0 引言

1965年Montroll和Weiss将无规则游走的固定时间间隔和固定跳跃距离都假设成随机变量, 提出了连续时间随机游走模型 (CTRW), 它能够有效地描述奇异扩散现象.在CTRW模型中, 如果粒子的随机跳跃步长$Y_i (i=1, 2, \cdots)$和随机等待时间$J_i (i=1, 2, \cdots)$相互独立, 称模型是非耦合的, 否则就称为耦合的. n次跳跃后粒子的位置是$S(n)=\sum_{i=1}^n Y_i$, 粒子在$t>0$时刻的跳跃次数是

$ {N_t} = \max \left\{ {n \in \mathbb{N}:T\left( n \right) \leqslant t} \right\}, $

(0.1)

其中$T(n)\!=\!\sum_{i=1}^n J_i$, $S(N_t)$是t时刻粒子的位置.在合适的条件下, 时空随机变量$(S(n), T(n))$的标准化过程收敛到$\mathbb{R}\times\mathbb{R}_+$上的Lévy过程$\{(A(u), D(u))\}_{u\geqslant 0}$, 其中$D(u)$是不减的Lévy过程 (也称为从属过程), 并且$S(N_t)$收敛到时变过程$\{A(E(t)-)\}_{t\geqslant 0}$, 其中时间过程

$ E\left( t \right) = \inf \left\{ {u > 0:D\left( u \right) > t} \right\},t \ge 0 $

(0.2)

是从属过程$D(u)$的逆过程, 也称为逆从属过程^[1].注意当$A(u)$和$D(u)$不独立时Lévy过程$A(E(t)-)$和$A(E(t))$是不同的^[2].

近年来关于耦合CTRW模型及其控制方程的研究开始兴起^[3-5].例如, 在统计物理中耦合CTRW模型被用于模拟奇异扩散^[6-7], 在金融领域里耦合CTRW理论被用来描述当跳跃步长是对数收益和等待时间推迟时对数价格的波动^[8].但是, 这些文章中考虑的$D(u)$都是$\alpha$-稳定从属过程$D_\alpha(u) (0< \alpha<1)$, 其Laplace变换为

$ \mathbb{E}\left[ {{{\text{e}}^{ - s{D_\alpha }\left( u \right)}}} \right] = {{\text{e}}^{ - u{s^\alpha }}}, $

(0.3)

并且耦合CTRW模型中跳跃步长关于等待时间的条件概率密度函数通常是Gauss型函数^{[4, 6]}

$ \lambda \left( {x\left| t \right.} \right) = \frac{1}{{\sqrt {2\pi g\left( t \right)} }}\exp \left[ { - \frac{{{x^2}}}{{2g\left( t \right)}}} \right],\;\;\;\;g\left( t \right) > 0. $

(0.4)

事实上, 除了逆$\alpha$-稳定从属过程, 其他的逆从属过程也可以用来模拟时间过程^[8-10].所以本文第一部分将从一般的从属过程$D(u)$出发构造一类耦合CTRW模型的极限过程$A(E(t)-)$, 并得出它的概率密度函数的Fourier-Laplace变换形式; 第二部分引入一个算子$\Phi$得出$A(E(t)-)$的控制方程; 第三部分讨论了当$E(t)$为三种不同的逆从属过程时$A(E(t)-)$的控制方程的具体形式, 这三种从属过程分别为$\alpha$-稳定过程, 温和稳定从属过程和Gamma过程.另外, 还简单讨论了这三种时变Lévy过程的各阶矩的敛散情况.

1 过程$A(E(t)-)$的构造

令$\widehat{f}(k)={\mathcal{F}}(f)(k)=\int {_\mathbb{R}} \textrm{e}^{\textrm{i}kx}f(x)\textrm{d}x$表示函数$f:\mathbb{R}\to \mathbb{R}$的Fourier变换, $\widetilde{g}(s)=\mathcal {L}(g)(s)= \int_0^\infty \textrm{e}^{-st}g(t)\textrm{d}t$表示函数$g:\mathbb{R}_+\to \mathbb{R}$的Laplace变换, $\overline{h}(k, s)=\mathcal {F}\mathcal {L}(h)(k, s)=\int_0^\infty\int_\mathbb{R} \textrm{e}^{-st+\textrm{i}kx}\times h(x, t)\textrm{d}x\textrm{d}t$表示函数$h:\mathbb{R}\times \mathbb{R}_+\to\mathbb{R}$的Fourier-Laplace变换, 这里${\mathbb{R}}_{+}=[0, +\infty)$.

根据文献[4, 11], 我们首先介绍Lévy过程$(A(u), D(u))$的Lévy-Khintchine公式, 即$(A(u)$, $D(u))$的Fourier-Laplace变换是

$ \mathbb{E}\left[ {{{\text{e}}^{ - sD\left( u \right) + {\text{i}}kA\left( u \right)}}} \right] = {{\text{e}}^{ - u\psi \left( {k,s} \right)}},\;\;\;\;k \in \mathbb{R},\;\;\;\;s \geqslant 0, $

(1.1)

其中Fourier-Laplace指数$\psi(k, s)$是

$ \psi \left( {k,s} \right) = {\text{i}}ak + \frac{1}{2}{\sigma ^2}{k^2} + \int_{\mathbb{R} \times \mathbb{R} + \backslash \left\{ {\left( {0,0} \right)} \right\}} {\left( {1 - {{\text{e}}^{{\text{i}}kx}}{{\text{e}}^{ - st}} + \frac{{{\text{i}}kx}}{{1 + {x^2}}}} \right)} \phi \left( {{\text{d}}x,{\text{d}}t} \right), $

(1.2)

这里$a\in \mathbb{R}$, $\sigma^2\geqslant 0$, 并且$\phi(\textrm{d}x, \textrm{d}t)$是$\mathbb{R}\times\mathbb{R}_ +\setminus\{(0, 0)\}$上的Lévy测度.

用$\phi_A(\textrm{d}x)=\phi(\textrm{d}x, \mathbb{R}_+)$表示$A(u)$的Lévy测度, 当$s=0$时表达式 (1.1) 变成

$ \mathbb{E}\left[ {{{\text{e}}^{{\text{i}}kA\left( u \right)}}} \right] = {{\text{e}}^{ - u\psi A\left( k \right)}}, $

(1.3)

其中

$ \psi A\left( k \right) = {\text{i}}ak + \frac{1}{2}{\sigma ^2}{k^2} + \int_{\mathbb{R}\backslash \left\{ 0 \right\}} {\left( {1 - {{\text{e}}^{{\text{i}}kx}} + \frac{{{\text{i}}kx}}{{1 + {x^2}}}} \right)} \phi A\left( {{\text{d}}x} \right) $

(1.4)

是$A(u)$的Fourier指数.类似地, 用$\phi_D(\textrm{d}t)=\phi(\mathbb{R}, \textrm{d}t)$表示$D(u)$的Lévy测度, 令表达式 (1.1) 中的$k=0$, 有

$ \mathbb{E}\left[ {{{\text{e}}^{ - sD\left( u \right)}}} \right] = {{\text{e}}^{ - u{\psi _D}\left( s \right)}}, $

(1.5)

其中$\psi_D(s)=\int\nolimits_0^\infty(1-\textrm{e}^{-st})\phi_D(\textrm{d}t)$是$D(u)$的Laplace指数.本文中我们要求$ \phi_D(0, \infty)=\infty $, 这样$D(u)$就严格增, 从而它的逆过程$E(t)$是连续的.

对于给定的Lévy过程$(X(u), D(u))$, 其中$D(u)$是一般从属过程, $X(u)$与$D(u)$独立, 给定参数$c>0$, 取独立同分布的随机变量$(Y_i^{(c)}, J_i^{(c)})=(X(D(c^{-1})), D(c^{-1}))$, 有

$ \left( {{S^{\left( c \right)}}\left( {cu} \right),{T^{\left( c \right)}}\left( {cu} \right)} \right) = \left( {\sum\limits_{i = 1}^{\left[ {cu} \right]} {Y_i^{\left( c \right)}} ,\sum\limits_{i = 1}^{\left[ {cu} \right]} {J_i^{\left( c \right)}} } \right) \Rightarrow \left( {A\left( u \right),D\left( u \right)} \right),\;\;\;当c \to \infty . $

(1.6)

这里$\Rightarrow$表示所有有限维边缘分布的收敛, 并且$A(u)=X(D(u))$^{[4, 11]}.因为$X(u)$和$D(u)$独立, 所以对于每个$k\in \mathbb{R}$, 有

$ \begin{gathered} \mathbb{E}\left[ {{{\text{e}}^{{\text{i}}kA\left( u \right)}}} \right] = \int_\mathbb{R} {{{\text{e}}^{{\text{i}}kx}}{f_{A\left( u \right)}}\left( x \right){\text{d}}x} = \int_\mathbb{R} {{{\text{e}}^{{\text{i}}kx}}} \int_0^\infty {{f_{D\left( t \right)}}\left( u \right){f_{X\left( u \right)}}\left( x \right){\text{d}}u{\text{d}}x} \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\; = \int_0^\infty {{f_{D\left( t \right)}}\left( u \right){\text{d}}u} \int_\mathbb{R} {{{\text{e}}^{{\text{i}}kx}}{f_{X\left( u \right)}}\left( x \right){\text{d}}x} = \int_0^\infty {{f_{D\left( t \right)}}\left( u \right){{\text{e}}^{ - u{\psi _X}\left( k \right)}}{\text{d}}u} \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\; = {{\text{e}}^{ - {\psi _D}\left( {{\psi _X}\left( k \right)} \right)}}, \hfill \\ \end{gathered} $

(1.7)

其中$f_{A(u)}(x), f_{D(t)}(u)$和$f_{X(u)}(x)$分别表示$A(u)$, $D(t)$和$X(u)$的概率密度函数.令$D\!=\!D(1)$, $A=A(1)$, 有

$ \mathbb{E}\left[ {{{\text{e}}^{ - sD}}{{\text{e}}^{{\text{i}}kA}}} \right] = \mathbb{E}\left[ {\mathbb{E}\left[ {{{\text{e}}^{ - sD}}{{\text{e}}^{{\text{i}}kX\left( D \right)}}\left| {D = t} \right.} \right]} \right] = \mathbb{E}\left[ {{{\text{e}}^{ - sD}}{{\text{e}}^{ - D{\psi _X}\left( k \right)}}} \right] = {{\text{e}}^{ - {\psi _D}\left( {s + {\psi _X}\left( k \right)} \right)}}, $

(1.8)

所以$(A(u), D(u))$的Fourier-Laplace指数是$\psi(k, s)=\psi_D(s+\psi_X(k))$.

根据文献[11]中的推论3.8和条件$\int\nolimits_0^1y|\ln y|\phi_D(\textrm{d}y)<\infty$, 我们得到$A(E(t)-)$的概率密度函数的Fourier-Laplace变换是

$ {\cal F}{\cal L}\left( h \right)\left( {k,s} \right) = \frac{1}{s} \cdot \frac{{{\psi _D}\left( s \right)}}{{{\psi _D}\left( {s + {\psi _X}\left( k \right)} \right)}}. $

(1.9)

条件$\int\nolimits_0^1y|\ln y|\phi_D(\textrm{d}y)<\infty$仅仅是^[11]中得到推论3.8的一个很弱的要求, 我们假设本文中涉及到的所有从属过程都满足此条件, 并且可以验证后文的例子均满足.

2 过程$A(E(t)-)$的控制方程

这部分我们将研究当$X(u)$为对称稳定过程时$A(E(t)-)$的控制方程.我们知道具有稳定分布的随机变量$Z$的Fourier变换具有以下形式

$ \mathbb{E}\left[ {{{\text{e}}^{{\text{i}}kZ}}} \right] = \exp \left( {{\text{i}}ka - b{{\left| k \right|}^\alpha }\left( {1 + {\text{i}}\beta {\text{sgn}}\left( k \right){\omega _\alpha }\left( k \right)} \right)} \right), $

(2.1)

其中参数$a\in\mathbb{R}$, $b\in[0, \infty)$, $\alpha\in(0, 2]$, [-1, 1].函数$\omega_\alpha(k)$的定义是

$ {\omega _\alpha }\left( k \right) = \left\{ \begin{array}{l} \tan \left( {\pi \alpha /2} \right),\;\;\;\;\;如果\;\;\alpha \ne 1,\\ \frac{2}{\pi }\ln \left| k \right|,\;\;\;如果\;\;\;\alpha = 1. \end{array} \right. $

在 (2.1) 中取参数$a=\beta=0$, $\alpha=b=1$, 选择这个特殊分布作为$X(1)$的分布.事实上此时Lévy过程$X(u)$具有Cauchy分布, 其概率密度函数为

$ {f_{X\left( u \right)}}\left( x \right) = \frac{u}{{\pi \left( {{u^2} + {x^2}} \right)}},\;\;\;\;u > 0,x \in \mathbb{R}, $

(2.2)

从而$X(u)$的Fourier指数为$\psi_X(k)=|k|$.

对任意的$\lambda>0$, 令$L_\lambda^1(\mathbb{R}\times\mathbb{R}_+)$表示$\mathbb{R}\times\mathbb{R}_+$上实值可测并且其模

$ {\left\| f \right\|_\lambda } = \int_0^\infty {\int_\mathbb{R} {{{\text{e}}^{ - \lambda t}}\left| {f\left( {x,t} \right)} \right|{\text{d}}x{\text{d}}t} } $

存在的全体函数. $L_\lambda^1(\mathbb{R}\times\mathbb{R}_+)$是个包含$L^1(\mathbb{R}\times\mathbb{R}_+)$的Banach空间.

定理2.1 令$\Phi$是定义在$L_\lambda^1(\mathbb{R}\times\mathbb{R}_+)$上的算子

$ \Phi \left( f \right)\left( {x,t} \right) = \int_{\mathbb{R} \times \mathbb{R} + \backslash \left\{ {\left( {0,0} \right)} \right\}} {\left( {h\left( {x,t} \right) - H\left( {t - \xi } \right)f\left( {x - \frac{\xi }{t}y,t - \xi } \right)} \right)} \frac{t}{{\pi \left( {{t^2} + {y^2}} \right)}}{\text{d}}y{\phi _D}\left( {{\text{d}}\xi } \right), $

(2.3)

其中$H(t)=I (t\geq 0)$是Heaviside阶跃函数, 则$A(E(t)-)$的概率密度函数$h(x, t)$满足

$ {\cal F}{\cal L}\left( {\Phi \left( h \right)} \right)\left( {k,s} \right) = \frac{{{\psi _D}\left( s \right)}}{s}. $

(2.4)

证明 当$t<0$, $h(x, t)=0$时, 有

$ \begin{array}{l} {\cal F}{\cal L}\left( {\Phi \left( h \right)} \right)\left( {k,s} \right)\\ = \int_0^\infty {{\rm{d}}t} \int_{ - \infty }^\infty {{\rm{d}}x} \int_0^\infty \\{{\phi _D}\left( {{\rm{d}}\xi } \right)} \int_{ - \infty }^\infty {\left( {{{\rm{e}}^{ - st}}{{\rm{e}}^{{\rm{i}}kx}}\left( {f\left( {x,t} \right) - H\left( {t - \xi } \right)f\left( {x - \frac{\xi }{t}y,t - \xi } \right)} \right)\frac{t}{{\pi \left( {{t^2} + {y^2}} \right)}}} \right)dy} \\ = \int_0^\infty {{\phi _D}\left( {{\rm{d}}\xi } \right){\cal F}{\cal L}\left( f \right)\left( {k,s} \right)} \\ \;\;\; - \int_0^\infty {{\rm{d}}t'} \int_{ - \infty }^\infty {{\rm{d}}x'} \int_0^\infty {{\phi _D}\left( {{\rm{d}}\xi } \right)} \int_{ - \infty }^\infty {{{\rm{e}}^{ - s\left( {t' + \xi } \right)}}{{\rm{e}}^{{\rm{i}}k\left( {x' + \frac{\xi }{{t' + \xi }}y} \right)}}f\left( {x',t'} \right)\frac{{t' + \xi }}{{\pi \left( {{{\left( {t' + \xi } \right)}^2} + {y^2}} \right)}}{\rm{d}}y} \\ = \int_0^\infty {{\phi _D}\left( {{\rm{d}}\xi } \right){\cal F}{\cal L}\left( f \right)\left( {k,s} \right)} - \int_0^\infty {{\rm{d}}t'} \int_0^\infty {{\phi _D}\left( {{\rm{d}}\xi } \right)\hat f\left( {k,t'} \right){{\rm{e}}^{ - s\left( {t' + \xi } \right)}}} \int_{ - \infty }^\infty \\{{{\rm{e}}^{{\rm{i}}k\frac{\xi }{{\xi + t'}}y}}\frac{{t' + \xi }}{{\pi \left( {{{\left( {t' + \xi } \right)}^2} + {y^2}} \right)}}{\rm{d}}y} \\ = \int_0^\infty {{\phi _D}\left( {{\rm{d}}\xi } \right){\cal F}{\cal L}\left( f \right)\left( {k,s} \right)} - \int_0^\infty {{\phi _D}\left( {{\rm{d}}\xi } \right)} \int_0^\infty {\hat f\left( {k,t'} \right){{\rm{e}}^{ - st'}}{{\rm{e}}^{ - s\xi }}{{\rm{e}}^{ - \xi \left| k \right|}}{\rm{d}}t'} \\ = \int_0^\infty {{\phi _D}\left( {{\rm{d}}\xi } \right){\cal F}{\cal L}\left( f \right)\left( {k,s} \right)} - \int_0^\infty {\bar f\left( {k,s} \right){{\rm{e}}^{ - \left( {s + \left| k \right|} \right)\xi }}{\phi _D}\left( {{\rm{d}}\xi } \right)} \\ = \int_0^\infty {\left( {{\rm{1}} - {{\rm{e}}^{ - \left( {s + \left| k \right|} \right)\xi }}} \right){\phi _D}\left( {{\rm{d}}\xi } \right){\cal F}{\cal L}\left( f \right)\left( {k,s} \right)} \\ = {\psi _D}\left( {s + \left| k \right|} \right){\cal F}{\cal L}\left( f \right)\left( {k,s} \right). \end{array} $

(2.5)

注意到在 (1.9) 中, 有$\psi_{X}(k)=|k|$, 故

$ {\cal F}{\cal L}\left( h \right)\left( {k,s} \right) = \frac{1}{s} \cdot \frac{{{\psi _D}\left( s \right)}}{{{\psi _D}\left( {s + \left| k \right|} \right)}}. $

(2.6)

因此

$ {\cal F}{\cal L}\left( {\Phi \left( h \right)} \right)\left( {k,s} \right) = \frac{{{\psi _D}\left( s \right)}}{s}. $

(2.7)

3 例子

我们将具体分析当$D(u)$分别为$\alpha$-稳定、温和稳定和Gamma过程时时变过程 $A(E(t)-)$的控制方程, 并且考察这些过程的各阶矩的情况, 希望这些矩的敛散性能在模拟奇异扩散中有所应用.

例3.1 $\alpha$-稳定从属过程$D_\alpha(u)$的Laplace指数是$\psi_{D}(s)=s^\alpha$.因为$s^\alpha=\int\nolimits_0^\infty (1-\textrm{e}^{-st})\times \frac{\alpha}{\Gamma(1-\alpha)}t^{-1-\alpha}\textrm{d}t$, 所以$D_\alpha(u)$的Lévy测度是$\phi_D(\textrm{d}t)=\frac{\alpha}{\Gamma(1-\alpha)}t^{-\alpha-1}\textrm{d}t$.由定理2.1的 (2.4) 和

$ {\cal L}_{t \to s}^{ - 1}\left\{ {\frac{{{\psi _D}\left( s \right)}}{s}} \right\}\left( t \right) = {\cal L}_{t \to s}^{ - 1}\left\{ {{s^{\alpha - 1}}} \right\}\left( t \right) = \frac{{{t^{ - \alpha }}}}{{\Gamma \left( {1 - \alpha } \right)}}, $

(3.1)

此时$A(E(t)-)$的控制方程为

$ \begin{gathered} \int_{\mathbb{R} \times \mathbb{R} + \backslash \left\{ {\left( {0,0} \right)} \right\}} {\left( {h\left( {x,t} \right) - H\left( {t - \xi } \right)h\left( {x - \frac{\xi }{t}y,t - \xi } \right)} \right)} \frac{t}{{\pi \left( {{t^2} + {y^2}} \right)}}\\{\text{d}}y\frac{\alpha }{{\Gamma \left( {1 - \alpha } \right)}}{\xi ^{ - \alpha - 1}}{\text{d}}\xi \hfill = \delta \left( x \right)\frac{{{t^{ - \alpha }}}}{{\Gamma \left( {1 - \alpha } \right)}}. \hfill \\ \end{gathered} $

(3.2)

并且

$ \int_0^1 {y\left| {\ln y} \right|{\phi _D}\left( {{\rm{d}}y} \right)} = - \int_0^1 {y\ln y\frac{\alpha }{{\Gamma \left( {1 - \alpha } \right)}}{y^{ - \alpha - 1}}{\rm{d}}y} = \frac{\alpha }{{\Gamma \left( {1 - \alpha } \right)}}\int_0^\infty {{\rm{t}}{{\rm{e}}^{ - \left( {1 - \alpha } \right)t}}{\rm{d}}t < \infty } , $

(3.3)

由 (1.9), 有

$ {\cal F}{\cal L}\left( h \right)\left( {k,s} \right) = \frac{{{s^{\alpha - 1}}}}{{{{\left( {s + \left| k \right|} \right)}^\alpha }}}. $

(3.4)

对 (3.4) 做逆Laplace变换

$ {\cal F}\left( h \right)\left( {k,t} \right) = \frac{1}{{\Gamma \left( \alpha \right)\Gamma \left( {1 - \alpha } \right)}}\int_0^t {{{\rm{e}}^{ - \left| k \right|u}}{u^{\alpha - 1}}{{\left( {t - u} \right)}^{ - \alpha }}{\rm{d}}u} , $

(3.5)

这里用到了公式$\mathcal {L}[\textrm{e}^{-ct}g(t)](s)=\mathcal {L}(g)(s+c)$和$\mathcal {L}(f\ast g)(s)=\widetilde{f}(s)\widetilde{g}(s)$, 再对其取逆Fourier变换

$ h\left( {x,t} \right) = \frac{1}{{x\Gamma \left( \alpha \right)\Gamma \left( {1 - \alpha } \right)}}\int_0^t {\frac{{{u^\alpha }{{\left( {t - u} \right)}^{ - \alpha }}}}{{{u^2} + {x^2}}}{\rm{d}}u} . $

(3.6)

因此

$ \begin{gathered} \mathbb{E}\left[ {{{\left| {A\left( {E\left( t \right) - } \right)} \right|}^\mu }} \right] = \int_\mathbb{R} {{{\left| x \right|}^\mu }h\left( {x,t} \right){\text{d}}x} \\= \frac{1}{{\pi \Gamma \left( \alpha \right)\Gamma \left( {1 - \alpha } \right)}}\int_\mathbb{R} {{{\left| x \right|}^\mu }\left( {\int_\mathbb{R} {\frac{{{u^\alpha }{{\left( {t - u} \right)}^{ - \alpha }}}}{{{u^2} + {x^2}}}} } \right){\text{d}}x} \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \frac{1}{{\pi \Gamma \left( \alpha \right)\Gamma \left( {1 - \alpha } \right)}}\int_0^t {{u^\alpha }{{\left( {t - u} \right)}^{ - \alpha }}} \int_\mathbb{R} {\frac{{{{\left| x \right|}^\mu }}}{{{u^2} + {x^2}}}} {\text{d}}x{\text{d}}u \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \frac{1}{{\pi \Gamma \left( \alpha \right)\Gamma \left( {1 - \alpha } \right)}}\int_0^t {{u^{\mu + \alpha - 1}}{{\left( {t - u} \right)}^{ - \alpha }}{\text{d}}u} \int_\mathbb{R} {\frac{{{{\left| x \right|}^\mu }}}{{1 + {x^2}}}} {\text{d}}x. \hfill \\ \end{gathered} $

(3.7)

容易看到$\int\nolimits_\mathbb{R} \frac{|x|^\mu}{1+x^2}\textrm{d}x$当$\mu<1$时收敛, 当$\mu\geqslant1$时发散.所以当$\mu<1$时$\mathbb{E}[|A(E(t)-)|^\mu]\sim t^\mu $, 当$\mu\geqslant 1$时过程$A(E(t)-)$的$\mu$阶矩无穷大.

例3.2 温和稳定从属过程$D_{\alpha, \lambda}(u) (0<\alpha<1, \lambda >0)$的Laplace变换为

$ \mathbb{E}\left[ {{{\text{e}}^{ - sD\left( u \right)}}} \right] = {{\text{e}}^{ - u\left( {{{\left( {s + \lambda } \right)}^\alpha } - {\lambda ^\alpha }} \right)}}. $

(3.8)

因为

$ {\left( {s + \lambda } \right)^\alpha } - {\lambda ^\alpha } = \int_0^\infty {\left( {1 - {{\rm{e}}^{ - st}}} \right)} \frac{1}{{\Gamma \left( {1 - \alpha } \right)}}{t^{ - 1 - \alpha }}{{\rm{e}}^{ - \lambda t}}{\rm{d}}t, $

(3.9)

所以其Lévy测度是$\phi_D(\textrm{d}t)=\frac{\alpha}{\Gamma(1-\alpha)}t^{-\alpha-1}\textrm{e}^{-\lambda t}\textrm{d}t$.此时控制方程 (2.4) 为

$ \begin{gathered} \int_{\mathbb{R} \times \mathbb{R} + \backslash \left\{ {\left( {0,0} \right)} \right\}}\\{\left( {h\left( {x,t} \right) - H\left( {t - \xi } \right)h\left( {x - \frac{\xi }{t}y,t - \xi } \right)} \right)\frac{t}{{\pi \left( {{t^2} + {y^2}} \right)}}{\text{d}}y} \frac{\alpha }{{\Gamma \left( {1 - \alpha } \right)}}{\xi ^{ - \alpha - 1}}{{\text{e}}^{ - \lambda \xi }}{\text{d}}\xi \hfill \\ = \delta \left( x \right)M\left( t \right). \hfill \\ \end{gathered} $

(3.10)

这里, 由文献[12], 有

$ \begin{array}{l} M\left( t \right) = {\cal L}_{t \to s}^{ - 1}\left\{ {\frac{{{{\left( {s + \lambda } \right)}^\alpha } - {\lambda ^\alpha }}}{s}} \right\}\left( t \right)\\ \;\;\;\;\;\;\;\;\;{\rm{ = }}\frac{\alpha }{{\Gamma \left( {1 - \alpha } \right)}}{t^{ - \alpha }}{{\rm{e}}^{ - \lambda t}}\left( {1 + \sum\limits_{n = 1}^\infty {\frac{{{{\left( {\lambda t} \right)}^n}}}{{\left( {1 - \alpha } \right)\left( {2 - \alpha } \right) \cdots \left( {n - \alpha } \right)}}} } \right) - {\lambda ^\alpha }. \end{array} $

(3.11)

并且

$ \int_0^1 {y\left| {\ln y} \right|{\phi _D}\left( {{\rm{d}}y} \right)} < - \int_0^1 {y\ln y\frac{\alpha }{{\Gamma \left( {1 - \alpha } \right)}}{y^{ - \alpha - 1}}{\rm{d}}y} = \frac{\alpha }{{\Gamma \left( {1 - \alpha } \right)}}\int_0^\infty {t{{\rm{e}}^{ - \left( {1 - \alpha } \right)t}}{\rm{d}}t} < \infty , $

(3.12)

从而

$ {\cal F}{\cal L}\left( h \right)\left( {k,s} \right) = \frac{{{{\left( {s + \lambda } \right)}^\alpha } - {\lambda ^\alpha }}}{s} \cdot \frac{1}{{{{\left( {s + \lambda + \left| k \right|} \right)}^\alpha } - {\lambda ^\alpha }}}. $

(3.13)

并且根据事实^[13]

$ {\cal L}_{t \to s}^{ - 1}\left\{ {\frac{{{{\left( {s + \lambda } \right)}^\alpha } - {\lambda ^\alpha }}}{s}} \right\}\left( t \right) = {{\rm{e}}^{ - \lambda t}}{t^{\alpha - 1}}{E_{\alpha ,\alpha }}\left( {{{\left( {\lambda t} \right)}^\alpha }} \right), $

(3.14)

其中

$ {E_{\alpha ,\beta }}\left( z \right) = \sum\limits_{k = 0}^\infty {\frac{{{z^k}}}{{\Gamma \left( {\alpha k + \beta } \right)}}} $

(3.15)

是一般Mittag-Leffler函数^[14], 仿照例3.1中关于矩的讨论可得这里的$A(E(t)-)$的任意阶矩的情况与上个例子的结论类似.

例3.3  设$D_{a, b}(u) (a>0, b>0)$是Gamma从属过程, 其概率密度函数是

$ {f_{D\left( u \right)}}\left( x \right) = \frac{{{b^{au}}}}{{\Gamma \left( {au} \right)}}{x^{au - 1}}{{\rm{e}}^{ - bx}}{I_{x > 0}}. $

(3.16)

通过简单计算可得当$s\geqslant 0$时,

$ \int_0^\infty {{{\rm{e}}^{ - sx}}{f_{D\left( u \right)}}\left( x \right){\rm{d}}x} = {\left( {1 + \frac{s}{b}} \right)^{ - au}} = \exp \left[ { - ua\log \left( {1 + \frac{s}{b}} \right)} \right], $

(3.17)

这说明$D_{a, b}(u)$的Laplace指数是$\psi_{D}(s)=a\log(1+\frac{s}{b}).$注意到

$ \begin{array}{l} \int_0^\infty {\left( {1 - {{\rm{e}}^{ - st}}} \right)a{t^{ - 1}}{{\rm{e}}^{ - bt}}{\rm{d}}t} = a\int_0^\infty {\frac{{{{\rm{e}}^{ - bt}} - {{\rm{e}}^{ - \left( {s + b} \right)t}}}}{t}{\rm{d}}t} = a\int_0^\infty {{\rm{d}}t} \int_b^{s + b} {{{\rm{e}}^{ - ty}}{\rm{d}}y} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = a\int_b^{s + b} {{\rm{d}}y} \int_0^\infty {{{\rm{e}}^{ - ty}}{\rm{d}}t} = a\int_b^{s + b} {\frac{1}{y}{\rm{d}}y} = a\log \left( {1 + \frac{s}{b}} \right), \end{array} $

(3.18)

$D_{a, b}(u)$的Lévy测度是$\phi_D(\textrm{d}t)=at^{-1}\textrm{e}^{-bt}\textrm{d}t$.因为

$ {\cal L}_{t \to s}^{ - 1}\left\{ {\frac{{a\log \left( {1 + \frac{s}{b}} \right)}}{s}} \right\}\left( t \right) = a\int_{bt}^\infty {\frac{{{{\rm{e}}^{ - x}}}}{x}{\rm{d}}x} , $

(3.19)

所以控制方程 (2.4) 变成

$ \begin{gathered} \int_{\mathbb{R} \times \mathbb{R} + \backslash \left\{ {\left( {0,0} \right)} \right\}} {\left( {h\left( {x,t} \right) - H\left( {t - \xi } \right)h\left( {x - \frac{\xi }{t}y,t - \xi } \right)} \right)\frac{t}{{\pi \left( {{t^2} + {y^2}} \right)}}{\text{d}}y} a{\xi ^{ - 1}}{{\text{e}}^{ - b\xi }}{\text{d}}\xi \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\; = \delta \left( x \right)a\int_{bt}^\infty {\frac{{{{\text{e}}^{ - x}}}}{x}{\text{d}}x} . \hfill \\ \end{gathered} $

(3.20)

概率密度函数$h(x, t)$的Fourier-Laplace变换是

$ {\cal F}{\cal L}\left( h \right)\left( {k,s} \right) = \frac{{a\log \left( {1 + \frac{s}{b}} \right)}}{s} \cdot \frac{1}{{a\log \left( {1 + \frac{{s + \left| k \right|}}{b}} \right)}}. $

(3.21)

关于矩的讨论和结论类似于例3.1.