经典的复合Poisson风险模型$R(t)=u+ct+\sum\limits_{i=1}^{N(t)}X_i$假定按照单位时间常速率取得保单, 且每张保单的保险费也相同, 但在实际应用中收到的保单数和每张保单的保费都是随机的, 所以近年来许多学者对其进行了推广, 一方面用复合Poisson过程来描述保费收入, 且用更一般的点过程（更新过程、Cox过程）来描述理赔次数；另一方面也考虑在经济环境中利率、通货膨胀率、随机干扰等因素的影响, 其中带利率的风险模型是当前风险理论研究的热点之一^[1]-[7].

吴荣^[1]讨论了保费在单位时间内以常数速率连续收取、理赔到达过程为一般更新过程、带常数利率的风险模型$R(t)=ue^{\delta t}+\frac{c(e^{\delta t}-1)}{\delta}-\sum\limits_{i=1}^{N(t)}X_ie^{\delta (t-T_i)}, $给出了几个重要的精算诊断量如破产概率、破产时余额的分布、破产前瞬间余额分布等的级数展式；朱柘琍^[2]对保费收取次数为Poisson过程、每次收取保费为常数C、理赔到达过程为一般更新过程、带常数利率的风险模型$U_\delta {(t)}=ue^{\delta t}+C\sum\limits_{i=1}^{M(t)}e^{\delta(t-K_i)}-\sum\limits_{i=1}^{N(t)}X_ie^{\delta(t-T_i)}$讨论了类似的问题.本文将研究带常数利率的保费收入为复合Poisson过程, 理赔到达过程为一般更新过程的风险模型$U_\delta(t)=ue^{\delta t}+\sum\limits_{i=1}^{M(t)}Y_ie^{\delta(t-K_i)}-\sum\limits_{i=1}^{N(t)}X_ie^{\delta(t-T_i)}, $推广吴荣^[1]和朱柘琍^[2]的相关结论.

设$(\Omega, {\cal{F}}, P)$为一个完备的概率空间, 以下所涉及到的随机变量均为该空间上的随机变量. ${\{N(t), t\geqslant0\}}$为一普通的更新过程, 它表示$(0, t]$时间间隔内的理赔次数, 相应的更新流为$0=T_0<T_1<T_2<\cdots<T_n<\cdots$, 它表示各次理赔时刻, 更新间隔$W_n=T_n-T_{n-1}, n=1, 2, \cdots$独立同分布, 其分布函数为$\gamma(x)$. ${\{M(t), t\geq0}\}$为一参数为$\lambda$的Poisson过程, 它表示$(0, t]$时间间隔内收取保费的次数, 相应的Poisson流为$0=K_0<K_1<K_2<\cdots<K_n<\cdots$, 它表示各次保费收取时刻. $\{X_n, n=1, 2, \cdots\}$, $\{Y_n, n=1, 2, \cdots\}$均为独立同分布的非负随机变量序列, 分别表示各次理赔额和各次投保额, 其分布函数分别为$F(x)$和$G(x)$.于是$X(t)=\sum\limits_{i=1}^{N(t)}X_i$表示$(0, t]$时间间隔内的理赔总额, $Y(t)=\sum\limits_{i=1}^{M(t)}X_i$表示$(0, t]$时间间隔内收取的保费总额.假定$\{X_n, n=1, 2, \cdots\}$, $\{Y_n, n=1, 2, \cdots\}$, $\{N_t, t\geq0\}$, $\{M_t, t\geq0\}$相互独立.

若保险公司除保费收入之外, 还以常数利率$\delta(\delta >0)$获取资产余额的利息, 于是考虑如下带常数利率的一般更新风险模型$dU_\delta(t)=\delta U_\delta(t)dt+dY(t)-dX(t), U_\delta (0)=u\geq0.$即

$ \begin{eqnarray*}U_\delta (t)&=&ue^{\delta t}+\int_0^te^{\delta(t-v)}dY(v)-\int_0^te^{\delta(t-v)}dX(v)\\ &=&ue^{\delta t}+\sum\limits_{i=1}^{M(t)}Y_ie^{\delta(t-K_i)}-\sum\limits_{i=1}^{N(t)}X_ie^{\delta(t-T_i)}.\end{eqnarray*} $

记$\psi_\delta(u)=P\{\bigcup\limits_{t\geq0}(U_\delta(t)<0)\mid U_\delta(0)=u\}$, $\psi_\delta(u)$表示保险公司初始准备金为$u$时风险过程$\{U_\delta(t), t\geq0\}$的破产概率.用$T$表示相应的破产时, 即$T=\inf\{t>0:U_\delta(t)<0\}$, 则$T$是一个停时, 也有$\psi_\delta(u)=P\{T<\infty \mid U_\delta(0)=u\}$.注意到破产只有可能在索赔时刻发生, 所以

$ \psi_\delta(u)=P\{\bigcup\limits_{n=1}^\infty(T=T_n)\mid U_\delta(0)=u\}=P\{\bigcup\limits_{n=1}^\infty(U_\delta(T_n)<0)\mid U_\delta(0)=u\}. $

初始准备金为$u$时风险过程$\{U_\delta(t), t\geq0\}$的生存概率记为$\Phi_\delta(u)$, 即$\Phi_\delta(u)=1-\psi_\delta(u)$.

首先我们准备几个引理.

引理 1 ^[7] 对所有的$0\leq s<t$及整数$m, n(m<n)$, 在$M(s)=m, M(t)=n$的条件下$(K_{m+1}-s, K_{m+2}-s, \cdots, K_n-s)$的条件分布密度为

$ f(s_1, s_2, \cdots, s_{n-m})=\frac{(n-m)!}{{(t-s)}^{n-m}} I_{0<s_1<s_2<\cdots<s_{n-m}\leq{t-s}}~~(s_1, s_2, \cdots, s_{n-m}). $

也就是说, 若$U_1, U_2, \cdots, U_{n-m}$为${[0, t-s]}$上$n-m$个均匀分布的独立的随机变量, $U_{(i)}, $ $i=1, 2, \cdots, n-m$为相应的顺序统计量, 则在$M(s)=m, M(t)=n$的条件下$(K_{m+1}-s, K_{m+2}-s, \cdots, K_n-s)$的条件分布密度与$(U_{(1)}, U_{(2)}, \cdots, U_{(n-m)})$的分布密度相同.特别地, 若令$s=0, m=0$, 则有在$M(t)=n$的条件下$(K_1, K_2, \cdots, K_n)$的条件分布密度为

$ f(s_1, s_2, \cdots, s_n)=\frac{n!}{t^n}I_{0<s_1<s_2<\cdots<s_n\leq t}(s_1, s_2, \cdots, s_n). $

即在$M(t)=n$的条件下$(K_1, K_2, \cdots, K_n)$的条件分布密度与$(U_{(1)}, U_{(2)}, \cdots, U_{(n)})$的分布密度相同, 其中$U_{(i)}, i=1, 2, \cdots, n$表示$[0, t]$上$n$个均匀分布的独立的随机变量$U_1, U_2, \cdots, U_n$的第$i$个次序统计量.

引理 2  $P(\sum\limits_{i=m+1}^nY_ie^{-\delta(K_i-t)}\leq z\mid {M(t)=m}, M(t+s)=n)=P(\sum\limits_{i=1}^{n-m}Y_ie^{-\delta U_{(i)}}\leq z)$, 其中$U_{(i)}, i=1, 2, \cdots, n-m$表示$[0, s]$上$n-m$个均匀分布的独立的随机变量$U_1, U_2, \cdots, U_{n-m}$的第$i$个次序统计量.特别地, 取$t=0, m=0$, 则有

$ P(\sum\limits_{i=1}^nY_ie^{-\delta K_i}\leq z\mid {M(s)=n})=P(\sum\limits_{i=1}^nY_ie^{-\delta U_{(i)}}\leq z), $

其中$U_{(i)}, i=1, 2, \cdots, n$表示$[0, s]$上$n$个均匀分布的独立的随机变量$U_1, U_2, \cdots, U_n$的第$i$个次序统计量.

证

$ \begin{array}{l} \;\;\;\;P(({Y_{m + 1}},{Y_{m + 2}}, \cdots ,{Y_n}) \in A,({K_{m + 1}} - t,{K_{m + 2}} - t,{K_n} - t) \in B\;\mid M(t) = m,\\\;\;\;\;M(t + s) = n)\\ = P(({K_{m + 1}} - t,{K_{m + 2}} - t,{K_n} - t) \in B\mid M(t) = m,\\\;\;\;\;M(t + s) = n)\\ \;\;\;\; \times P(({Y_{m + 1}},{Y_{m + 2}}, \cdots ,{Y_n}) \in A\mid ({K_{m + 1}} - t,{K_{m + 2}} - t,{K_n} - t) \in B,M(t) = m,\\\;\;\;\;M(t + s) = n). \end{array} $

注意到$\{{Y_n, n=1, 2, \cdots}\}$与$\{{M_t, t\geq0}\}, \{{K_n, n=1, 2, \cdots}\}$均相互独立, 所以

$ \begin{array}{l} \;\;\;\;P(({Y_{m + 1}},{Y_{m + 2}}, \cdots ,{Y_n}) \in A\mid ({K_{m + 1}} - t,{K_{m + 2}} - t,{K_n} - t) \in B,{\rm{ }}M(t) = m,\\\;\;\;\;{\rm{ }}M(t + s) = n)\\ = \;\;P(({Y_{m + 1}},{Y_{m + 2}}, \cdots ,{Y_n}) \in A). \end{array} $

从而由引理1知

$ \begin{array}{l} \;\;\;\;P(({Y_{m + 1}},{Y_{m + 2}}, \cdots ,{Y_n}) \in A,({K_{m + 1}} - t,{K_{m + 2}} - t,{K_n} - t) \in B\;\mid M(t) = m,\\\;\;\;\;M(t + s) = n)\\ = \;P(({U_{(1)}},{U_{(2)}}, \cdots ,{U_{(n - m)}}) \in B)P(({Y_{m + 1}},{Y_{m + 2}}, \cdots ,{Y_n}) \in A), \end{array} $

其中$U_{(i)}, i=1, 2, \cdots, n-m$表示$[0, s]$上$n-m$个均匀分布的独立的随机变量$U_1, U_2, \cdots, U_{n-m}$的第$i$个次序统计量.注意到$\{{Y_n, n=1, 2, \cdots}\}$独立同分布, 则有

$ \begin{eqnarray*}&&P((Y_{m+1}, Y_{m+2}, \cdots, Y_n)\in A, \\ &&(K_{m+1}-t, K_{m+2}-t, K_n-t)\in B\mid M(t)=m, M(t+s)=n)\\ &=&P((U_{(1)}, U_{(2)}, \cdots, U_{(n-m)})\in B) P((Y_1, Y_2, \cdots, Y_{n-m})\in A), \end{eqnarray*} $

于是

$ P(\sum\limits_{i=m+1}^nY_ie^{-\delta(K_i-t)}\leq z\mid M(t)=m, M(t+s)=n)=P(\sum\limits_{i=1}^{n-m}Y_ie^{-\delta U_{(i)}}\leq z). $

引理 3  保费总额折现过程$\{L_\delta(t)=\sum\limits_{i=1}^{M(t)}Y_ie^{-\delta K_i}, t\geq 0\}$满足

$ L_\delta(T_n)-L_\delta(T_{n-1})\stackrel{d}{=}e^{-\delta T_{n-1}}L_\delta(W_n), $

其中$X\stackrel{d}{=}Y$表示随机变量$X$、$Y$有相同的分布函数.

证  首先证明下面$(2.1)$式

$ \begin{eqnarray} &&P(L_\delta(t+s)-L_\delta(t)\leq z\mid M(t)=m, M(t+s)=n)\nonumber\\ &=&\;P(e^{-\delta t}\sum\limits_{i=1}^{n-m}Y_ie^{-\delta K_i}\leq z\mid M(s)=n-m), \\ &&P(L_\delta(t+s)-L_\delta(t)\leq z\mid M(t)=m, M(t+s)=n)\nonumber\\ &=&\;P(\sum\limits_{i=M(t)+1}^{M(t+s)}Y_ie^{-\delta K_i}\leq z\mid M(t)=m, M(t+s)=n)\nonumber\\ &=&\;P(\sum\limits_{i=m+1}^nY_ie^{-\delta(K_i-t)}\leq ze^{\delta t}\mid M(t)=m, M(t+s)=n) =P(\sum\limits_{i=1}^{n-m}Y_ie^{-\delta U_{(i)}}\leq ze^{\delta t}) \nonumber \end{eqnarray} $

(2.1)

$ \begin{eqnarray} & =&\;P(\sum\limits_{i=1}^{n-m}Y_ie^{-\delta K_i}\leq ze^{\delta t}\mid M(s)=n-m) \end{eqnarray} $

(2.2)

$ \begin{eqnarray} &=&\;P(e^{-\delta t}\sum\limits_{i=1}^{n-m}Y_ie^{-\delta K_i}\leq z\mid M(s)=n-m), \end{eqnarray} $

(2.3)

其中式$(2.2), (2.3)$由引理$2$得到, 于是$(2.1)$式得证.

记$F(s, t)=P(W_n\leq s, T_{n-1}\leq t)$, 则有

$ \begin{eqnarray*} &&P(L_\delta(T_n)-L_\delta(T_{n-1})\leq z)=P(L_\delta(T_{n-1}+W_n)-L_\delta(T_{n-1})\leq z)\\ &=&\int P(L_\delta(T_{n-1}+W_n)-L_\delta(T_{n-1})\leq z\mid W_n=s, T_{n-1}=t)F(ds, dt)\\ &=&\int P(L_\delta(t+s)-L_\delta(t)\leq z\mid W_n=s, T_{n-1}=t)F(ds, dt)\\ &=&\int P(\sum\limits_{i=M(t)+1}^{M(t+s)}Y_ie^{-\delta K_i}\leq z\mid W_n=s, T_{n-1}=t)F(ds, dt)\\ &=&\int P(\sum\limits_{i=M(t)+1}^{M(t+s)}Y_ie^{-\delta K_i}\leq z)F(ds, dt) =\int P(L_\delta(t+s)-L_\delta(t)\leq z)F(ds, dt)\\ &=&\int \sum\limits_{m=0}^\infty \sum\limits_{n=m}^\infty P(L_\delta(t+s)-L_\delta(t)\leq z\mid M(t)=m, M(t+s)=n)\\ &&P(M(t)=m, M(t+s)=n)F(ds, dt)\\ &=&\int \sum\limits_{m=0}^\infty \sum\limits_{n=m}^\infty P(e^{-\delta t}\sum\limits_{i=1}^{n-m}Y_ie^{-\delta K_i}\leq z\mid M(s)=n-m)\\ &&P(M(t)=m, M(t+s)=n)F(ds, dt)\mbox{(由(2.1)式知)}\\ &=&\int \sum\limits_{m=0}^\infty \sum\limits_{n=m}^\infty P(e^{-\delta t}\sum\limits_{i=1}^{M(s)}Y_ie^{-\delta K_i}\leq z\mid M(s)=n-m)\\ &&P(M(t)=m)P(M(s)=n-m)F(ds, dt)\end{eqnarray*} $

$ \begin{eqnarray*} &=&\;\int P(e^{-\delta t}\sum\limits_{i=1}^{M(s)}Y_ie^{-\delta K_i}\leq z)F(ds, dt)\\ &=&\int P (e^{-\delta t}\sum\limits_{i=1}^{M(s)}Y_ie^{-\delta K_i}\leq z\mid W_n=s, T_{n-1}=t)F(ds, dt)\\ &=&\;\int P(e^{-\delta T_{n-1}}\sum\limits_{i=1}^{M(W_n)}Y_ie^{-\delta K_i}\leq z\mid W_n=s, T_{n-1}=t)F(ds, dt)\\ &=&\;P(e^{-\delta T_{n-1}}\sum\limits_{i=1}^{M(W_n)}Y_ie^{-\delta K_i}\leq z) =\;P(e^{-\delta T_{n-1}}L_\delta(W_n)\leq z), \end{eqnarray*} $

从而引理3得证.

注意到$N(T_i)=i, i=1, 2, \cdots$, 则有

$ \begin{eqnarray*} U_\delta(T_n)&\;=&\;ue^{\delta T_n}+\sum\limits_{i=1}^{M(T_n)}Y_ie^{\delta(T_n-K_i)}-\sum\limits_{i=1}^nX_ie^{\delta(T_n-T_i)}\\ &\;=&\;e^{\delta W_n}U_\delta(T_{n-1})+e^{\delta W_n}e^{\delta T_{n-1}} \sum\limits_{i=M(T_{n-1})+1}^{M(T_n)}Y_ie^{-\delta K_i}-X_n. \end{eqnarray*} $

用与引理3类似的方法可以证明

$ \begin{equation}U_\delta(T_n)\stackrel{d}{=} e^{\delta W_n}U_\delta(T_{n-1})+e^{\delta W_n}L_\delta(W_n)-X_n. \end{equation} $

(2.4)

引理 4  记$\{L_\delta(t)=\sum\limits_{i=1}^{M(t)}Y_ie^{-\delta K_i}, t\geq 0\}$的分布函数为$F_t(x)$, 则

$ \begin{eqnarray*}F_t(x)&\;=&\;e^{-\lambda t}\sum\limits_{n=1}^\infty\{\lambda^n\int_0^tds_1\int_{s_1}^tds_2\cdots\int_{s_{n-1}}^tds_n\int _0^\infty\int_0^\infty\\ &&\cdots\int_0^\infty I_{(0, x]}(\sum\limits_{i=1}^ny_ie^{-\delta s_i})dG(y_1)dG(y_2)\cdots dG(y_n)\}.\end{eqnarray*} $

证  由引理1及$\{Y_n, n=1, 2, \cdots\}$与$\{K_n, n=1, 2, \cdots\}$和$\{M_t, t\geq 0\}$互相独立知

$ \begin{eqnarray*} &&P(L_\delta(t)\leq x\mid M(t)=n) =P(\sum\limits_{i=1}^nY_ie^{-\delta K_i}\leq x\mid M(t)=n)\\ &\;=&\;\int_{-\infty}^\infty ds_1\int_{-\infty}^\infty ds_2 \cdots\int_{-\infty}^\infty ds_nP(\sum\limits_{i=1}^nY_ie^{-\delta s_i}\leq x\mid M(t)=n, K_i=s_i, i=1, 2, \cdots, n) \frac{n!}{t^n}, \\ &&I_{0<s_1<s_2<\cdots<s_n\leq t}(s_1, s_2, \cdots, s_n)\\ &=&\;\frac{n!}{t^n}\int_0^tds_1\int_{s_1}^tds_2\cdots\int_{s_{n-1}}^tds_nP(\sum\limits_{i=1}^nY_i e^{-\delta s_i}\leq x)\\ &=&\;\frac{n!}{t^n}\int_0^tds_1\int_{s_1}^tds_2\cdots\int_{s_{n-1}}^tds_nEI_{(0, x]}(\sum\limits_{i=1} ^nY_ie^{-\delta s_i}), \end{eqnarray*} $

于是

$ \begin{eqnarray*} F_t(x)&=&\;P(L_\delta(t)\leq x) =\sum\limits_{n=1}^\infty P(L_\delta(t)\leq x\mid M(t)=n)P(M(t)=n)\\ &=&\;\sum\limits_{n=1}^\infty\frac{n!}{t^n}\int_0^tds_1\int_{s_1}^tds_2\cdots\int_{s_{n-1}} ^tds_n\int_0^\infty\int_0^\infty\\ &&\cdots\int_0^\infty I_{(0, x]}(\sum\limits_{i=1}^ny_ie ^{-\delta s_i})dG(y_1)dG(y_2) \cdots dG(y_n)\frac{(\lambda t)^ne^{-\lambda t}}{n!}\\ &=&\;e^{-\lambda t}\sum\limits_{n=1}^\infty\{\lambda^n\int_0^tds_1\int_{s_1}^tds_2\cdots\int_{s_{n-1}} ^tds_n\int_0^\infty\int_0^\infty\\ &&\cdots\int_0^\infty I_{(0, x]}(\sum\limits_{i=1}^ny_ie ^{-\delta s_i})dG(y_1)dG(y_2)\cdots dG(y_n)\}. \end{eqnarray*} $

记$\Im_n=\sigma\{U_\delta(T_i), i\leq n\}, n=1, 2, \cdots$, 则对任意Borel可测集$B\subset R$, 有

$ \begin{eqnarray*} &&P(U_\delta(T_n)\epsilon B\mid \Im_{n-1})\\ &=&\;P(e^{\delta W_n}U_\delta(T_{n-1})+ e^{\delta W_n}e^{\delta T_{n-1}} \sum\limits_{i=M(T_{n-1})+1}^{M(T_n)}Y_ie^{-\delta K_i}-X_n\epsilon B\mid \Im_{n-1})\\ &=&\;P(e^{\delta W_n}U_\delta(T_{n-1})+e^{\delta W_n}e^{\delta T_{n-1}} \sum\limits_{i=M(T_{n-1})+1}^{M(T_n)}Y_ie^{-\delta K_i}-X_n\epsilon B\mid U_\delta(T_{n-1}))\\ &=&\;P(U_\delta(T_n)\epsilon B\mid U_\delta(T_{n-1})), \end{eqnarray*} $

上述第二个等式成立是因为$e^{\delta W_n}, \sum\limits_{i=M(T_{n-1})}^{M(T_n)}Y_ie^{-\delta K_i}$和$X_n$关于$\Im_{n-1}$独立, 而$U_\delta(T_{n-1})$和$e^{\delta T_{n-1}}$关于$\sigma(U_\delta(T_{n-1}))$可测, 且$\sigma(U_\delta(T_{n-1}))\subset \Im_{n-1}$.

于是$\{U_\delta(T_n), n=1, 2, \cdots\}$为一马氏过程, 下面考虑其转移概率.由(2.4) 式, 有

$ \begin{eqnarray*} &&Q(n-1, x, n, A)=P(U_\delta(T_n)\epsilon A\mid U_\delta(T_{n-1})=x)\\ &=&\;P(e^{\delta W_n}U_\delta(T_{n-1})+e^{\delta W_n}e^{\delta T_{n-1}} \sum\limits _{i=M(T_{n-1})+1}^{M(T_n)}Y_ie^{-\delta K_i}-X_n\epsilon A\mid U_\delta(T_{n-1})=x)\\ &=&\;P(e^{\delta W_n}U_\delta(T_{n-1})+e^{\delta W_n}L_\delta(W_n)-X_n\epsilon A \mid U_\delta(T_{n-1})=x)\\ &=&\;P(xe^{\delta W_n}+e^{\delta W_n}L_\delta(W_n)-X_n\epsilon A \mid U_\delta(T_{n-1})=x)\\ &=&\;P(xe^{\delta W_n}+e^{\delta W_n}L_\delta(W_n)-X_n\epsilon A)\\ &=&\;P(xe^{\delta W_1}+e^{\delta W_1}L_\delta(W_1)-X_1\epsilon A). \end{eqnarray*} $

由于转移概率与时间无关, 于是可记$Q(n-1, x, n, A)=Q(x, A)$.综上, 有

定理 1  $\{{U_\delta(T_n), n=0, 1, 2, \cdots}\}$为一时齐马氏链, 其转移概率为

$ Q(x, A)=P(xe^{\delta W_1}+e^{\delta W_1}L_\delta (W_1)-X_1\in A). $

下面利用转移概率$Q(x, A)$刻画破产概率、破产时余额分布及破产前瞬间余额的分布, 给出其级数展开式.

定理 2 (1) 破产概率$\psi_\delta(u)$有如下展式:

$ \begin{eqnarray*}\psi_\delta(u)&\;=&\;\int_{-\infty}^0Q(u, dx)+\sum\limits_{n=2}^\infty\int_0^\infty Q(u, dx_1)\int_0^\infty Q(x_1, dx_2)\\ &&\cdots\int_0^\infty Q(x_{n-2}, dx_{n-1})\int_{-\infty}^0Q(x_{n-1}, dx_n);\end{eqnarray*} $

(2) 生存概率$\Phi_\delta(u)$满足$\Phi_\delta(u)=\int_0^\infty\Phi_\delta(x)Q(u, dx);$

(3) 生存概率$\Phi_\delta(u)$满足

$ \Phi_\delta(u)=\int_0^\infty d\gamma(t)\int_0^\infty dF_t(y)\int_0^{(u+y)e^{\delta t}}\Phi_\delta((u+y)e^{\delta t}-x)dF(x). $

证 (1)

$ \begin{eqnarray*}&& \psi_\delta(u)\\ &=&\;P\{\bigcup\limits_{n=1}^\infty (T=T_n)|U_\delta(0)=u\} =\sum\limits _{n=1}^\infty P\{T=T_n|U_\delta(0)=u\} =P\{U_\delta(T_1)<0|U_\delta(0)=u\}\\ &&+\sum\limits_{n=2}^\infty P\{U_\delta(T_1)>0, U_\delta(T_2)>0, \cdots, U_\delta(T_{n-1})>0, U_\delta(T_n)<0|U_\delta(0)=u\}\\ &=&\;\int_{-\infty}^0Q(u, dx)+\sum\limits_{n=2}^\infty\int_0^\infty Q(u, dx_1)\int_0^\infty Q(x_1, dx_2)\\ &&\cdots\int_0^\infty Q(x_{n-2}, dx_{n-1})\int_{-\infty}^0Q(x_{n-1}, dx_n). \end{eqnarray*} $

(2) 由定理1知

$ P(T=T_n|U_\delta(0)=u)=\int_0^\infty P(T=T_{n-1}|U_\delta(0)=x)Q(u, dx), n\geq2, $

两边对$n\geq2$求和并注意到$\psi_\delta(u)=1-\Phi_\delta(u)$可得$\Phi_\delta(u)=\int_0^\infty\Phi_\delta(x)Q(u, dx).$

(3) 注意到$\Phi_\delta(y)=0, y<0$, 由定理1知

$ \begin{aligned} \Phi_\delta(u)=&E\Phi_\delta(U_\delta(T_1))\\ =&E\Phi_\delta(ue^{\delta T_1}+e^{\delta T_1}L_\delta (T_1)-X_1)\\ =&\int_0^\infty d\gamma(t)\int_0^\infty dF_t(y)\int_0^{(u+y)e^{\delta t}}\Phi_\delta((u+y)e^{\delta t}-x)dF(x) \end{aligned} $

Gerber, Goovaerts和Kass^[8]引入破产时余额的分布函数亦即破产时赤字的分布函数$G_\delta(u, y)$来描述“破产”的严重程度, 即

$ G_\delta(u, y)=P(T<\infty, -y\leq U_\delta(T)<0|U_\delta(0)=u), y>0. $

定理 3  破产时余额的分布$G_\delta(u, y)$有如下展式

$ \begin{eqnarray*}G_\delta(u, y)&\;=&\;\int_{-y}^0Q(u, dx)+\sum\limits_{n=2}^\infty\int_0^\infty Q(u, dx_1)\int_0^\infty Q(x_1, dx_2)\\ &&\cdots\int_0^\infty Q(x_{n-2}, dx_{n-1})\int_{-y}^0Q(x_{n-1}, dx_n).\end{eqnarray*} $

证

$ \;\;\;\;G_\delta(u, y)=P(T<\infty, -y\leq U_\delta(T)<0|U_\delta(0)=u)\\ =\;P(\bigcup\limits_{n=1}^\infty(T=T_n), -y\leq U_\delta(T)<0|U_\delta(0)=u)\\ =\;\sum\limits_{n=1}^\infty P(T=T_n, -y\leq U_\delta(T)<0|U_\delta(0)=u)\\ =\;P\{{-y\leq U_\delta(T_1)<0|U_\delta(0)=u}\}\\ \;\;\;\;+\sum\limits_{n=2}^\infty P(U_\delta(T_1)>0, U_\delta(T_2)>0, \cdots, U_\delta(T_{n-1})>0, -y\leq U_\delta(T_n)<0|U_\delta(0)=u)\\ =\;\int_{-y}^0Q(u, dx)+\sum\limits_{n=2}^\infty\int_0^\infty Q(u, dx_1)\int_0^\infty Q(x_1, dx_2)\cdots\int_0^\infty Q(x_{n-2}, dx_{n-1})\\\\;\;\;\;int_{-y}^0Q(x_{n-1}, dx_n). $

在Gerber, Goovaerts和Kass^[8]工作的基础上, Dufresne和Gerber^[9]引入了如下破产前瞬间余额分布函数

$ F_\delta(u, y)=P(T<\infty, 0<U_\delta(T-)<y|U_\delta(0)=u), y>0. $

定理 4  破产前瞬间余额的分布$F_\delta(u, y)$有如下展式

$ \;\;\;\;F_\delta(u, y)\\ =\;I_{(u<y)}[\int_0^{\frac{\ln y-\ln u}{\delta}}d\gamma(t)\int_{ue^{\delta t}}^y F_t(xe^{-\delta t}-u)dF(x)+\int_0^{\frac{\ln y-\ln u}{\delta}}(1-F(y))\\\;\;\;\;F_t(ye^{-\delta t}-u)d\gamma(t)]\\ \;\;\;\;+\sum\limits_{n=2}^\infty\int_0^\infty Q(u, dx_1)\int_0^\infty Q(x_1, dx_2)\\ \;\;\;\;\cdots\int_0^yQ(x_{n-2}, dx_{n-1})\int_0^{\frac{\ln y-\ln x_{n-1}}{\delta}}d\gamma(t)\int_{x_{n-1}e^{\delta t}}^yF_t(xe^{-\delta t}-x_{n-1})dF(x)\\ \;\;\;\;+\sum\limits_{n=2}^\infty\int_0^\infty Q(u, dx_1)\int_0^\infty Q(x_1, dx_2)\\ \;\;\;\;\cdots\int_0^yQ(x_{n-2}, dx_{n-1})\int_0^{\frac{\ln y-\ln x_{n-1}}{\delta}}(1-F(y))F_t(ye^{-\delta t}-x_{n-1})d\gamma(t). $

证

$ \;\;\;\;F_\delta(u, y)=P(T<\infty, 0<U_\delta(T-)\leq y|U_\delta(0)=u)\\ =\;I_{(u<y)}P(T=T_1, 0<U_\delta(T_1-)\leq y|U_\delta(0)=u)\\ \;\;\;\;+\sum\limits_{n=2}^\infty P(T=T_n, 0<U_\delta(T_n-)\leq y|U_\delta(0)=u)\\ =\;I_{(u<y)}P(U_\delta(T_1)<0, 0<U_\delta(T_1-)\leq y|U_\delta(0)=u)\\ \;\;\;\;+\sum\limits_{n=2}^\infty P(U_\delta(T_1)>0, U_\delta(T_2)>0, \cdots, U_\delta(T_{n-1})>0, U_\delta(T_n)<0, \\\;\;\;\;0<U_\delta(T_n-)\leq y|U_\delta(0)=u), $

而

$ \begin{eqnarray*} &&I_{(u<y)}P(T=T_1, 0<U_\delta(T_1-)\leq y|U_\delta(0)=u)\\ &=&\;I_{(u<y)}P(ue^{\delta T_1}+e^{\delta T_1}L_\delta(T_1)-X_1<0, 0<ue^{\delta T_1}+e^{\delta T_1}L_\delta(T_1)\leq y)\\ &=&\;I_{(u<y)}\int_0^{\frac{\ln y-\ln u}{\delta}}P(ue^{\delta t}+e^{\delta t}L_\delta(t)-X_1<0, 0<ue^{\delta t}+e^{\delta t}L_\delta(t)\leq y)d\gamma(t)\\ &=&\;I_{(u<y)}\int_0^{\frac{\ln y-\ln u}{\delta}}P(L_\delta(t)<X_1e ^{-\delta t}-u, L_\delta(t)\leq ye^{-\delta t}-u)d\gamma(t)\\ &=&\;I_{(u<y)}\int_0^{\frac{\ln y-\ln u}{\delta}}P(ue^{\delta t}<X_1\leq y, L_\delta(t)<X_1e ^{-\delta t}-u)d\gamma(t)\\ &&+I_{(u<y)}\int_0^{\frac{\ln y-\ln u}{\delta}}P(X_1>y, L_\delta(t)< ye^{-\delta t}-u)d\gamma(t)\\ &=&\;I_{(u<y)}\int_0^{\frac{\ln y-\ln u}{\delta}}d\gamma(t)\int_{ue^{\delta t}}^yF_t(xe^{-\delta t}-u)dF(x)\\ &&+I_{(u<y)}\int_0^{\frac{\ln y-\ln u}{\delta}}(1-F(y))F_t(ye^{-\delta t}-u)d\gamma(t), \end{eqnarray*} $

又由定理1知

$ \begin{eqnarray*}&&P(U_\delta(T_1)>0, U_\delta(T_2)>0, \cdots, U_\delta(T_{n-1})>0, U_\delta(T_n)<0, 0<U_\delta(T_n-)\leq y|U_\delta(0)=u)\\ &=&\;\int_0^\infty Q(u, dx_1)\int_0^\infty Q(x_1, dx_2)\cdots\int_0^yQ(x_{n-2}, dx_{n-1})\\ &&P(U_\delta(T_n)<0, 0<U_\delta(T_n-)\leq y|U_\delta(T_{n-1})=x_{n-1})\\ &=&\;\int_0^\infty Q(u, dx_1)\int_0^\infty Q(x_1, dx_2)\\&&\cdots\int_0^yQ(x_{n-2}, dx_{n-1})\int_0^{\frac{\ln y-\ln x_{n-1}}{\delta}}d\gamma(t)\int_{x_{n-1}e^{\delta t}}^yF_t(xe^{-\delta t}-x_{n-1})dF(x)\\ &&+\int_0^\infty Q(u, dx_1)\int_0^\infty Q(x_1, dx_2)\\ &&\cdots\int_0^yQ(x_{n-2}, dx_{n-1})\int_0^{\frac{\ln y-\ln x_{n-1}}{\delta}}(1-F(y))F_t(ye^{-\delta t}-x_{n-1})d\gamma(t).\end{eqnarray*} $

于是定理得证.