近十几年, Parisian破产问题受到了越来越多的风险领域专家学者的关注. Parisian破产是在经典破产问题中引入了金融领域的Parisian延迟概念.即当股票价格连续地位于一个特定水平以上或以下的时间超过特定时间才做出决策.如在风险模型中,盈余过程持续位于0水平下的连续时间超过长度$X$时,才做出破产的决策. Dassios和Wu(2008)^[1]首次在保险风险模型中介绍了Parisian延迟,并且定义Parisian破产时间为盈余过程在第一次位于0水平下的连续时间超过长度$r\, (>0)$的时刻. Parisian延迟有两种情况:一种是延迟期为确定,即延迟期$X=r(>0)$为一非负常数. Czarna和Palmowski(2011)^[2], Loeffen等(2013)^[3], Wong和Cheung(2015)^[4]等研究均是这种类型的Parisian破产问题;另一种是延迟期为一非负的随机变量,即$X$为非负的随机变量. Baurdoux等(2015)^[5], Landrianlt等(2011, 2014)^[6-7]讨论的就是这种随机延迟的情形. Parisian破产时间被广泛认可是因为:现实中的破产概率通常很小,即使盈余为负值,公司也可以运行一段时间,且很大可能会迅速恢复到正盈余水平.因此,它比经典破产更具有应用价值.同时,可以将Parisian延迟期看作公司的负盈余水平的恢复期. Parisian延迟的引入可以在企业偿付能力与收益性之间找到一个平衡.

本文研究的折射Lévy风险过程是一个保费根据公司的经营状况作出调整的风险过程.即,当公司的盈余值超过某一个临界值$b$时,公司可以对保费进行调整,降低保费率;但一旦公司的盈余值回到临界值以下,公司保费的收取就会回到原来水平.这种现象就是常见的折射.本文研究的风险过程就是具有确定Parisian延迟($r>0$)的折射Lévy风险过程.

假设$b\ (\geq0)$是临界值,当公司盈余值超过$b$时,保费就会做出调整,若公司盈余值回到了$b$以下时保费会恢复到原水平.

在本文中,令$U=\{U_t, t \geq 0\}$表示公司的盈余过程,其满足下面的随机微分方程

(1.1) ${\rm d}U_t={\rm d}X_t-\delta I_{\{U_t>b\}}{\rm d}t, \qquad t\geq0, \quad \delta\geq0.$

其中,过程$X=\{X_t, t \geq0\}$为一谱负Lévy过程,其Laplace指数为

$\Psi(s)=\gamma s+\frac{1}{2}s^2\sigma^2+\int_0^{\infty}({\rm e}^{-sz}-1+szI_{\{0<z\leq1\}})\pi({\rm d}z).$

$\gamma$为谱负Lévy过程$X$的漂移系数; $\sigma\geq0$为扩散系数;测度$\pi$是在$(0, \infty)$上的$\sigma$有限测度,满足

$\int_0^{\infty}(1\wedge z^2)\pi({\rm d}z)<\infty.$

若$\sigma=0$, $\int_0^{\infty}z\pi({\rm d}z)<\infty, $谱负Lévy过程$\{X_t:t\geq0\}$为有界变差过程,可简记为$X_t=ct-S_t.$其中, $c:=\gamma+\int_0^{\infty}z\pi({\rm d}z)$称为$X$的漂移系数, $S_t=\{S(t), t\geq0\}$为复合泊松过程.为了保证风险过程正的保费收入,假定

$0<\delta<c=\gamma+\int_0^{\infty}z\pi({\rm d}z).$

谱负Lévy过程$X$刻画了在临界水平$b$下方公司盈余值变化;令$Y_t=X_t-\delta t, \ \ t\geq0, $则过程$Y=\{Y_t, t \geq 0\}$也是谱负Lévy过程,它刻画了在临界水平$b$上方公司盈余的变化.即,当公司状况良好,其盈余超过临界水平$b$时,公司保费率减少$\delta.$为了模型中安全负荷为正,要求$E[U_1]>0$或$E[Y_1]>0.$

在本文中,主要研究了Parisian破产时刻$k_r^U:$

$k_r^U=\inf\{t>0:t-g_t^U>r\}, $

其中, $g_t^U=\sup\{0\leq s\leq t:U_s\geq 0\}.$许多学者研究过不同风险模型的Parisian破产问题.如, Dassios和Wu(2008)^[1]研究了经典复合Poisson风险模型的Parisian破产时间的Laplace变换; Czarna和Palmowski(2011)^[2]和Loeffen等(2013)^[3]在谱负Lévy风险模型下研究了Parisian破产时间的Laplace变换; Wong和Cheung(2015)^[4]讨论了对偶风险模型的Parisian破产问题. Lkabous等(2017)^[8]讨论了在$0$水平折射的Lévy风险过程的Parisian破产问题.本文推广了Lkabous等(2017)^[8]的模型,风险过程的折射水平由$0$变为任意的有限非负数$b\ (\geq0).$在本文中,给出了Parisian破产概率的表达式.

本文结构如下:在第二章中,介绍了本文所涉及到的谱负Lévy过程的尺度函数(scale function)概念及波动性(fluctuation)理论;第三章给出了本文的主要定理及其证明;在第四章中,给出了两个例子.

尺度函数是Lévy过程中一个强有力的工具,在本节中主要介绍了相关知识.对于谱负Lévy过程$X$,其$q$-尺度函数$W^{(q)}(x)$的定义为: $W^{(q)}(x)$为定义在$[0, +\infty)$上的连续函数,其拉普拉斯变换为

(2.1) $\begin{equation}\int_0^{\infty}{\rm e}^{-\lambda y}W^{(q)}(y){\rm d}y=\frac{1}{\Psi(\lambda)-q}, \ \ \lambda>\Phi(q), \ \ q\geq0, \end{equation}$

其中, $\Psi$为$X$的Laplace指数; $\Phi(q)=\sup\{s\geq0\mid\Psi(s)=q\}.$显然, $\Psi(\Phi(q))=q.$由Lévy过程的理论可知,尺度函数$W^{(q)}(x)$为关于$x$单调不减的非负函数且关于参数$q\, (\geq0)$也连续.当$x<0$时,令$W^{(q)}(x)=0, $则$W^{(q)}(x)$的定义域变为整个实数域${\Bbb R}.$由谱负Lévy过程的理论可知

$W^{(q)}(0)=\lim\limits_{x \to 0}W^{(q)}(x)=\left\{\begin{array}{ll} \frac{1}{c}, & \sigma=0 \ \textrm{且}\ \int_0^{1}z\pi({\rm d}z)<\infty;\\[2mm]0, &\textrm{其他.}\end{array}\right.$

除了$q$-尺度函数$W^{(q)}(x)$以外,尺度函数$Z^{(q)}(x)$也是非常重要的,其定义为

(2.2) $\begin{equation}Z^{(q)}(x)=1+q\int_0^xW^{(q)}(y){\rm d}y, \ \ x\geq 0.\end{equation}$

类似地,对于谱负Lévy过程$Y=\{Y_t=X_t-\delta t, t\geq0\}, $可以给出其相应的尺度函数, ${\Bbb W}^{(q)}$和${\Bbb Z}^{(q)}$:

(2.3) $\begin{equation}\int_0^{\infty}{\rm e}^{-\lambday}{\Bbb W}^{(q)}(y){\rm d}y=\frac{1}{\Psi(\lambda)-\delta \lambda-q}, \\\lambda>\varphi(q), \label{2.3}\end{equation}$

(2.4) $\begin{equation}{\Bbb Z}^{(q)}(x)=1+q\int_0^x{\Bbb W}^{(q)}(y){\rm d}y, \\ x\geq0.\label{2.4}\end{equation}$

其中, $\varphi(q)=\sup\{\lambda \geq 0:\Psi(\lambda)-\delta \lambda=q\}.$

对于折射谱负Lévy过程$U$定义$w^{(q)}$和$z^{(q)}$:

(2.5) $\begin{equation}w^{(q)}(x;z)=W^{(q)}(x-z)+\delta I_{\{x\geq b\}}\int_b^x{\Bbb W}^{(q)}(x-y)W^{(q)'}(y-z){\rm d}y, \label{2.5}\end{equation}$

(2.6) $\begin{equation}z^{(q)}(x;z)=Z^{(q)}(x-z)+\delta qI_{\{x\geq b\}}\int_b^x{\Bbb W}^{(q)}(x-y)W^{(q)}(y-z){\rm d}y, \label{2.6}\end{equation}$

其中, $W^{(q)'}(x)$为尺度函数$W^{(q)}(x)$的关于$x$的导函数. (2.1)-(2.6)式可参见文献Kyprianou和Loeffen(2010)^[9], Lkabous等(2017)^[8]或Renaud(2013)^[10].

注2.1  当$x<b$时, $w^{(q)}(x;z)=W^{(q)}(x-z);\ \ \ z^{(q)}(x;z)=Z^{(q)}(x-z).$

注2.2  当$q=0$时,记$W^{(0)}(x)=W(x), \ \ {\Bbb W}^{(0)}(x)={\Bbb W}(x), \ \ w^{(0)}(x;z)=w(x;z)$.

下面给出谱负Lévy过程$X, Y$以及折射谱负Lévy过程溢出时的定义及相关结论.对任意的常数$a, c\in {\Bbb R}$,令

$\tau_a^-=\inf\{t>0:X_t<a\}, \qquad \tau_c^+=\inf\{t>0:X_t\geqc\};\\\nu_a^-=\inf\{t>0:Y_t<a\}, \qquad \nu_c^+=\inf\{t>0:Y_t\geqc\};\\k_a^-=\inf\{t>0:U_t<a\}, \qquad k_c^+=\inf\{t>0:U_t\geqc\}.$

在本文以后的部分,令$P_x$表示过程的初始值为$x$时相对应的概率测度, $E_x$表示过程的初始值为$x$时相对应的数学期望.若$x=0$时,分别利用$P, \ E$表示概率测度和数学期望.谱负Lévy过程$X, \ Y$及$U$的双边溢出问题已经被许多学者研究讨论过,如Kyprianou(2006)^[11], Kyprianou和Loeffen(2010)^[9].以下的结论在这些论文中均可找到.设$a\leq x \leq c$,则有

(2.7) $\begin{equation}E_x[{\rm e}^{-q\tau_c^+};\tau_c^+<\tau_a^-]=\frac{W^{(q)}(x-a)}{W^{(q)}(c-a)}, \label{2.7}\end{equation}$

(2.8) $\begin{equation}E_x[{\rm e}^{-q\tau_a^-};\tau_a^-<\tau_c^+]=Z^{(q)}(x-a)-\frac{Z^{(q)}(c-a)}{W^{(q)}(c-a)}W^{(q)}(x-a), \label{2.8}\end{equation}$

(2.9) $\begin{equation}E_x[{\rm e}^{-q\nu_c^+};\nu_c^+<\nu_a^-]=\frac{{\Bbb W}^{(q)}(x-a)}{{\Bbb W}^{(q)}(c-a)}, \label{2.9}\end{equation}$

(2.10) $\begin{equation}E_x[{\rm e}^{-qk_c^+};k_c^+<k_a^-]=\frac{w^{(q)}(x;a)}{w^{(q)}(c;a)}, \label{2.10}\end{equation}$

(2.11) $\begin{equation}E_x[{\rm e}^{-qk_a^-};k_a^-<k_c^+]=z^{(q)}(x;a)-\frac{z^{(q)}(c;a)}{w^{(q)}(c;a)}w^{(q)}(x;a).\label{2.11}\end{equation}$

在本节最后我们给出论文后半部分涉及到的关于谱负Lévy过程相关结论.谱负Lévy过程$X$的相关量如前所示.

命题2.1^[11]  对于$x\leq a, \lambda\geq0, $有

(2.12) $\begin{equation}E_x[{\rm e}^{-\lambda\tau_a^+}]={\rm e}^{\Phi(\lambda)(x-a)}, \end{equation}$

其中, $\Phi(\lambda)=\sup\{s\geq0\mid\Psi(s)=\lambda\}.$

命题2.2^[7]  对任意的$0<x\leq a$,有

(2.13) $\begin{eqnarray}E_x\big[{\rm e}^{-q\tau_0^-+\alphaX_{\tau_0^-}}I_{\{\tau_0^-\tau_a^+\}}\big] ={\rm e}^{\alphax}+\big(q-\Psi(\alpha)\big){\rm e}^{\alpha x}\int_0^x{\rm e}^{-\alphaz}W^{(q)}(z){\rm d}z\nonumber\\-\frac{W^{(q)}(x)}{W^{(q)}(a)}\Big({\rm e}^{\alpha a}+\big(q-\Psi(\alpha)\big){\rm e}^{\alphaa}\int_0^a{\rm e}^{-\alpha z}W^{(q)}(z){\rm d}z\Big).\label{2.13} \end{eqnarray}$

命题2.3^[12]  对于$p, q\geq0, \ y\leq a\leq x\leq b$,有

(2.14) $\begin{eqnarray}&& E_x\big[{\rm e}^{-p\tau_a^-}W^{(q)}(X_{\tau_a^-}-y)I_{\{\tau_a^-<\tau_b^+\}}\big]\\ &=&W^{(q)}(x-y)-(q-p)\int_a^xW^{(p)}(x-z)W^{(q)}(z-y){\rm d}z\\ &&-\frac{W^{(p)}(x-a)}{W^{(p)}(b-a)}\Big(W^{(q)}(b-y)-(q-p)\int_a^bW^{(p)}(b-z)W^{(q)}(z-y){\rm d}z\Big).\label{2.14}\end{eqnarray}$

命题2.4^[3]  对于$\lambda>0, \ y\geq0, $有

(2.15) $\begin{equation}\int_0^\infty {\rm e}^{-\lambdar}\int_y^\infty\frac{z}{r}P(X_r\in{\rm d}z){\rm d}r=\frac{1}{\Phi(\lambda)}{\rm e}^{-\Phi(\lambda)y}, \label{2.15}\end{equation}$

(2.16) $\begin{equation}\int_0^\infty{\rm e}^{-\lambda r}\int_0^\infty W^{(q)}(z-y)\frac{z}{r}P(X_r\in{\rm d}z){\rm d}r=\frac{{\rm e}^{-\Phi(\lambda)y}}{\lambda-q}, \label{2.16}\end{equation}$

其中, $\Phi(\lambda)=\sup\{s\geq0\mid\Psi(s)=\lambda\}.$

由命题2.4易得下面结论

(2.17) $\int_0^\infty W^{(q)}(z)\frac{z}{r}P(X_r\in{\rm d}z)={\rm e}^{q\, r}.$

对于谱负Lévy过程$X, \ Y, $由论文Loeffen等(2014)^[12]可得如下命题.

命题2.5^[12]  对任意的$p, q\geq0, \ b\leq x\leq c$,有

(2.17) $\begin{equation}E_x\big[{\rm e}^{-p\nu_b^-}W^{(q)}(Y_{\nu_b^-})I_{\{\nu_b^-<\nu_c^+\}}\big]=\\W^{(q)}(x)-{\cal H}(x) -\frac{{\Bbb W}^{(p)}(x-b)}{{\Bbb W}^{(p)}(c-b)}\Big( W^{(q)}(c)-{\cal H}(c)\Big), \label{2.18}\end{equation}$

其中, ${\cal H}(x)=\int_0^{x-b}\Big((q-p)W^{(q)}(x-z)-\delta W^{(q)'}(x-z)\Big){\Bbb W}^{(p)}(z){\rm d}z.$

注2.3  当$p=q$时, $E_x\big[{\rm e}^{-q\nu_b^-}W^{(q)}(Y_{\nu_b^-})I_{\{\nu_b^-<\nu_c^+\}}\big]=w(x;0)-\frac{{\Bbb W}^{(q)}(x-b)}{{\Bbb W}^{(q)}(c-b)}w(c;0).$

在本节中,借助尺度函数以及谱负Lévy过程的波动性理论,给出折射Lévy风险过程的Parisian破产概率表达式.

定理3.1  折射谱负Lévy风险过程$U$的Parisian破产概率具有如下的表达式

(3.1) $\begin{eqnarray}P_x(k_r^U<\infty)=1-(E[X_1]-\delta)_+\frac{\int_0^\infty w(x;-z)zP(X_r\in{\rm d}z)}{\int_0^\infty \big(1-\delta W(b+z)\big)zP(X_r\in{\rm d}z)}, \ \ x\in{\Bbb R}.\end{eqnarray}$

注3.1  在折射谱负Lévy风险过程$U$中,若令$\delta=0$,则过程$U$就变为非折射的谱负Lévy风险过程, (3.1)式和Loeffen等(2013)^[3]的定理1的(2)式相同;若$b=0, $ (3.1)式和Lkabous等(2017)^[8]的定理2的(14)式一样.

为了证明我们的定理3.1,首先证明以下引理.

引理3.1  对于任意的$x\in {\Bbb R}, \ \ 0\leq b<c, $有

(3.2) $\begin{equation}E_x\Big[P_{X_{\tau_0^-}}(\tau_0^+\leq r)I_{\{\tau_0^-<\tau_b^+\}}\Big]=\int_0^\infty\Big(W(x+z)-\frac{W(x)}{W(b)}W(b+z)\Big)\frac{z}{r}P(X_r\in{\rm d}z), \label{3.2}\end{equation}$

(3.3) $\begin{equation}E_x\Big[P_{Y_{\nu_b^-}}(k_0^-\geq k_b^+)I_{\{\nu_b^-<\infty\}}\Big]=\frac{w(x;0)-{\Bbb W}(x-b)(1-\delta W(b))}{W(b)}.\label{3.3}\end{equation}$

证  由Fubini定理,空间时齐性以及(2.14)式可得

(3.4) $\begin{eqnarray}&&E_x\Big[{\rm e}^{-q\tau_0^-}E_{X_{\tau_0^-}}[{\rm e}^{-q\tau_0^+} I_{\{\tau_0^+\leq r\}}]I_{\{\tau_0^-<\tau_b^+\}}\Big]\nonumber\\ &=&E_x\Big[{\rm e}^{-q\tau_0^-}\int_0^\infty {\rm e}^{-qr}W^{(q)}(X_{\tau_0^-}+z)\frac{z}{r}P(X_r\in{\rm d}z)I_{\{\tau_0^-<\tau_b^+\}}\Big]\nonumber\\ &=&\int_0^\infty {\rm e}^{-qr}E_x \Big[{\rm e}^{-q\tau_0^-}W^{(q)}(X_{\tau_0^-}+z)I_{\{\tau_0^-<\tau_b^+\}}\Big]\frac{z}{r}P(X_r\in{\rm d}z)\nonumber\\ &=&\int_0^\infty {\rm e}^{-qr}E_{x+z} \Big[{\rm e}^{-q\tau_z^-}W^{(q)}(X_{\tau_z^-}+z)I_{\{\tau_z^-<\tau_{b+z}^+\}}\Big]\frac{z}{r}P(X_r\in{\rm d}z)\nonumber\\ &=&\int_0^\infty {\rm e}^{-qr}\Big(W^{(q)}(x+z)-\frac{W^{(q)}(x)}{W^{(q)}(b)}W^{(q)}(b+z)\Big)\frac{z}{r}P(X_r\in{\rm d}z), \label{3.4}\end{eqnarray}$

其中

$E_{X_{\tau_0^-}}\left[{\rm e}^{-q\tau_0^+} I_{\{\tau_0^+\leq r\}}\right]=\int_0^\infty {\rm e}^{-qr}W^{(q)}(X_{\tau_0^-}+z)\frac{z}{r}P(X_r\in {\rm d}z), $

可详见Lkabous等(2017)^[8]的引理8.而

$\begin{eqnarray*} &E_x\left[P_{X_{\tau_0^-}}(\tau_0^+\leq r)I_{\{\tau_0^-<\tau_b^+\}}\right]\\=&\lim\limits_{q\to 0}{\rm e}^{qr}E_x\Big[{\rm e}^{-q\tau_0^-}E_{X_{\tau_0^-}}[{\rm e}^{-q\tau_0^+} I_{\{\tau_0^+\leq r\}}]I_{\{\tau_0^-<\tau_b^+\}}\Big]\nonumber\\ =&\int_0^\infty\Big(W(x+z)-\frac{W(x)}{W(b)}W(b+z)\Big)\frac{z}{r}P(X_r\in{\rm d}z).\nonumber\end{eqnarray*}$

所以, (3.2)式成立.下证(3.3)式成立.由(2.5), (2.10)及(2.18)式可得

(3.5) $\begin{eqnarray}E_x\Big[{\rm e}^{-q\nu_b^-}E_{Y_{\nu_b^-}}[{\rm e}^{-qk_b^+}I_{\{k_b^+\leqk_0^-\}}]I_{\{\nu_b^-<\nu_c^+\}}\Big]&=&E_x\left[{\rm e}^{-q\nu_b^-}\frac{w^{(q)}(Y_{\nu_b^-};0)}{w^{(q)}(b;0)}I_{\{\nu_b^-<\nu_c^+\}}\right]\nonumber\\&=&\frac{E_x[{\rm e}^{-q\nu_b^-}W^{(q)}(Y_{\nu_b^-})I_{\{\nu_b^-<\nu_c^+\}}]}{W^{(q)}(b)}\nonumber\\&=&\frac{w^{(q)}(x;0)-\frac{{\Bbb W}^{(q)}(x-b)}{{\Bbb W}^{(q)}(c-b)}w^{(q)}(c;0)}{W^{(q)}(b)}.\label{3.5}\end{eqnarray}$

又因为

$\lim\limits_{c \to \infty}\frac{W^{(q)}(c)}{{\Bbb W}^{(q)}(c-b)}=0, ~~~~~ \lim\limits_{c \to \infty}\frac{{\Bbb W}^{(q)}(c-y)}{{\Bbb W}^{(q)}(c-b)}={\rm e}^{-\varphi(q)(y-b)}.$

所以

(3.6) $\begin{eqnarray}&&\lim\limits_{q \to 0}\lim\limits_{c \to\infty}\frac{w^{(q)}(c;0)}{{\Bbb W}^{(q)}(c-b)}=\lim\limits_{q \to 0}\lim\limits_{c \to \infty}\frac{W^{(q)}(c)+\delta\int_b^c {\Bbb W}^{(q)}(c-y)W^{(q)'}(y){\rm d}y}{{\Bbb W}^{(q)}(c-b)}\nonumber\\&=&\lim\limits_{q \to 0}-\delta W^{(q)}(b)+\delta {\rm e}^{\varphi(q)(b)}\bigg(\frac{1}{\delta}-\varphi(q)\int_{0}^{b}{\rm e}^{-\varphi(q)(y)}W^{(q)}(y){\rm d}y\bigg)\nonumber\\&=&1-\delta W(b).\label{3.6}\end{eqnarray}$

借助(3.6)式,在(3.5)式中分别令$c\to\infty, \ \ q\to0$立得(3.3)式.

引理3.2  对任意的$x\in {\Bbb R}, \ \ b\geq0, $有

(3.7) $\begin{eqnarray}&&E_x\Big[E_{Y_{\nu_b^-}}[P_{X_{\tau_0^-}}(\tau_0^+\leq r)I_{\{\tau_0^-<\tau_b^+\}}]I_{\{\nu_b^-<\infty\}}\Big]\nonumber\\&=&\int_0^\infty \Big[w(x;-z)-{\Bbb W}(x-b)\big(1-\delta W(b+z)\big)\Big]\frac{z}{r}P(X_r\in{\rm d}z)\nonumber\\ &&-\int_0^\infty \Big[w(x;0)-{\Bbb W}(x-b)\big(1-\delta W(b)\big)\Big]\frac{W(b+z)}{W(b)}\frac{z}{r}P(X_r\in{\rm d}z).\label{3.7}\end{eqnarray}$

证  为证明(3.7)式,首先需要计算下面的极限

(3.8) $\begin{eqnarray}&&E_x\Big[E_{Y_{\nu_b^-}}[P_{X_{\tau_0^-}}(\tau_0^+\leq r)I_{\{\tau_0^-<\tau_b^+\}}]I_{\{\nu_b^-<\infty\}}\Big]\nonumber\\&=&\lim\limits_{q \to 0}\lim\limits_{c \to \infty}{\rm e}^{qr}E_x\bigg[{\rm e}^{-q\nu_b^-}E_{Y_{\nu_b^-}}\Big[{\rm e}^{-q\tau_0^-}E_{X_{\tau_0^-}}[{\rm e}^{-q\tau_0^+}I_{\{\tau_0^+\leqr\}}]I_{\{\tau_0^-<\tau_b^+\}}\Big]I_{\{\nu_b^-<\nu_c^+\}}\bigg].\label{3.8}\end{eqnarray}$

利用(3.4)式,空间时齐性及(2.18)式可得

(3.9) $\begin{eqnarray}&&E_x\bigg[{\rm e}^{-q\nu_b^-}E_{Y_{\nu_b^-}}\Big[{\rm e}^{-q\tau_0^-}E_{X_{\tau_0^-}}[{\rm e}^{-q\tau_0^+}I_{\{\tau_0^+\leq r\}}]I_{\{\tau_0^-<\tau_b^+\}}\Big]I_{\{\nu_b^-<\nu_c^+\}}\bigg]\nonumber\\&=&E_x\bigg[{\rm e}^{-q\nu_b^-}\int_0^\infty {\rm e}^{-qr}\Big(W^{(q)}(Y_{\nu_b^-}+z)\\&&-\frac{W^{(q)}(Y_{\nu_b^-})}{W^{(q)}(b)}W^{(q)}(b+z)\Big)\frac{z}{r}P(X_r\in{\rm d}z)I_{\{\nu_b^-<\nu_c^+\}}\bigg]\nonumber\\&=&\int_0^\infty {\rm e}^{-qr} E_{x+z}\Big[{\rm e}^{-q\nu_{b+z}^-}W^{(q)}(Y_{\nu_{b+z}^-})I_{\{\nu_{b+z}^-<\nu_{c+z}^+\}}\Big]\frac{z}{r}P(X_r\in{\rm d}z)\nonumber\\&&-\int_0^\infty {\rm e}^{-qr} E_x\Big[{\rm e}^{-q\nu_b^-}W^{(q)}(Y_{\nu_b^-})I_{\{\nu_b^-<\nu_c^+\}}\Big]\frac{W^{(q)}(b+z)}{W^{(q)}(b)}\frac{z}{r}P(X_r\in{\rm d}z)\nonumber\\&=&\int_0^\infty {\rm e}^{-qr}\Big(w^{(q)}(x;-z)-\frac{{\Bbb W}^{(q)}(x-b)}{{\Bbb W}^{(q)}(c-b)}w^{(q)}(c;-z)\Big)\frac{z}{r}P(X_r\in{\rm d}z)\nonumber\\&&-\int_0^\infty{\rm e}^{-qr}\Big(w^{(q)}(x;0)-\frac{{\Bbb W}^{(q)}(x-b)}{{\Bbb W}^{(q)}(c-b)}w^{(q)}(c;0)\Big)\\&&\frac{W^{(q)}(b+z)}{W^{(q)}(b)}\frac{z}{r}P(X_r\in{\rm d}z).\label{3.9}\end{eqnarray}$

借助于极限

$\lim\limits_{c \to \infty}\frac{W^{(q)}(c+z)}{{\Bbb W}^{(q)}(c-b)}=0,~~~~~~~\lim\limits_{c \to \infty}\frac{{\Bbb W}^{(q)}(c-y)}{{\Bbb W}^{(q)}(c-b)}={\rm e}^{-\varphi(q)(y-b)}, $

可得

(3.10) $\begin{eqnarray}&&\lim\limits_{q \to 0}\lim\limits_{c \to\infty}\frac{w^{(q)}(c;-z)}{{\Bbb W}^{(q)}(c-b)}\\&=&\lim\limits_{q \to 0}\lim\limits_{c \to \infty}\frac{W^{(q)}(c+z)+\delta\int_b^c {\Bbb W}^{(q)}(c-y)W^{(q)'}(y+z){\rm d}y}{{\Bbb W}^{(q)}(c-b)}\nonumber\\&=&\lim\limits_{q \to 0}-\delta W^{(q)}(b+z)+\delta {\rm e}^{-\varphi(q)(z+b)}\Big(\frac{1}{\delta}-\varphi(q)\int_{0}^{b+z}{\rm e}^{-\varphi(q)(y)}W^{(q)}(y){\rm d}y\Big)\nonumber\\&=&1-\delta W(b+z).\label{3.10}\end{eqnarray}$

将(3.9)及(3.10)式代入(3.8)式即可得结论(3.7).

引理3.3  对任意的$x\in {\Bbb R}, \ b\geq0$及$\epsilon>0, $有

(3.11) $\begin{equation}E_x[P_{X_{\tau_0^-}}(\tau_\epsilon^+\leq r)I_{\{\tau_0^-<\tau_{b+\epsilon}^+\}}]=\\\int_\epsilon^\infty\Big(W(z+x-\epsilon)-W(x)\frac{W(z+b)}{W(b+\epsilon)}\Big)\frac{z}{r}P(X_r\in{\rm d}z), \label{3.11}\end{equation}$

(3.12) $\begin{equation}E_x[P_{Y_{\nu_b^-}}(k_0^->k_{b+\epsilon}^+)I_{\{\nu_b^-<\infty\}}]=\frac{w(x;0)-{\Bbb W}(x-b)(1-\deltaW(b))}{w(b+\epsilon;0)}.\label{3.12}\end{equation}$

证  由(2.12), (2.13)及(2.15)式,同时利用Fubini定理和分部积分可得下面的Laplace变换

$\begin{eqnarray*}&&\int_0^\infty {\rm e}^{-\theta r}E_x[P_{X_{\tau_0^-}}(\tau_\epsilon^+\leqr)I_{\{\tau_0^-<\tau_{b+\epsilon}^+\}}]{\rm d}r\\&=&E_x\Big[\int_0^\infty {\rm e}^{-\theta r}P_{X_{\tau_0^-}}(\tau_\epsilon^+\leq r){\rm d}rI_{\{\tau_0^-<\tau_{b+\epsilon}^+\}}\Big]\nonumber\\&=&\frac{1}{\theta}E_x\Big[E_{X_{\tau_0^-}}[{\rm e}^{-\theta\tau_\epsilon^+}]I_{\{\tau_0^-<\tau_{b+\epsilon}^+\}}\Big]=\frac{{\rm e}^{-\Phi(\theta)\epsilon}}{\theta}E_x[{\rm e}^{\Phi(\theta)X_{\tau_0^-}}I_{\{\tau_0^-<\tau_{b+\epsilon}^+\}}]\nonumber\\&=&\frac{{\rm e}^{-\Phi(\theta)\epsilon}}{\theta}\Big(\theta\int_0^\infty{\rm e}^{-\Phi(\theta)y}W(y+x){\rm d}y-\\&&\frac{W(x)}{W(b+\epsilon)}\theta \int_0^\infty {\rm e}^{-\Phi(\theta)y}W(y+b+\epsilon){\rm d}y\Big)\nonumber\\&=&\frac{1}{\Phi(\theta)}\int_0^\infty {\rm e}^{-\Phi(\theta)(y+\epsilon)}W'(y+x){\rm d}y-\\&&\frac{W(x)}{W(b+\epsilon)}\frac{1}{\Phi(\theta)}\int_0^\infty {\rm e}^{-\Phi(\theta)(y+\epsilon)}W'(y+b+\epsilon){\rm d}y\nonumber\\&=&\int_0^\infty {\rm e}^{-\thetar}\int_\epsilon^\infty\int_0^{z-\epsilon}W'(y+x)-\\&&\frac{W(x)}{W(b+\epsilon)}W'(y+b+\epsilon){\rm d}y\frac{z}{r}P(X_r\in{\rm d}z){\rm d}r.\nonumber\end{eqnarray*}$

所以,由上式及Laplace变换的性质可得

$\begin{eqnarray*}E_x[P_{X_{\tau_0^-}}(\tau_\epsilon^+\leq r)I_{\{\tau_0^-<\tau_{b+\epsilon}^+\}}]=\\\int_\epsilon^\infty\int_0^{z-\epsilon}W'(y+x)-\frac{W(x)}{W(b)}W'(y+b){\rm d}y\frac{z}{r}P(X_r\in{\rm d}z)\nonumber\\=\int_\epsilon^\infty\Big(W(z+x-\epsilon)-\frac{W(x)}{W(b+\epsilon)}W(z+b)\Big)\frac{z}{r}P(X_r\in{\rm d}z).\nonumber \end{eqnarray*}$

下证(3.12)式.由(2.10)式及命题2.5可得

$\begin{eqnarray*}E_x[P_{Y_{\nu_b^-}}(k_0^->k_{b+\epsilon}^+)I_{\{\nu_b^-<\infty\}}]&=&\lim\limits_{c\to\infty}E_x[P_{Y_{\nu_b^-}}(k_0^->k_{b+\epsilon}^+)I_{\{\nu_b^-<\nu_c^+\}}]\nonumber\\&=&\lim\limits_{c\to\infty}\frac{E_x[W(Y_{\nu_b^-})I_{\{\nu_b^-<\nu_c^+\}}]}{w(b+\epsilon;0)}\\&=&\lim\limits_{c\to\infty}\frac{w(x;0)-\frac{{\Bbb W}(x-b)}{{\Bbb W}(c-b)}w(c;0)}{w(b+\epsilon;0)}\nonumber\\&=&\frac{w(x;0)-{\Bbb W}(x-b)(1-\deltaW(b))}{w(b+\epsilon;0)}.\nonumber\end{eqnarray*}$

证毕.

引理3.4  对于任意的$x\in {\Bbb R}, \ 0\leq b<c$及$\epsilon>0, $有

(3.13) $\begin{eqnarray}&&E_x\Big[E_{Y_{\nu_b^-}}[P_{X_{\tau_0^-}}(\tau_\epsilon^+\leq r)I_{\{\tau_0^-<\tau_{b+\epsilon}^+\}}]I_{\{\nu_b^-<\infty\}}\Big]\nonumber\\&=&\int_\epsilon^\infty\Big(w(x;-z+\epsilon)-{\Bbb W}(x-b)\big(1-\delta W(b+z-\epsilon)\big)\Big) \frac{z}{r}P(X_r\in{\rm d}z)\nonumber\\&&-\int_\epsilon^\infty \Big(w(x;0)-{\Bbb W}(x-b)\big(1-\deltaW(b)\big)\Big) \frac{W(z+b)}{W(b+\epsilon)}\frac{z}{r}P(X_r\in{\rm d}z).\label{3.13}\end{eqnarray}$

证  根据(3.11)式且利用空间齐次性和命题2.5可得

$\begin{eqnarray*}&&E_x\Big[E_{Y_{\nu_b^-}}[P_{X_{\tau_0^-}}(\tau_\epsilon^+\leq r)I_{\{\tau_0^-<\tau_{b+\epsilon}^+\}}]I_{\{\nu_b^-<\infty\}}\Big]\nonumber\\&=&\lim\limits_{c \to \infty}E_x\Big[E_{Y_{\nu_b^-}}[P_{X_{\tau_0^-}}(\tau_\epsilon^+\leq r)I_{\{\tau_0^-<\tau_{b+\epsilon}^+\}}]I_{\{\nu_b^-<\nu_c^+\}}\Big]\nonumber\\&=&\lim\limits_{c \to \infty}\int_\epsilon^\infty\Big(E_x[W(z+Y_{\nu_b^-}-\epsilon)I_{\{\nu_b^-<\nu_c^+\}}]-\\&&E_x[W(Y_{\nu_b^-})I_{\{\nu_b^-<\nu_c^+\}}]\frac{W(z+b)}{W(b+\epsilon)}\Big)\frac{z}{r}P(X_r\in{\rm d}z)\nonumber\\&=&\int_\epsilon^\infty \Big(w(x;-z+\epsilon)-{\Bbb W}(x-b)\big(1-\delta W(b+z-\epsilon)\big)\Big)\frac{z}{r}P(X_r\in{\rm d}z)\nonumber\\&&-\int_\epsilon^\infty \Big(w(x;0)-{\Bbb W}(x-b)\big(1-\deltaW(b)\big)\Big) \frac{W(z+b)}{W(b+\epsilon)}\frac{z}{r}P(X_r\in{\rm d}z).\nonumber\end{eqnarray*}$

证毕.

定理3.1的证明  只要求Parisian破产时刻等于无穷的概率$P_x(k_r^U=\infty), $由$P_x(k_r^U<\infty)=1-P_x(k_r^U=\infty)$立得(3.1)式.设$x$为过程的出发位置,则

(Ⅰ)当$x<0$时,因为谱负Lévy过程只有负的跳跃,所以$U_{k_0^+}=0, $再由强马氏性可得

$P_x(k_r^U=\infty)=E_x\left[P_x(k_r^U=\infty|{\cal F}_{K_0^+})I_{\{k_0^+<\infty\}}\right]=P_x(k_0^+\leqr)\, P_0(k_r^U=\infty).$

又由于在$x<0$条件下, $\{X_t, t<\tau_b^+\}$和$\{U_t, t<k_b^+\}$在$P_x$下有相同的分布.所以

(3.14) $\begin{eqnarray}P_x(k_r^U=\infty)=P_x(\tau_0^+\leq r)\, P_0(k_r^U=\infty).\end{eqnarray}$

(Ⅱ)当$0\leq x<b$时, $\{X_t, t<\tau_b^+\}$和$\{U_t, t<k_b^+\}$在$P_x$下有相同的分布.由强马氏性及(3.14)式,可得

(3.15) $\begin{eqnarray}P_x(k_r^U=\infty)&=&P_x(k_r^U=\infty, k_0^-> k_b^+)+P_x(k_r^U=\infty, k_0^-<k_b^+)\nonumber\\&=&E_x[P_b(k_r^U=\infty)I_{\{k_0^-> k_b^+\}}]+E_x[P_{U_{k_0^-}}(k_r^U=\infty)I_{\{k_0^-< k_b^+\}}]\nonumber\\&=&P_x(k_0^->k_b^+)P_b(k_r^U=\infty)+P_0(k_r^U=\infty)\\&&E_x[P_{X_{\tau_0^-}}(\tau_0^+\leqr)I_{\{\tau_0^-< \tau_b^+\}}].\label{3.15} \end{eqnarray}$

同时,可以得到(3.15)式在$x<0$时恰为(3.14)式.

(Ⅲ)当$x\geq b$时, $\{Y_t, t<\nu_b^-\}$和$\{U_t, t<k_b^-\}$在$P_x$下有相同的分布.同样利用强马氏性,双期望公式及(3.15)式,可得

(3.16) $\begin{eqnarray}P_x(k_r^U=\infty)&=&P_x(k_b^-=\infty)+E_x[P_x(k_r^U=\infty |{\cal F}_{k_b^-})I_{\{k_b^-<\infty\}}]\nonumber\\&=&P_x(\nu_b^-=\infty)+P_b(k_r^U=\infty)E_x[P_{Y_{\nu_b^-}}(k_0^-> k_b^+)I_{\{\nu_b^-<\infty\}}]\nonumber\\&&+P_0(k_r^U=\infty)E_x[E_{Y_{\nu_b^-}}[P_{X_{\tau_0^-}}(\tau_0^+\leq r)I_{\{\tau_0^-<\tau_b^+\}}]I_{\{\nu_b^-< \infty\}}]. \label{3.16}\end{eqnarray}$

与(Ⅱ)中一样, (3.16)式在$x<b$时与(3.14)和(3.15)式相同.

下面将分两种情形来证明定理3.1.

(Pi) 假设谱负Lévy过程$X$具有有界变差(BV)的样本轨道.在(3.14)式中令$x=0$,可得

(3.17) $\begin{equation}P_0(k_r^U=\infty)=\frac{P_0(k_0^-> k_b^+)P_b(k_r^U=\infty)}{1-E_0[P_{X_{\tau_0^-}}(\tau_0^+\leq r)I_{\{\tau_0^-<\tau_b^+\}}]}.\end{equation}$

同样在(3.16)式中令$x=b$,可得

(3.18) $\begin{equation}P_b(k_r^U=\infty)=\frac{P_b(\nu_b^-=\infty)+P_0(k_r^U=\infty)E_b\Big[E_{Y_{\nu_b^-}}[P_{X_{\tau_0^-}}(\tau_0^+\leqr)I_{\{\tau_0^-< \tau_b^+\}}]I_{\{\nu_b^-< \infty\}}\Big]}{1-E_b\Big[P_{Y_{\nu_b^-}}(k_0^->k_b^+)I_{\{\nu_b^-<\infty\}}\Big]}. \end{equation}$

联立(3.17)及(3.18)式可解得

(3.19) $\begin{equation}P_0(k_r^U=\infty)=P_b(\nu_b^-=\infty)P_0(k_0->k_b^+){\cal J}^{-1}_{0, b}, \label{3.19}\end{equation}$

(3.20) $\begin{equation}P_b(k_r^U=\infty)=P_b(\nu_b^-=\infty)\big(1-E_0[P_{X_{\tau_0^-}}(\tau_0^+\leqr)I_{\{\tau_0^-\!<\tau_b^+\}}]\big){\cal J}^{-1}_{0, b}, \label{3.20}\end{equation}$

其中

$ \begin{eqnarray*}{\cal J}_{0, b}&=&\big(1-E_b[P_{Y_{\nu_b^-}}(k_0^-\geqk_b^+)I_{\{\nu_b^-<\infty\}}]\big)\times\big(1-E_0[P_{X_{\tau_0^-}}(\tau_0^+\leq r)I_{\{\tau_0^-\!<\tau_b^+\}}]\big)\nonumber\\&&-P_0(k_0^-> k_b^+)E_b\Big[E_{Y_{\nu_b^-}}[P_{X_{\tau_0^-}}(\tau_0^+\leqr)I_{\{\tau_0^-< \tau_b^+\}}]I_{\{\nu_b^-<\infty\}}\Big].\nonumber\end{eqnarray*}$

对于谱负Lévy风险过程$Y, $由Kyprianou(2006)^[11]中的(8.7)式,可知

(3.21) $\begin{equation}P_x(\nu_0^-<\infty)=1-(E[X_1]-\delta)_+{\Bbb W}(x), \ \ x\geq0.\end{equation}$

由上式及空间齐次性可得

(3.22) $\begin{equation}P_b(\nu_b^-=\infty)=P_0(\nu_0^-=\infty)=(E[X_1]-\delta)_+{\Bbb W}(0).\end{equation}$

在(3.2)式中令$x=0$,得到

(3.23) $\begin{equation}E_0[P_{X_{\tau_0^-}}(\tau_0^+\leq r)I_{\{\tau_0^-<\tau_b^+\}}]=1-\frac{W(0)}{W(b)}\int_0^\infty W(b+z)\frac{z}{r}P(X_r\in{\rm d}z).\end{equation}$

同样,在(3.3)和(3.7)式中,令$x=b$可得

(3.24) $\begin{equation}E_b\left[P_{Y_{\nu_b^-}}(k_0^-> k_b^+)I_{\{\nu_b^-<\infty\}}\right]=1-{\Bbb W}(0)\left(\frac{1}{W(b)}-\delta\right), \label{3.24}\end{equation}$

(3.25) $\begin{equation}E_b\Big[E_{Y_{\nu_b^-}}[P_{X_{\tau_0^-}}(\tau_0^+\leqr)I_{\{\tau_0^-<\tau_b^+\}}]I_{\{\nu_b^-<\infty\}}\Big]=\\{\Bbb W}(0)\int_0^\infty\Big(\frac{W(b+z)}{W(b)}-1\Big)\frac{z}{r}P(X_r\in{\rm d}z).\label{3.25}\end{equation}$

在(2.10)式中令$q=0, \ a=0, \ c=b, \ x=0$,可得

(3.26) $\begin{equation}P_0(k_0^-> k_b^+)=\frac{W(0)}{W(b)}.\end{equation}$

将(3.22)-(3.26)式分别代入(3.19)及(3.20)式,化简可得

(3.27) $\begin{equation}P_b(k_r^U=\infty)=\frac{(E[X_1]-\delta)_+\int_0^\inftyW(b+z)\frac{z}{r}P(X_r\in{\rm d}z)} {\int_0^\infty \big(1-\deltaW(b+z)\big)\frac{z}{r}P(X_r\in{\rm d}z)}, \label{3.27}\end{equation}$

(3.28) $\begin{equation}P_0(k_r^U=\infty)=\frac{(E[X_1]-\delta)_+} {\int_0^\infty\big(1-\delta W(b+z)\big)\frac{z}{r}P(X_r\in{\rm d}z)}.\label{3.28}\end{equation}$

(Pⅱ)假设谱负Lévy风险过程$X$是无界变差过程(UBV).本部分利用与Loeffen等(2013)^[3]类似的逼近方法来研究相关问题.为此,引入停时$k_{r, \epsilon}^U.$对任意的$\epsilon\geq0, $有

$k_{r, \epsilon}^U=\inf\{t>r:t-g_t^\epsilon >r, U_{t-r}<0\}, $

其中, $g_t^\epsilon=\sup\{0\leq s\leq t:U_s\geq\epsilon\}.$停时$k_{r, \epsilon}^U$表示过程$U$从零水平下方出发到首次回到$\epsilon$水平上方这段时间大于$r$的停时.类似于有界变差,有

(Ⅳ)当$x<0$时,有

(3.29) $\begin{equation}P_x(k_{r, \epsilon}^U=\infty)=P_x(\tau_\epsilon^+\leq r)P_\epsilon(k_{r, \epsilon}^U=\infty).\end{equation}$

(Ⅴ)当$0\leq x<b$时,有

(3.30) $\begin{eqnarray}P_x(k_{r, \epsilon}^U=\infty)&=&P_x(k_0^-> k_{b+\epsilon}^+)P_{b+\epsilon}(k_{r, \epsilon}^U=\infty)\nonumber\\&&+P_\epsilon(k_{r, \epsilon}^U=\infty)E_x[P_{X_{\tau_0^-}}(\tau_\epsilon^+\leqr)I_{\{\tau_0^-< \tau_{b+\epsilon}^+\}}].\label{3.30}\end{eqnarray}$

(Ⅵ)当$x\geq b$时,有

(3.31) $\begin{eqnarray}P_x(k_{r, \epsilon}^U=\infty)&=&P_x(\nu_b^-=\infty)+P_{b+\epsilon}(k_{r, \epsilon}^U=\infty)E_x[P_{Y_{\nu_b^-}}(k_0^-> k_{b+\epsilon}^+)I_{\{\nu_b^-<\infty\}}]\nonumber\\&&+P_\epsilon(k_{r, \epsilon}^U=\infty)E_x\Big[E_{Y_{\nu_b^-}}[P_{X_{\tau_0^-}}(\tau_\epsilon^+\leqr)I_{\{\tau_0^-<\tau_{b+\epsilon}^+\}}]I_{\{\nu_b^-<\infty\}}\Big].\label{3.31}\end{eqnarray}$

分别在(3.30)及(3.31)式中令$x=\epsilon, \ \ x=b+\epsilon$,可得

(3.32) $\begin{equation}P_\epsilon(k_{r, \epsilon}^U=\infty)=\frac{P_\epsilon(k_0^->k_{b+\epsilon}^+)P_{b+\epsilon}(k_{r, \epsilon}^U=\infty)}{1-E_\epsilon[P_{X_{\tau_0^-}}(\tau_\epsilon^+\leqr)I_{\{\tau_0^-<\tau_{b+\epsilon}^+\}}]}, \label{3.32}\end{equation}$

(3.33) $\begin{eqnarray}&&P_{b+\epsilon}(k_{r, \epsilon}^U=\infty)=\\&&\frac{P_{b+\epsilon}(\nu_b^-=\infty)+P_\epsilon(k_{r, \epsilon}^U=\infty)E_{b+\epsilon}[E_{Y_{\nu_b^-}}[P_{X_{\tau_0^-}}(\tau_\epsilon^+\leqr)I_{\{\tau_0^-< \tau_{b+\epsilon}^+\}}]I_{\{\nu_b^-< \infty\}}]}{1-E_{b+\epsilon}[P_{Y_{\nu_b^-}}(k_0^->k_{b+\epsilon}^+)I_{\{\nu_b^-<\infty\}}]}.\label{3.33}\end{eqnarray}$

联立(3.32)及(3.33)式,可解得

(3.34) $\begin{equation}P_{b+\epsilon}(k_{r, \epsilon}^U=\infty)=P_{b+\epsilon}(\nu_b^-=\infty)\big(1-E_\epsilon[P_{X_{\tau_0^-}}(\tau_\varepsilon^+\leqr)I_{\{\tau_0^-< \tau_{b+\epsilon}^+\}}]\big){\cal J}^{-1}_{\epsilon, b}, \label{3.34}\end{equation}$

(3.35) $\begin{equation}P_\epsilon(k_{r, \epsilon}^U=\infty)=P_{b+\epsilon}(\nu_b^-=\infty)P_\epsilon(k_0^->k_{b+\epsilon}^+){\cal J}^{-1}_{\epsilon, b}, \label{3.35}\end{equation}$

其中

$\begin{eqnarray*}{\cal J}_{\epsilon, b}&=&(1-E_{b+\epsilon}[P_{Y_{\nu_b^-}}(k_0^-> k_{b+\epsilon}^+)I_{\{\nu_b^-<\infty\}}])\times(1-E_\epsilon[P_{X_{\tau_0^-}}(\tau_\epsilon^+\leq r)I_{\{\tau_0^-\!< \tau_{b+\epsilon}^+\}}])\nonumber\\&&-P_\epsilon(k_0^-> k_{b+\epsilon}^+)E_{b+\epsilon}\Big[E_{Y_{\nu_b^-}}[P_{X_{\tau_0^-}}(\tau_\epsilon^+\leqr)I_{\{\tau_0^-< \tau_{b+\epsilon}^+\}}]I_{\{\nu_b^-<\infty\}}\Big].\nonumber\end{eqnarray*}$

为了求解(3.34)和(3.35)式,首先来计算以下几个式子.由空间齐次性及(3.21)式,可得

(3.36) $\begin{equation}P_{b+\epsilon}(\nu_b^-=\infty)=P_{\epsilon}(\nu_0^-=\infty)=(E[X_1]-\delta)_+{\Bbb W}(\epsilon).\end{equation}$

分别在(3.11)-(3.13)及(2.10)式中,令$x=\epsilon, \ x=b+\epsilon, \ x=b+\epsilon$及$q=\epsilon, a=0, $$ c=b+\epsilon, $$ x=\epsilon, $可得

(3.37) $\begin{equation}E_\epsilon [P_{X_{\tau_0^-}}(\tau_\epsilon^+\leq r)I_{\{\tau_0^-<\tau_{b+\epsilon}^+\}}]=\int_\epsilon^\infty\Big(W(z)-W(\epsilon)\frac{W(z+b)}{W(b+\epsilon)}\Big)\frac{z}{r}P(X_r\in{\rm d}z), \label{3.37}\end{equation}$

(3.38) $\begin{equation}E_{b+\epsilon}[P_{Y_{\nu_b^-}}(k_0^->k_{b+\epsilon}^+)I_{\{\nu_b^-<\infty\}}]=1-\frac{{\Bbb W}(\epsilon)(1-\deltaW(b))}{w(b+\epsilon;0)}, \label{3.38}\end{equation}$

(3.39) $\begin{eqnarray}&&E_{b+\epsilon}\Big[E_{Y_{\nu_b^-}}[P_{X_{\tau_0^-}}(\tau_\epsilon^+\leq r)I_{\{\tau_0^-<\tau_{b+\epsilon}^+\}}]I_{\{\nu_b^-<\infty\}}\Big]\nonumber\\&=&\int_\epsilon^\infty\Big(w(b+\epsilon;-z+\epsilon)-{\Bbb W}(\epsilon)\big(1-\delta W(b+z-\epsilon)\big)\Big) \frac{z}{r}P(X_r\in{\rm d}z)\nonumber\\&&-\int_\epsilon^\infty\Big(w(b+\epsilon;0)-{\Bbb W}(\epsilon)\big(1-\deltaW(b)\big)\Big) \frac{W(z+b)}{W(b+\epsilon)}\frac{z}{r}P(X_r\in{\rm d}z), \label{3.39}\end{eqnarray}$

(3.40) $\begin{equation}P_\epsilon (k_0^->k_{b+\epsilon}^+)=\frac{W(\epsilon)}{w(b+\epsilon;0)}.\label{3.40}\end{equation}$

将(3.36)-(3.40)式代入(3.35)式,化简可得

(3.41) $\begin{equation}P_\epsilon (k_{r, \epsilon}^U=\infty)={\cal M}^{-1}_{\epsilon, 0}(E[X_1]-\delta)_+\; , \end{equation}$

其中

$\begin{eqnarray*}{\cal M}_{\epsilon, 0}&=&\int_\epsilon^\infty\big(1-\deltaW(b+z-\epsilon)\big)\frac{z}{r}P(X_r\in{\rm d}z)\nonumber\\&&+\frac{1}{W(\epsilon)}\bigg[\Big(1-\int_\epsilon^\infty W(z)\frac{z}{r}P(X_r\in{\rm d}z)\Big)\big(1-\deltaW(b)\big)\bigg]\nonumber\\&&+\frac{1}{{\Bbb W}(\epsilon)}\int_\epsilon^\infty\Big(w(b+\epsilon;0)\frac{W(z+b)}{W(b+\epsilon)}-w(b+\epsilon;-z+\epsilon)\Big)\frac{z}{r}P(X_r\in{\rm d}z).\nonumber\end{eqnarray*}$

类似地,将(3.36)-(3.40)式代入(3.34)式,化简可得

(3.42) $\begin{eqnarray}P_{b+\epsilon}(k_{r, \epsilon}^U=\infty)&=&{\cal M}^{-1}_{\epsilon, b}\, (E[X_1]-\delta)_+\nonumber\\&&\times\Big[\frac{W(b+\epsilon)}{W(\epsilon)}\int_0^\epsilon W(z)\frac{z}{r}P(X_r\in{\rm d}z)\\&&+\int_\epsilon^\infty W(z+b)\frac{z}{r}P(X_r\in{\rm d}z)\Big], \label{3.42} \end{eqnarray}$

其中

$\begin{eqnarray*}{\cal M}_{\epsilon, b}&=&\frac{1-\delta W(b)}{w(b+\epsilon;0)W(\epsilon)}\int_0^\epsilon W(z)\frac{z}{r}P(X_r\in{\rm d}z)\nonumber\\&&-\frac{\delta\, W(b+\epsilon)}{r\cdotw(b+\epsilon;0)\, {\Bbb W}(\epsilon)}\int_\epsilon^\infty\int_b^{b+\epsilon}{\Bbb W}(b+\epsilon-y)W'(y+z-\epsilon)\cdot\\&&z\, dy\, P(X_r\in{\rm d}z)\nonumber\\&&+\frac{\delta}{r\cdotw(b+\epsilon;0)\, {\Bbb W}(\epsilon)}\int_\epsilon^\infty\int_b^{b+\epsilon}{\Bbb W}(b+\epsilon-y)W'(y)W(z+b)\cdot\\&& z{\rm d}yP(X_r\in{\rm d}z)\nonumber\\&&+\frac{W(b+\epsilon)}{r\cdot\, w(b+\epsilon;0)}\int_\epsilon^\infty\big(1-\deltaW(b+z-\epsilon)\big)\cdot z \, P(X_r\in{\rm d}z).\nonumber\end{eqnarray*}$

为了求出$P_0(k_r^U=\infty)$及$P_b(k_r^U=\infty)$的表达式,引入与$k_{r, \epsilon}^U$类似的停时$\tilde{k}_{r, \epsilon}^U.$对任意的$\epsilon\geq0, $有

$\tilde{k}_{r, \epsilon}^U=\inf\{t>r:t-g_t >r, U_{t-r}<-\epsilon\}, $

其中, $g_t=\sup\{0\leq s\leq t:U_{s}\geq0\}.$

对任意的$0<\epsilon'<\epsilon$,显然有, $\{\tilde{k}_{r, \epsilon}^U<\infty\}\subset\{\tilde{k}_{r, \epsilon'}^U<\infty\}$且$\cup_{\epsilon>0}\{\tilde{k}_{r, \epsilon}^U<\infty\}=\{k_{r}^U<\infty\}, $由谱负Lévy过程的空间时齐性,可得

$\lim\limits_{\epsilon \to 0}P_\epsilon(k_{r, \epsilon}^U=\infty)=\lim\limits_{\epsilon \to 0}P_0(\tilde{k}_{r, \epsilon}^U=\infty)=P_0(k_{r}^U=\infty), $

$\lim\limits_{\epsilon \to 0}P_{b+\epsilon}(k_{r, \epsilon}^U=\infty)=\lim\limits_{\epsilon \to 0}P_b(\tilde{k}_{r, \epsilon}^U=\infty)=P_b(k_{r}^U=\infty).$

所以只要在(3.41)及(3.42)式两边同时令$\epsilon \to 0$即得$P_0(k_r^U=\infty)$及$P_b(k_r^U=\infty)$的表达式.而

(3.43) $\begin{eqnarray}0\leq \lim\limits_{\epsilon \to0}\frac{1}{W(\epsilon)}\left(1-\int_\epsilon^\inftyW(z)\frac{z}{r}P(X_r\in{\rm d}z)\right)\leq\lim\limits_{\epsilon \to0}\int_0^\epsilon \frac{z}{r}P(X_r\in{\rm d}z)=0, \label{3.43}\end{eqnarray}$

且

(3.44) $\begin{eqnarray}\lim\limits_{\epsilon \to 0}\int_\epsilon^\infty\big(1-\deltaW(b+z-\epsilon)\big)\frac{z}{r}P(X_r\in{\rm d}z)=\\\int_0^\infty\big(1-\delta W(b+z)\big)\frac{z}{r}P(X_r\in{\rm d}z).\label{3.44}\end{eqnarray}$

对任意$z\in(\epsilon, \infty), y\in(b, b+\epsilon), $利用尺度函数的单调性及非负性,可得

$\begin{eqnarray*}0&\leq&\int_b^{b+\epsilon}\frac{{\Bbb W}(b+\epsilon-y)}{{\Bbb W}(\epsilon)}\bigg(W'(y)\frac{W(z+b)}{W(b+\epsilon)}-W'(y+z-\epsilon)\bigg){\rm d}y\nonumber\\&\leq&\int_b^{b+\epsilon}\left(W'(y)-W'(y+z-\epsilon)\right){\rm d}y=\\&&W(b+\epsilon)-W(b)-W(b+z)+W(b+z-\epsilon), \nonumber\end{eqnarray*}$

所以

(3.45) $\begin{eqnarray}0&\leq& \lim\limits_{\epsilon \to0}\frac{1}{{\Bbb W}(\epsilon)}\int_\epsilon^\infty\left(w(b+\epsilon;0)\frac{W(z+b)}{W(b+\epsilon)}-w(b+\epsilon;-z+\epsilon)\right)\frac{z}{r}P(X_r\in{\rm d}z)\nonumber\\&=&\frac{\delta}{{\Bbb W}(\epsilon)}\int_\epsilon^\infty\int_b^{b+\epsilon}{\Bbb W}(b+\epsilon-y)\\&&\bigg(W'(y)\frac{W(z+b)}{W(b+\epsilon)}-W'(y+z-\epsilon)\bigg){\rm d}y\frac{z}{r}\, P(X_r\in{\rm d}z)\nonumber\\&\leq&\lim\limits_{\epsilon \to 0}\delta \int_\epsilon^\infty\Big(W(b+\epsilon)-\\&&W(b)-W(b+z)+W(b+z-\epsilon)\Big)\frac{z}{r}P(X_r\in{\rm d}z)=0.\label{3.45}\end{eqnarray}$

在(3.41)式两边同时令$\epsilon \to 0, $同时借助(3.43)-(3.45)式,可得

$P_0(k_r^U=\infty)=\frac{(E[X_1-\delta])_+}{\int_0^\infty \big(1-\delta W(b+z)\big)\frac{z}{r}P(X_r\in{\rm d}z)}.$

对任意的$y\in(b, b+\epsilon)$,有

$0\leq\int_b^{b+\epsilon}\frac{{\Bbb W}(b+\epsilon-y)W'(y+z-\epsilon)}{{\Bbb W}(\epsilon)}{\rm d}y\leq W(b+z)-W(b+z-\epsilon).$

所以

(3.46) $\begin{equation}0\leq\lim\limits_{\epsilon \to0}\int_\epsilon^\infty\int_b^{b+\epsilon}\frac{{\Bbb W}(b+\epsilon-y)W'(y+z-\epsilon)}{{\Bbb W}(\epsilon)}{\rm d}y\frac{z}{r}P(X_r\in{\rm d}z)\leq0.\label{3.46}\end{equation}$

其次对任意的$y\in(b, b+\epsilon)$,有

(3.47) $\begin{equation}0\leq\lim\limits_{\epsilon \to 0}\frac{\int_b^{b+\epsilon}{\Bbb W}(b+\epsilon-y)W'(y){\rm d}y}{{\Bbb W}(\epsilon)}\leq\lim\limits_{\epsilon \to 0}W(b+\epsilon)-W(b)=0.\end{equation}$

另外

(3.48) $\begin{equation}\lim\limits_{\epsilon \to 0}\int_\epsilon^\infty\frac{W(b+\epsilon)(1-\delta W(b+z-\epsilon))}{w(b+\epsilon;0)}\frac{z}{r}P(X_r\in{\rm d}z)=\\\int_0^\infty \big(1-\delta W(b+z)\big)\frac{z}{r}P(X_r\in{\rm d}z).\end{equation}$

在(3.42)式两边同时令$\epsilon \to 0, $同时借助(3.46)-(3.48)式,可得

$P_b(k_r^U=\infty)=\frac{(E[X_1]-\delta)_+\int_0^\infty W(b+z)\frac{z}{r}P(X_r\in{\rm d}z)} {\int_0^\infty \big(1-\delta W(b+z)\big)\frac{z}{r}P(X_r\in {\rm d}z)}.$

$P_0(k_r^U=\infty)$及$P_b(k_r^U=\infty)$的表达式和有界变差的情形具有相同的表达形式.

最后,无论谱负Lévy过程的样本路径是有界变差(BV)还是无界变差(UBV),均可化简(3.16)式.由谱负Lévy过程的空间时齐性及(3.21)式可得

(3.49) $P_x(\nu_b^-=\infty)=P_{x-b}(\nu_0^-=\infty)=(E[X_1]-\delta)^+{\Bbb W}(x-b).$

将(3.3), (3.7), (3.27), (3.28)及(3.49)式代入(3.16)式中,便得到结论(3.1)式.

本节对两个特殊谱负Lévy风险过程,给出了Parisian破产概率的解析表达式.

假设$X$和$Y$都是带有指数索赔的Cramer-Lundberg风险过程,它们的数学表达式分别为

$X_t=ct-\sum\limits_{i=1}^{N_t}C_{i}~~~~~~Y_t=(c-\delta)t-\sum\limits_{i=1}^{N_t}C_{i}, $

其中$N=\{N_t, t\geq0\}$是强度为$\lambda$的泊松过程; $\{C_1, C_2, C_3, \cdots\}$是相互独立且同分布的随机变量,服从参数为$\alpha$的指数分布; $N$与$C_i$是相互独立的.在这种情况下$X$的拉普拉斯指数为

$\Psi(s)=cs+\lambda\left(\frac{\alpha}{s+\alpha}-1\right), ~~~~~~ s>-\alpha, $

安全负荷为正的条件用数学表达为

$E[Y_1]=c-\delta-\frac{\lambda}{\alpha}\geq0, $

过程$X$的尺度函数为

$W(x)=\frac{1}{c-\frac{\lambda}{\alpha}}\bigg(1-\frac{\lambda}{c\alpha}{\rm e}^{(\frac{\lambda}{c}-\alpha)x}\bigg), ~~ x\geq0, $

过程$Y$的尺度函数为

${\Bbb W}(x)=\frac{1}{c-\delta-\frac{\lambda}{\alpha}}\bigg(1-\frac{\lambda}{(c-\delta)\alpha}{\rm e}^{(\frac{\lambda}{c-\delta}-\alpha)x}\bigg), ~~x\geq0, $

则$U$的尺度函数为

$w(x;-z)=\frac{1}{c-\frac{\lambda}{\alpha}}\bigg(1-\frac{\lambda}{c\alpha}{\rm e}^{(\frac{\lambda}{c}-\alpha)(x+z)}\bigg)+{\rm e}^{(\frac{\lambda}{c}-\alpha)z}K(x, \delta, \alpha, \lambda, c, b), ~~x\in \mathbb{R}, $

其中

$\begin{eqnarray*}&&K(x, \delta, \alpha, \lambda, c, b)\\&&=\frac{\delta\lambda}{(c-\delta-\frac{\lambda}{\alpha})c}\bigg(\frac{1}{\lambda-c\alpha}\\&&({\rm e}^{(\frac{\lambda}{c}-\alpha)x}-{\rm e}^{(\frac{\lambda}{c}-\alpha)b})-\frac{1}{\delta\alpha}{\rm e}^{(\frac{\lambda}{c-\delta}-\alpha)x}({\rm e}^{\frac{-\lambda\delta b}{c(c-\delta)}}-{\rm e}^{\frac{-\lambda\deltax}{c(c-\delta)}})\bigg).\end{eqnarray*}$

Loeffen等(2013)^[3]给出了下式

$ P\bigg(\sum\limits_{i=1}^{N_r}C_{i}\in{\rm d}y\bigg)={\rm e}^{-\lambda r} \bigg(\delta_0({\rm d}y)+{\rm e}^{-\alpha y}\sum\limits_{m=0}^{\infty}\frac{(\alpha\lambda r)^{m+1}}{m!(m+1)!}y^m{\rm d}y\bigg), $

其中, $\delta_0({\rm d}y)$是在零点的狄拉克函数

$ \int_0^\infty zP(X_r\in{\rm d}z)=\\{\rm e}^{-\lambda r}\bigg(cr+\sum\limits_{m=0}^\infty\frac{(\lambda r)^{m+1}}{m!(m+1)!}\Big[cr\Gamma(m+1, cr\alpha)-\frac{1}{\alpha}\Gamma(m+2, cr\alpha)\Big]\bigg), $

其中, $\Gamma(a, x)=\int_0^x{\rm e}^{-t}t^{a-1}{\rm d}t$是下不完全伽马函数.另外

$\int_0^\infty {\rm e}^{(\frac{\lambda}{c}-\alpha)z}zP(X_r\in{\rm d}z)=\int_0^\infty zP(X_r\in{\rm d}z)-(c-\frac{\lambda}{\alpha})r.$

将上述式子代入(3.1)式,可得

(4.1) $\begin{eqnarray} P_x (k_r^U<\infty)&=&1-\left(c-\delta-\frac{\lambda}{\alpha}\right)\\&&\left[\frac{{\cal N}_1} {c-\lambda/\alpha-\alpha+\frac{\lambda}{c}{\rm e}^{(\lambda/c-\alpha)b}}+{\cal N}_1 \times{\cal N}_2^{-1}\frac{\lambda r}{c}{\rm e}^{(\lambda/c-\alpha)b+\lambda r}\right] \nonumber\\& &-\left(c-\delta-\frac{\lambda}{\alpha}\right)\\&& \left(\frac{\lambda}{c^2\alpha-c\lambda}{\rm e}^{(\lambda/c-\alpha)x}-K(x, \delta, \alpha, \lambda, c, b)\right) r{\rm e}^{\lambda r}\times{\cal N}_3^{-1}, \label{4.1}\end{eqnarray}$

其中

${\cal N}_1=1-\frac{\lambda}{c\alpha}{\rm e}^{(\frac{\lambda}{\alpha}-\alpha)x}+(c-\lambda/\alpha)K(x, \delta, \alpha, \lambda, c, b), $

$\begin{eqnarray*}{\cal N}_2&=&\Big(c-\lambda/\alpha-\alpha+\frac{\lambda}{c}{\rm e}^{(\frac{\lambda}{c}-\alpha)b}\Big)\bigg(\Big(1-\frac{\alpha}{c-\lambda/\alpha}+\frac{1}{c-\lambda/\alpha}\frac{\lambda}{c}{\rm e}^{(\lambda/c-\alpha)b}\Big) \\ &&\times\bigg(cr+\sum\limits_{m=0}^\infty\frac{(\lambda r)^{m+1}}{m!(m+1)!}\\&& \Big[cr\Gamma(m+1, cr\alpha)-\frac{1}{\alpha}\Gamma(m+2, cr\alpha)\Big]\bigg)-\frac{\lambda r}{c}{\rm e}^{(\lambda/c-\alpha)b+\lambda r}\bigg), \nonumber\end{eqnarray*}$

$\begin{eqnarray*} {\cal N}_3&=&\Big(c-\alpha-\lambda/\alpha+\frac{\lambda}{c}{\rm e}^{(\lambda/c-\alpha)b}\Big) \bigg(cr+\sum\limits_{m=0}^\infty\frac{(\lambda r)^{m+1}}{m!(m+1)!}\\ &&\times\Big[cr\Gamma(m+1, cr\alpha)-\frac{1}{\alpha}\Gamma(m+2, cr\alpha)\Big]\bigg)-r\Big(\lambda-\frac{\lambda^2}{c\alpha}\Big){\rm e}^{(\frac{\lambda}{c}-\alpha)b+\lambdar}.\end{eqnarray*}$

假设$X$和$Y$都是Brownian风险过程,则

$X_t=ct+\sigma B_t, ~~~~~~ Y_t=(c-\delta)t+\sigma B_t, $

其中$c, \sigma>0, B=\{B_t, t\geq0\}$为标准布朗运动.则$X$的拉普拉斯指数为

$\Psi(s)=cs+\frac{1}{2}\sigma^2s^2.$

过程$X, \ Y$的尺度函数分别为

$W(x)=\frac{1}{c}(1-{\rm e}^{-2\frac{c}{\sigma^2}x}), ~~~~~~x\geq0, $

${\Bbb W}(x)=\frac{1}{c-\delta}(1-{\rm e}^{-2\frac{c-\delta}{\sigma^2}x}), ~~~~~~x\geq0.$

并且有

$w(x;-z)=\frac{1}{c}(1-{\rm e}^{-2\frac{c}{\sigma^2}(x+z)})+{\rm e}^{-2\frac{c}{\sigma^2}z}M(x, \delta, \sigma, c, b), ~~~~~~ x\in \mathbb{R}, $

其中

$M(x, \delta, \sigma, c, b)=\frac{\delta}{c-\delta}\Big[\frac{{\rm e}^{-2\frac{c}{\sigma^2}b}-{\rm e}^{-2\frac{c}{\sigma^2}x}}{c}-\frac{1}{\delta}({\rm e}^{-2\frac{c-\delta}{\sigma^2}x-2\frac{\delta}{\sigma^2}b}-{\rm e}^{-2\frac{c}{\sigma^2}x})\Big].$

由Loeffen等(2013)^[3]可知

$\int_0^\infty {\rm e}^{-2\frac{c}{\sigma^2}z}zP(X_r\in{\rm d}z)=\int_0^\infty zP(X_r\in{\rm d}z)-cr, $

$ \int_0^\infty zP(X_r\in{\rm d}z)=\frac{\sigma\sqrt{r}}{\sqrt{2\pi}}{\rm e}^{-c^2r/(2\sigma^2)}+cr{\cal N}\Big(\frac{c\sqrt{r}}{\sigma}\Big), $

其中${\cal N}$是标准正态分布的分布函数.

将以上式子代入(3.1)式可得

(4.2) $\begin{eqnarray} P_x(k_r^U<\infty)&=&1-(c-\alpha)\frac{1-{\rm e}^{\frac{-2c}{\sigma^2}x}+cM(x, \delta, \sigma, c, b)}{c-\delta+\delta {\rm e}^{\frac{-2cb}{\sigma^2}}} \\& &-\frac{c-\delta}{c}\frac{\delta r{\rm e}^{-\frac{2cb}{\sigma^2}}(1-{\rm e}^{\frac{-2c}{\sigma^2}x}+cM(x, \delta, \sigma, c, b))} {(1-\delta/c+\delta/c {\rm e}^{\frac{-2cb}{\sigma^2}})^2 (\frac{\sigma\sqrt{r}}{\sqrt{2\pi}}{\rm e}^{-c^2r/(2\sigma^2)}+cr{\cal N}(\frac{c\sqrt{r}}{\sigma}))}\\&&-\frac{\frac{(c-\delta)^2r}{c}( {\rm e}^{\frac{-2cb}{\sigma^2}}-c M(x, \delta, \sigma, c, b) )} {(1-\delta/c+\delta/c {\rm e}^{\frac{-2cb}{\sigma^2}}) (\frac{\sigma\sqrt{r}}{\sqrt{2\pi}}{\rm e}^{-c^2r/(2\sigma^2)}+cr{\cal N}(\frac{c\sqrt{r}}{\sigma}))-\delta r{\rm e}^{\frac{-2cb}{\sigma^2}}}.\label{4.2}\end{eqnarray}$

注4.1  令$b=0$, (4.1)式与(4.2)式分别与Lkabous等(2017)^[8]中的例4.1以及例4.2中的$P_x (k_r^U<\infty)$表达式相同.