在实际当中, 保险公司内部有巨大的现金流量, 很多学者考虑了将保险公司资产投资于资本市场的情况, 如文献[1-5], 得到了很多有实用意义的结果.文献[6-8]考虑了通过比例再保险来控制保险公司的收益和风险, 在假定资本市场变动与保险公司资产收益变动相互独立的情况下, 求解出了最优再保险策略.但是, 实证结果表明, 资本市场变动与保险公司资产变动存在相关性.例如, 保险公司承保事业险, 当经济低迷时, 资本市场回报不尽人意, 同时出现大量的失业理赔现象, 这两者之间存在很强的相关性.本文在考虑了保险公司收益与资本市场回报之间存在相关性的前提下, 讨论了这种相关性对保险公司最优再保险策略、最小破产概率的影响.

经典的风险模型是Cramér-Lunderberg提出的.在这个模型中, 不考虑再保险、投资和分红, 保险公司的资产变动服从

$ {X_t} = x + pt - \sum\limits_{i = 1}^{{N_t}} {{U_i}}, $

(1.1)

其中${X_t}$为保险公司$t$时刻的盈余资金, $x$是初始盈余资金, ${N_t}$是泊松过程, 用于记录$t$时刻之前的索赔次数, ${U_t}$是独立同分布随机变量, 表示为第$i$次索赔金额, 记$E{N_t} = {\lambda _0}t, E{U_i} = {\mu _0}, EU_i^2 = \sigma _0^2$. $p$是保险收益率, 由溢价率和期望赔付价值的乘积得到. $p = (1 + \beta ){\lambda _0}{\mu _0}$, 其中$\beta \ge 0$是风险溢价率.保险公司在通过连续购买比例再保险来控制自身的业务风险后, 在保险方面的资本盈余过程为

$ {X_t} = x + {p^{{a_t}}}t - \sum\limits_{i = 1}^{{N_t}} {U_i^{{a_t}}}, $

(1.2)

这里${a_t} \in [0, 1]$为$t$时刻的存留(保持)比例, 记$1 - {a_t}$为$t$时刻的再保险比例.因此, 任意时刻保险公司的保险盈余过程满足

$ \left\{\begin{array}{l} d{X_t} = (\mu - (1 - a)\lambda )dt + \sigma dB_t^{(0)}, \\ {X_0} = x, \end{array}\right. $

(1.3)

其中$\mu = p - {\lambda _0}{\mu _0}, \sigma = \sqrt {\mu _0^2 + \sigma _0^2} $, $B_t^{(0)}$是标准的布朗运动.再保险公司的风险收益率$\lambda $与保险公司的风险收益率$\mu $不同, 且有$\lambda > \mu $, 这是由于再保险公司和保险公司收取的风险溢价率不同造成的.如果$\lambda = \mu $, 保险公司选择全部再保险, 可以避免破产.

注  (1.3) 式推导过程可参考文献[2].

假定在$t$时刻, 保险公司的资产用于投资到风险资本市场.在资本市场上风险证券价格满足下面随机微分方程

$ dp(t) = p(t)(rdt + {\sigma _p}dB_t^{(1)}), $

(1.4)

其中$r, {\sigma _p}$分别表示风险资产的回报率与波动率, 且$r > 0, {\sigma _p} > 0$, $B_t^{(1)}$为标准布朗运动.

为了便于讨论, 定义完备的概率空间$(\Omega, F, {F_t}, P)$, 其中$F = \{ {F_t}, t \ge 0\} $, ${F_t}$是$t$时刻的信息空间. ${B_t}$关于${F_t}$是适应的.所有可行策略过程的集合$\Pi = \{ \pi \} $, $\pi = ({a_\pi }(t), {\tau _\pi })$, ${a_\pi }(t) \in [0, 1]$是关于${F_t}$适应的.因此, 在给定的控制策略$\pi \in \Pi $下, 保险公司总盈余过程满足下面微分方程

$ dR_t^\pi = [\mu-(1-{a_\pi }(t))\lambda + rR_t^\pi]dt + \sigma {a_\pi }(t)dB_t^{(0)} + {\sigma _p}R_t^\pi dB_t^{(1)}, $

(1.5)

其中$R_0^\pi = x$为初始资本盈余.令${M_t} = (B_t^{(0)}, B_t^{(1)})$, ${M_t}$为${\{ {F_t}\} _{t \ge 0}}$ -鞅.式(1.5) 可以写成

$ dR_t^\pi = [\mu-(1-{a_\pi }(t))\lambda + rR_t^\pi]dt + (\sigma {a_\pi }(t), {\sigma _p}R_t^\pi )d{M_t}. $

(1.6)

假设$B_t^{(0)}$和$B_t^{(1)}$的相关性满足$E\{ B_t^{(0)}B_t^{(1)}\} = \rho (s \wedge t)$, 对任意的$s, t \ge 0$和$\left| \rho \right| \le 1$成立.

注  本文只讨论$\rho \ne \pm 1$的情况.若$\rho = \pm 1$时, 讨论的方法同$\rho \ne \pm 1$的情形.

定义破产时刻为${\tau _\pi } = \inf \{ t > 0:R_t^\pi \le 0\} $, 这里${\tau _\pi }$为${F_t}$停时.在策略$\pi $下的破产概率为${J_\pi }(x) = P({\tau _\pi } < \infty )$.因此, 最优控制策略${\pi ^ * }$满足

$ V(x) = {J_{{\pi ^ * }}}(x) = \mathop {\inf }\limits_{\pi \in \Pi } \{ {J_\pi }(x)\}. $

(1.7)

引理 2.1^[2]  假定$V \in {C^2}(0, + \infty )$是递减凸函数, 则$V$满足Hamiliton-Jacobi-Bellman (HJB)变分不等方程

$ \begin{equation}\mathop {\inf }\limits_{0 \le {a_t} \le 1} \{ \frac{1}{2}({a^2}{\sigma ^2} + 2\rho a\sigma {\sigma _p}x + \sigma _p^2{x^2})V''(x) + [\mu-(1-a)\lambda + rx]V'(x)\} = 0\end{equation} $

(2.1)

满足的边界条件为

$ \begin{equation}\begin{aligned} V(0) = 1, ~~~~ V(\infty ) = 0 . \end{aligned}\end{equation} $

(2.2)

引理 2.2^[2]  假定$f \in {C^2}(0, + \infty )$, $V(x)$由(1.7) 式给出, 且都为方程(2.1), (2.2) 的解, 任意的$x$, ${a^ * }(x)$满足下列方程

$ \frac{1}{2}({a^ * }{(x)^2}{\sigma ^2} + 2\rho {a^ * }(x)\sigma {\sigma _p}x + \sigma _p^2{x^2})V''(x) + [\mu-(1-{a^ * }(x))\lambda + rx]V'(x) = 0, $

那么对最优策略${\pi ^ * }$有$f(x) = V(x) = {V_{{\pi ^ * }}}(x)$, 其中${a^ * }(R_s^{{\pi ^ * }}) = {a_{{\pi ^ * }}}(s)$, $R_s^{{\pi ^ * }}$满足(1.5) 式.

下面求解HJB方程(2.1).已知$V''(x) > 0$, 对任意的$x \ge 0$, 有

$ \begin{eqnarray}&& a(x) = - \frac{{\lambda V'}}{{{\sigma ^2}V''}} - \frac{{\rho {\sigma _p}x}}{\sigma }, \end{eqnarray} $

(2.3)

$ \begin{eqnarray} && - \frac{1}{2}\frac{{{\lambda ^2}}}{{{\sigma ^2}}}{(\frac{{V'}}{{V''}})^2} + (\mu - \lambda + rx - \frac{{\rho \lambda {\sigma _p}x}}{\sigma })\frac{{V'}}{{V''}} + \frac{1}{2}\sigma _p^2{x^2}(1 - {\rho ^2}) = 0. \end{eqnarray} $

(2.4)

令$\frac{{V'(x)}}{{V''(x)}} = g(x)$, 有

$ \begin{eqnarray}&& g(x) = \frac{{(\mu - \lambda + rx - \frac{{\rho \lambda {\sigma _p}x}}{\sigma }) - \sqrt {{{(\mu - \lambda + rx - \frac{{\rho \lambda {\sigma _p}x}}{\sigma })}^2} + \frac{{{\lambda ^2}\sigma _p^{^2}{x^2}(1 - {\rho ^2})}}{{{\sigma ^2}}}} }}{{\frac{{{\lambda ^2}}}{{{\sigma ^2}}}}}, \end{eqnarray} $

(2.5)

$ \begin{eqnarray} && a(x) = \frac{1}{\lambda }(\lambda - \mu - rx + \sqrt {{{(\mu - \lambda + rx - \frac{{\rho \lambda {\sigma _p}x}}{\sigma })}^2} + \frac{{{\lambda ^2}\sigma _p^2{x^2}(1 - {\rho ^2})}}{{{\sigma ^2}}}} ).\end{eqnarray} $

(2.6)

由$a(x) = 0$得

$ \begin{equation}{x_0} = \frac{{ - 2\rho \sigma (\lambda - \mu )}}{{\lambda {\sigma _p} - 2r\rho \sigma }};\end{equation} $

(2.7)

由$a(x) = 1$得

$ \begin{equation}(\lambda \sigma _p^2 - 2r\rho \sigma {\sigma _p}){x^2} + (2\rho \lambda \sigma {\sigma _p} - 2\mu \rho \sigma {\sigma _p} - 2r{\sigma ^2})x + {\sigma ^2}(\lambda - 2\mu ) = 0.\end{equation} $

(2.8)

取$\nabla = {(2\rho \lambda \sigma {\sigma _p} - 2\mu \rho \sigma {\sigma _p} - 2r{\sigma ^2})^2} - 4{\sigma ^2}(\lambda \sigma _p^2 - 2r\rho \sigma {\sigma _p})(\lambda - 2\mu )$.若$\lambda {\sigma _p} - 2r\rho \sigma \ne 0$且$\nabla \ge 0$, 有

$ \begin{equation}{x_{1, 2}} = \frac{\sigma }{{{\sigma _p}}}\frac{{(r\sigma + \rho \mu {\sigma _p} - \rho \lambda {\sigma _p}) \mp \sqrt {{{(r\sigma + \rho \mu {\sigma _p} - \rho \lambda {\sigma _p})}^2} - (\lambda \sigma _p^2 - 2r\rho \sigma {\sigma _p})(\lambda - 2\mu )} }} {{\lambda {\sigma _p} - 2r\rho \sigma }}.\end{equation} $

(2.9)

由$a(x) = 1$, 带入(2.1) 式得$(\mu + rx)V' + \frac{1}{2}({\sigma ^2} + 2\rho \sigma {\sigma _p} + \sigma _p^2{x^2})V'' = 0$, 即

$ \begin{eqnarray} \ln \left| {V'(x)} \right|& =& \ln \left| {{{(\sigma _p^2{x^2} + 2\rho \sigma {\sigma _p}x + {\sigma ^2})}^{\frac{{ - r}}{{\sigma _p^2}}}}} \right| \nonumber\\ && + (\frac{{2r\rho \sigma }}{{{\sigma _p}}} - 2\mu )\int {\frac{1}{{\sigma _p^2{x^2} + 2\rho \sigma {\sigma _p}x + {\sigma ^2}}} + C}, \end{eqnarray} $

(2.10)

$ \begin{eqnarray} V(x) &=& - {C_1}\int_0^x {{{[\frac{{\sigma _p^2}}{{{\sigma ^2}(1 -{\rho ^2})}}{{(t + \frac{{\rho \sigma }}{{{\sigma _p}}})}^2} + 1]}^{\frac{{ - r}}{{\sigma _p^2}}}}} \nonumber\\ && \cdot \exp \{ (\frac{{r\rho }}{{\sigma _p^2}} - \frac{\mu }{{\sigma {\sigma _p}}})\frac{2}{{\sqrt {1 - {\rho ^2}} }}\arctan \frac{{{\sigma _p}t + \rho \sigma }}{{\sigma \sqrt {1 - {\rho ^2}} }}\} dt + {C_2}. \end{eqnarray} $

(2.11)

令

$ h(t) = {[\frac{{\sigma _p^2}}{{{\sigma ^2}(1-{\rho ^2})}}{(t + \frac{{\rho \sigma }}{{{\sigma _p}}})^2} + 1]^{\frac{{ -r}} {{\sigma _p^2}}}} \cdot {e^{(\frac{{r\rho }}{{\sigma _p^2}} - \frac{\mu }{{\sigma {\sigma _p}}})\frac{2}{{\sqrt {1 -{\rho ^2}} }} \arctan \frac{{{\sigma _p}t -\rho \sigma }}{{\sigma \sqrt {1 - {\rho ^2}} }}}}. $

(2.12)

因此, 当$a(x) = 1$时, 有

$ V(x) = - {C_1}\int_0^x {h(t)dt + {C_2}}. $

(2.13)

根据边界条件(2.2) 可得${C_1} = (\int_0^\infty {h(t)dt{)^{ - 1}}} $.因为当$r \le \frac{{\sigma _p^2}}{2}$时, 积分$\int_0^\infty {h(t)dt} $不收敛, 于是令

$ {V_N}(x) = 1 - \frac{{\int_0^x {h(t)dt} }}{{\int_0^N {h(t)dt} }}, $

(2.14)

${V_N}(x)$满足(2.1) 式, 且${V_N}(0) = 1, {V_N}(N) = 0$.当$N \to \infty $时, ${V_N}(x) \to 1$.根据引理2.2可知, 对任何的再保险策略都有${V_N}(x) = V(x) = 1$, $x \in (0, \infty )$.

(1) 若$0 < {x_1} < {x_2}$.在$x \in (0, {x_1}]$时, $a(x) \ge 1$, 取${a^ * }(x) = 1$; 在$x \in ({x_1}, {x_2})$时, $0 < a(x) < 1$, 取${a^ * }(x) = a(x)$, 这里$a(x)$由(2.6) 定义; 在$x \in [{x_2}, \infty )$时, $a(x) \ge 1$, 取${a^ * }(x) = 1$.因此最优策略为

$ ~~~~~~~~~~{a^ * }(x) = \left\{ \begin{array}{l} 1, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~0 \le x \le {x_1}, \\ \frac{1}{\lambda }(\lambda- \mu- rx + \sqrt {{{(\mu- \lambda + rx - \frac{{\rho \lambda {\sigma _p}x}}{\sigma })}^2} + \frac{{{\lambda ^2}\sigma _p^2(1 - {\rho ^2})}}{{{\sigma ^2}}}} ), ~~~~~{x_1} < x < {x_2}, \\ 1, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x \ge {x_2}. \end{array} \right. $

定理 2.1.1  若$r > \frac{{\sigma _p^2}}{2}$时, 最小破产概率为

$ ~~~~~~~~~~~V(x) = \left\{ \begin{array}{l} \displaystyle - {C_1}\int_0^x {h(t)dt + {C_2}}, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~0 \le x \le {x_1}, \\ \displaystyle - {C_3}\int_{{x_1}}^x {\exp (\int_{{x_1}}^t {\frac{1}{{g(y)}}dy)dt + {C_4}} }, ~~~~~~~~~~~~~{x_1} < x < {x_2}, \\ \displaystyle - {C_5}\int_{{x_2}}^x {h(t)dt + {C_6}}, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x \ge {x_2}, \end{array} \right. $

这里$g, h$, 分别由(2.5), (2.12) 式定义, ${C_i}$由(2.15), (2.16) 式给出.

证  若${a^ * }(x) = a(x)$, 即${x_1} < x < {x_2}$时, 由$\frac{{V'(x)}}{{V''(x)}} = g(x)$得

$ V(x) = - {C_3}\int_{{x_1}}^x {\exp (\int_{{x_1}}^t {\frac{1}{{g(y)}}dy)dt + {C_4}}}. $

若${a^ * }(x) = 1$时, $V(x)$由(2.13) 式定义.因此在$0 \le x \le {x_1}$时, $V(x) = - {C_1}\int_0^x {h(t)dt + {C_2}} $; 在$x \ge {x_2}$时, $V(x) = - {C_5}\int_{{x_2}}^x {h(t)dt + {C_6}} $.根据$V(x)$和$V'(x)$在${x_1}, {x_2}$的光滑性以及边界条件(2.2) 可知

$ \begin{aligned} {C_2} = 1, ~{C_4} = 1 - {K_1}{C_1}, ~{C_6} = {C_4} - {C_3}{K_3}, ~{C_3} = {C_1}{K_2}, ~{C_5} = {C_3}{K_4}, ~{C_6} = {C_5}{K_5}, \end{aligned} $

(2.15)

其中

$ \begin{eqnarray*} &&{K_1} = \int_0^{x{}_1} {h(t)dt}, {K_2} = h({x_1}), {K_3} = \int_{{x_1}}^{{x_2}} {\exp (\int_{{x_1}}^t {\frac{1}{{g(y)}}} } dy)dt, \\ && {K_4} = \frac{{\exp (\int_{{x_1}}^{{x_2}} {\frac{1}{{g(y)}}dy)} }}{{h({x_2})}}, {K_5} = \int_{{x_2}}^\infty {h(t)dt}. \end{eqnarray*} $

因此,

$ \begin{aligned} & {C_1} = \frac{1}{{{K_1} + {K_2}{K_3} + {K_2}{K_4}{K_5}}}, {C_3} = \frac{{{K_2}}}{{{K_1} + {K_2}{K_3} + {K_2}{K_4}{K_5}}}, \\ & {C_4} = \frac{{{K_2}{K_3} + {K_2}{K_4}{K_5}}}{{{K_1} + {K_2}{K_3} + {K_2}{K_4}{K_5}}}, \\ & {C_5} = \frac{{{K_2}{K_4}}}{{{K_1} + {K_2}{K_3} + {K_2}{K_4}{K_5}}}, {C_6} = \frac{{{K_2}{K_4}{K_5}}}{{{K_1} + {K_2}{K_3} + {K_2}{K_4}{K_5}}}, \end{aligned} $

(2.16)

这里${K_5} = \int_{{x_2}}^\infty {h(t)dt} $, 当$x \to \infty $时, $\arctan (x) \to \pi /2$, 在$r > \frac{{\sigma _p^2}}{2}$的情形下, 积分$\int_{{x_2}}^\infty {h(t)dt} $是收敛的, 即${K_5}$是有界的.于是, 当$r > \frac{{\sigma _p^2}}{2}$时, 所有${C_i}$为正值, 因此定理得证.

(2) 若${x_1} \le 0 < {x_2}$.

$ ~~~~~~~~{a^ * }(x) = \left\{ \begin{array}{l} \frac{1}{\lambda }(\lambda - \mu - rx + \sqrt {{{(\mu - \lambda + rx - \frac{{\rho \lambda {\sigma _p}x}}{\sigma })}^2} + \frac{{{\lambda ^2}\sigma _p^{^2}{x^2}(1 - {\rho ^2})}}{{{\sigma ^2}}}} ), ~~~0 \le x < {x_2}, \\ 1, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x \ge {x_2}. \end{array} \right. $

定理 2.1.2  若$r > \frac{{\sigma _p^2}}{2}$时, 最小破产概率为

$ ~~~~~~~~~V(x) = \left\{ \begin{array}{l} \displaystyle - {c_1}\int_0^x {\exp (\int_0^t {\frac{1}{{g(y)}}dy)dt} + {c_2}}, ~~~~~0 \le x < {x_2}, \\ \displaystyle - {c_3}\int_{{x_2}}^x {h(t)dt + {c_4}}, ~~~~~~~~~~~~~~~~~~~~~~x \ge {x_2}, \end{array} \right. $

这里$g$由(2.5) 式定义, $h$由(2.12) 式定义, ${c_i}$由(2.17) 式给出.

$ \begin{aligned} {c_1} = \frac{1}{{{k_1} + {k_2}{k_3}}}, ~~~~~{c_2} = 1, ~~~~~{c_3} = \frac{{{k_2}}}{{{k_1} + {k_2}{k_3}}}, ~~~~~{c_4} = \frac{{{k_2}{k_3}}}{{{k_1} + {k_2}{k_3}}}, \end{aligned} $

(2.17)

其中

$ {k_1} = \int_0^{{x_2}} {\exp (\int_0^t {\frac{1}{{g(y)}}} } dy)dt, {k_2} = \frac{{\int_0^{{x_2}} {\frac{1}{{g(y)}}dy} }}{{h({x_2})}}, {k_3} = \int_{{x_2}}^\infty {h(t)dt}. $

(3) 若${x_1} < {x_2} \le 0$或$\nabla \le 0$.

当${x_1} < {x_2} \le 0$或$\nabla \le 0$时, 对任意$x \in (0, \infty )$, 都有$a(x) \ge 1$, 因此取最优再保险策略${a^ * }(x) = 1$.

定理 2.1.3  若$r >\frac{{\sigma _p^2}}{2}$时, 最小破产概率$V(x) = - {c_1}\int_0^x {h(t)dt + {c_2}}$, 其中${c_1} = \frac{1}{{\int_0^\infty {h(t)dt} }}$, ${c_2} = 1$.

若$\lambda {\sigma _p} = 2r\rho \sigma $时, 由(2.6) 式可知$a(x)$为单调递减曲线, 且恒有$a(x) > 0$.根据(2.8) 式得

$ {x_3} = \frac{{\sigma (\lambda - 2\mu )}}{{2(r\sigma + \rho \mu {\sigma _p} - \rho \lambda {\sigma _p})}}. $

(2.18)

因此, 当$x \ge {x_3}$时, 必有$0 < a(x) \le 1$.相应的破产概率

$ V(x) = - {c_1}\int_{{x_3}}^x {\exp (\int_{{x_3}}^t {\frac{1}{{g(y)}}dy)dt + {c_2}} } . $

根据边界条件(2.2) 可得${c_1} = \displaystyle(\int_{{x_3}}^\infty {\exp (\int_{{x_3}}^t {\frac{1}{{g(y)}}} } dy)dt{)^{ - 1}}$.当$y \to \infty $时, $\frac{1}{{g(y)}} \to p\frac{1}{y}$, 其中

$ p = \frac{{{\lambda ^2}}}{{{\sigma ^2}(r - \frac{{\rho \lambda {\sigma _p}}}{\sigma } - \sqrt {{{(r - \frac{{\rho \lambda {\sigma _p}}}{\sigma })}^2} + \frac{{{\lambda ^2}\sigma _p^2(1 - {\rho ^2})}}{{{\sigma ^2}}}} )}}, $

(2.19)

因为$p < 0$, 所以当$0 < - p \le 1$时, $\displaystyle\int_{{x_3}}^\infty {\exp (\int_{{x_3}}^t {\frac{1}{{g(y)}}} } dy)dt$不收敛, 于是根据引理2.2知, 在$\lambda {\sigma _p} = 2r\rho \sigma $且$0 < - p \le 1$时, 对任何再保险策略都有$V(x) = 1$, $x \in (0, \infty )$.

(1) 若${x_3} > 0$.

$ ~~~~~~~~{a^ * }(x) = \left\{ \begin{array}{l} 1, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~0 \le x < {x_3}, \\ \frac{1}{\lambda }(\lambda - \mu - rx + \sqrt {{{(\mu - \lambda + rx - \frac{{\rho \lambda {\sigma _p}x}}{\sigma })}^2} + \frac{{{\lambda ^2}\sigma _p^{^2}{x^2}(1 - {\rho ^2})}}{{{\sigma ^2}}}} ), ~~x \ge {x_3}. \end{array} \right. $

定理 2.2.1  若$ - p > 1$, 则最小破产概率为

$ ~~~~~~~~~V(x) = \left\{ \begin{array}{l} \displaystyle- {c_1}\int_0^x {h(t)dt + {c_2}}, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~0 \le x < {x_3}, \\ \displaystyle- {c_3}\int_{{x_3}}^x {\exp (\int_{{x_3}}^t {\frac{1}{{g(y)}}dy)dt + {c_4}} }, ~~~~~~~~~~x \ge {x_3}, \end{array} \right. $

这里$g$由(2.5) 式定义, $h$由(2.12) 式定义, ${c_i}$由(2.20) 式给出.

$ \begin{aligned} {c_1} = \frac{1}{{{k_1} + {k_2}{k_3}}}, ~~~~~{c_2} = 1, ~~~~~{c_3} = \frac{{{k_2}}}{{{k_1} + {k_2}{k_3}}}, ~~~~~{c_4} = \frac{{{k_2}{k_3}}}{{{k_1} + {k_2}{k_3}}}, \end{aligned} $

(2.20)

其中

$ \begin{aligned} ~~~~ {k_1} = \int_0^{{x_3}} {h(t)dt}, ~~~~~~~{k_2} = h({x_3}), ~~~~~~~ {k_3} = \int_{{x_3}}^\infty {\exp (\int_{{x_3}}^t {\frac{1}{{g(y)}}} } dy)dt . \end{aligned} $

(2) 若${x_3} \le 0$.

$ {a^ * }(x) = \frac{1}{\lambda }(\lambda - \mu - rx + \sqrt {{{(\mu - \lambda + rx - \frac{{\rho \lambda {\sigma _p}x}}{\sigma })}^2} + \frac{{{\lambda ^2}\sigma _p^2{x^2}(1 - {\rho ^2})}}{{{\sigma ^2}}}} ). $

定理 2.2.2  若$ - p > 1$, 则最小破产概率为

$ V(x) = - {c_1}\int_0^x {\exp (\int_0^t {\frac{1}{{g(y)}}} } dy)dt + {c_2}, $

其中${c_1} = (\int_0^\infty {\exp (\int_0^t {\frac{1}{{g(y)}}} } dy)dt{)^{ - 1}}$, ${c_2} = 1$.

若$\lambda {\sigma _p} < 2r\rho \sigma $时, 根据$a(x)$曲线的单调性知${x_2}$是增根, 应舍去. ${x_1}$是$a(x) = 1$的根.

$ \begin{equation}{x_1} = \frac{\sigma }{{{\sigma _p}}}\frac{{(r\sigma + \rho \mu {\sigma _p} - \rho \lambda {\sigma _p}) - \sqrt {{{(r\sigma + \rho \mu {\sigma _p} - \rho \lambda {\sigma _p})}^2} - (\lambda \sigma _p^2 - 2r\rho \sigma {\sigma _p})(\lambda - 2\mu )} }}{{\lambda {\sigma _p} - 2r\rho \sigma }}.\end{equation} $

(2.21)

因为当$x \ge {x_0}$时, 必有$a(x) \le 0$, 所以取${a^ * }(x) = 0$, 由(2.1) 式可得

$ \begin{eqnarray}&& (\mu - \lambda + rx)V' + \frac{1}{2}\sigma _p^2{x^2}V'' = 0, \nonumber\\ && V(x) = - {c_1}\int_0^x {{t^{\frac{{ - 2r}}{{\sigma _p^2}}}}} \exp \{ \frac{2}{{\sigma _p^2}}(\mu - \lambda )\frac{1}{t}\} dt + {c_2}. \end{eqnarray} $

(2.22)

令

$ \begin{eqnarray}&& f(t) = {t^{\frac{{ - 2r}}{{\sigma _p^2}}}}\exp \{ \frac{2}{{\sigma _p^2}}(\mu - \lambda )\frac{1}{t}\}, \end{eqnarray} $

(2.23)

$ \begin{eqnarray} && V(x) = - {c_1}\int_0^x {f(t)} dt + {c_2}. \end{eqnarray} $

(2.24)

根据边界条件(2.2) 可得${c_1} = (\int_0^\infty {f(t)dt{)^{ - 1}}} $.当$r \le \frac{{\sigma _p^2}}{2}$时, 积分$\int_0^\infty {f(t)dt} $不收敛, 于是根据引理2.2可知, 在$\lambda {\sigma _p} < 2r\rho \sigma $且$r \le \frac{{\sigma _p^2}}{2}$时, 对任何再保险策略都有$V(x) = 1$, $x \in (0, \infty )$.

(1) 若${x_1} \le 0$.

$ ~~~~~~~{a^ * }(x) = \left\{ \begin{array}{l} \displaystyle \frac{1}{\lambda }(\lambda - \mu - rx\\ + \sqrt {{{(\mu - \lambda + rx - \frac{{\rho \lambda {\sigma _p}x}}{\sigma })}^2} + \frac{{{\lambda ^2}\sigma _p^{^2}{x^2}(1 - {\rho ^2})}}{{{\sigma ^2}}}} ), ~~~~0 \le x < {x_0}, \\ 0, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x \ge {x_0}. \end{array} \right. $

定理 2.3.1  当$\frac{{2r}}{{\sigma _p^2}} > 1$时, 最小破产概率为

$ ~~~~~~~V(x) = \left\{ \begin{array}{l} \displaystyle- {c_1}\int_0^x {\exp (\int_0^t {\frac{1}{{g(y)}}dy)dt} + {c_2}}, ~~~~~0 \le x < {x_0}, \\ \displaystyle - {c_3}\int_{{x_0}}^x {f(t)dt + {c_4}}, ~~~~~~~~~~~~~~~~~~~~~~x \ge {x_0}, \end{array} \right. $

这里$g, f$分别由(2.5) 和(2.23) 式定义, ${c_i}$由(2.25) 式确定.

$ \begin{eqnarray} {c_1} = \frac{1}{{{k_1} + {k_2}{k_3}}}, ~~~~~~~~{c_2} = 1, ~~~~~~ {c_3} = \frac{{{k_2}}}{{{k_1} + {k_2}{k_3}}}, ~~~~~~~ {c_4} = \frac{{{k_2}{k_3}}}{{{k_1} + {k_2}{k_3}}}, \end{eqnarray} $

(2.25)

其中

$ \begin{aligned} ~~~~~~~{k_1} = \int_0^{{x_0}} {\exp (\int_0^t {\frac{1}{{g(y)}}dy)dt} }, ~~~~~~~{k_2} = \frac{{\exp (\int_0^{{x_0}} {\frac{1}{{g(y)}}dy)} }}{{f({x_0})}}, ~~~~~~~ {k_3} = \int_{{x_0}}^\infty {f(t)dt} . \end{aligned} $

(2) 若${x_1} > 0$.

$ ~~~~~~~~{a^ * }(x) = \left\{ \begin{array}{l} 1, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~0 \le x \le {x_1}, \\ \frac{1}{\lambda }(\lambda - \mu - rx + \sqrt {{{(\mu - \lambda + rx - \frac{{\rho \lambda {\sigma _p}x}}{\sigma })}^2} + \frac{{{\lambda ^2}\sigma _p^2(1 - {\rho ^2})}}{{{\sigma ^2}}}} ), ~~~{x_1} < x < {x_0}, \\ 0, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x \ge {x_0}.\\ \end{array} \right. $

定理 2.3.2  当$\frac{{2r}}{{\sigma _p^2}} > 1$时, 最小破产概率为

$ ~~~~~~~~V(x) = \left\{ \begin{array}{l} - {c_1}\int_0^x {h(t)dt + {c_2}}, ~~~~~~~~~~~~~~~~~~~~~~~~0 \le x \le {x_1}, \\ - {c_3}\int_{{x_1}}^x {\exp (\int_{{x_1}}^t {\frac{1}{{g(y)}}dy)dt + {c_4}} }, ~~~~~{x_1} < x < {x_0}, \\ - {c_5}\int_{{x_0}}^x {f(t)dt + {c_6}}, ~~~~~~~~~~~~~~~~~~~~~~~~~x \ge {x_0}, \end{array} \right. $

这里$g, h, f$分别由(2.5), (2.12), (2.23) 式定义, ${c_i}$由(2.26) 式给出

$ \begin{eqnarray} && {c_1} = \frac{1}{{{K_1} + {K_2}{K_3} + {K_2}{K_4}{K_5}}}, {c_2} = 1, {c_3} = \frac{{{K_2}}}{{{K_1} + {K_2}{K_3} + {K_2}{K_4}{K_5}}}, \nonumber\\ && {c_4} = \frac{{{K_2}{K_3} + {K_2}{K_4}{K_5}}}{{{K_1} + {K_2}{K_3} + {K_2}{K_4}{K_5}}}, \nonumber\\ && {c_5} = \frac{{{K_2}{K_4}}}{{{K_1} + {K_2}{K_3} + {K_2}{K_4}{K_5}}}, {c_6} = \frac{{{K_2}{K_4}{K_5}}}{{{K_1} + {K_2}{K_3} + {K_2}{K_4}{K_5}}}, \end{eqnarray} $

(2.26)

其中

$ \begin{eqnarray*} && {K_1} = \int_0^{x{}_1} {h(t)dt}, {K_2} = h({x_1}), {K_3} = \int_{{x_1}}^{{x_0}} {\exp (\int_{{x_1}}^t {\frac{1}{{g(y)}}} } dy)dt, \\ && {K_4} = \frac{{\exp (\int_{{x_1}}^{{x_0}} {\frac{1}{{g(y)}}dy)} }}{{f({x_0})}} , {K_5} = \int_{{x_0}}^\infty {f(t)dt}. \end{eqnarray*} $

通过上述讨论表明在$\rho \ne \pm 1$和$\lambda {\sigma _p} \ne 2r\rho \sigma $的情况下, 当$r \le \sigma _p^2/2$时, 对任何的再保险策略都有$V(x) = 1$.这是因为当${\sigma _p}$越大, 资本市场的波动就越强, 保险公司承受的风险也就越大.在$r \le \sigma _p^2/2$时, 资本市场的波动会导致投资盈余趋于0, 此时保险盈余只能有限度的调整资本市场带来的不利影响, 但不能阻止其破产.

以下通过数值例子来说明布朗运动${B^{(0)}}, {B^{(1)}}$的相关系数$\rho $对最小破产概率$V(x)$和相应的最优再保险策略${a^ * }(x)$的影响.

图 1显示$\rho $对$a(x)$的影响.其中, $\mu = 0.3$, $\lambda = 0.6$, $r = 0.21$, $\sigma = 0.5$, ${\sigma _p} = 0.25$.在$\lambda {\sigma _p} - 2r\rho \sigma > 0$时, $a(x)$的图形为抛物线, 但随着$\rho $逐渐的增大, $a(x)$的开口也随之增大.在$\rho = 0.75$时, $\lambda {\sigma _p} - 2r\rho \sigma = 0$, 此时随着$x$的增大, $a(x)$的值逐渐趋向0, 并最终为一大于0的恒定值.当$\rho > 0.75$时, $\lambda {\sigma _p} - 2r\rho \sigma < 0$, 此时$a(x)$随$x$的增大而减小.

图 2显示最优再保险策略${a^ * }(x)$与相应的最小破产概率$V(x)$的图形, 其中$\mu = 0.24$, $\lambda = 0.55$, $r = 0.18$, $\sigma = 0.66$, ${\sigma _p} = 0.3$, $\rho = 0.15$.经过计得${x_1} = 0.2359, {x_2} = 3.3304$. ${C_1} = 1.6998$, ${C_2} = 1$, ${C_3} = 0.9322$, ${C_4} = 0.7458$, ${C_5} = 1.5796$, ${C_6} = 0.0195$.

图 3显示$\rho $对$V(x)$的影响, 其中$\mu = 0.25$, $\lambda = 0.53$, $r = 0.2$, $\sigma = 0.65$, ${\sigma _p} = 0.4$. $V(x)$为关于$\rho $的单调递增函数. $\rho $越大, 表示保险收益和资本市场回报波动的相关性越强, 此时, 总体的波动相应变大, 从而增加了保险公司风险, 同时削弱了公司的盈利能力, 因此破产的概率随之增大.