_{1}\left(s\right)-{\tau }_{\lambda ,1}\right],$

$⋮$

${\rho }_{n-1}\left(s\right)={\left(-1\right)}^{n-1}\left(\underset{i=1}{\overset{n}{\sum }}\frac{{\lambda }_{i}+\delta }{c}-s\right),$

${\rho }_{n}\left(s\right)={\left(-1\right)}^{n},$

${\tau }_{\lambda ,0}=\underset{i=1}{\overset{n}{\prod }}\frac{{\lambda }_{i}+\delta }{c}$

${\tau }_{\lambda ,i}=\frac{{\tau }_{\lambda ,i-1}\left(s\right)}{\frac{{\lambda }_{j}+\delta }{c}},\text{\hspace{0.17em}}i\ne j$

$\begin{array}{l}{\stackrel{^}{y}}_{1,\delta }\left(s\right)=-\frac{1}{s}\left[\underset{i=1}{\overset{n}{\prod }}\left(\frac{{\lambda }_{i}+\delta }{c}-s\right)-\underset{i=1}{\overset{n}{\prod }}\frac{{\lambda }_{i}+\delta }{c}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdot {\left[\underset{i=1}{\overset{n}{\prod }}\left(\frac{{\lambda }_{i}+\delta }{c}-s\right)-\underset{i=1}{\overset{n}{\sum }}\frac{{\lambda }_{i}}{c}\underset{j=1,j\ne i}{\overset{n}{\prod }}\left(\frac{{\lambda }_{j}-\delta }{c}-s\right){\stackrel{^}{f}}_{i}\left(s\right)\right]}^{-1},\end{array}$ (13)

${y}_{1,\delta }\left(0\right)=0,{y}_{2,\delta }\left(0\right)=1,\cdots ,{y}_{n,\delta }\left(0\right)=0$

$\begin{array}{l}{\stackrel{^}{y}}_{2,\delta }\left(s\right)=\frac{{\left(-1\right)}^{2}}{s}\left[{\rho }_{1}\left(s\right)-\underset{i=1}{\overset{n}{\sum }}\underset{j=i,j\ne i}{\overset{n}{\prod }}\frac{{\lambda }_{j}+\delta }{c}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdot {\left[\underset{i=1}{\overset{n}{\prod }}\left(\frac{{\lambda }_{i}+\delta }{c}-s\right)-\underset{i=1}{\overset{n}{\sum }}\frac{{\lambda }_{i}}{c}\underset{j=1,j\ne i}{\overset{n}{\prod }}\left(\frac{{\lambda }_{j}-\delta }{c}-s\right){\stackrel{^}{f}}_{i}\left(s\right)\right]}^{-1},\end{array}$ (14)

$⋮$

${y}_{1,\delta }\left(0\right)=0,{y}_{2,\delta }\left(0\right)=0,\cdots ,{y}_{n,\delta }\left(0\right)=1$

${\stackrel{^}{y}}_{2,\delta }\left(s\right)={\left[\underset{i=1}{\overset{n}{\prod }}\left(\frac{{\lambda }_{i}+\delta }{c}-s\right)-\underset{i=1}{\overset{n}{\sum }}\frac{{\lambda }_{i}}{c}\underset{j=1,j\ne i}{\overset{n}{\prod }}\left(\frac{{\lambda }_{j}-\delta }{c}-s\right){\stackrel{^}{f}}_{i}\left(s\right)\right]}^{-1}.$ (15)

${m}_{b,\delta }\left(u\right)={m}_{\infty ,\delta }\left(u\right)+\underset{i=1}{\overset{n}{\sum }}{\xi }_{i}{y}_{i,\delta }\left(u\right),\text{\hspace{0.17em}}0\le u\le b,$

$\underset{i=1}{\overset{n}{\sum }}{\xi }_{i}{{y}^{\prime }}_{i,\delta }\left(b\right)=-{{m}^{\prime }}_{\infty ,\delta }\left(b\right),$ (16)

$\begin{array}{l}\underset{j=1}{\overset{n}{\sum }}{\xi }_{j}{y}_{j,\delta }^{\left(n\right)}\left(b\right)+\underset{j=1}{\overset{n}{\sum }}{\xi }_{j}\underset{i=1}{\overset{n}{\sum }}\frac{{\lambda }_{i}}{c}{D}^{n-1}{\int }_{0}^{u}{y}_{j,\delta }\left(u-x\right){f}_{i}\left(x\right)\text{d}x|{}_{u=b}\\ =-\left[{m}_{b,\delta }^{\left(n\right)}\left(b\right)+\underset{i=1}{\overset{n}{\sum }}\frac{{\lambda }_{i}}{c}{D}^{n-1}{\int }_{0}^{u}{m}_{b,\delta }\left(u\right){f}_{i}\left(x\right)\text{d}x|{}_{u=b}+\underset{i=1}{\overset{n}{\sum }}\frac{{\lambda }_{i}}{c}{\omega }_{i}^{\left(n-1\right)}\left(b\right)\right].\end{array}$ (17)

${m}_{b,\delta }\left(u\right)={m}_{\infty ,\delta }\left(u\right)+\underset{i=1}{\overset{n}{\sum }}{\xi }_{i}{y}_{i,\delta }\left(u\right),$ (18)

${{m}^{\prime }}_{\infty ,\delta }\left(u\right)+\underset{i=1}{\overset{n}{\sum }}{\xi }_{i}{{y}^{\prime }}_{i,\delta }\left(b\right)=0,$ (19)

${m}_{\infty ,\delta }^{\left(n\right)}\left(u\right)+\underset{i=1}{\overset{n}{\sum }}{\xi }_{i}{y}_{i,\delta }^{\left(n\right)}\left(b\right)=-\underset{i=1}{\overset{n}{\sum }}\frac{{\lambda }_{i}}{c}{\sigma }_{i}^{\left(n-1\right)}\left(b\right).$ (20)

(19)即为(16)式，利用 ${m}_{b,\delta }\left(u\right)$ 的结构形式(17)式，对(5)式中u求 $n-1$ 阶导，得到

${D}^{n-1}\left({\sigma }_{i}\left(u\right)\right)=\underset{i=1}{\overset{n}{\sum }}{\xi }_{i}{D}^{n-1}{\int }_{0}^{u}{y}_{i,\delta }\left(u-x\right){f}_{i}\left(x\right)\text{d}x+{D}^{n-1}{\omega }_{i}\left(u\right).$ (21)

$u=b$ 处将(21)式带入(20)右手边，得出(17)式。

${\stackrel{^}{y}}_{1,\delta }\left(s\right)=\left[\underset{i=1}{\overset{n}{\sum }}{\rho }_{1}\left(s\right)\frac{\underset{i=1}{\overset{n}{\prod }}\left(s-{s}_{i}\right)}{\left(s-{s}_{i}\right){\left[\underset{i=1}{\overset{n}{\prod }}\left(s-{s}_{i}\right)\right]}^{\prime }}\right]\cdot {\left[\underset{i=1}{\overset{n}{\prod }}\left(s-{s}_{i}\right)\right]}^{-1}{\left[1-{T}_{s}{T}_{{s}_{n}}\cdots {T}_{{s}_{2}}{T}_{{s}_{1}}{h}_{2,\delta }\left(0\right)\right]}^{-1},$

${\stackrel{^}{y}}_{2,\delta }\left(s\right)=\left[\underset{i=1}{\overset{n-1}{\sum }}{\rho }_{2}\left(s\right)\frac{\underset{i=1}{\overset{n-1}{\prod }}\left(s-{s}_{i}\right)}{\left(s-{s}_{i}\right){\left(\underset{i=1}{\overset{n-1}{\prod }}\left(s-{s}_{i}\right)\right)}^{\prime }}\right]\cdot {\left[\underset{i=1}{\overset{n}{\prod }}\left(s-{s}_{i}\right)\right]}^{-1}{\left[1-{T}_{s}{T}_{{s}_{n}}\cdots {T}_{{s}_{2}}{T}_{{s}_{1}}{h}_{2,\delta }\left(0\right)\right]}^{-1},$

$⋮$

${\stackrel{^}{y}}_{n,\delta }\left(s\right)={\left[\underset{i=1}{\overset{n}{\prod }}\left(s-{s}_{i}\right)\right]}^{-1}{\left[1-{T}_{s}{T}_{{s}_{n}}\cdots {T}_{{s}_{2}}{T}_{{s}_{1}}{h}_{2,\delta }\left(0\right)\right]}^{-1},$

${\stackrel{^}{h}}_{2,\delta }\left(s\right)=\underset{i=1}{\overset{n}{\sum }}\frac{{\lambda }_{i}}{c}\underset{j=1,j\ne i}{\overset{n}{\prod }}\left(\frac{{\lambda }_{j}+\delta }{c}-s\right){\stackrel{^}{f}}_{i}\left(s\right).$

${y}_{1,\delta }\left(u\right)={k}_{\delta }{\int }_{0}^{u}{y}_{1,\delta }\left(u-y\right){g}_{\delta }\left(y\right)\text{d}y+\underset{i=1}{\overset{n}{\sum }}\frac{{\rho }_{1}\left({s}_{i}\right){\text{e}}^{{s}_{i}u}}{\underset{j=1,j\ne i}{\overset{n}{\prod }}\left({s}_{i}-{s}_{j}\right)},$

${y}_{2,\delta }\left(u\right)={k}_{\delta }{\int }_{0}^{u}{y}_{2,\delta }\left(u-y\right){g}_{\delta }\left(y\right)\text{d}y+\underset{i=1}{\overset{n-1}{\sum }}\frac{{\rho }_{2}\left({s}_{i}\right)\left({\text{e}}^{{s}_{i}u}-{\text{e}}^{{s}_{n}u}\right)}{\underset{j=1,j\ne i}{\overset{n}{\prod }}\left({s}_{i}-{s}_{j}\right)},$

$⋮$

${y}_{n,\delta }\left(u\right)={k}_{\delta }{\int }_{0}^{u}{y}_{n,\delta }\left(u-y\right){g}_{\delta }\left(y\right)\text{d}y+\underset{i=1}{\overset{n-1}{\sum }}\frac{{\text{e}}^{{s}_{i}u}}{\underset{j=1,j\ne i}{\overset{n}{\prod }}\left({s}_{i}-{s}_{j}\right)}.$

${g}_{\delta }\left(y\right)=\frac{{T}_{{s}_{n}}\cdots {T}_{{s}_{2}}{T}_{{s}_{1}}}{{k}_{\delta }}$

${\tau }_{\delta ,0}\left(s\right)=\underset{i=1}{\overset{n}{\prod }}\left(\frac{{\lambda }_{i}+\delta }{c}-{s}_{i}\right)$

${\tau }_{\delta ,i}\left(s\right)=\frac{{\tau }_{\delta ,i-1}\left(s\right)}{\frac{{\lambda }_{j}+\delta }{c}-{s}_{j}},\text{\hspace{0.17em}}i\ne j$

${L}_{\delta }\left(u\right)=1-\underset{n=1}{\overset{\infty }{\sum }}\left(1-{k}_{\delta }\right){\left({k}_{\delta }\right)}^{n}{\stackrel{¯}{G}}_{\delta }^{\ast n}\left(y\right),$

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