﻿ 二维对偶模型的最优分红 Optimal Dividend in the Two-Dimensional Dual Model

Statistics and Application
Vol.06 No.05(2017), Article ID:23123,10 pages
10.12677/SA.2017.65058

Optimal Dividend in the Two-Dimensional Dual Model

Mi Han, Shixia Ma, Tong Li

Hebei University of Technology, Tianjin

Received: Dec. 1st, 2017; accepted: Dec. 15th, 2017; published: Dec. 22nd, 2017

ABSTRACT

We study the two-dimensional dual model with diffusion under a threshold dividend strategy. We obtain a group of integro-differential equations satisfied by the expected total sum of discount dividends until ruin. And explicit results when the gains of the two projects are exponentially distributed are derived. By applying the method of the Laplace transform, we solve the case where gains follow general distributions. We also illustrate how to calculate the optimal dividend threshold.

Keywords:Two-Dimensional Dual Model, Threshold Strategy, Diffusion, Proportional Transaction Costs, Laplace Transform

1. 引言

2. 模型的建立

${X}_{1}\left(t\right)={x}_{1}-{c}_{11}t+\sum _{n=1}^{{N}_{1}\left(t\right)}{U}_{n},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{X}_{2}\left(t\right)={x}_{2}-{c}_{12}t+\sum _{n=1}^{{N}_{2}\left(t\right)}{V}_{n},$

$X\left(t\right)=x-{c}_{1}t+{S}_{1}\left(t\right)+{S}_{2}\left(t\right)+\sigma W\left(t\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}t\ge 0$

$\text{d}X\left(t\right)=-C\left(X\left(t\right)\right)\text{d}t+\text{d}{S}_{1}\left(t\right)+\text{d}{S}_{2}\left(t\right)+\sigma \text{d}W\left(t\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}X\left(0\right)=x,$

$C\left(x\right)=\left\{\begin{array}{ll}{c}_{1},\hfill & 0b.\hfill \end{array}$ (1)

$V\left(x,b\right)=E\left[{\int }_{0}^{T}\eta \left({c}_{2}-{c}_{1}\right){\text{e}}^{-\delta t}I\left(X\left(t\right)>b\right)\text{d}t\right],$

3. 积分–微分方程

$0\le x 时，

$\begin{array}{l}\frac{{\sigma }^{2}}{2}{V}_{xx}\left(x,b\right)-{c}_{1}{V}_{x}\left(x,b\right)-\left({\lambda }_{1}+{\lambda }_{2}+\delta \right)V\left(x,b\right)\\ +{\lambda }_{1}{\int }_{0}^{\infty }V\left(x+u,b\right)\text{d}{F}_{U}\left(u\right)+{\lambda }_{2}{\int }_{0}^{\infty }V\left(x+v,b\right)\text{d}{F}_{V}\left(v\right)=0,\end{array}$ (2)

$x>b$ 时，

$\begin{array}{l}\frac{{\sigma }^{2}}{2}{V}_{xx}\left(x,b\right)-{c}_{2}{V}_{x}\left(x,b\right)-\left({\lambda }_{1}+{\lambda }_{2}+\delta \right)V\left(x,b\right)\\ +{\lambda }_{1}{\int }_{0}^{\infty }V\left(x+u,b\right)\text{d}{F}_{U}\left(u\right)+{\lambda }_{2}{\int }_{0}^{\infty }V\left(x+v,b\right)\text{d}{F}_{V}\left(v\right)+\eta \left({c}_{2}-{c}_{1}\right)=0,\end{array}$ (3)

(1) 在到时刻 $h$ 前，两类项目均无收益；

(2) 在到时刻 $h$ 前，第一类项目有收益第二类项目无收益；

(3) 在到时刻 $h$ 前，第一类项目无收益第二类项目有收益；

(4) 在到时刻 $h$ 前，两类项目都有收益；

$\begin{array}{c}V\left(x,b\right)=\eta \left({c}_{2}-{c}_{1}\right){\int }_{0}^{h}{\text{e}}^{-\delta t}\text{d}t+\left(1-\delta h\right)\left\{\left(1-{\lambda }_{1}h\right)\left(1-{\lambda }_{2}h\right)E\left[V\left(x-{c}_{2}h+\sigma W\left(h\right),b\right)\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\lambda }_{1}h\left(1-{\lambda }_{2}h\right){\int }_{0}^{\infty }E\left[V\left(x+u-{c}_{2}h+\sigma W\left(h\right),b\right)\right]\text{d}{F}_{U}\left(u\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left(1-{\lambda }_{1}h\right){\lambda }_{2}h{\int }_{0}^{\infty }E\left[V\left(x+v-{c}_{2}h+\sigma W\left(h\right),b\right)\right]\text{d}{F}_{V}\left(v\right)\right\}+o\left(h\right).\end{array}$ (4)

$V\left(x-{c}_{2}h+\sigma W\left(h\right),b\right)=V\left(x,b\right)+{V}_{x}\left(x,b\right)\left(-{c}_{2}h+\sigma W\left(h\right)\right)+{V}_{xx}\left(x,b\right)\frac{{\left(-{c}_{2}h+\sigma W\left(h\right)\right)}^{2}}{2}+o\left(h\right).$

$E\left[V\left(x-{c}_{2}h+\sigma W\left(h\right),b\right)\right]=V\left(x,b\right)-{c}_{2}h{V}_{x}\left(x,b\right)+\frac{{\sigma }^{2}}{2}h{V}_{xx}\left(x,b\right)+o\left(h\right).$ (5)

$\begin{array}{l}\frac{{\sigma }^{2}}{2}h{V}_{xx}\left(x,b\right)+\eta h\left({c}_{2}-{c}_{1}\right)-{c}_{2}h{V}_{x}\left(x,b\right)-\left({\lambda }_{1}+{\lambda }_{2}+\delta \right)hV\left(x,b\right)\\ +{\lambda }_{1}h{\int }_{0}^{\infty }V\left(x+u,b\right)\text{d}{F}_{U}\left(u\right)+{\lambda }_{2}h{\int }_{0}^{\infty }V\left(x+v,b\right)\text{d}{F}_{V}\left(v\right)+o\left(h\right)=0.\end{array}$ (6)

$\begin{array}{l}\frac{{\sigma }^{2}}{2}{V}_{xx}\left(b-,b\right)-{c}_{1}{V}_{x}\left(b-,b\right)-\left({\lambda }_{1}+{\lambda }_{2}+\delta \right)V\left(b-,b\right)\\ +{\lambda }_{1}{\int }_{0}^{\infty }V\left(b+u,b\right)\text{d}{F}_{U}\left(u\right)+{\lambda }_{2}{\int }_{0}^{\infty }V\left(b+v,b\right)\text{d}{F}_{V}\left(v\right)=0,\end{array}$

$\begin{array}{l}\frac{{\sigma }^{2}}{2}{V}_{xx}\left(b+,b\right)-{c}_{2}{V}_{x}\left(b+,b\right)-\left({\lambda }_{1}+{\lambda }_{2}+\delta \right)V\left(b+,b\right)\\ +{\lambda }_{1}{\int }_{0}^{\infty }V\left(b+u,b\right)\text{d}{F}_{U}\left(u\right)+{\lambda }_{2}{\int }_{0}^{\infty }V\left(b+v,b\right)\text{d}{F}_{V}\left(v\right)+\eta \left({c}_{2}-{c}_{1}\right)=0.\end{array}$

$\frac{{\sigma }^{2}}{2}\left[{V}_{xx}\left(b-,b\right)-{V}_{xx}\left(b+,b\right)\right]+{c}_{2}{V}_{x}\left(b+,b\right)-{c}_{1}{V}_{x}\left(b-,b\right)=\eta \left({c}_{2}-{c}_{1}\right),$

$V\left(x,b\right)$ 是有界的。

$V\left(x,b\right)=\left\{\begin{array}{ll}{V}_{1}\left(x\right),\hfill & 0b.\hfill \end{array}$ (7)

$0\le x 时，

$\begin{array}{l}\frac{{\sigma }^{2}}{2}{{V}^{″}}_{1}\left(x\right)-{c}_{1}{{V}^{\prime }}_{1}\left(x\right)-\left(\lambda +\delta \right){V}_{1}\left(x\right)+{\lambda }_{1}{\int }_{0}^{b-x}{V}_{1}\left(x+u\right)\text{d}{F}_{U}\left(u\right)\\ +{\lambda }_{1}{\int }_{b-x}^{\infty }{V}_{2}\left(x+u\right)\text{d}{F}_{U}\left(u\right)+{\lambda }_{2}{\int }_{0}^{b-x}{V}_{2}\left(x+v\right)\text{d}{F}_{V}\left(v\right)+{\lambda }_{2}{\int }_{b-x}^{\infty }{V}_{2}\left(x+v\right)\text{d}{F}_{V}\left(v\right)=0,\end{array}$ (8)

$x>b$ 时，

$\begin{array}{l}\frac{{\sigma }^{2}}{2}{{V}^{″}}_{2}\left(x\right)-{c}_{2}{{V}^{\prime }}_{2}\left(x\right)-\left(\lambda +\delta \right){V}_{2}\left(x\right)+{\lambda }_{1}{\int }_{0}^{\infty }{V}_{2}\left(x+u\right)\text{d}{F}_{U}\left(u\right)\\ +{\lambda }_{2}{\int }_{0}^{\infty }{V}_{2}\left(x+v\right)\text{d}{F}_{V}\left(v\right)+\eta \left({c}_{2}-{c}_{1}\right)=0,\end{array}$ (9)

3.1. 收益服从指数分布

$x>b$ 时，

$\begin{array}{l}\frac{{\sigma }^{2}}{2}{{V}^{″}}_{2}\left(x\right)-{c}_{2}{{V}^{\prime }}_{2}\left(x\right)-\left(\lambda +\delta \right){V}_{2}\left(x\right)+{\lambda }_{1}{\int }_{x}^{\infty }{V}_{2}\left(u\right)\alpha {\text{e}}^{-\alpha \left(u-x\right)}\text{d}u\\ +{\lambda }_{2}{\int }_{x}^{\infty }{V}_{2}\left(v\right)\beta {\text{e}}^{-\beta \left(v-x\right)}\text{d}v+\eta \left({c}_{2}-{c}_{1}\right)=0.\end{array}$ (10)

$\mathcal{I}$ 表示值函数 ${V}_{2}\left(x\right)$ 的恒等算子， $\mathcal{Q}$ 表示值函数 ${V}_{2}\left(x\right)$ 的微分算子。对(10)两侧进行 $\mathcal{Q}\left(\mathcal{Q}-\alpha \mathcal{I}\right)-\beta \left(\mathcal{Q}-\alpha \mathcal{I}\right)$ 运算得到

$\begin{array}{l}\frac{{\sigma }^{2}}{2}{V}_{2}^{\left(4\right)}\left(x\right)-\left[\frac{{\sigma }^{2}}{2}\left(\alpha +\beta \right)+{c}_{2}\right]{V}_{2}^{\left(3\right)}\left(x\right)+\left[{c}_{2}\left(\alpha +\beta \right)-\left(\lambda +\delta \right)+\alpha \beta \frac{{\sigma }^{2}}{2}\right]{V}_{2}^{\left(2\right)}\left(x\right)\\ +\left[\left(\alpha +\beta \right)\left(\delta +\lambda \right)-{\lambda }_{1}\alpha -{\lambda }_{2}\beta -\alpha \beta {c}_{2}\right]{{V}^{\prime }}_{2}\left(x\right)-\alpha \beta \delta {V}_{2}\left(x\right)+\alpha \beta \eta \left({c}_{2}-{c}_{1}\right)=0.\end{array}$

$\begin{array}{l}\frac{{\sigma }^{2}}{2}{r}^{4}-\left[\frac{{\sigma }^{2}}{2}\left(\alpha +\beta \right)+{c}_{2}\right]{r}^{3}+\left[{c}_{2}\left(\alpha +\beta \right)-\left(\lambda +\delta \right)+\alpha \beta \frac{{\sigma }^{2}}{2}\right]{r}^{2}\\ +\left[\left(\alpha +\beta \right)\left(\delta +\lambda \right)-{\lambda }_{1}\alpha -{\lambda }_{2}\beta -\alpha \beta {c}_{2}\right]r-\alpha \beta \delta =0,\end{array}$

${V}_{2}\left(x\right)={D}_{1}{\text{e}}^{{r}_{1}x}+{D}_{2}{\text{e}}^{{r}_{2}x}+{D}_{3}{\text{e}}^{{r}_{3}x}+{D}_{4}{\text{e}}^{{r}_{4}x}+\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta },$ (11)

${{V}^{\prime }}_{2}\left(x\right)={D}_{1}{r}_{1}{\text{e}}^{{r}_{1}x}+{D}_{2}{r}_{2}{\text{e}}^{{r}_{2}x}+{D}_{3}{r}_{3}{\text{e}}^{{r}_{3}x}+{D}_{4}{r}_{4}{\text{e}}^{{r}_{4}x}.$

${D}_{4}<0$ ，当 $x$ 足够大时有 ${{V}^{\prime }}_{2}\left(x\right)<0$ ，这与 ${V}_{2}\left(x\right)$ 是递增的事实相矛盾，所以 ${D}_{4}=0$ 。同样可证得 ${D}_{3}={D}_{2}=0$${D}_{1}<0$ ，故

${V}_{2}\left(x\right)={D}_{1}{\text{e}}^{{r}_{1}x}+\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }.$

$0\le x\le b$ 时，

$\begin{array}{l}\frac{{\sigma }^{2}}{2}{{V}^{″}}_{1}\left(x\right)-{c}_{1}{{V}^{\prime }}_{1}\left(x\right)-\left(\lambda +\delta \right){V}_{1}\left(x\right)\\ +\alpha {\lambda }_{1}{\text{e}}^{\alpha x}\left[{\int }_{x}^{b}{V}_{1}\left(u\right){\text{e}}^{-\alpha u}\text{d}u+{D}_{1}{\int }_{b}^{\infty }{\text{e}}^{\left({r}_{1}-\alpha \right)u}\text{d}u+\eta \frac{{c}_{2}-{c}_{1}}{\delta }{\int }_{b}^{\infty }{\text{e}}^{-\alpha u}\text{d}u\right]\\ +\beta {\lambda }_{2}{\text{e}}^{\beta x}\left[{\int }_{x}^{b}{V}_{1}\left(v\right){\text{e}}^{-\beta v}\text{d}v+{D}_{1}{\int }_{b}^{\infty }{\text{e}}^{\left({r}_{1}-\beta \right)v}\text{d}v+\eta \frac{{c}_{2}-{c}_{1}}{\delta }{\int }_{b}^{\infty }{\text{e}}^{-\beta v}\text{d}v\right]=0,\end{array}$ (12)

$\begin{array}{l}\frac{{\sigma }^{2}}{2}{V}_{1}^{\left(4\right)}\left(x\right)-\left[\frac{{\sigma }^{2}}{2}\left(\alpha +\beta \right)+{c}_{1}\right]{V}_{1}^{\left(3\right)}\left(x\right)+\left[{c}_{1}\left(\alpha +\beta \right)-\left(\lambda +\delta \right)+\alpha \beta \frac{{\sigma }^{2}}{2}\right]{V}_{1}^{\left(2\right)}\left(x\right)\\ +\left[\left(\alpha +\beta \right)\left(\delta +\lambda \right)-{\lambda }_{1}\alpha -{\lambda }_{2}\beta -\alpha \beta {c}_{1}\right]{{V}^{\prime }}_{1}\left(x\right)-\alpha \beta \delta {V}_{1}\left(x\right)=0.\end{array}$

$\begin{array}{l}\frac{{\sigma }^{2}}{2}{s}^{4}-\left[\frac{{\sigma }^{2}}{2}\left(\alpha +\beta \right)+{c}_{2}\right]{s}^{3}+\left[{c}_{1}\left(\alpha +\beta \right)-\left(\lambda +\delta \right)+\alpha \beta \frac{{\sigma }^{2}}{2}\right]{s}^{2}\\ +\left[\left(\alpha +\beta \right)\left(\delta +\lambda \right)-{\lambda }_{1}\alpha -{\lambda }_{2}\beta -\alpha \beta {c}_{1}\right]s-\alpha \beta \delta =0.\end{array}$

${E}_{1}+{E}_{2}+{E}_{3}+{E}_{4}=0$ (13)

${D}_{1}{\text{e}}^{{r}_{1}b}+\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }={E}_{1}{\text{e}}^{{s}_{1}b}+{E}_{2}{\text{e}}^{{s}_{2}b}+{E}_{3}{\text{e}}^{{s}_{3}b}+{E}_{4}{\text{e}}^{{s}_{4}b}.$ (14)

${D}_{1}{r}_{1}{\text{e}}^{{r}_{1}b}={E}_{1}{s}_{1}{\text{e}}^{{s}_{1}b}+{E}_{2}{s}_{2}{\text{e}}^{{s}_{2}b}+{E}_{3}{s}_{3}{\text{e}}^{{s}_{3}b}+{E}_{4}{s}_{4}{\text{e}}^{{s}_{4}b}.$ (15)

${V}_{1}$${V}_{2}$ 的表达式代入(12)，可得

$\begin{array}{l}\frac{{\sigma }^{2}}{2}\left({E}_{1}{s}_{1}^{2}{\text{e}}^{{s}_{1}b}+{E}_{2}{s}_{2}^{2}{\text{e}}^{{s}_{2}b}+{E}_{3}{s}_{3}^{2}{\text{e}}^{{s}_{3}b}+{E}_{4}{s}_{4}^{2}{\text{e}}^{{s}_{4}b}\right)-{c}_{1}\left({E}_{1}{s}_{1}{\text{e}}^{{s}_{1}b}+{E}_{2}{s}_{2}{\text{e}}^{{s}_{2}b}+{E}_{3}{s}_{3}{\text{e}}^{{s}_{3}b}+{E}_{4}{s}_{4}{\text{e}}^{{s}_{4}b}\right)\\ -\left(\lambda +\delta \right)\left({E}_{1}{\text{e}}^{{s}_{1}x}+{E}_{2}{\text{e}}^{{s}_{2}x}+{E}_{3}{\text{e}}^{{s}_{3}x}+{E}_{4}{\text{e}}^{{s}_{4}x}\right)+\alpha {\lambda }_{1}{\text{e}}^{\alpha x}\left[{\int }_{x}^{b}\left({E}_{1}{\text{e}}^{{s}_{1}x}+{E}_{2}{\text{e}}^{{s}_{2}x}+{E}_{3}{\text{e}}^{{s}_{3}x}+{E}_{4}{\text{e}}^{{s}_{4}x}\right){\text{e}}^{-\alpha u}\text{d}u\stackrel{}{}\\ +{D}_{1}{\int }_{b}^{\infty }{\text{e}}^{\left({r}_{1}-\alpha \right)u}\text{d}u+\eta \frac{{c}_{2}-{c}_{1}}{\delta }{\int }_{b}^{\infty }{\text{e}}^{-\alpha u}\text{d}u\right]+\beta {\lambda }_{2}{\text{e}}^{\beta x}\left[{\int }_{x}^{b}\left({E}_{1}{\text{e}}^{{s}_{1}x}+{E}_{2}{\text{e}}^{{s}_{2}x}+{E}_{3}{\text{e}}^{{s}_{3}x}+{E}_{4}{\text{e}}^{{s}_{4}x}\right){\text{e}}^{-\beta v}\text{d}v\stackrel{}{}\\ +{D}_{1}{\int }_{b}^{\infty }{\text{e}}^{\left({r}_{1}-\beta \right)v}\text{d}v+\eta \frac{{c}_{2}-{c}_{1}}{\delta }{\int }_{b}^{\infty }{\text{e}}^{-\beta v}\text{d}v\right]=0.\end{array}$

$\frac{{E}_{1}}{{s}_{1}-\alpha }{\text{e}}^{{s}_{1}b}+\frac{{E}_{2}}{{s}_{2}-\alpha }{\text{e}}^{{s}_{2}b}+\frac{{E}_{3}}{{s}_{3}-\alpha }{\text{e}}^{{s}_{3}b}+\frac{{E}_{4}}{{s}_{4}-\alpha }{\text{e}}^{{s}_{4}b}-\frac{{D}_{1}}{{r}_{1}-\alpha }{\text{e}}^{{r}_{1}b}+\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\alpha \delta }=0,$ (16)

$\frac{{E}_{1}}{{s}_{1}-\beta }{\text{e}}^{{s}_{1}b}+\frac{{E}_{2}}{{s}_{2}-\beta }{\text{e}}^{{s}_{2}b}+\frac{{E}_{3}}{{s}_{3}-\beta }{\text{e}}^{{s}_{3}b}+\frac{{E}_{4}}{{s}_{4}-\beta }{\text{e}}^{{s}_{4}b}-\frac{{D}_{1}}{{r}_{1}-\beta }{\text{e}}^{{r}_{1}b}+\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\beta \delta }=0.$ (17)

$V\left(x,b\right)=\left\{\begin{array}{ll}{E}_{1}{\text{e}}^{{s}_{1}x}+{E}_{2}{\text{e}}^{{s}_{2}x}+{E}_{3}{\text{e}}^{{s}_{3}x}+{E}_{4}{\text{e}}^{{s}_{4}x},\hfill & 0\le x\le b,\hfill \\ {D}_{1}{\text{e}}^{{r}_{1}x}+\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta },\hfill & x>b.\hfill \end{array}$ (18)

3.2. 用 ${V}_{1}\left(x\right)$ 求解 $V\left(x,b\right)$

${V}_{2}\left(x\right)={D}_{1}{\text{e}}^{{r}_{1}x}+\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta },$ (19)

$\stackrel{˜}{X}\left(t\right)=x-ct+{S}_{1}\left(t\right)+{S}_{2}\left(t\right)+\sigma W\left(t\right).$

$\psi \left(x\right)=E\left[{\text{e}}^{-\delta \stackrel{˜}{T}}|\stackrel{˜}{X}\left(0\right)=x\right]={\text{e}}^{{R}_{\delta }x}$ (20)

$-\delta -c\theta +{\lambda }_{1}\left({M}_{U}\left(\theta \right)-1\right)+{\lambda }_{2}\left({M}_{V}\left(\theta \right)-1\right)+\frac{{\sigma }^{2}{\theta }^{2}}{2}=0,$

${\text{e}}^{-\delta t+\theta x-ct\theta +{\lambda }_{1}t\left({M}_{U}\left(\theta \right)-1\right)+{\lambda }_{2}t\left({M}_{V}\left(\theta \right)-1\right)+t\frac{{\sigma }^{2}{\theta }^{2}}{2}}={\text{e}}^{\theta x}.$

$-\delta -c\theta +{\lambda }_{1}\left({M}_{U}\left(\theta \right)-1\right)+{\lambda }_{2}\left({M}_{V}\left(\theta \right)-1\right)+\frac{{\sigma }^{2}{\theta }^{2}}{2}=0.$

$\varphi \left(\theta \right):=-c\theta +{\lambda }_{1}\left({M}_{U}\left(\theta \right)-1\right)+{\lambda }_{2}\left({M}_{V}\left(\theta \right)-1\right)+\frac{{\sigma }^{2}{\theta }^{2}}{2}$ ，可得到

$\underset{\theta \to -\infty }{\mathrm{lim}}\varphi \left(\theta \right)=+\infty ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\varphi \left(0\right)=0,$

${\varphi }^{″}\left(\theta \right)={\lambda }_{1}{\int }_{0}^{\infty }{u}^{2}{\text{e}}^{-\theta u}\text{d}{F}_{U}\left(u\right)+{\lambda }_{2}{\int }_{0}^{\infty }{v}^{2}{\text{e}}^{-\theta v}\text{d}{F}_{V}\left(v\right)+{\sigma }^{2}>0,$

$V\left(x,b\right)=\eta \left({c}_{2}-{c}_{1}\right)\frac{1-{\text{e}}^{{\overline{R}}_{\delta }\left(x-b\right)}}{\delta }+{\text{e}}^{{\overline{R}}_{\delta }\left(x-b\right)}V\left(b,b\right),$ (21)

${\overline{R}}_{\delta }$ 是以下拓展的Lundberg基本方程的唯一非正根

$\frac{{\sigma }^{2}{\theta }^{2}}{2}+{\lambda }_{1}\left({M}_{U}\left(\theta \right)-1\right)+{\lambda }_{2}\left({M}_{V}\left(\theta \right)-1\right)=\delta +{c}_{2}\theta .$

$b$ 后到破产时的分红现值期望为

$\begin{array}{c}V\left(b+\chi \right)=V\left(x,b\right)\\ =E\left[\eta \left({c}_{2}-{c}_{1}\right){\int }_{0}^{{\overline{T}}_{-\chi }}{\text{e}}^{-\delta t}|X\left(0\right)-b=\chi \right]+{\text{e}}^{\overline{{R}_{\delta }}\chi }V\left(b,b\right)\\ =-\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\left({\text{e}}^{\overline{{R}_{\delta }}\chi }-1\right)+{\text{e}}^{\overline{{R}_{\delta }}\chi }V\left(b,b\right)\\ =\eta \left({c}_{2}-{c}_{1}\right)\frac{1-{\text{e}}^{{\overline{R}}_{\delta }\left(x-b\right)}}{\delta }+{\text{e}}^{{\overline{R}}_{\delta }\left(x-b\right)}V\left(b,b\right).\end{array}$

$\begin{array}{l}\frac{{\sigma }^{2}}{2}{V}_{xx}\left(x,b\right)-{c}_{1}{V}_{x}\left(x,b\right)-\left(\lambda +\delta \right)V\left(x,b\right)+{\lambda }_{1}{\int }_{0}^{b-x}V\left(x+u,b\right)\text{d}{F}_{U}\left(u\right)\\ +{\lambda }_{1}\left\{{\int }_{b-x}^{\infty }\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\text{d}{F}_{U}\left(u\right)+\left[V\left(b,b\right)-\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\right]{\int }_{b-x}^{\infty }{\text{e}}^{{\overline{R}}_{\delta }\left(x+u-b\right)}\text{d}{F}_{U}\left(u\right)\right\}\\ +{\lambda }_{2}{\int }_{0}^{b-x}V\left(x+v,b\right)\text{d}{F}_{V}\left(v\right)+{\lambda }_{2}\left\{{\int }_{b-x}^{\infty }\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\text{d}{F}_{V}\left(v\right)\\ +\left[V\left(b,b\right)-\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\right]{\int }_{b-x}^{\infty }{\text{e}}^{{\overline{R}}_{\delta }\left(x+v-b\right)}\text{d}{F}_{V}\left(v\right)\right\}=0.\end{array}$ (22)

4. Laplace变换

$H\left(z,b\right)=V\left(b-z,b\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le z\le b.$ (23)

$\begin{array}{l}\frac{{\sigma }^{2}}{2}{H}_{zz}\left(z,b\right)-{c}_{1}{H}_{z}\left(z,b\right)-\left(\lambda +\delta \right)H\left(z,b\right)+{\lambda }_{1}{\int }_{0}^{z}H\left(y,b\right){P}_{U}\left(z-y\right)\text{d}y\\ +{\lambda }_{1}\left\{\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\left(1-{F}_{U}\left(u\right)\right)+\left[H\left(0,b\right)-\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\right]{\int }_{z}^{\infty }{\text{e}}^{{\overline{R}}_{\delta }\left(u-z\right)}\text{d}{F}_{U}\left(u\right)\right\}\\ +{\lambda }_{2}{\int }_{0}^{z}H\left(y,b\right){P}_{V}\left(z-y\right)\text{d}y+{\lambda }_{2}\left\{\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\left(1-{F}_{V}\left(v\right)\right)\\ +\left[H\left(0,b\right)-\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\right]{\int }_{z}^{\infty }{\text{e}}^{{\overline{R}}_{\delta }\left(u-z\right)}\text{d}{F}_{V}\left(v\right)\right\}=0.\end{array}$ (24)

$\begin{array}{l}\frac{{\sigma }^{2}}{2}{w}^{″}\left(z\right)-{c}_{1}{w}^{\prime }\left(z\right)-\left(\lambda +\delta \right)w\left(z\right)+{\lambda }_{1}{\int }_{0}^{z}w\left(y\right){P}_{U}\left(z-y\right)\text{d}y\\ +{\lambda }_{1}\left\{\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\left(1-{F}_{U}\left(u\right)\right)+\left[w\left(0\right)-\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\right]{\int }_{z}^{\infty }{\text{e}}^{{\overline{R}}_{\delta }\left(u-z\right)}\text{d}{F}_{U}\left(u\right)\right\}\\ +{\lambda }_{2}{\int }_{0}^{z}w\left(y\right){P}_{V}\left(z-y\right)\text{d}y+{\lambda }_{2}\left\{\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\left(1-{F}_{V}\left(v\right)\right)\\ +\left[w\left(0\right)-\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\right]{\int }_{z}^{\infty }{\text{e}}^{{\overline{R}}_{\delta }\left(u-z\right)}\text{d}{F}_{V}\left(v\right)\right\}=0.\end{array}$ (25)

$\stackrel{^}{w}$${\stackrel{^}{P}}_{U}$${\stackrel{^}{P}}_{V}$ 分别表示 $w$${P}_{U}$${P}_{V}$ 的Laplace变换。对上式中 $w\left(z\right)$${P}_{U}\left(u\right)$${P}_{V}\left(v\right)$ 均做Laplace变换，可得

$\begin{array}{l}\frac{{\sigma }^{2}{\xi }^{2}}{2}\stackrel{^}{w}\left(\xi \right)-\frac{{\sigma }^{2}\xi }{2}w\left(0\right)-\frac{{\sigma }^{2}}{2}{w}^{\prime }\left(0\right)-{c}_{1}\xi \stackrel{^}{w}\left(\xi \right)+{c}_{1}w\left(0\right)\\ -\left(\lambda +\delta \right)\stackrel{^}{w}\left(z\right)+{\lambda }_{1}\stackrel{^}{w}\left(\xi \right){\stackrel{^}{P}}_{U}\left(\xi \right)+\frac{{\lambda }_{1}\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\frac{1-{\stackrel{^}{P}}_{U}\left(\xi \right)}{\xi }\\ +{\lambda }_{1}\left[w\left(0\right)-\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\right]\frac{{\stackrel{^}{P}}_{U}\left(-{\overline{R}}_{\delta }\right)-{\stackrel{^}{P}}_{U}\left(\xi \right)}{\xi +{\overline{R}}_{\delta }}+{\lambda }_{2}\stackrel{^}{w}\left(\xi \right){\stackrel{^}{P}}_{v}\left(\xi \right)\\ +\frac{{\lambda }_{2}\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\frac{1-{\stackrel{^}{P}}_{v}\left(\xi \right)}{\xi }+{\lambda }_{2}\left[w\left(0\right)-\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\right]\frac{{\stackrel{^}{P}}_{v}\left(-{\overline{R}}_{\delta }\right)-{\stackrel{^}{P}}_{v}\left(\xi \right)}{\xi +{\overline{R}}_{\delta }}=0,\end{array}$

$\stackrel{^}{w}\left(\xi \right)=\frac{f\left(\xi \right)}{\frac{{\sigma }^{2}{\xi }^{2}}{2}-{c}_{1}\xi -\left(\lambda +\delta \right)+{\lambda }_{1}{\stackrel{^}{P}}_{U}+{\lambda }_{2}{\stackrel{^}{P}}_{V}},$ (26)

$\begin{array}{c}f\left(\xi \right)=-\frac{{\lambda }_{1}\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\frac{1-{\stackrel{^}{P}}_{U}\left(\xi \right)}{\xi }-{\lambda }_{1}\left[w\left(0\right)-\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\right]\frac{{\stackrel{^}{P}}_{U}\left(-{\overline{R}}_{\delta }\right)-{\stackrel{^}{P}}_{U}\left(\xi \right)}{\xi +{\overline{R}}_{\delta }}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-\frac{{\lambda }_{2}\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\frac{1-{\stackrel{^}{P}}_{v}\left(\xi \right)}{\xi }-{\lambda }_{2}\left[w\left(0\right)-\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }\right]\frac{{\stackrel{^}{P}}_{v}\left(-{\overline{R}}_{\delta }\right)-{\stackrel{^}{P}}_{v}\left(\xi \right)}{\xi +{\overline{R}}_{\delta }}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{{\sigma }^{2}\xi }{2}w\left(0\right)+\frac{{\sigma }^{2}}{2}{w}^{\prime }\left(0\right)-{c}_{1}w\left(0\right).\end{array}$

$w\left(0\right)=V\left(b,b\right)$${w}^{\prime }\left(0\right)={V}_{x}\left(b,b\right)$ 可解得 $V\left(x,b\right)$$b$ 。过程如下：对(26)式求逆得到 $w\left(z\right)$ ，由 $w\left(b\right)=0$ 可解得 $b$ 值，最后代换得到当 $0\le x\le b$ 时， $V\left(x,b\right)=w\left(b-x\right)$

5. 求解最优策略

${\frac{\partial V\left(x,b\right)}{\partial b}|}_{b={b}^{*}}=0.$ (27)

${V}_{xx}\left(b-,b\right)+{\frac{{\partial }^{2}V\left(x,b\right)}{\partial x\partial b}|}_{b=b-}={V}_{xx}\left(b+,b\right)+{\frac{{\partial }^{2}V\left(x,b\right)}{\partial x\partial b}|}_{b=b+}.$

${V}_{xx}\left({b}^{*}-,{b}^{*}\right)={V}_{xx}\left({b}^{*}+,{b}^{*}\right).$

${V}_{x}\left({b}^{*},{b}^{*}\right)={V}_{x}\left({b}^{*}-,{b}^{*}\right)={V}_{x}\left({b}^{*}+,{b}^{*}\right)=\eta .$

${V}_{x}\left(x,b\right)=-\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }{\overline{R}}_{\delta }{\text{e}}^{{\overline{R}}_{\delta }\left(x-b\right)}+{\overline{R}}_{\delta }{\text{e}}^{{\overline{R}}_{\delta }\left(x-b\right)}V\left(b,b\right),$

${V}_{x}\left({b}^{*},{b}^{*}\right)=-\frac{\eta \left({c}_{2}-{c}_{1}\right)}{\delta }{\overline{R}}_{\delta }+{\overline{R}}_{\delta }V\left({b}^{*},{b}^{*}\right)=\eta ,$

$V\left({b}^{*},{b}^{*}\right)=\eta \left(\frac{{c}_{2}-{c}_{1}}{\delta }+\frac{1}{{\overline{R}}_{\delta }}\right).$ (28)

(28)式是进行Laplace变换非常关键的一个条件，它等价于 $H\left(0,{b}^{*}\right)=\eta \left(\frac{{c}_{2}-{c}_{1}}{\delta }+\frac{1}{{\overline{R}}_{\delta }}\right)$ 。在(26)式中给出了 $w\left(0\right)=\eta \left(\frac{{c}_{2}-{c}_{1}}{\delta }+\frac{1}{{\overline{R}}_{\delta }}\right)$${w}^{\prime }\left(0\right)=\eta$ 这两个条件，进而可由 $\stackrel{^}{w}\left(\xi \right)$ 求逆得 $w\left(z\right)$${b}^{*}$$w\left(z\right)=0$ 可解得，且当 $0\le x\le {b}^{*}$ 时有 $V\left(x,{b}^{*}\right)=w\left({b}^{*}-x\right)$ ，又由定理2可求当 $x\ge b$$V\left(x,b\right)$ 的表达式。

Optimal Dividend in the Two-Dimensional Dual Model[J]. 统计学与应用, 2017, 06(05): 516-525. http://dx.doi.org/10.12677/SA.2017.65058

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