文献[1]考虑双险种的理赔量具有二维一阶自回归结构的常利率二维离散时间风险模型, 通过随机过程中的鞅方法得到了最终破产概率的伦德贝格不等式进而导出了破产概率的指数上界, 并结合数值分析了参数对上界影响; 文献[2]在利率满足马尔科夫性的条件下, 通过数学递推归纳方法得到了破产前盈余和破产后赤字联合分布及最终破产概率满足的递推公式; 文献[3]在利率服从$m$阶自回归的离散时间风险模型中讨论了保费收入分别期初和期末缴纳时破产前最大盈余分布等破产指标; 文献[4]考虑理赔总量具有一阶自回归模型的离散时间风险模型, 利用随机过程中鞅方法得到了最终破产概率的伦德贝格上界; 文献[5]利用无穷小方法研究了具有相依结构的风险模型，得到了罚金折现期望函数所满足的积分-微分方程, 并给出了一些破产问题在积分-微分方程的应用; 文献[6]在文献[4]的基础上继续讨论理赔量满足一阶自回归结构修正的离散时间风险模型中破产前的盈余等一些的破产分布; 文献[7]在折现率因素下对离散时间风险模型中各破产指标进行讨论; 文献[8]考虑一类索赔达到计数过程相关的两险种风险模型给出了分类破产概率的渐进结果, 并在指数索赔条件下得到分类破产概率的具体表达式; 文献[9]在允许理赔支出可延迟的情况下对索赔过程为复合二项过程的离散型风险模型采用递推算法推导给出罚金折现期望函数, 进而给出破产概率、破产赤字的概率以及联合概率, 最后结合数值插图说明其实际意义; 文献[10]研究双险种离散时间风险模型, 当索赔过程服从复合二项过程时得到罚金期望函数及有时间内破产概率等结果.本文则在文献[4, 7, 10]的基础上研究两种广义的风险模型:理赔量满足一阶自回归的离散时间风险模型及具有相依理赔量带有双险种和折现率的离散时间风险模型, 通过数学归纳递推方法分别得到了破产持续时间的分布以及盈余首次穿过给定水平$x$的时刻分布所满足的积分方程, 并将两类模型进行对比, 得出相依理赔量带有双险种和折现率的离散时间风险模型更加贴合实际, 最后当理赔量为指数分布时给出了数值分析.

假定保险公司只经营险种$A$, 修正的离散风险模型^[4]

$ \begin{equation} \label{eq:1} \overline{U}_n=\overline{u}+nc-\overline{S}_n, n = 0, 1, 2, \cdots, \end{equation} $

(1.1)

其中$\overline{u} = \overline{U}_0 = u-\dfrac{a}{1-a}\omega$为修正初始盈余; $c$单位时间区间$(n-1, n]$内收取险种$A$的保费; $\overline{S}_n = \sum \limits_{i=1}^{n}\overline{X_i}$前$n$个单位时期的修正理赔总量; 这里理赔量满足一阶自回归结构AR$(1)$, $\overline{X}_i = \dfrac{W_i}{1-a}$第$i$个理赔时期的修正理赔量$(i=1, 2, \cdots)$, 这里$W_0=\omega$, 且$\overline{X_1}, \overline{X_2}, \cdots$, 相互独立同分布, 记$H(w)=P\{\dfrac{W_1}{1-a} \leq w\}$.为了便于讨论将此离散时间风险模型记为模型(Ⅰ).

假定保险公司同时经营两种相关险种$A$和$B$, 在所考虑的单位时期内有固定的折现率$\delta$, 而$u \ge 0$表示初始盈余, 保费在每个时期期初收取而理赔在每个时期的期末支付, 则在时刻$n~(n = 0, 1, 2, \cdots)$保险公司累积盈余折现到初始时刻的盈余为

$ \begin{equation} \label{eq:2} U_n^\delta = u+\sum\limits_{i=1}^{n}e^{-\delta(i-1)}(c_1+c_2)-\sum\limits_{i=1}^{n}e^{-\delta i}(X_i+Y_i), \end{equation} $

(1.2)

其中$c_1$和$c_2$分别表示$A$和$B$两个险种的单位时间内保费收入量, $\{X_i, i = 1, 2, \cdots\}$, $\{Y_i, i = 1, 2, \cdots\}$是两个独立同分布的随机变量(r.v.)序列, 分别表示在$(i-1, i]$时间内险种$A$和$B$的理赔支出, 并假定$X_i$和$Y_i$的期望值$E(X_i) < \infty$, $E(Y_i) < \infty$.

此外考虑$A$和$B$两个险种的理赔量满足$AR(1)$模型

$ \begin{eqnarray} \label{eq:3} &&X_i = W_i+aX_{i-1}, \end{eqnarray} $

(1.3)

$ \begin{eqnarray} \label{eq:4} &&Y_i = V_i+bY_{i-1}, \end{eqnarray} $

(1.4)

其中$|a| < 1$和$|b| < 1$, 且$\{W_i, i = 1, 2, \cdots\}$, $\{V_i, i = 1, 2, \cdots\}$分别是独立同分布的随机变量序列, 这里记$W_0=\omega$, $V_0=\nu$.为了后面讨论的方便, 把满足(1.2), (1.3) 和(1.4) 式的离散时间破产模型记为模型(Ⅱ).

通过(1.3) 和(1.4) 式对模型(1.2) 进行修正, 得到另一修正的离散时间风险模型

$ \begin{equation} \label{eq:5} \begin{aligned} \hat{U}_n^\delta &= \hat{u}+\sum\limits_{i=1}^n e^{-(i-1)\delta}(c_1+c_2)-\sum\limits_{i=1}^n e^{-i\delta}(\hat{X}_i+\hat{Y}_i)\\ &= \hat{u}+\sum\limits_{i=1}^n e^{-(i-1)\delta}(c_1+c_2) -\hat{S}_n, \end{aligned} \end{equation} $

(1.5)

其中$\hat{u} = u-\dfrac{ae^{-\delta}}{1-ae^{-\delta}}\omega-\dfrac{be^{-\delta}} {1-be^{-\delta}}\nu$表示修正初始盈余, 随机变量$\hat{X}_i=\dfrac{W_i}{1-ae^{-\delta}}$和$\hat{Y}_i=\dfrac{V_i}{1-be^{-\delta}}$分别表示$A$和$B$险种在$(i-1, i]$时间区间内的修正理赔支出, $\{\hat{X}_i\}$和$\{\hat{Y}_i\}$分别是独立同分布的随机变量序列$(i=1, 2, \cdots)$.记$k_n = \sum\limits_{i=1}^n e^{-\delta(i-1)} = \dfrac{1-e^{-n\delta}}{1-e^{-\delta}}, n = 0, 1, 2, \cdots$, 表示保费收入在前$n$个时段内的折现权重.特别地, $k_0 = 0$，$k_1 = 1$.这里$\hat{S}_n = \sum\limits_{i=1}^n e^{-\delta i}(\hat{X}_i+\hat{Y}_i)$是前$n$个时段内修正总理赔.

记$Z_i = \hat{X}_i+\hat{Y}_i-e^\delta(c_1+c_2)$称为时刻$i$的净亏损额, 则$\{Z_i, i=1, 2, \cdots\}$是独立同分布的随机变量, 这里要求$E(Z_i) < 0$是为了保证保险公司的经常运作所必须附加的风险负荷.记$F_{\hat{X}}(x)$和$F_{\hat{Y}}(y)$分别为随机变量$\hat{X}$和$\hat{Y}$的分布函数, 可得$Z$的分布函数$G(z)=\displaystyle\int_0^\infty F_{\hat{Y}}(z+e^{\delta}(c_1+c_2)-x)d F_{\hat{X}}(x)$.若令$\hat{L}_n = \sum\limits_{i=1}^n e^{-i\delta}Z_i$, 此时(1.5) 式的模型即化为

$ \begin{equation} \label{eq:6} \hat{U}_n^{\delta} = \hat{u}-\hat{L}_n. \end{equation} $

(1.6)

注  在离散时间风险模型(Ⅱ)中令$\delta=0$且只考虑一个险种即为离散时间风险模型(Ⅰ).

破产时刻即保险公司的首次盈余小于零的时刻在模型(Ⅰ)下定义为

$ \begin{equation} \label{eq:7} T(\overline{u})=\inf\limits_{n}\{n:n>0, \overline{U}_n < 0\} = \inf\limits_{n}\{n:n>0, \overline{S}_n>\overline{u}+nc\}. \end{equation} $

(2.1)

在模型(Ⅱ)下定义为

$ T(\hat u) = \mathop {\inf }\limits_n \{ n:n > 0,\hat U_n^\delta < 0\} = \mathop {\inf }\limits_n \{ n:n > 0,{\hat S_n} > \hat u + {k_n}({c_1} + {c_2})\} . $

(2.2)

下面讨论在离散时间模型(Ⅰ)和(Ⅱ)下破产持续时间的分布和首达某一水平$x$的时刻分布.

为了表明停时对初始准备金的依赖关系, 将破产后保险公司的盈余首次回为正的时刻定义为^[3]

$ \begin{equation} \label{eq:9} \Gamma(\overline{u}) = \inf\limits_{n}\{n:n>T(\overline{u}), \overline{U}_n>0\}. \end{equation} $

(2.3)

于是, 破产持续时间定义为^[3]

$ \begin{equation} \label{eq:10} \tilde{T}(\overline{u}) = \left\{ \begin{aligned} &\Gamma(\overline{u})-T(\overline{u}), &T(\overline{u}) < \infty, \\ &0, &T(\overline{u})=\infty. \end{aligned} \right. \end{equation} $

(2.4)

定理1  这是定理内容.在离散时间风险模型(Ⅰ)下破产持续时间为$n$期的概率为

$ \begin{equation} \label{eq:11} \Psi_n(\overline{u}) = P\{\tilde{T}(\overline{u})=n\} = \sum\limits_{k=1}^\infty M_k^{(n)}(\overline{u}), \end{equation} $

(2.5)

其中$M_1^{(n)}(\overline{u})=\displaystyle\int_{\overline{u}+c}^\infty M_1^{(n-1)}(\overline{u}+c-w)dH(w)$, $M_1^{(1)}(\overline{u})=\displaystyle\int_{\overline{u}+c}^\infty H(\overline{u}+2c-w)dH(w)$, $M_k^{(n)}(\overline{u})=\displaystyle\int_0^{\overline{u}+c} M_{k-1}^{(n)}(\overline{u}+c-w)dH(w)(k\ge2)$.

证  当$\tilde{T}(\overline{u}) = 1$时,

$ \begin{eqnarray} \label{eq:12} \Psi_1(\overline{u}) &=&P\{\tilde{T}(\overline{u}) = 1\} = P\{\Gamma(\overline{u}) = T(\overline{u})+1\}\nonumber\\ &=&\sum\limits_{k=1}^\infty P\{\overline{U}_0\ge0, \cdots, \overline{U}_{k-1}\ge0, \overline{U}_k < 0, \overline{U}_{k+1}\ge0\}\nonumber\\ &=&\sum\limits_{k=1}^\infty M_k^{(1)}(\overline{u}), \end{eqnarray} $

(2.6)

$ \begin{eqnarray} \label{eq:13} M_1^{(1)}(\overline{u}) &=&P\{\overline{U}_0\ge0, \overline{U}_1 < 0, \overline{U}_2\ge0\}\nonumber\\ &=&\displaystyle\int_{\overline{u}+c}^{+\infty}H(\overline{u}+2c-w)dH(w)\nonumber\\ &=&B_1(\overline{u}), \nonumber\\ M_2^{(1)}(\overline{u}) &=&\displaystyle\int^{\overline{u}+c}_{-\infty}P\{\frac{W_2}{1-a}>\overline{u}+2c-w, \frac{W_2}{1-a}+\frac{W_3}{1-a}\le\overline{u}+3c\}dH(w)\nonumber\\ &=&\displaystyle\int^{\overline{u}+c}_{-\infty}M_1^{(1)}(\overline{u}+c-w)dH(w). \end{eqnarray} $

(2.7)

由数学归纳得到, 破产持续1期且在$k~ (k\ge2)$时刻破产概率为

$ \begin{equation} \label{eq:14} M_k^{(1)}(\overline{u}) = \displaystyle\int^{\overline{u}+c}_{-\infty}M_{k-1}^{(1)}(\overline{u}+c-w)dH(w). \end{equation} $

(2.8)

所以破产持续1期的概率为

$ \begin{equation} \label{eq:15} \Psi_1(\overline{u}) = P\{\tilde{T}(\overline{u})=1\} = \sum\limits_{k=1}^{+\infty} M_k^{(1)}(\overline{u}). \end{equation} $

(2.9)

当$\tilde{T}(\overline{u}) = 2$时, 同理可得

$ \begin{equation} \label{eq:16} \Psi_2(\overline{u})= \sum\limits_{k=1}^{+\infty} M_k^{(2)}(\overline{u}), \end{equation} $

(2.10)

这里记$B_2(\overline{u}) = \displaystyle\int_{\overline{u}+c}^{+\infty} B_1(\overline{u}+c-w)dH(w)$, 则在(2.10) 式中

$ \begin{equation} \label{eq:17} \begin{aligned} M_1^{(2)}(\overline{u}) &=\displaystyle\int_{\overline{u}+c}^{+\infty}B_1(\overline{u}+c-w)dH(w)=B_2(\overline{u}), \\ M_2^{(2)}(\overline{u}) &=P\{\overline{U}_0\ge0, \overline{U}_1\ge0, \overline{U}_2 < 0, \overline{U}_3 < 0, \overline{U}_4\ge0\}\\ &=\displaystyle\int_{-\infty}^{\overline{u}+c}M_1^{(2)}(\overline{u}+c-w)dH(w). \end{aligned} \end{equation} $

(2.11)

由数学归纳法得到, 对于$k\ge2$有

$ \begin{equation} \label{eq:18} M_k^{(2)}(\overline{u}) = \displaystyle\int^{\overline{u}+c}_{-\infty}M_{k-1}^{(2)}(\overline{u}+c-w)dH(w). \end{equation} $

(2.12)

类似地, 记$B_n(\overline{u}) = \displaystyle\int_{\overline{u}+c}^{+\infty} B_{n-1}(\overline{u}+c-w)dH(w)$, 当$\tilde{T}(\overline{u}) = n$时, 同理有

$ \begin{equation} \nonumber \begin{aligned} M_1^{(n)}(\overline{u}) &=B_n(\overline{u}), \\ M_k^{(n)}(\overline{u}) &=\displaystyle\int_{-\infty}^{\overline{u}+c}M_{k-1}^{(n)}(\overline{u}+c-w)dH(w)~(k\ge2). \end{aligned} \end{equation} $

故破产持续$n$期的概率为

$ \begin{equation} \label{eq:19} \Psi_n(\overline{u}) = P\{\tilde{T}(\overline{u})=n\} = \sum\limits_{k=1}^\infty M_k^{(n)}(\overline{u}). \end{equation} $

(2.13)

定理2  对离散时间风险模型(Ⅱ)其破产持续时间为$n$期的概率为

$ \begin{equation} \label{eq:20} \Phi_n(\hat{u}) = P\{\tilde{T}(\hat{u}) = n\} = \sum\limits_{k=1}^\infty \eta_k^{(n)}(\hat{u}) \end{equation} $

(2.14)

其中$\eta_1^{(n)}(\hat{u}) = \displaystyle\int_{\hat{u}e^{\delta}}^\infty \eta_1^{(n-1)}(\hat{u}e^{\delta}-z)dG(z)$, $\eta_1^{(1)}(\hat{u}) = \displaystyle\int_{\hat{u}e^{\delta}}^\infty G(e^{\delta}(\hat{u}e^{\delta}-z))dG(z)$; 当$k\ge2$时, $\eta_k^{(n)}(\hat{u}) = \displaystyle\int^{\hat{u}e^{-\delta}}_\infty \eta_{k1}^{(n)}(\hat{u}e^{\delta}-z)dG(z)$.

证  当$\tilde{T}(\hat{u}) = 1$时, 即当破产持续时间为1期的破产概率为

$ \begin{equation} \nonumber \begin{aligned} \Phi_1(\hat{u}) &=P\{\Gamma(\hat{u})=T(\hat{u})+1\} =\sum\limits_{k=1}^\infty P\{\Gamma(\hat{u})=k+1|T(\hat{u})=k\}P\{T(\hat{u})=k\}\\ &=\sum\limits_{k=1}^{\infty}P\{\hat{U}_0^\delta\ge0, \hat{U}_1^\delta\ge0, \cdots, \hat{U}_{k-1}^\delta\ge0, \hat{U}_{k}^\delta < 0, \hat{U}_{k+1}^\delta\ge0\}\\ &=\sum\limits_{k=1}^\infty \eta_k^{(1)}(\hat{u}), \\ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \eta_1^{(1)}(\hat{u}) &=P\{\hat{U}_0^\delta\ge0, \hat{U}_1^\delta < 0, \hat{U}_2^\delta\ge0\} =P\{\hat{L}_1>\hat{u}, \hat{L}_2\le\hat{u}\}\\ &=P\{Z_1>\hat{u}e^{\delta}, Z_1+\frac{Z_2}{e^{\delta}}\le \hat{u}e^{\delta}\}=\displaystyle\int_{\hat{u}e^{\delta}}^\infty G(e^{\delta}(\hat{u}e^{\delta}-z))dG(z), \\ %%%%%%%%%%%%%%%%%%%%%%%%%%%% \eta_2^{(1)}(\hat{u}) &=P\{\hat{U}_1^\delta\ge0, \hat{U}_2^\delta < 0, \hat{U}_3^\delta\ge0\}= P\{\hat{L}_1\le\hat{u}, \hat{L}_2>\hat{u}, \hat{L}_3\le \hat{u}\}\\ &=\displaystyle\int^{\hat{u}e^{\delta}}_{-\infty} P\{Z_2>(\hat{u}e^{\delta}-z)e^{\delta}, Z_2+\frac{Z_3}{e^{\delta}}\le (\hat{u}e^{\delta}-z)e^{\delta}\}dG(z)\\ &=\displaystyle\int^{\hat{u}e^{\delta}}_{-\infty} \eta_1^{(1)}(\hat{u}e^{\delta}-z)dG(z). \end{aligned} \end{equation} $

由数学归纳法, 对$k\ge2$有

$ \begin{equation} \label{eq:21} \eta_k^{(1)}(\hat{u})=\displaystyle\int^{\hat{u}e^{\delta}}_{-\infty} \eta_{k-1}^{(1)}(\hat{u}e^{\delta}-z)dG(z), \end{equation} $

(2.15)

$\eta_k^{(1)}(\hat{u})$表示当$k$时刻破产持续时间为1期的概率, 故得到破产持续时间为1期的概率为$\Phi_1(\hat{u}) = P\{\tilde{T}(\hat{u}) = 1\}=\sum\limits_{k=1}^\infty \eta_k^{(1)}(\hat{u})$.

当$\tilde{T}(\hat{u}) = 2$时, 类似地有

$ \begin{align} & {{\Phi }_{2}}(\hat{u})=P\{\tilde{T}(\hat{u})=2\}=\sum\limits_{k=1}^{\infty }{P}\{\Gamma (\hat{u})=k+2|T(\hat{u})=k\}P\{T(\hat{u})=k\} \\ & =\sum\limits_{k=1}^{\infty }{P}\{\hat{U}_{0}^{\delta }\ge 0,\hat{U}_{1}^{\delta }\ge 0,\cdots ,\hat{U}_{k-1}^{\delta }\ge 0,\hat{U}_{k}^{\delta } < 0,\hat{U}_{k+1}^{\delta } < 0,\hat{U}_{k+2}^{\delta }\ge 0\} \\ & =\sum\limits_{k=1}^{\infty }{\eta _{k}^{(2)}}(\hat{u}), \\ \end{align} $

(2.16)

$ \begin{array}{l} \eta _1^{(2)}(\hat u) = P\{ \hat U_0^\delta \ge 0,\hat U_1^\delta < 0,\hat U_2^\delta < 0,\hat U_3^\delta \ge 0\} = P\{ {{\hat L}_1} > \hat u,{{\hat L}_2} > \hat u,{{\hat L}_3} \le \hat u\} \\ = \int_{\hat u{e^\delta }}^\infty {\eta _1^{(1)}(\hat u{e^\delta } - z)dG(z),} \\ \eta _2^{(2)}(\hat u)n = P\{ \hat U_1^\delta \ge 0,\hat U_2^\delta < 0,\hat U_3^\delta < 0,\hat U_4^\delta \ge 0\} = \int_{ - \infty }^{\hat u{e^\delta }} {\eta _1^{(2)}(\hat u{e^\delta } - z)dG(z).} \end{array} $

对$k\ge 2$, 由数学归纳法可得

$ \begin{equation} \nonumber \eta_k^{(2)}(\hat{u})=\displaystyle\int^{\hat{u}e^{\delta}}_{-\infty} \eta_{k-1}^{(2)}(\hat{u}e^{\delta}-z)dG(z), \end{equation} $

所以破产持续时间为2期的破产概率为

$ \begin{equation} \nonumber \Phi_2(\hat{u}) = P\{\tilde{T}(\hat{u}) = 2\}=\sum\limits_{k=1}^\infty \eta_k^{(2)}(\hat{u}). \end{equation} $

同理, 由数学归纳法知当$\tilde{T}(\hat{u}) = n$时, 有

$ \begin{equation} \nonumber \begin{aligned} &\eta_1^{(n)}(\hat{u})=\displaystyle\int_{\hat{u}e^{\delta}}^{\infty} \eta_1^{(n-1)}(\hat{u}e^{\delta}-z)dG(z), \\ &\eta_k^{(n)}(\hat{u})=\displaystyle\int^{\hat{u}e^{\delta}}_{-\infty} \eta_{k-1}^{(n)}(\hat{u}e^{\delta}-z)dG(z)~~(k\ge2) \end{aligned} \end{equation} $

且破产持续时间为$n$期的概率为$\Phi_n(\hat{u}) = P\{\tilde{T}(\hat{u}) = n\} = \sum\limits_{k=1}^\infty \eta_k^{(n)}(\hat{u}).$

风险理论研究的一个主要问题是破产发生的概率, 但是“破产”实际发生的可能性还是很小的, 因而我们多数关注在什么时候保险公司的盈余将达到一个给定的水平$x$^[10].

对任意的$x>0$, 定义盈余过程$\{U_n\}_{n=0}^{+\infty}$首次穿过水平$x$的时刻为

$ \begin{equation} \label{eq:23} T_x=\left\{ \begin{aligned} &k, &\inf\{n:U_n(u)\ge x|U_0 = u\}=k, \\ &\infty, &\inf\{n:U_n(u)\ge x|U_0 = u\}=\infty, \end{aligned} \right. \end{equation} $

(2.17)

则称$T_x$为初始准备金为$u$时盈余$U_n(u)$首次穿过水平$x$的时刻.当初始盈余为$\overline{u}$, 则记为$\overline{T}_x$; 初始盈余为$\hat{u}$, 则记为$\hat{T}_x$.

定理3   这是定理内容.对修正的离散时间风险模型(Ⅰ)其首次穿过水平$x$的时刻满足下式

$ \begin{equation} \label{eq:24} \rho(\overline{u}, x) = \sum\limits_{n=1}^\infty P\{\overline{T}_x = n\} = H(\overline{u}+c-x)+\displaystyle\int_{\overline{u}+c-x}^\infty \rho(\overline{u}+c-w, x)dH(w). \end{equation} $

(2.18)

证  由(2.17) 式容易看出, 当$x\le \overline{u}$时, $\overline{T}_x = 0$.不失一般性, 以下假定$x>\overline{u}$.记$t_n(\overline{u}, x) = P\{\overline{T}_x = n\}$表示盈余过程$\{\overline{U}_n\}_{n=0}^{+\infty}$最终能穿过水平$x$的概率.用$\rho(\overline{u}, x)$描述首次穿过给定水平$x$的时刻的分布, 则$\rho(\overline{u}, x)$可以做如下分解

$ \begin{equation} \label{eq:25} \begin{aligned} \rho(\overline{u}, x) &=\sum\limits_{n=1}^\infty P\{\overline{U}_n \ge x, \overline{U}_{n-1} < x, \cdots, \overline{U}_1 < x\} \\ &=\sum\limits_{n=1}^\infty \{\overline{S}_n\le \overline{u}+nc-x, \overline{S}_{n-1}>\overline{u}+(n-1)c-x, \cdots, \overline{S}_{1}>\overline{u}+c-x\}\\ &=\sum\limits_{n=1}^\infty t_n(\overline{u}, x), \end{aligned} \end{equation} $

(2.19)

这里

$ \begin{equation} \label{eq:26} \begin{aligned} t_1(\overline{u}, x) &=P\{\overline{U}_1\ge x\}=P\{\overline{S}_1 \le \overline{u}+c-x\}=P\{\frac{W_1}{1-a} \le \overline{u}+c-x\}\\ &=H(\overline{u}+c-x), \\ t_2(\overline{u}, x) &=P\{\overline{U}_2 \ge x, \overline{U}_1 < x\}\\ &=P\{\frac{W_1}{1-a}+\frac{W_2}{1-a} \le \overline{u}+2c-x, \frac{W_1}{1-a}>\overline{u}+c-x\}\\ &=\displaystyle\int_{\overline{u}+c-x}^{\infty} P\{\frac{W_2}{1-a} \le \overline{u}+2c-x-w\}dH(w)\\ &=\displaystyle\int_{\overline{u}+c-x}^{\infty} t_1(\overline{u}+c-w, x)dH(w). \end{aligned} \end{equation} $

(2.20)

对于$n\ge 2$, 有

$ \begin{equation} \label{eq:27} \begin{aligned} t_n(\overline{u}, x) &=H(\overline{u}+c-x)+\sum\limits_{n=2}^{\infty} \displaystyle\int_{\overline{u}+c-x}^{\infty} t_{n-1}(\overline{u}+c-w, x)dH(w), \end{aligned} \end{equation} $

(2.21)

由(2.19), (2.20) 和(2.21) 式可得保险公司盈余首次穿过给定水平$x$的时刻分布满足积分方程

$ \begin{equation} \nonumber \rho(\overline{u}, x)=H(\overline{u}+c-x)+\displaystyle\int_{\overline{u}+c-x}^\infty \rho(\overline{u}+c-w, x)dH(w). \end{equation} $

定理4  对修正的离散时间风险模型(Ⅱ)其首次穿过水平$x$的时刻满足下面的式子

$ \begin{equation} \label{eq:28} P(\hat{T}_x < \infty) = \rho(\hat{u}, x) = G((\hat{u}-x)e^\delta)+\displaystyle\int_{(\hat{u}-x)e^\delta}^\infty \rho(\hat{u}e^\delta-z, xe^\delta)dG(z). \end{equation} $

(2.22)

证  由定义知当$x\le \hat{u}$时$\hat{T}_x = 0$, $P(\hat{T}=n) = t_n(\hat{u}, x)$; 当$ x > \hat{u}$时, $P(\hat{T}_x < \infty) = \rho(\hat{u}, x)$.按照上述$\hat{T}_x$的定义并不难推算得出

$ \begin{eqnarray} \label{eq:29} &&t_1(\hat{u}, x) = P\{\hat{U}_1^\delta \ge x\} = P\{Z_1 \le (\hat{u}-x)e^\delta\} = G((\hat{u}-x)e^\delta), \end{eqnarray} $

(2.23)

$ \begin{eqnarray} \label{eq:30} &&\begin{aligned} t_2(\hat{u}, x) = P\{\hat{U}_1^\delta < x, \hat{U}_2^\delta \ge x\} = \displaystyle\int_{(\hat{u}-x)e^\delta}^\infty t_1(\hat{u}e^\delta-z, xe^\delta)dG(z), \end{aligned} \end{eqnarray} $

(2.24)

由数学归纳法对$n \ge 2$有

$ \begin{equation} \label{eq:31} \begin{aligned} t_n(\hat{u}, x) = P\{\hat{U}_1^\delta < x, \cdots, \hat{U}_{n-1}^\delta < x, \hat{U}_n^\delta \ge x\} = \displaystyle\int_{(\hat{u}-x)e^\delta}^\infty t_{n-1}(\hat{u}e^\delta-z, xe^\delta)dG(z). \end{aligned} \end{equation} $

(2.25)

综上可得保险公司折现盈余$\hat{U}_n^{\delta}$首次穿过水平$x$的时刻分布

$ \begin{equation} \label{eq:32} \begin{aligned} \rho(\hat{u}, x) = \sum\limits_{n=1}^{\infty} P\{\hat{T}_x = n\} = G((\hat{u}-x)e^\delta)+\displaystyle\int_{(\hat{u}-x)e^\delta}^\infty \rho(\hat{u}e^\delta-z, xe^\delta)dG(z). \end{aligned} \end{equation} $

(2.26)

假设两个险种$A$和$B$的理赔额$\hat{X}$, $\hat{Y}$均服从指数分布, 那么$F_{\hat{X}}(x) = 1-e^{-\lambda_1x}$, $\lambda_1>0$, $x \ge 0$, $F_{\hat{Y}}(y) = 1-e^{-\lambda_2y}$, $\lambda_2>0$, $y \ge 0$, $\lambda_1 \ge \lambda_2$, 且记$c=c_1+c_2$, 那么

$ G(z)=\displaystyle\int_o^\infty F_{\hat{y}}(z+e^\delta(c_1+c_2)-x)dF_{\hat{X}}(x) = 1-\dfrac{\lambda_1e^{-\lambda_2(z+e^\delta c)}} {(\lambda_1-\lambda_2)}. $

由定理2可知

$ \eta_1^{(1)}(\hat{u}) = \displaystyle\int_{\hat{u}e^\delta}^{\infty}G(e^\delta(\hat{u}e^\delta-z))dG(z) = \dfrac{\lambda_1}{\lambda_1-\lambda_2}e^{-\lambda_2e^\delta(\hat{u}+c)}(\lambda_2- \dfrac{\lambda_1e^{-\lambda_2e^\delta c}}{(\lambda_1-\lambda_2)(e^\delta-1)}), $

进而

$ \begin{eqnarray*}&&\eta_1^{(2)}(\hat{u}) = \displaystyle\int_{\hat{u}e^\delta}^{\infty}\eta_1^{(1)}(\hat{u}e^\delta-z)dG(z) = \dfrac{\lambda_1^2}{(\lambda_1-\lambda_2)^2}(\lambda_2- \dfrac{\lambda_1e^{-\lambda_2e^\delta c}} {(\lambda_1-\lambda_2)(e^\delta-1)})\dfrac{e^{-\lambda_2e^\delta(\hat{u}+2c)}}{(e^\delta-1)}, \\ &&\eta_1^{(3)}(\hat{u}) = \displaystyle\int_{\hat{u}e^\delta}^{\infty}\eta_1^{(2)}(\hat{u}e^\delta-z)dG(z) = \dfrac{\lambda_1^3} {(\lambda_1-\lambda_2)^3}(\lambda_2- \dfrac{\lambda_1e^{-\lambda_2e^\delta c}} {(\lambda_1-\lambda_2)(e^\delta-1)})\dfrac{e^{-\lambda_2e^\delta(\hat{u}+3c)}}{(e^\delta-1)^2}.\end{eqnarray*} $

通过数学归纳可得

$ \eta_1^{(n)}(\hat{u}) = \dfrac{\lambda_1^n}{(\lambda_1-\lambda_2)^n} (\lambda_2- \dfrac{\lambda_1e^{-\lambda_2e^\delta c}}{(\lambda_1-\lambda_2)(e^\delta-1)}) \dfrac{e^{-\lambda_2e^\delta(\hat{u}+nc)}}{(e^\delta-1)^{n-1}}. $

考虑到净亏损到达无穷大是不贴合实际的, 所以可以选定积分下限为$m$, 进而可得

$ \begin{eqnarray*}\eta_2^{(n)}(\hat{u}) &=&\displaystyle\int^{\hat{u}e^{\delta}}_m \eta_1^{(n)}(\hat{u}e^{\delta}-z)dG(z)\\ &=&\dfrac{\lambda_1^{n+1}}{(\lambda_1-\lambda_2)^{n+1}}(\lambda_2- \dfrac{\lambda_1e^{-\lambda_2e^\delta c}}{(\lambda_1-\lambda_2)(e^\delta-1)})\dfrac{e^{-\lambda_2e^\delta(\hat{u}+(n+1)c)}}{(e^\delta-1)^n} (e^{\lambda_2(e^\delta-1)m}-e^{\lambda_2(e^\delta-1)\hat{u}e^\delta}), \\ \eta_3^{(n)}(\hat{u}) &=&\displaystyle\int^{\hat{u}e^{\delta}}_m \eta_2^{(n)}(\hat{u}e^{\delta}-z)dG(z)\\ &=&\dfrac{\lambda_1^{n+2}}{(\lambda_1-\lambda_2)^{n+2}}(\lambda_2- \dfrac{\lambda_1e^{-\lambda_2e^\delta c}}{(\lambda_1-\lambda_2)(e^\delta-1)}) \dfrac{e^{-\lambda_2e^\delta(\hat{u}+(n+2)c)}}{(e^\delta-1)^{n+1}} (e^{\lambda_2(e^\delta-1)m}-e^{\lambda_2(e^\delta-1)\hat{u}e^\delta})^2.\end{eqnarray*} $

通过数学归纳法可得当$k \ge 2$时, 由定理2知对离散时间风险模型(Ⅱ)其破产持续时间为$n$的概率

$ \begin{equation} \label{eq:33} \begin{aligned} \eta_k^{(n)}(\hat{u}) =&\frac{\lambda_1^{n+k-1}}{(\lambda_1-\lambda_2)^{n+k-1}}(\lambda_2- \frac{\lambda_1e^{-\lambda_2e^\delta c}}{(\lambda_1-\lambda_2)(e^\delta-1)})\\ &\frac{e^{-\lambda_2e^\delta(\hat{u}+(n+k-1)c)}}{(e^\delta-1)^{n+k-2}} (e^{\lambda_2(e^\delta-1)m}-e^{\lambda_2(e^\delta-1)\hat{u}e^\delta})^{k-1}. \end{aligned} \end{equation} $

(3.1)

又由定理4对离散时间风险模型(Ⅱ)其首次穿过水平$x$的时刻满足下面的式子

$ \begin{equation} \nonumber \begin{aligned} &t_1(\hat{u}, x)=1-\frac{\lambda_1e^{-\lambda_2e^\delta((\hat{u}-x)+c)}}{(\lambda_1-\lambda_2)}, \\ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% &t_2(\hat{u}, x)=e^{-\lambda_2e^\delta((\hat{u}-x)+c)}\frac{\lambda_1}{(\lambda_1-\lambda_2)} [\frac{\lambda_1}{(\lambda_1-\lambda_2)(e^\delta-1)}+1], \\ %%%%%%%%%%%%%%%%%%%%%%%%%%%% &t_3(\hat{u}, x)=e^{-\lambda_2e^\delta((\hat{u}-x)+2c)}\frac{(\lambda_1)^2}{(\lambda_1-\lambda_2)^2(e^\delta-1)} [\frac{\lambda_1}{(\lambda_1-\lambda_2)(e^\delta-1)}+1]. \\ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \end{aligned} \end{equation} $

由数学归纳法对$n\ge 2$有

$ \begin{equation} \nonumber t_n(\hat{u}, x)=e^{-\lambda_2e^\delta((\hat{u}-x)+(n-1)c)}\frac{(\lambda_1)^{n-1}} {(\lambda_1-\lambda_2)^{n-1}(e^\delta-1)^{n-2}} [\frac{\lambda_1}{(\lambda_1-\lambda_2)(e^\delta-1)}+1]. \end{equation} $

进而可得首次穿过水平$x$的时刻分布

$ \begin{equation} \label{eq:34} \begin{aligned} \rho(\hat{u}, x) &= \sum\limits_{n=1}^\infty P\{\hat{T}_x = n\} = \sum\limits_{n=1}^\infty t_n(\hat{u}, x) \\ &= 1+\frac{e^{-\lambda_2e^\delta((\hat{u}-x)+c)}[e^{-\lambda_2e^\delta c}+1]\lambda_1^2} {(\lambda_1-\lambda_2)^2(e^{\delta}-1)-\lambda_1(\lambda_1-\lambda_2)e^{-\lambda_2e^\delta c}}, \end{aligned} \end{equation} $

(3.2)

在(2.27) 和(2.28) 式中取一种险种且$\delta = 0$可得离散时间风险模型(Ⅰ)其破产持续时间为$n$的概率和首次穿过水平$x$的时刻分布.

本文研究了具有一阶自回归结构理赔量的修正的离散时间风险模型和引入折现率和双险种因素且理赔相依的修正的离散时间风险模型, 利用递推方法得到了破产持续时间$n$的概率以及盈余首次穿过水平$x$所满足的积分表达式, 最后结合理赔服从指数分布给出了具体的结果, 因而可以根据保险公司给出的数据做进一步的数据分析.后期可以讨论理赔量是$n~(n \ge 2)$阶自回归结构, 此外现实生活中除了双险种, 折现率和理赔量的三个因素的影响还可以考虑受到投资和通货膨胀等因素影响, 双险种还可以推广到多险种, 比如财产保险、健康保险、人身保险、意外险等.