西南大学学报 (自然科学版)  2019, Vol. 41 Issue (11): 54-63.  DOI: 10.13718/j.cnki.xdzk.2019.11.008
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  • 带借贷利率和干扰的双Poisson-Geometric风险过程模型    [PDF全文]
    王月明1, 魏广华2, 郭楠1, 高艳艳1    
    1. 南京工程学院 数理部, 南京 211167;
    2. 金陵科技学院 理学院, 南京 211169
    摘要:考虑了带借贷利率及干扰的双复合Poisson-Geometric风险过程,借助全期望公式、微分和伊藤积分等知识,并综合引起破产的原因得到无限时破产概率积分微分方程和有限时破产概率的积分偏微分方程.
    关键词借贷    复合Poisson-Geometric过程    布朗运动    破产概率    积分微分方程    积分偏微分方程    

    近年来,破产概率是风险理论研究中的热门话题,越来越多学者开始研究各种风险模型下的破产概率、利率等问题[1-11].文献[9]考虑带借贷的理赔为单复合泊松模型的风险模型,文献[2]研究带借贷及干扰的理赔为双复合泊松模型的风险过程,获得积分微分方程. Poisson分布的重要特征是方差等于均值,但是索赔事件和风险事件不是等价的,实际生活中,保险公司和投保人提高了风险意识,保险公司采用回避风险机制,如免赔制度、无赔款折扣(NCD)制度,所以事故发生时,投保人会权衡其利益得失而决定是否进行索赔,从而理赔次数小于事故发生次数,因此索赔次数并不完全遵循Poisson分布,文献[3-5]对该问题进行了研究,考虑索赔次数为复合Poisson-Geometric风险过程,得到破产概率的积分方程.本文推广上述风险模型,引入双复合Poisson-Geometric风险过程,并用布朗运动描述不确定的付款和收益的影响,同时考虑借贷因素,即研究带借贷及干扰的双复合Poisson-Geometric模型.

    $ \begin{array}{l} U\left( t \right) = u + ct - \sum\limits_{i = 1}^{{N_1}\left( t \right)} {{X_i}} - \sum\limits_{j = 1}^{{N_2}\left( t \right)} {{Y_j}} + \sigma B\left( t \right) = \\ \;\;\;\;\;\;\;\;\;\;u + ct - {S_1}\left( t \right) - {S_2}\left( t \right) + \sigma B\left( t \right)\;\;\;\;\;\;\;t \ge 0 \end{array} $ (1)

    其中:u是初始准备金,c是保险公司单位时间征收的保险费率;{XXii=1,2,…}是期望为μ1的独立同分布的非负随机变量序列,X表示A险种的理赔额,其分布为G(x),密度函数为g(x),且G*k(x),g*k(x)分别为G(x),g(x)的k重卷积;{N1(t),t≥0}是参数为(λ1ρ1)的Poisson-Geometric过程,表示到时刻t为止A险种理赔发生的次数;{YYjj=1,2,…}是期望为μ2的独立同分布的非负随机变量序列,Y表示B险种的理赔额,其分布为F(x),密度函数为f(x),且F*k(x),f*k(x)分别为F(x),f(x)的k重卷积;{N2(t),t≥0}是参数为(λ2ρ2)的Poisson-Geometric过程,表示到时刻t为止B险种理赔发生的次数;{B(t),t≥0}是一标准布朗运动,表示不确定的付款和收入,σ是一常数;由于各个保险过程和理赔是相互独立的,设{Xii=1,2,…},{Yjj=1,2,…},{N1(t),t≥0},{N2(t),t≥0}以及{B(t),t≥0}是相互独立的.

    为了使模型更具有实际意义,假设当保险公司财政赤字时,即盈余是负的,可以允许其以利息力δ>0进行借贷并继续经营其业务,保险公司通过它的保费收入来偿还其债务,然而当公司的盈余低于$ - \frac{c}{\delta }$时,绝对破产出现.

    带借贷的双复合Poisson-Geometric模型风险过程,即

    $ {\rm{d}}{U_\delta }\left( t \right) = \left( {c + \delta {U_\delta }\left( t \right)I\left( {{U_\delta }\left( t \right) < 0} \right)} \right){\rm{d}}t + \sigma {\rm{d}}B\left( t \right) - {\rm{d}}{S_1}\left( t \right) - {\rm{d}}{S_2}\left( t \right)\;\;\;\;{U_\delta }\left( 0 \right) = u $ (2)

    其中:I(A)是集合A上的示性函数,设Tδ表示风险模型(2)的破产时刻,${T_\delta } = \inf \left\{ {t \ge 0:{U_\delta }(t) < - \frac{c}{\delta }} \right\}$且约定infϕ=∞;ψ(u)表示模型(2)中初始资本为u的无限时破产概率,即

    $ \psi \left( u \right) = \mathit{Pr}\left\{ {{T_\delta } < \infty |{U_\delta }\left( 0 \right) = u} \right\},u > - \frac{c}{\delta },\psi \left( { - \frac{c}{\delta }} \right) = 1 $

    当盈余处于不同水平时,破产概率符合不同的积分微分方程,为了研究问题的方便,当u≥0时,记ψ(u)=ψ+(u);当$ - \frac{c}{\delta }$u<0时,记ψ(u)=ψ-(u).

    T表示风险模型(1)的破产时刻,T= inf{t≥0:U(t)<0};ϕ(u)表示风险模型(1)中初始资本为u的无限时破产概率,即

    $ \phi \left( u \right) = \mathit{Pr}\left\{ {T < \infty |{U_0} = u} \right\},u > 0 $

    显然当u≥0时,TTδψ+(u)≤ϕ(u)且$\mathop {\lim }\limits_{u \to + \infty } $ ψ+(u)=0及0<ψ+(u)≤ϕ(u)<1.

    由[5,13]可知,导致破产出现有两种可能.一种是由于理赔引起的,ψs(u)表示破产是由理赔引起的破产概率;另一种是由于扰动引起的,ψd(u)表示破产是由扰动引起的破产概率.因此无限时破产概率有以下分解:

    $ \psi \left( u \right) = {\psi _s}\left( u \right) + {\psi _d}\left( u \right)\;\;\;u \ge - \frac{c}{\delta } $

    而且,

    $ {\psi _d}\left( { - \frac{c}{\delta }} \right) = \psi \left( { - \frac{c}{\delta }} \right) = 1,{\psi _s}\left( { - \frac{c}{\delta }} \right) = 0 $

    类似地,当u≥0时,记ψs+=ψs(u),ψd+=ψd(u);当$ - \frac{c}{\delta }$u<0时,记ψs-=ψs(u),ψd-=ψd(u).

    同样地,定义有限时破产所有变量ψ(ut),ψs(ut),ψd(ut),ψs-(ut),ψs+(ut),ψd-(ut),ψd+(ut).

    为了保证保险公司的稳定经营,假定保费收入的期望值大于总的理赔的均值,由此定义安全负荷条件为

    $ c > \frac{{{\lambda _1}}}{{1 - {\rho _1}}}{\mu _1} + \frac{{{\lambda _2}}}{{1 - {\rho _2}}}{\mu _2} $
    1 相关知识简要回顾

    定义1  称母函数$G(t) = {{\rm{e}}^{\frac{{\lambda (t - 1)}}{{1 - \rho t}}}}$所对应的分布为复合Poisson-Geometric分布,记为PG(λρ),其中λ>0,0≤ρ<1.

    引理1[6]  当ρ=0时,PG(λρ)是参数为λ的Poisson分布.

    引理2[6]  对t>0,若N(t)服从PG(λtρ)分布,则

    $ E\left( {N\left( t \right)} \right) = \frac{{\lambda t}}{{1 - \rho }}, {\rm{var}}\left( {N\left( t \right)} \right) = \frac{{\lambda t\left( {1 + \rho } \right)}}{{{{\left( {1 - \rho } \right)}^2}}} $

    引理3[6]  若{Ni(t);t≥0},是参数为λiρi的Poisson-Geometric过程,记${\alpha _i} = \frac{{{\lambda _i}\left( {1 - {\rho _i}} \right)}}{{{\rho _i}}}$(若ρi=0,则取αi=λi),则当t足够小时有

    $ P\left( {{N_i}\left( t \right) = 0} \right) = {{\rm{e}}^{ - {\lambda _i}t}} = 1 - {\lambda _i}t + o\left( t \right),P\left( {{N_i}\left( t \right) = k} \right) = {\alpha _i}\rho _i^kt + A_k^i\left( t \right)o\left( t \right)\;\;\;\;k = 1,2,3, \cdots $

    其中

    $ A_k^i\left( t \right) = \rho _i^k + \left( {k - 1} \right){\left( {{\rho _i}\left( {1 + {\alpha _i}t} \right)} \right)^{k - 2}} $

    $\sum\limits_{k = 0}^\infty {A_k^i} (t)$一致收敛,i=1,2,o(t)与k无关.

    2 主要结果

    定理1  假设ψs(u)是二次连续可微的,当u≥0时,则ψs(u)符合下面积分微分方程:

    $ \begin{array}{l} \left( {{\lambda _1} + {\lambda _2}} \right){\psi _{s + }}\left( u \right) = c{{\psi '}_{s + }}\left( u \right) + \frac{1}{2}{\sigma ^2}{{\psi ''}_{s + }}\left( u \right) + \int_0^u {{\psi _{s + }}\left( {u - x} \right){g_{{\rho _1}}}\left( x \right){\rm{d}}x} + {B_{{s_1}}}\left( u \right) + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\int_0^u {{\psi _{s + }}\left( {u - y} \right){f_{{\rho _2}}}\left( y \right){\rm{d}}y} + {B_{{s_2}}}\left( u \right) \end{array} $ (3)

    $ - \frac{c}{\delta }$u<0时,则ψs(u)符合下面积分微分方程:

    $ \begin{array}{l} \left( {{\lambda _1} + {\lambda _2}} \right){\psi _{s - }}\left( u \right) = \left( {u\delta + c} \right){{\psi '}_{s - }}\left( u \right) + \frac{1}{2}{\sigma ^2}{{\psi ''}_{s - }}\left( u \right) + \int_0^{\mu + \frac{c}{\delta }} {{\psi _{s - }}} \left( {u - x} \right){g_{{\rho _1}}}\left( x \right){\rm{d}}x + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{k = 1}^\infty {\overline {{G^{*k}}} \left( {u + \frac{c}{\delta }} \right){\alpha _1}\rho _1^k} + \int_0^{u + \frac{c}{\delta }} {{\psi _{s - }}\left( {u - y} \right){f_{{\rho _2}}}\left( y \right){\rm{d}}y} + \sum\limits_{k = 1}^\infty {\overline {{F^{*k}}} \left( {u + \frac{c}{\delta }} \right){\alpha _2}\rho _2^k} \end{array} $ (4)

    边界条件为

    $ \left\{ {\begin{array}{*{20}{l}} {{\psi _{s - }}\left( { - \frac{c}{\delta }} \right) = 0}\\ {\frac{1}{2}{\sigma ^2}{{\psi ''}_{s - }}\left( { - \frac{c}{\delta }} \right) = - \sum\limits_{k = 1}^\infty {{\alpha _1}\rho _1^k} - \sum\limits_{k = 1}^\infty {{\alpha _2}\rho _2^k} }\\ {\mathop {\lim }\limits_{u \to \infty } {\psi _{s + }}\left( u \right) = 0} \end{array}} \right. $ (5)

    其中

    $ {B_{{s_1}}}\left( u \right) = \int_u^{u + \frac{c}{\delta }} {{\psi _{s - }}\left( {u - x} \right){g_{{\rho _1}}}\left( x \right){\rm{d}}x} + \sum\limits_{k = 1}^\infty {\overline {{G^{*k}}} \left( {u + \frac{c}{\delta }} \right){\alpha _1}\rho _1^k} $
    $ {B_{{s_2}}}\left( u \right) = \int_u^{u + \frac{c}{\delta }} {{\psi _{s - }}\left( {u - y} \right){f_{{\rho _2}}}\left( y \right){\rm{d}}y} + \sum\limits_{k = 1}^\infty {\overline {{F^{*k}}} \left( {u + \frac{c}{\delta }} \right){\alpha _2}\rho _2^k} \;\;\;\;u \ge 0 $

      当u≥0时,令

    $ H\left( t \right) = u + ct + \sigma B\left( t \right) $

    $ H\left( 0 \right) = u $

    $ {\rm{d}}H\left( t \right) = c{\rm{d}}t + \sigma {\rm{d}}{B_t} $

    由伊藤积分公式有

    $ {\rm{d}}{\psi _{s + }}\left( {H\left( t \right)} \right) = \left( {c{{\psi '}_{s + }}\left( {H\left( t \right)} \right) + \frac{1}{2}{\sigma ^2}{{\psi ''}_{s + }}\left( {H\left( t \right)} \right)} \right){\rm{d}}t + \sigma {{\psi '}_{s + }}\left( {H\left( t \right)} \right){\rm{d}}{B_t} $

    $ {\psi _{s + }}\left( {H\left( t \right)} \right) - {\psi _{s + }}\left( u \right) = \int_0^t {\left( {c{{\psi '}_{s + }}\left( {H\left( x \right)} \right) + \frac{1}{2}{\sigma ^2}{{\psi ''}_{s + }}\left( {H\left( x \right)} \right)} \right){\rm{d}}x} + \int_0^t {\sigma {{\psi '}_{s + }}\left( {H\left( x \right)} \right){\rm{d}}{B_x}} $

    所以

    $ E\left( {{\psi _{s + }}\left( {H\left( t \right)} \right)} \right) - {\psi _{s + }}\left( u \right) = \int_0^\tau {\left( {cE\left[ {{{\psi '}_{s + }}\left( {H\left( x \right)} \right)} \right] + \frac{1}{2}{\sigma ^2}E\left[ {{{\psi ''}_{s + }}\left( {H\left( x \right)} \right)} \right]} \right){\rm{d}}x} $ (6)

    在充分小的时间段(0,t]内,考虑(2)式定义的风险过程Uδ(t).既然N1(t)和N2(t)都是Poisson-Geometric过程,则在(0,t]有以下4种可能情况:

    1) N1(t)和N2(t)都没有跳跃,即没有保单到达,也没有索赔发生,其发生的概率为(1-λ1 t+o(t))(1-λ2 t+o(t)).

    2) N1(t)至少有一个跳跃,且N2(t)没有跳跃,其发生的概率为(1-λ2 t+o(t))$\sum\limits_{k = 1}^\infty$[α1ρ1k t+Ak1(t)o(t)].

    3) N1(t)没有跳跃,且N2(t)至少有一跳跃,其发生的概率为(1-λ1 t+o(t))$\sum\limits_{k = 1}^\infty$[α2ρ2k t+Ak2(t)o(t)].

    4) N1(t)(或者N2(t))至少有两个跳跃或者N1(t)和N2(t)同时有跳跃,其发生的概率为o(t).

    因此

    $ \begin{array}{l} {\psi _{s + }}\left( u \right) = \left( {1 - {\lambda _1}t + o\left( t \right)} \right)\left( {1 - {\lambda _2}t + o\left( t \right)} \right)E\left[ {{\psi _{s + }}\left( {H\left( t \right)} \right)} \right] + \left( {1 - {\lambda _2}t + o\left( t \right)} \right)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{k = 1}^\infty {E\left[ {\int_0^{H\left( t \right)} {{\psi _{s + }}\left( {H\left( t \right) - x} \right){\rm{d}}{G^{*k}}\left( x \right)} + \int_{H\left( t \right)}^{H\left( t \right) + \frac{c}{\delta }} {{\psi _{s - }}\left( {H\left( t \right) - x} \right){\rm{d}}{G^{*k}}\left( x \right)} + \int_{H\left( t \right) + \frac{c}{\delta }}^\infty {{\rm{d}}{G^{*k}}\left( x \right)} } \right]} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\left( {{\alpha _1}\rho _1^kt + A_k^1\left( t \right)o\left( t \right)} \right) + \left( {1 - {\lambda _1}t + o\left( t \right)} \right)\sum\limits_{k = 1}^\infty E \left[ {\int_0^{H\left( t \right)} {{\psi _{s + }}\left( {H\left( t \right) - y} \right){\rm{d}}{F^{*k}}\left( y \right)} + } \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\int_{H\left( t \right)}^{H\left( t \right) + \frac{c}{\delta }} {{\psi _{s - }}\left( {H\left( t \right) - y} \right){\rm{d}}{F^{*k}}\left( y \right)} + \int_{H\left( t \right) + \frac{c}{\delta }}^\infty {\left. {{\rm{d}}{F^{*k}}\left( y \right)} \right]\left( {{\alpha _2}\rho _2^kt + A_k^2\left( t \right)o\left( t \right)} \right) + o\left( t \right)} \end{array} $

    整理可得

    $ \begin{array}{l} \left( {{\lambda _1} + {\lambda _2}} \right)tE\left[ {{\psi _{s + }}\left( {H\left( t \right)} \right)} \right] = E\left[ {{\psi _{s + }}\left( {H\left( t \right)} \right)} \right] - {\psi _{s + }}\left( u \right) + \sum\limits_{k = 1}^\infty E \left[ {\int_0^{H(h)} {{\psi _{s + }}\left( {H\left( t \right) - x} \right){g^{*k}}\left( x \right){\rm{d}}x} + } \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left. {\int_{H\left( t \right)}^{H\left( t \right) + \frac{c}{\delta }} {{\psi _{s - }}\left( {H\left( t \right) - x} \right){g^{*k}}\left( x \right){\rm{d}}x} + \overline {{G^{*k}}} \left( {H\left( t \right) + \frac{c}{\delta }} \right)} \right]{\alpha _1}\rho _1^kt + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{k = 1}^\infty E \left[ {\int_0^{H\left( t \right)} {{\psi _{s + }}\left( {H\left( t \right) - y} \right){f^{*k}}\left( y \right){\rm{d}}y} + } \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left. {\int_{H\left( t \right)}^{H\left( t \right) + \frac{c}{\delta }} {{\psi _{s - }}\left( {H\left( t \right) - y} \right){f^{*k}}\left( y \right){\rm{d}}x} + \overline {{F^{*k}}} \left( {H\left( t \right) + \frac{c}{\delta }} \right)} \right]{\alpha _2}\rho _2^kt + o\left( t \right) \end{array} $ (7)

    在(7)式两边同时除以t,令t→0,同时利用(6)式则有

    $ \begin{array}{l} \left( {{\lambda _1} + {\lambda _2}} \right){\psi _{s + }}\left( u \right) = c{{\psi '}_{s + }}\left( u \right) + \frac{1}{2}{\sigma ^2}{{\psi ''}_{s + }}\left( u \right) + \int_0^u {{\psi _{s + }}\left( {u - x} \right){g_{{\rho _1}}}\left( x \right){\rm{d}}x} + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\int_u^{u + \frac{c}{\delta }} {{\psi _{s - }}\left( {u - x} \right){g_{{\rho _1}}}\left( x \right){\rm{d}}x} + \sum\limits_{k = 1}^\infty {\overline {{G^{*k}}} \left( {u + \frac{c}{\delta }} \right){\alpha _1}\rho _1^k} + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\int_0^u {{\psi _{s + }}\left( {u - y} \right){f_{{\rho _2}}}\left( y \right){\rm{d}}y} + \int_u^{u + \frac{c}{\delta }} {{\psi _{s - }}\left( {u - y} \right){f_{{\rho _2}}}\left( y \right){\rm{d}}y} + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{k = 1}^\infty {\overline {{F^{*k}}} \left( {u + \frac{c}{\delta }} \right){\alpha _2}\rho _2^k} \end{array} $

    其中

    $ \begin{array}{*{20}{c}} {{g_{{\rho _1}}}\left( x \right) = \sum\limits_{k = 1}^\infty {{\alpha _1}\rho _1^k{g^{*k}}\left( x \right)} }&{{f_{{\rho _2}}}\left( y \right) = \sum\limits_{k = 1}^\infty {{\alpha _2}\rho _2^k{f^{*k}}\left( y \right)} } \end{array} $

    所以(3)式成立.

    $ - \frac{c}{\delta } < u < 0$时,令Y(t)=ueδt+ct+σBt,则Y(0)=u及dY(t)=(eδt+c)dt+σdBt.

    由伊藤积分公式有:

    $ {\rm{d}}{\psi _{s - }}\left( {Y\left( t \right)} \right) = \left( {\left( {u\delta {{\rm{e}}^{\delta t}} + c} \right){\psi _{s - }}\left( {Y\left( t \right)} \right) + \frac{1}{2}{\sigma ^2}{{\psi ''}_{s - }}\left( {Y\left( t \right)} \right)} \right){\rm{d}}t + \sigma {{\psi '}_{s - }}\left( {Y\left( t \right)} \right){\rm{d}}{B_t} $

    $ {\psi _{s - }}\left( {Y\left( t \right)} \right) - {\psi _{s - }}\left( u \right) = \int_0^t {\left( {\left( {u\delta {{\rm{e}}^{\delta x}} + c} \right){\psi _{s - }}\left( {Y\left( x \right)} \right) + \frac{1}{2}{\sigma ^2}{{\psi ''}_{s - }}\left( {Y\left( x \right)} \right)} \right){\rm{d}}x} + \int_0^t {\sigma {{\psi '}_{s - }}\left( {Y\left( x \right)} \right){\rm{d}}{B_x}} $

    所以

    $ E\left( {{\psi _{s - }}\left( {Y\left( t \right)} \right)} \right) - {\psi _{s - }}\left( u \right) = \int_0^t {\left( {\left( {u\delta {{\rm{e}}^{\delta x}} + c} \right)E\left[ {{\psi _{s - }}\left( {Y\left( x \right)} \right)} \right] + \frac{1}{2}{\sigma ^2}E\left[ {{{\psi ''}_{s - }}\left( {Y\left( x \right)} \right)} \right]} \right){\rm{d}}x} $ (8)

    利用同u≥0时的讨论方法可得:

    $ \begin{array}{l} {\psi _{s - }}\left( u \right) = \left( {1 - {\lambda _1}t + o\left( t \right)} \right)\left( {1 - {\lambda _2}t + o\left( t \right)} \right)E\left[ {{\psi _{s - }}\left( {Y\left( t \right)} \right)} \right] + \left( {1 - {\lambda _2}t + o\left( t \right)} \right)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{k = 1}^\infty E \left( {\int_0^{Y\left( t \right) + \frac{c}{\delta }} {{\psi _{s - }}\left( {Y\left( t \right) - x} \right){\rm{d}}{G^{*k}}\left( x \right)} + \int_{Y\left( t \right) + \frac{c}{\delta }}^\infty {{\rm{d}}{G^{*k}}\left( x \right)} } \right)\left( {{\alpha _1}\rho _1^kt + A_k^1\left( t \right)o\left( t \right)} \right) + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\left( {1 - {\lambda _1}t + o\left( t \right)} \right)\sum\limits_{k = 1}^\infty E \left( {\int_0^{Y\left( t \right) + \frac{c}{\delta }} {{\psi _{s - }}\left( {Y\left( t \right) - y} \right){\rm{d}}{F^{*k}}\left( y \right)} + \int_{Y\left( t \right) + \frac{c}{\delta }}^\infty {{\rm{d}}{F^{*k}}\left( y \right)} } \right)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\left( {{\alpha _2}\rho _2^kt + A_k^2\left( t \right)o\left( t \right)} \right) + o\left( t \right) \end{array} $

    整理可得

    $ \begin{array}{l} \left( {{\lambda _1} + {\lambda _2}} \right)tE\left( {{\psi _{s - }}\left( {Y\left( t \right)} \right)} \right) = E\left( {{\psi _{s - }}\left( {Y\left( t \right)} \right)} \right) - {\psi _{s - }}\left( u \right) + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{k = 1}^\infty E \left( {\int_0^{Y\left( t \right) + \frac{c}{\delta }} {{\psi _{s - }}\left( {Y\left( t \right) - x} \right){g^{*k}}\left( x \right){\rm{d}}x} + \overline {{G^{*k}}} \left( {Y\left( t \right) + \frac{c}{\delta }} \right)} \right){\alpha _1}\rho _1^kt + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{k = 1}^\infty E \left( {\int_0^{Y\left( t \right) + \frac{c}{\delta }} {{\psi _{s - }}\left( {Y\left( t \right) - y} \right){f^{*k}}\left( y \right){\rm{d}}y} + \overline {{F^{*k}}} \left( {Y\left( t \right) + \frac{c}{\delta }} \right)} \right){\alpha _2}\rho _2^kt + o\left( t \right) \end{array} $ (9)

    在(9)式两边同时除以t,令t→0,同时利用(8)式则有(4)式成立.

    在(4)式中,令$u \to - \frac{c}{\delta }$,得边界条件(5)式中的第2式.

    注1  当ρi=0时,αi=λii=1,2,(3),(4)式分别退化为文献[2]中的(3),(4)式.

    定理2  假设ψd(u)是二次连续可微的,当u≥0时,则ψd(u)符合下面积分微分方程:

    $ \begin{array}{l} \left( {{\lambda _1} + {\lambda _2}} \right){\psi _{d + }}\left( u \right) = c{{\psi '}_{d + }}\left( u \right) + \frac{1}{2}{\sigma ^2}{{\psi ''}_{d + }}\left( u \right) + \int_0^u {{\psi _{d + }}\left( {u - x} \right){g_{{\rho _1}}}\left( x \right){\rm{d}}x} + {B_{{d_1}}}\left( u \right) + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\int_0^u {{\psi _{d + }}\left( {u - y} \right){f_{{\rho _2}}}\left( y \right){\rm{d}}y} + {B_{{d_2}}}\left( u \right) \end{array} $ (10)

    $ - \frac{c}{\delta } < u < 0$时,ψd(u)符合下面积分微分方程:

    $ \begin{array}{l} \left( {{\lambda _1} + {\lambda _2}} \right){\psi _{d - }}\left( u \right) = \left( {u\delta + c} \right){{\psi '}_{d - }}\left( u \right) + \frac{1}{2}{\sigma ^2}{{\psi ''}_{d - }}\left( u \right) + \int_0^{u + \frac{c}{\delta }} {{\psi _{d - }}} \left( {u - x} \right){g_{{\rho _1}}}\left( x \right){\rm{d}}x + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\int_0^{u + \frac{c}{\delta }} {{\psi _{s - }}} (u - y){f_{{\rho _2}}}(y){\rm{d}}y \end{array} $ (11)

    边界条件为

    $ \left\{ {\begin{array}{*{20}{l}} {{\psi _{d - }}\left( { - \frac{c}{\delta }} \right) = 1}\\ {\frac{1}{2}{\sigma ^2}{{\psi ''}_{d - }}\left( { - \frac{c}{\delta }} \right) = {\lambda _1} + {\lambda _2}}\\ {\mathop {\lim }\limits_{u \to \infty } {\psi _{d + }}\left( u \right) = 0} \end{array}} \right. $ (12)

    其中

    $ {B_{{d_1}}}\left( u \right) = \int_u^{u + \frac{c}{\delta }} {{\psi _{d - }}\left( {u - x} \right){g_{{\rho _1}}}\left( x \right){\rm{d}}x} $
    $ {B_{{d_2}}}\left( u \right) = \int_u^{u + \frac{c}{\delta }} {{\psi _{d - }}\left( {u - y} \right){f_{{\rho _2}}}\left( y \right){\rm{d}}y} ,u \ge 0 $

      类似于定理1.

    注2  当ρi=0时,αi=λii=1,2,(10),(11)式分别退化为文献[2]中的(10),(11)式.

    推论1  在定理1和定理2的条件下,当u>0时,ψ(u)符合下面的积分微分方程:

    $ \begin{array}{l} \left( {{\lambda _1} + {\lambda _2}} \right)\psi \left( u \right) = c\psi '\left( u \right) + \frac{1}{2}{\sigma ^2}\psi ''\left( u \right) + \int_0^u {\psi \left( {u - x} \right){g_{{\rho _1}}}\left( x \right){\rm{d}}x} + {B_1}\left( u \right) + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\int_0^u {\psi \left( {u - y} \right){f_{{\rho _2}}}\left( y \right){\rm{d}}y} + {B_2}\left( u \right) \end{array} $ (13)

    $ - \frac{c}{\delta } < u < 0$时,则ψ(u)符合下面积分微分方程:

    $ \begin{array}{l} \left( {{\lambda _1} + {\lambda _2}} \right)\psi \left( u \right) = (u\delta + c)\psi '\left( u \right) + \frac{1}{2}{\sigma ^2}\psi ''\left( u \right) + \int_0^{u + \frac{c}{\delta }} \psi (u - x){g_{{\rho _1}}}(x){\rm{d}}x + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{k = 1}^\infty {\overline {{G^{*k}}} } \left( {u + \frac{c}{\delta }} \right){\alpha _1}\rho _1^k + \int_0^{u + \frac{c}{\delta }} \psi (u - y){f_{{\rho _2}}}(y){\rm{d}}y + \sum\limits_{k = 1}^\infty {\overline {{F^{*k}}} } \left( {u + \frac{c}{\delta }} \right){\alpha _2}\rho _2^k \end{array} $ (14)

    边界条件为

    $ \left\{ {\begin{array}{*{20}{l}} {{\psi _ - }\left( { - \frac{c}{\delta }} \right) = 1}\\ {\frac{1}{2}{\sigma ^2}{{\psi ''}_ - }\left( { - \frac{c}{\delta }} \right) = {\lambda _1} + {\lambda _2} - \sum\limits_{k = 1}^\infty {{\alpha _1}} \rho _1^k - \sum\limits_{k = 1}^\infty {{\alpha _2}} \rho _2^k}\\ {\mathop {\lim }\limits_{u \to \infty } {\psi _{d + }}\left( u \right) = 0} \end{array}} \right. $ (15)

    其中

    $ {B_1}\left( u \right) = \int_u^{u + \frac{c}{\delta }} {\psi \left( {u - x} \right){g_{{\rho _1}}}\left( x \right){\rm{d}}x} + \sum\limits_{k = 1}^\infty {\overline {{G^{*k}}} \left( {u + \frac{c}{\delta }} \right){\alpha _1}\rho _1^k} $
    $ {B_2}\left( u \right) = \int_u^{u + \frac{c}{\delta }} {\psi \left( {u - y} \right){f_{{\rho _2}}}\left( y \right){\rm{d}}y} + \sum\limits_{k = 1}^\infty {\overline {{F^{*k}}} } \left( {u + \frac{c}{\delta }} \right){\alpha _2}\rho _2^k,u \ge 0 $

    $ {\psi _d}\left( u \right) + {\psi _s}\left( u \right) = \psi \left( u \right),{{\psi '}_d}\left( u \right) + {{\psi '}_s}\left( u \right) = \psi '\left( u \right),{{\psi ''}_d}\left( u \right) + {{\psi ''}_s},\left( u \right) = \psi ''\left( u \right) $

    因此,根据(3)式和(10)式得(13)式,根据(4)式和(11)式得(14)式,根据(5)式和(12)式得(15)式.

    注3  当ρi=0时,αi=λii=1,2,(13),(14)及(15)式分别退化为文献[2]中的(13),(14)及(15)式.

    定理3  假设ψs(ut)对u是二次连续可微的,对t是一次连续可微的.当u≥0时,ψs(ut)符合下面的偏微分积分方程:

    $ \begin{array}{l} \left( {{\lambda _1} + {\lambda _2}} \right){\psi _{s + }}\left( {u,t} \right) = c\frac{{\partial {\psi _{{s^ + }}}\left( {u,t} \right)}}{{\partial u}} + \frac{{\partial {\psi _{{s^ + }}}\left( {u,t} \right)}}{{\partial t}} + \frac{1}{2}{\sigma ^2}\frac{{{\partial ^2}{\psi _{{s^ + }}}\left( {u,t} \right)}}{{\partial {u^2}}} + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\int_0^u {{\phi _{s + }}} \left( {u - x,t} \right){g_{{\rho _1}}}\left( x \right){\rm{d}}x + {B_{{s_1}}}\left( {u,t} \right) + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\int_0^u {{\phi _{s + }}} \left( {u - y,t} \right){f_{{\rho _2}}}\left( y \right){\rm{d}}y + {B_{{s_2}}}\left( {u,t} \right) \end{array} $ (16)

    $ - \frac{c}{\delta } < u < 0$时,则ψs(ut)符合下面积分微分方程:

    $ \begin{array}{l} \left( {u\delta + c} \right)\frac{{\partial {\psi _{s - }}\left( {u,t} \right)}}{{\partial u}} + \frac{{\partial {\psi _{s - }}\left( {u,t} \right)}}{{\partial t}} + \frac{1}{2}{\sigma ^2}\frac{{{\partial ^2}{\psi _{s - }}\left( {u,t} \right)}}{{\partial {u^2}}} - \left( {{\lambda _1} + {\lambda _2}} \right){\psi _{s - }}\left( {u,t} \right) = \\ - \int_0^{u + \frac{c}{\delta }} {{\psi _{s - }}\left( {u - x,t} \right){g_{{\rho _1}}}\left( x \right){\rm{d}}x} - \sum\limits_{k = 1}^\infty {\overline {{G^{*k}}} } \left( {u + \frac{c}{\delta }} \right){\alpha _1}\rho _1^k - \\ \int_0^{u + \frac{c}{\delta }} {{\psi _{s - }}} (u - y,t){f_{{\rho _2}}}(y){\rm{d}}y - \overline {{F^{*k}}} \left( {u + \frac{c}{\delta }} \right){\alpha _2}\rho _2^k \end{array} $ (17)

    边界条件为

    $ \left\{ {\begin{array}{*{20}{l}} {{\psi _{s + }}\left( { + \infty ,t} \right) = 0}\\ {{\psi _s}\left( {u, + \infty } \right) = {\psi _s}\left( u \right)} \end{array}} \right. $ (18)

    其中

    $ {B_{{s_1}}}\left( {u,t} \right) = \int_u^{u + \frac{c}{\delta }} {{\psi _{s - }}} (u - x,t){g_{{\rho _1}}}(x){\rm{d}}x + \sum\limits_{k = 1}^\infty {\overline {{G^{*k}}} } \left( {u + \frac{c}{\delta }} \right){\alpha _1}\rho _1^k $
    $ {B_{{s_2}}}\left( {u,t} \right) = \int_u^{u + \frac{c}{\delta }} {{\psi _{s - }}} (u - y,t){f_{{\rho _2}}}(y){\rm{d}}y + \sum\limits_{k = 1}^\infty {\overline {{F^{*k}}} } \left( {u + \frac{c}{\delta }} \right){\alpha _2}\rho _2^k,u \ge 0 $

    注4  当ρi=0时,αi=λii=1,2,(16),(17)式分别退化为文献[2]中的(23),(24)式.

      当u≥0时,令H(t)=u+ct+σB(t),则H(0)=u且dH(Δ)=cdΔ+σdBΔ.

    由伊藤积分公式有

    $ \begin{array}{l} {\psi _{s + }}\left( {H\left( \Delta \right),t - \Delta } \right) - {\psi _{s + }}\left( {u,t} \right) = \int_0^\Delta {\left( {c\frac{{\partial {\psi _{s + }}\left( {H\left( x \right),t - x} \right)}}{{\partial u}} + \frac{{\partial {\psi _{s + }}\left( {H\left( x \right),t - x} \right)}}{{\partial t}} + } \right.} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left. {\frac{1}{2}{\sigma ^2}\frac{{{\partial ^2}{\psi _{s + }}\left( {H\left( x \right),t - x} \right)}}{{\partial {u^2}}}} \right){\rm{d}}x + \int_0^\Delta \sigma \frac{{\partial {\psi _{s + }}\left( {H\left( x \right),t - x} \right)}}{{\partial u}}{\rm{d}}{B_x} \end{array} $

    所以

    $ \begin{array}{l} E\left[ {{\psi _{s + }}(H(\Delta ),t - \Delta )} \right] - {\psi _{s + }}(u,t) = \int_0^\Delta {\left( {cE\left[ {\frac{{\partial {\psi _{s + }}(H(x),t - x)}}{{\partial u}}} \right] + E\left[ {\frac{{\partial {\psi _{s + }}(H(x),t - x)}}{{\partial t}}} \right] + } \right.} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left. {\frac{1}{2}{\sigma ^2}E\left[ {\frac{{{\partial ^2}{\psi _{s + }}(H(x),t - x)}}{{\partial {u^2}}}} \right]} \right){\rm{d}}x \end{array} $ (19)

    在充分小的时间段(0,Δ]内,考虑(2)式定义的风险过程.首先研究索赔引起的有限时破产概率ψs(ut),由全概公式整理可得

    $ \begin{array}{l} \left( {{\lambda _1} + {\lambda _2}} \right)\Delta E\left[ {{\psi _{s + }}(H(\Delta ),t - \Delta )} \right] = E\left[ {{\psi _{s + }}(H(\Delta ),t - \Delta )} \right] - {\psi _{s + }}(u,t) + \\ \sum\limits_{k = 1}^\infty E \left[ {\int_0^{H(\Delta )} {{\psi _{s + }}} (H(\Delta ) - x,t - \Delta ){g^{*k}}(x){\rm{d}}x + \int_{H(\Delta )}^{H(\Delta ) + \frac{c}{\delta }} {{\psi _{s - }}} (H(\Delta ) - x,t - \Delta ){g^{*k}}(x){\rm{d}}x + } \right.\\ \left. {\overline {{G^{*k}}} \left( {H(\Delta ) + \frac{c}{\delta }} \right)} \right]{\alpha _1}\rho _1^k\Delta + \sum\limits_{k = 1}^\infty E \left[ {\int_0^{H(\Delta )} {{\psi _{s + }}} (H(\Delta ) - y,t - \Delta ){f^{*k}}(y){\rm{d}}y + } \right.\\ \left. {\int_{H(\Delta )}^{H(\Delta ) + \frac{c}{\delta }} {{\psi _{s - }}(H(\Delta ) - y,t - \Delta ){f^{*k}}(y){\rm{d}}x} + \overline {{F^{*k}}} \left( {H(\Delta ) + \frac{c}{\delta }} \right)} \right]{\alpha _2}\rho _2^k\Delta + o(\Delta ) \end{array} $ (20)

    在(20)式两边同时除以Δ,令Δ→0,同时利用(19)式得(16)式.

    $ - \frac{c}{\delta } < u < 0$时,令Y(t)=ueδt+ct+σBt,则Y(0)=u及dY(Δ)=(uδeδΔ+c)dΔ+σdBΔ.由伊藤积分公式可得

    $ \begin{array}{l} E\left[ {{\psi _{s - }}(Y(\Delta ),t - \Delta )} \right] - {\psi _{s - }}(u,t) = \int_0^\Delta {\left( {\left( {u\delta {{\rm{e}}^{\delta x}} + c} \right)E\left[ {\frac{{\partial {\psi _{s - }}(Y(x),t - x)}}{{\partial u}}} \right] + } \right.} \\ \left. {E\left[ {\frac{{\partial {\psi _{s - }}(Y(x),t - x)}}{{\partial t}}} \right] + \frac{1}{2}{\sigma ^2}E\left[ {\frac{{{\partial ^2}{\psi _{s - }}(Y(x),t - x)}}{{\partial {u^2}}}} \right]} \right){\rm{d}}x \end{array} $ (21)

    利用同u≥0时讨论的方法得

    $ \begin{array}{l} \left( {{\lambda _1} + {\lambda _2}} \right)\Delta E\left[ {{\psi _{s - }}(Y(\Delta ),t - \Delta )} \right] = E\left[ {{\psi _{s - }}(Y(\Delta ),t - \Delta )} \right] - {\psi _{s - }}(u,t) + \\ \sum\limits_{k = 1}^\infty E \left[ {\int_0^{Y(\Delta ) + \frac{c}{\delta }} {{\psi _{s - }}} (Y(\Delta ) - x,t - \Delta ){g^{*k}}(x){\rm{d}}x + \overline {{G^{*k}}} \left( {Y(\Delta ) + \frac{c}{\delta }} \right)} \right]{\alpha _1}\rho _1^k\Delta + \\ \sum\limits_{k = 1}^\infty E \left[ {\int_0^{Y(\Delta ) + \frac{c}{\delta }} {{\psi _{s - }}} (Y(\Delta ) - y,t - \Delta ){f^{*k}}(y){\rm{d}}y + \overline {{F^{*k}}} \left( {Y(\Delta ) + \frac{c}{\delta }} \right)} \right]{\alpha _2}\rho _2^k\Delta + o(\Delta ) \end{array} $ (22)

    在(22)式两边同时除以Δ,令Δ→0,同时利用(21)式得(17)式.

    注5  ${\left. {\frac{{\partial {\psi _s}(u, t)}}{{\partial t}}} \right|_{t = \infty }}$,当t→∞时,(16)式即为(3)式,(17)式即为(4)式.

    定理4  假设ψd(ut)对u是二次连续可微的,对t是一次连续可微的.当u≥0,ψs(ut)符合下面的偏微分积分方程:

    $ \begin{array}{l} \left( {{\lambda _1} + {\lambda _2}} \right){\psi _{d + }}(u,t) = c\frac{{\partial {\psi _{{d^ + }}}(u,t)}}{{\partial u}} + \frac{{\partial {\psi _{{d^ + }}}(u,t)}}{{\partial t}} + \frac{1}{2}{\sigma ^2}\frac{{{\partial ^2}{\psi _{{d^ + }}}(u,t)}}{{\partial {u^2}}} + \int_0^u {{\psi _{d + }}} \left( {u - x,t} \right)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{g_{{\rho _1}}}(x){\rm{d}}x + {B_{{d_1}}}(u,t) + \int_0^u {{\psi _{d + }}} (u - y,t){f_{{\rho _2}}}(y){\rm{d}}y + {B_{{d_2}}}(u,t) \end{array} $ (23)

    $ - \frac{c}{\delta } < u < 0$时,则ψd(ut)符合下面积分微分方程:

    $ \begin{array}{l} \left( {u\delta + c} \right)\frac{{\partial {\psi _{d - }}(u,t)}}{{\partial u}} + \frac{{\partial {\psi _{d - }}(u,t)}}{{\partial t}} + \frac{1}{2}{\sigma ^2}\frac{{{\partial ^2}{\psi _{d - }}(u,t)}}{{\partial {u^2}}} - \left( {{\lambda _1} + {\lambda _2}} \right){\psi _{d - }}(u,t) = \\ \int_0^{u + \frac{c}{\delta }} {{\psi _{d - }}} (u - x,t){g_{{\rho _1}}}(x){\rm{d}}x - \int_0^{u + \frac{c}{\delta }} {{\psi _{d - }}} (u - y,t){f_{{\rho _2}}}(y){\rm{d}}y \end{array} $ (24)

    边界条件为

    $ \left\{ {\begin{array}{*{20}{l}} {{\psi _{d + }}( + \infty ,t) = 0}\\ {{\psi _d}(u, + \infty ) = {\psi _d}(u)} \end{array}} \right. $

    其中

    $ {B_{{d_1}}}(u,t) = \int_u^{u + \frac{c}{\delta }} {{\psi _{d - }}} (u - x,t){g_{{\rho _1}}}(x){\rm{d}}x $
    $ {B_{{d_2}}}(u,t) = \int_u^{u + \frac{c}{\delta }} {{\psi _{d - }}} (u - y,t){f_{{\rho _2}}}(y){\rm{d}}y,u \ge 0 $

    注6  ${\left. {\frac{{\partial {\psi _d}(u, t)}}{{\partial t}}} \right|_{t = \infty }}$=0,当t→∞时,(23)式即为(10)式,(24)式即为(11)式.

    注7  当ρi=0时,αi=λii=1,2,(23),(24)式分别退化为文献[2]中的(30),(31)式.

    推论2  在定理3和定理4条件下,当u≥0,ψ(ut)符合下面的偏微分积分方程:

    $ \begin{array}{l} \left( {{\lambda _1} + {\lambda _2}} \right)\psi (u,t) = c\frac{{\partial \psi (u,t)}}{{\partial u}} + \frac{{\partial \psi (u,t)}}{{\partial t}} + \frac{1}{2}{\sigma ^2}\frac{{{\partial ^2}\psi (u,t)}}{{\partial {u^2}}} + \int_0^u \psi (u - x,t){g_{{\rho _1}}}(x){\rm{d}}x + \\ {B_1}(u,t) + \int_0^u \psi (u - y,t){f_{{\rho _2}}}(y){\rm{d}}y + {B_2}(u,t) \end{array} $ (25)

    $ - \frac{c}{\delta } < u < 0$时,则ψ(ut)符合下面积分微分方程:

    $ \begin{array}{l} (u\delta + c)\frac{{\partial \psi (u,t)}}{{\partial u}} + \frac{{\partial \psi (u,t)}}{{\partial t}} + \frac{1}{2}{\sigma ^2}\frac{{{\partial ^2}\psi (u,t)}}{{\partial {u^2}}} - \left( {{\lambda _1} + {\lambda _2}} \right)\psi (u,t) = \\ - \int_0^{u + \frac{c}{\delta }} \psi (u - x,t){g_{{\rho _1}}}(x){\rm{d}}x - \sum\limits_{k = 1}^\infty {\overline {{G^{*k}}} } \left( {u + \frac{c}{\delta }} \right){\alpha _1}\rho _1^k - \\ \int_0^{u + \frac{c}{\delta }} \psi (u - y,t){f_{{\rho _2}}}(y){\rm{d}}y - \overline {{F^{*k}}} \left( {u + \frac{c}{\delta }} \right){\alpha _2}\rho _2^k \end{array} $ (26)

    边界条件为

    $ \left\{ {\begin{array}{*{20}{l}} {\psi ( + \infty ,t) = 0}\\ {\psi (u, + \infty ) = \psi (u)} \end{array}} \right. $

    其中

    $ {B_1}(u,t) = \int_u^{u + \frac{c}{\delta }} {{\psi _{s - }}} (u - x,t){g_{{\rho _1}}}(x){\rm{d}}x + \sum\limits_{k = 1}^\infty {\overline {{G^{*k}}} } \left( {u + \frac{c}{\delta }} \right){\alpha _1}\rho _1^k $
    $ {B_2}(u,t) = \int_u^{u + \delta } {{\psi _{s - }}} (u - y,t){f_{{\rho _2}}}(y){\rm{d}}y + \sum\limits_{k = 1}^\infty {\overline {{F^{*k}}} } \left( {u + \frac{c}{\delta }} \right){\alpha _2}\rho _2^k,u \ge 0 $

    注8  ${\left. {\frac{{\partial \psi (u, t)}}{{\partial t}}} \right|_{t = \infty }}$=0,当t→∞时,(25)式即为(13)式,(26)式即为(14)式.

    注9  当ρi=0时,αi=λii=1,2,(25),(26)式分别退化为文献[2]中的(32),(33)式.

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    A Perturbed Double Compound Poisson-Geometric Risk Process Model with Debit Interest
    WANG Yue-ming1, WEI Guang-hua2, GUO Nan1, GAO Yan-yan1    
    1. Department of Mathematics and Physics, Nanjing Institute of Technology, Nanjing 211167, China;
    2. School of Science, Jinling Institute of Technology, Nanjing 211169, China
    Abstract: In this paper, we consider a risk model which is a perturbed double compound Poisson-Geometric process with debit interest. We obtain the Integro-differential equation for infinite time ruin probability and then derive the integral partial differential equation for finite time ruin probability by total expectation formula, differential calculus, Ito formula and other knowledge and the causes of ruin.
    Key words: debit    compound Poisson-Geometric process    Brown motion    ruin probability    integro-differential equation    integral partial differential equation    
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