由α-稳定过程驱动的线性自排斥扩散过程的渐近行为和参数估计

Asymptotic Behavior and Parameter Estimation of the Linear Self-Repelling Diffusion Driven by α-Stable Motion

摘要: 设为一维-稳定模型且。本文主要研究如下线性自排斥扩散的长时间行为和参数估计：，其中、是两个未知参数且。当且t趋于无穷大时，对任意，我们有和几乎处处成立，其中。在连续观测条件下，建立和的最小二乘估计讨论其相合性与渐近分布。

Abstract: Let be an -stable motion of one-dimension with . In this paper, we consider large time behaviors and parameter estimation of the linear self-repelling diffusion of the forms where and are two unknown parameters. When and t tends to infinity, we show that the convergence and hold almost surely for all , where . The least squares estimates of and are established to discuss their coincidence and asymptotic distributions under continuous observation conditions.

1. 引言

在1995年，Cranston和Le Jan [1] 引入一个与路径相关的随机微分方程的特例(线性自吸引扩散)

$X_t=B_t-\theta {\displaystyle {\int }_0^t{\displaystyle {\int }_0^s\left(X_s-X_u\right)\text{d}u\text{d}s+\nu t,t\ge 0}},$

其中 $\theta >0$ 且B是一维标准布朗运动。他们证明了当t趋于无穷大时， $X_t$ 过程在 $L^2$ 上存在且几乎必然收敛。这种依赖于路径的随机微分方程是由Durrett和Rogers [2] 在1992年首次提出的，是描述了一种增长聚合物的模型，具体如下：

$X_t=X_0+B_t+{\displaystyle {\int }_0^t\text{d}s{\displaystyle {\int }_0^{s}f\left(X_s-X_u\right)\text{d}u}},$ (1)

其中 $B=\left\{B_t,t\ge 0\right\}$ 是d维标准布朗运动且f是Lipschitz连续的。 $X_t$ 对应于聚合物末端在时间t时的位置。在某些条件下，他们又建立了随机微分方程的渐近形态，并提出一些猜想和问题。该模型是自交互随机行走概念的连续模型。若 $f\left(x\right)=g\left(x\right)x/\Vert x\Vert$ ， $g\left(x\right)\ge 0$ ，则此时 $X_t$ 就是由Pemantle [3] 所提出的过程的一个连续类似物。我们可以将这种与过去的轨迹相互作用的布朗运动称为自交互运动。一般地，方程(1)在不对f做任何假设时可定义自交互扩散过程。如果对所有 $x\in R^d$ 均有 $x\cdot f\left(x\right)\ge 0\left(resp.\le 0\right)$ ，则称这个解为自吸引(或自排斥)。2002年，Benaïm [4] 等人还引入了一种依赖(卷积)经验测度的自交互扩散。这些扩散和布朗聚合物之间一个很大的区别是漂移项除了t。重要的是要注意，在f的许多情况下，相互作用势具有足够的吸引力，使得自交互扩散过程可近似等同于O-U过程，这使得其遍历行为可能存在并值得进行研究。更多的可参考Cranston和Mountford [5] ，Gauthier [6] ，Hermann和Scheutzow [7] ，Mountford和Tarr [8] ，Sun和Yan [9] 及文内其他文献。

另一方面，方程(1)可以重写为

$X_t=B_t-\theta {\displaystyle {\int }_0^t{\displaystyle {\int }_0^{s}u\text{d}{X}_u}}\text{d}s+\nu t,t\ge 0.$

受这一结论的启发，在这篇文章我们可以考虑以下方程的长时间行为和参数估计：

$X_t^{\alpha }=M_t^{\alpha }+\theta {\displaystyle {\int }_0^t{\displaystyle {\int }_0^s{\left(1+u\right)}^2\text{d}{X}_u^{\alpha }}}\text{d}s+\nu t,t\ge 0,$ (2)

其中 $M_t^{\alpha }$ 是一维 $\alpha$ 稳定运动且 $0<\alpha <2$ ， $\theta$ 和 $\nu$ 是两个实参数且 $\theta >0$ 。这篇文章的组织结构如下。

在第二节，我们简要回顾了 $\alpha$ 稳定运动相关的定义、性质以及公式命题。第三节研究了解的长时间行为。第四节研究了参数的最小二乘估计及参数估计量的渐近分布。

2. 预备知识

在这一节中，我们简要地介绍了 $\alpha$ -稳定运动的一些定义和性质。在本文，我们假设 $0<\alpha <2$ 是任意且固定的，并设 $M^{\alpha }=\left\{M_t^{\alpha },t\ge 0\right\}$ 是一个 $\left\{{\mathcal{F}}_t\right\}$ -适定过程。令参数 $\alpha ,\beta ,\lambda ,\mu$ 满足 $\alpha \in \left(0,2\right],\lambda \in \left(0,+\infty \right),\beta \in \left[-1,1\right],\mu \in \left(-\infty ,+\infty \right),$ 且定义 ${\varphi }_{\alpha }\left(u\right)=\{\begin{array}{l}-{\lambda }^{\alpha }{\left|u\right|}^{\alpha }\left(1-i\beta \mathrm{sgn}\left(u\right)\tan \frac{\alpha \pi }{2}\right)+i\mu u,\alpha \ne 1,\\ -\lambda \left|u\right|\left(1+i\beta \frac{2}{\pi }\mathrm{sgn}\left(u\right)\log \left|u\right|\right)+i\mu u,\alpha =1\end{array}$ 其中 $u\in \left(-\infty ,+\infty \right)$ 且 $i^2=-1$ 。如果随机变量 $\eta$ 具有特征函数 $E{\text{e}}^{iu\eta }={\text{e}}^{-{\varphi }_{\alpha }\left(u\right)}$ ，则称其具有 $\alpha$ -稳定分布，表示为 $\eta ~S_{\alpha }\left(\lambda ,\beta ,\mu \right)$ 。当 $\mu =0$ 和 $\beta =0$ 时，我们称 $\eta$ 是对称的 $\alpha$ -稳定过程。

进一步地，若对任意 $t>s>0$ 有 $E\left[{\text{e}}^{iu\left(M_t^{\alpha }-M_s^{\alpha }\right)}|{\mathcal{F}}_s\right]={\text{e}}^{-\left(t-s\right){\varphi }_{\alpha }\left(u\right)},$ 则称 $M^{\alpha }$ 是一个 $\alpha$ -稳定过程。在本文中，我们假定 $M^{\alpha }$ 是对称的 $\alpha$ -稳定过程，并且 $\lambda =1$ ， $0<\alpha <2$ ，即 ${\varphi }_{\alpha }\left(u\right)={\left|u\right|}^{\alpha }$ 。

命题2.1：方程(2)有唯一的解，且可表示为： $X_t^{\alpha }={\displaystyle {\int }_0^t{h}_{\theta }\left(t,s\right)\text{d}{M}_s^{\alpha }}+\nu {\displaystyle {\int }_0^t{h}_{\theta }\left(t,s\right)\text{d}s},t\ge \text{0,}$ 其中 $h_{\theta }\left(t,s\right)=\left(1+\theta {\left(1+s\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+s\right)}^3-1\right)}{\displaystyle {\int }_s^t{\text{e}}^{3^{-1}\theta \left({\left(1+u\right)}^3-1\right)}\text{d}u}\right),t\ge s\ge 0$ 。

证明：我们可以用常数变易法证明这个命题。

3. 长时间行为

设 $X_t^{\alpha }$ 是方程(2)的解且 $\theta >0$ 。这节我们主要讨论解的长时间行为。

引理3.1：设 $0<\alpha <2$ 且 $\theta >0$ ，定义过程 ${\xi }_t^{\alpha }:={\displaystyle {\int }_0^t{\left(1+s\right)}^2}{\text{e}}^{-3^{-1}\theta \left({\left(1+s\right)}^3-1\right)}\text{d}{M}_s^{\alpha },t\ge 0$ 。则随机变量 ${\xi }_{\infty }^{\alpha }$ 在 $L^p$ 上适定，且

${\left(1+t\right)}^{3n}\phi \left(t\right)\left({\xi }_{\infty }^{\alpha }-{\xi }_t^{\alpha }\right)\to 0$ (3)

在 $L^p$ 上适定且几乎必然收敛，其中 $\phi \left(t\right)={\left(1+t\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}{\displaystyle {\int }_0^t{\text{e}}^{3^{-1}\theta \left({\left(1+u\right)}^3-1\right)}}\text{d}u$ 。

证明：明显地，我们有

$E{\left|{\xi }_t^{\alpha }\right|}^p\le C_{p,\alpha }{\left({\displaystyle {\int }_0^t{\left(1+s\right)}^{2\alpha }{\text{e}}^{-\frac{\alpha \theta }{3}\left({\left(1+s\right)}^3-1\right)}\text{d}s}\right)}^{p/\alpha }\le C_{p,\alpha }{\left({\displaystyle {\int }_0^{\infty }{\left(1+s\right)}^{2\alpha }{\text{e}}^{-\frac{\alpha \theta }{3}\left({\left(1+s\right)}^3-1\right)}\text{d}s}\right)}^{p/\alpha }<\infty ,$

即随机变量 ${\xi }_{\infty }^{\alpha }$ 在 $L^p$ 上适定。相似地，我们有

$E{\left|{\left(1+t\right)}^{3n}\left({\xi }_{\infty }^{\alpha }-{\xi }_t^{\alpha }\right)\right|}^p\le C_{p,\alpha }{\left({\left(1+t\right)}^{3n\alpha }{\displaystyle {\int }_t^{\infty }{\left(1+s\right)}^{2\alpha }{\text{e}}^{-\frac{\alpha \theta }{3}\left({\left(1+s\right)}^3-1\right)}\text{d}s}\right)}^{p/\alpha }\to 0\left(t\to \infty \right).$

结合当t趋于无穷大时， $\phi \left(t\right)\to 1/\theta$ ，则这个收敛在 $L^p$ 上适定。

再考虑(3)以概率1收敛。运用分部积分和洛必达法则可得

$\begin{array}{l}{\left(1+t\right)}^{3n}\left({\xi }_{\infty }^{\alpha }-{\xi }_t^{\alpha }\right)={\left(1+t\right)}^{3n}{\displaystyle {\int }_t^{\infty }{\left(1+s\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+s\right)}^3-1\right)}}\text{d}{M}_s^{\alpha }\\ =-{\left(1+t\right)}^{3n+2}{\text{e}}^{-3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}{M}_t^{\alpha }-{\left(1+t\right)}^{3n}{\displaystyle {\int }_t^{\infty }\left(2\left(1+s\right)-\theta {\left(1+s\right)}^4\right){\text{e}}^{-3^{-1}\theta \left({\left(1+s\right)}^3-1\right)}}\text{d}{M}_s^{\alpha }\to 0\left(t\to \infty \right)\end{array}$

几乎必然收敛，且

${\left(1+t\right)}^{3n}\phi \left(t\right)\left({\xi }_{\infty }^{\alpha }-{\xi }_t^{\alpha }\right)~{\theta }^{-1}{\left(1+t\right)}^{3n}\left({\xi }_{\infty }^{\alpha }-{\xi }_t^{\alpha }\right)\to 0\left(t\to \infty \right)$

几乎必然收敛。因此，该引理得到证明。

引理3.2：设 $\theta >0$ 和 $\frac{1}{3}<\alpha <2$ 。定义函数 $I_{\theta }\left(t\right)=\theta {\left(1+t\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}{\displaystyle {\int }_0^t{\text{e}}^{3^{-1}\theta \left({\left(1+u\right)}^3-1\right)}}\text{d}u-1$ ，则随机变量 ${\displaystyle {\int }_0^{\infty }{I}_{\theta }}\left(s\right)\text{d}{M}_s^{\alpha }$ 在 $L^p$ 上适定，且

${\displaystyle {\int }_0^t{I}_{\theta }}\left(s\right)\text{d}{M}_s^{\alpha }\to {\displaystyle {\int }_0^{\infty }{I}_{\theta }}\left(s\right)\text{d}{M}_s^{\alpha }\left(t\to \infty \right)$ (4)

在 $L^p$ 上适定且几乎必然收敛， $0<p<\alpha$ 。

证明：定义 ${\eta }_t^{\alpha }:={\displaystyle {\int }_0^{\infty }{I}_{\theta }}\left(s\right)\text{d}{M}_s^{\alpha }-{\displaystyle {\int }_0^t{I}_{\theta }}\left(s\right)\text{d}{M}_s^{\alpha }\text{=}{\displaystyle {\int }_t^{\infty }{I}_{\theta }\left(s\right)\text{d}{M}_s^{\alpha }}$ ， $t\ge 0$ 。明显地，我们有

${\displaystyle {\int }_0^t{\text{e}}^{3^{-1}\theta \left({\left(1+u\right)}^3-1\right)}\left({\left(1+t\right)}^2-{\left(1+u\right)}^2\right)\text{d}u}={\displaystyle {\int }_0^{t}2\left(1+s\right)\text{d}s{\displaystyle {\int }_0^s{\text{e}}^{3^{-1}\theta \left({\left(1+u\right)}^3-1\right)}\text{d}u}}\to \infty \left(t\to \infty \right).$

紧接着，根据洛必达法则可得

$\begin{array}{c}\underset{t\to \infty }{\lim }{I}_{\theta }\left(t\right){\left(1+t\right)}^3=\underset{t\to \infty }{\lim }{\left(1+t\right)}^3{\text{e}}^{-3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}\left(\theta {\left(1+t\right)}^2{\displaystyle {\int }_0^t{\text{e}}^{3^{-1}\theta \left({\left(1+u\right)}^3-1\right)}\text{d}u}-{\text{e}}^{3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}\right)\\ =\underset{t\to \infty }{\lim }{\left(1+t\right)}^3{\text{e}}^{-3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}\left(\theta {\left(1+t\right)}^2{\displaystyle {\int }_0^t{\text{e}}^{3^{-1}\theta \left({\left(1+u\right)}^3-1\right)}\text{d}u}-{\displaystyle {\int }_0^t\theta {\left(1+u\right)}^2{\text{e}}^{3^{-1}\theta \left({\left(1+u\right)}^3-1\right)}\text{d}u-1}\right)\\ =\theta \underset{t\to \infty }{\lim }\frac{1}{{\left(1+t\right)}^{-3}{\text{e}}^{3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}}{\displaystyle {\int }_0^t{\text{e}}^{3^{-1}\theta \left({\left(1+u\right)}^3-1\right)}\left({\left(1+t\right)}^2-{\left(1+u\right)}^2\right)\text{d}u}=\frac{2}{\theta }.\end{array}$

根据 ${\displaystyle {\int }_0^{\infty }{\left(1+t\right)}^{-3}}\text{d}t=-1/2$ ，可得

$E{\left|{\displaystyle {\int }_0^t{I}_{\theta }}\left(s\right)\text{d}{M}_s^{\alpha }\right|}^p\le C_{p,\alpha }{\left({\displaystyle {\int }_0^t{\left|I_{\theta }\left(s\right)\right|}^{\alpha }\text{d}s}\right)}^{p/\alpha }\le C_{p,\alpha }{\left({\displaystyle {\int }_0^{\infty }{\left|I_{\theta }\left(s\right)\right|}^{\alpha }\text{d}s}\right)}^{p/\alpha }<\infty .$

因此，随机变量 ${\displaystyle {\int }_0^{\infty }{I}_{\theta }}\left(s\right)\text{d}{M}_s^{\alpha }$ 在 $L^p$ 上适定。同理，根据以下估计可知(4)在 $L^p$ 上适定：

$E{\left|{\eta }_t^{\alpha }\right|}^p\le C_{p,\alpha }{\left({\displaystyle {\int }_t^{\infty }{\left|I_{\theta }\left(s\right)\right|}^{\alpha }\text{d}s}\right)}^{p/\alpha }~\frac{1}{{\left(1+t\right)}^{3p-p/\alpha }}\left(t\to \infty \right),$

对所有 $\frac{1}{3}<\alpha <2$ ， $0<p<\alpha$ 都成立。最后，考虑(4)以概率1收敛。通过一个简单的计算可得

$\begin{array}{c}{I^{\prime }}_{\theta }\left(t\right)=2\theta \left(1+t\right){\text{e}}^{-3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}{\displaystyle {\int }_0^t{\text{e}}^{3^{-1}\theta \left({\left(1+u\right)}^3-1\right)}\text{d}u}-{\theta }^2{\left(1+t\right)}^4{\text{e}}^{-3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}{\displaystyle {\int }_0^t{\text{e}}^{3^{-1}\theta \left({\left(1+u\right)}^3-1\right)}\text{d}u}+\theta {\left(1+t\right)}^2\\ =\theta \left(1+t\right){\text{e}}^{-3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}\left({\displaystyle {\int }_0^{t}2{\text{e}}^{3^{-1}\theta \left({\left(1+u\right)}^3-1\right)}\text{d}u}-\theta {\left(1+t\right)}^3{\displaystyle {\int }_0^t{\text{e}}^{3^{-1}\theta \left({\left(1+u\right)}^3-1\right)}\text{d}u}+\left(1+t\right){\text{e}}^{3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}\right)\\ =\theta \left(1+t\right){\text{e}}^{-3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}{\displaystyle {\int }_0^t{\text{e}}^{3^{-1}\theta \left({\left(1+u\right)}^3-1\right)}\left(2-\theta {\left(1+t\right)}^3+\theta \left(1+t\right){\left(1+u\right)}^2\right)\text{d}u}+\theta {\left(1+t\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}.\end{array}$

定义 $f_{\theta }\left(t\right)={\displaystyle {\int }_0^t{\text{e}}^{3^{-1}\theta \left({\left(1+u\right)}^3-1\right)}\left(2-\theta {\left(1+t\right)}^3+\theta \left(1+t\right){\left(1+u\right)}^2\right)\text{d}u}$ ，则可得 $f_{\theta }\left(0\right)=0$ ，

${f^{\prime }}_{\theta }\left(t\right)=-3\theta {\displaystyle {\int }_0^{t}2\left(1+s\right)\text{d}s}{\displaystyle {\int }_0^t{\text{e}}^{3^{-1}\theta \left({\left(1+u\right)}^3-1\right)}}\text{d}u+2\to -\infty \left(t\to \infty \right)$

和 ${f^{\prime }}_{\theta }\left(t\right)\le 2$ 。从而有 $f_{\theta }\left(t\right)={\displaystyle {\int }_0^t{f^{\prime }}_{\theta }\left(s\right)\text{d}s}\to -\infty \left(t\to \infty \right)$ 。紧接着，运用洛必达法则可得

$\underset{t\to \infty }{\lim }{\left(1+t\right)}^4{I^{\prime }}_{\theta }\left(t\right)=-6{\theta }^{-1},$

即可证(4)以概率1收敛。

引理3.3：设 $\theta >0$ 和 $\frac{1}{3}<\alpha <2$ 。定义如下递归公式：

$I_{\theta }\left(t;1\right)=\theta {\left(1+t\right)}^3{I}_{\theta }\left(t\right),I_{\theta }\left(t;n+1\right)=\theta {\left(1+t\right)}^3\left[I_{\theta }\left(t;n\right)-{\displaystyle \underset{i=1}{\overset{n}{\prod }}\left(3i-1\right)}\right],$

则对所有 $n\ge 1$ 我们有

$\underset{t\to \infty }{\lim }{I}_{\theta }\left(t;n\right)={\displaystyle \underset{i=1}{\overset{n}{\prod }}\left(3i-1\right)}.$

证明：证明过程可参考文献Sun和Yan [10] 引理4.3。

定理3.1：设 $\theta >0$ 和 $\frac{1}{3}<\alpha <2$ 。则对任意 $0<p<\alpha$ 有

$J_t^{\alpha }\left(0,\theta \right):={\left(1+t\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}{X}_t^{\alpha }\to {\xi }_{\infty }^{\alpha }+\frac{\nu }{\theta }\left(t\to \infty \right)$

在 $L^p$ 上适定且几乎必然收敛。

证明：通过一个简单的计算可得

$\begin{array}{c}{J}_t^{\alpha }\left(0,\theta \right)={\left(1+t\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}{\displaystyle {\int }_0^t{h}_{\theta }\left(t,s\right)\text{d}{M}_s^{\alpha }}+\nu {\left(1+t\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}{\displaystyle {\int }_0^t{h}_{\theta }\left(t,s\right)\text{d}s}\\ ={\left(1+t\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}{M}_t^{\alpha }-\theta {\left(1+t\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}{\displaystyle {\int }_0^t{\text{e}}^{3^{-1}\theta \left({\left(1+u\right)}^3-1\right)}\left({\xi }_{\infty }^{\alpha }-{\xi }_u^{\alpha }\right)\text{d}u}\\ +\left({\xi }_{\infty }^{\alpha }+\frac{\nu }{\theta }\right)\theta {\left(1+t\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}{\displaystyle {\int }_0^t{\text{e}}^{3^{-1}\theta \left({\left(1+u\right)}^3-1\right)}\text{d}u}\\ =-{\left(1+t\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}{\displaystyle {\int }_0^t{I}_{\theta }\left(s\right)\text{d}{M}_s^{\alpha }}-\theta \phi \left(t\right)\left({\xi }_{\infty }^{\alpha }-{\xi }_t^{\alpha }\right)\\ +\left({\xi }_{\infty }^{\alpha }+\frac{\nu }{\theta }\right)\theta {\left(1+t\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}{\displaystyle {\int }_0^t{\text{e}}^{3^{-1}\theta \left({\left(1+u\right)}^3-1\right)}\text{d}u}.\end{array}$

从而有

$J_t^{\alpha }\left(0,\theta \right)-\left({\xi }_{\infty }^{\alpha }+\frac{\nu }{\theta }\right)=-{\left(1+t\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}{\displaystyle {\int }_0^t{I}_{\theta }\left(s\right)\text{d}{M}_s^{\alpha }}-\theta \phi \left(t\right)\left({\xi }_{\infty }^{\alpha }-{\xi }_t^{\alpha }\right)+\left({\xi }_{\infty }^{\alpha }+\frac{\nu }{\theta }\right)I_{\theta }\left(t\right).$ (5)

根据引理3.1，引理3.2和引理3.3，我们可证该定理成立。

定理3.2：设 $\theta >0$ 和 $\frac{1}{3}<\alpha <2$ 。定义递归过程： $J_t^{\alpha }\left(n,\theta \right):=\theta {\left(1+t\right)}^3\left(J_t^{\alpha }\left(n-1,\theta \right)-{\displaystyle \underset{i=1}{\overset{n-1}{\prod }}\left(3i-1\right)}\left({\xi }_{\infty }^{\alpha }+\frac{\nu }{\theta }\right)\right)$ 。则对所有 $n=\left\{0,1,2,\cdots \right\}$ ， $0<p<\alpha$ 有

$J_t^{\alpha }\left(n,\theta \right)\to {\displaystyle \underset{i=1}{\overset{n}{\prod }}\left(3i-1\right)}\left({\xi }_{\infty }^{\alpha }+\frac{\nu }{\theta }\right)\left(t\to \infty \right)$

在 $L^p$ 上适定且几乎必然收敛。

证明：参考(5)，可得

$J_t^{\alpha }\left(n,\theta \right)=-{\theta }^n{\left(1+t\right)}^{3n+2}{\text{e}}^{-3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}{\displaystyle {\int }_0^t{I}_{\theta }\left(s\right)\text{d}{M}_s^{\alpha }}-{\theta }^{n+1}{\left(1+t\right)}^{3n}\phi \left(t\right)\left({\xi }_{\infty }^{\alpha }-{\xi }_t^{\alpha }\right)+\left({\xi }_{\infty }^{\alpha }+\frac{\nu }{\theta }\right)I_{\theta }\left(t;n\right).$

根据引理3.1，引理3.2，引理3.3和定理3.1可证。

4. 参数估计与渐近分布

在这一节，我们主要研究两个实参 $\theta$ 和 $\nu$ 的相合性和渐近分布。记

$Y_t^{\alpha }:={\displaystyle {\int }_0^t{\left(1+s\right)}^2}\text{d}{X}_s^{\alpha },t\ge 0,$

且 $\theta$ 和 $\nu$ 的最小二乘估计量可以由以下比较函数的最小值求出：

$\rho \left(\theta ,\nu \right)={\displaystyle {\int }_0^T{\left|{\stackrel{\dot{}}{X}}_t^{\alpha }-\left(\theta Y_t^{\alpha }+\nu \right)\right|}^2}\text{d}t.$

则可得最小二乘估计量为

${\hat{\theta }}_T=\frac{T{\displaystyle {\int }_0^T{Y}_t^{\alpha }\text{d}{X}_t^{\alpha }-X_T^{\alpha }{\displaystyle {\int }_0^T{Y}_t^{\alpha }\text{d}t}}}{T{\displaystyle {\int }_0^T{\left(Y_t^{\alpha }\right)}^2\text{d}t-{\left({\displaystyle {\int }_0^T{Y}_t^{\alpha }\text{d}t}\right)}^2}},{\hat{\nu }}_T=\frac{1}{T}\left(X_T^{\alpha }-{\hat{\theta }}_T{\displaystyle {\int }_0^T{Y}_t^{\alpha }\text{d}t}\right)\text{.}$

从而有

${\hat{\theta }}_T-\theta =\frac{T{\displaystyle {\int }_0^T{Y}_t^{\alpha }\text{d}{M}_t^{\alpha }-M_T^{\alpha }{\displaystyle {\int }_0^T{Y}_t^{\alpha }\text{d}t}}}{T{\displaystyle {\int }_0^T{\left(Y_t^{\alpha }\right)}^2\text{d}t-{\left({\displaystyle {\int }_0^T{Y}_t^{\alpha }\text{d}t}\right)}^2}},$ (6)

${\hat{\nu }}_T-\nu =\frac{1}{T}\left(M_T^{\alpha }-\left({\hat{\theta }}_T-\theta \right){\displaystyle {\int }_0^T{Y}_t^{\alpha }\text{d}t}\right)\text{.}$

通过常数变易法可得： $Y_t^{\alpha }={\text{e}}^{3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}{\xi }_t^{\alpha }+\frac{\nu }{\theta }\left({\text{e}}^{3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}-1\right)$ ，其中 ${\xi }_t^{\alpha }={\displaystyle {\int }_0^t{\left(1+s\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+s\right)}^3-1\right)}}\text{d}{M}_s^{\alpha }$ 。当T趋于无穷大时，

$\begin{array}{l}{\left(1+T\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+T\right)}^3-1\right)}{\displaystyle {\int }_0^T{\text{e}}^{3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}{\xi }_t^{\alpha }\text{d}t}\to \frac{1}{\theta }{\xi }_{\infty }^{\alpha }\\ {\left(1+T\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+T\right)}^3-1\right)}{\displaystyle {\int }_0^T\frac{\nu }{\theta }\left({\text{e}}^{3^{-1}\theta \left({\left(1+t\right)}^3-1\right)}-1\right)\text{d}t}\to \frac{\nu }{{\theta }^2},\end{array}$

则有

${\left(1+T\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+T\right)}^3-1\right)}{\displaystyle {\int }_0^T{Y}_t^{\alpha }}\text{d}t\to \frac{1}{\theta }\left({\xi }_{\infty }^{\alpha }+\frac{\nu }{\theta }\right)\left(a.s.\right).$ (7)

定理4.1：设 $0<\alpha <2$ 且 $\theta >0$ ，则当T趋向于无穷时有

${\hat{\theta }}_T\to \theta ,{\hat{\nu }}_T\to \nu \left(a.s.\right).$

证明：定义 ${\psi }_T:=T{\displaystyle {\int }_0^T{\left(Y_t^{\alpha }\right)}^2\text{d}t}-{\left({\displaystyle {\int }_0^T{Y}_t^{\alpha }\text{d}t}\right)}^2$ 。根据洛必达法则，当T趋于无穷大时可得

$T^{-1}{\left(1+T\right)}^2{\text{e}}^{-\frac{2\theta }{3}\left({\left(1+T\right)}^3-1\right)}{\psi }_T\to \frac{1}{2\theta }{\left({\xi }_{\infty }^{\alpha }+\frac{\nu }{\theta }\right)}^2\left(a.s.\right).$

依据洛必达法则和公式(6)，当T趋于无穷大时有

$\begin{array}{l}{T}^{-2}{\text{e}}^{-3^{-1}\theta \left({\left(1+T\right)}^3-1\right)}\left(T{Y}_T^{\alpha }{M}_T^{\alpha }\right)\to 0\\ T^{-2}{\text{e}}^{-3^{-1}\theta \left({\left(1+T\right)}^3-1\right)}\left(T{\displaystyle {\int }_0^T{\stackrel{\dot{}}{Y}}_t^{\alpha }{\stackrel{\dot{}}{M}}_t^{\alpha }\text{d}t}\right)\to 0\\ T^{-2}{\text{e}}^{-3^{-1}\theta \left({\left(1+T\right)}^3-1\right)}\left(M_T^{\alpha }{\displaystyle {\int }_0^T{Y}_t^{\alpha }\text{d}t}\right)=T^{-1}{M}_T^{\alpha }\left({\left(1+T\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+T\right)}^3-1\right)}{\displaystyle {\int }_0^T{Y}_t^{\alpha }\text{d}t}\right)T^{-1}{\left(1+T\right)}^{-2}\to 0\left(a.s.\right).\end{array}$

从而得到当 $T\to \infty$ 时 ${\hat{\theta }}_T\to \theta$ 几乎必然收敛于0。同理可得，当 $T\to \infty$ 时有

${\hat{\nu }}_T-\nu =\frac{M_T^{\alpha }}{T}\text{-}\frac{T^{-2}{\text{e}}^{-3^{-1}\theta \left({\left(1+T\right)}^3-1\right)}\left({\hat{\theta }}_T-\theta \right)}{T^{-1}{\left(1+T\right)}^2{\text{e}}^{-\frac{2\theta }{3}\left({\left(1+T\right)}^3-1\right)}{\psi }_T}\left({\left(1+T\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+T\right)}^3-1\right)}{\displaystyle {\int }_0^T{Y}_t^{\alpha }\text{d}t}\right)\to \text{0}\left(a.s.\right)\text{.}$

定理4.2：设 $1<\alpha <2$ 且 $\theta >0$ ，则当T趋向于无穷时有

$\begin{array}{l}{\left(1+T\right)}^{2\left(1/\alpha -1\right)}{\text{e}}^{3^{-1}\theta \left({\left(1+T\right)}^3-1\right)}\left({\hat{\theta }}_T-\theta \right)~2{\theta }^{1-1/\alpha }{\alpha }^{-1/\alpha }{\left|{\xi }_{\infty }^{\alpha }+\frac{\nu }{\theta }\right|}^{-1}{\tilde{M}}_1\\ T{\left(1+T\right)}^{2/\alpha }\left({\hat{\nu }}_T-\nu -\frac{M_T^{\alpha }}{T}\right)~-2{\left(\theta \alpha \right)}^{-1/\alpha }\mathrm{sgn}\left({\xi }_{\infty }^{\alpha }+\frac{\nu }{\theta }\right){\tilde{M}}_1\left(a.s.\right).\end{array}$

证明：参考Rosinski-Woyczynski [10] ，定义 ${\displaystyle {\int }_0^T{Y}_t^{\alpha }}\text{d}{M}_t^{\alpha }:={\tilde{M}}_{{\tau }_{\alpha }\left(T\right)}^{\alpha },T\ge \text{0,}$ 其中 ${\tau }_{\alpha }\left(T\right)={\displaystyle {\int }_0^T{\left|Y_t^{\alpha }\right|}^{\alpha }}\text{d}t$ 。根据收敛(7)式和洛必达法则，当T趋向于无穷时有 $\begin{array}{l}{\left(1+T\right)}^{2/\alpha }{\text{e}}^{-3^{-1}\theta \left({\left(1+T\right)}^3-1\right)}{\left({\displaystyle {\int }_0^T{\left|Y_t^{\alpha }\right|}^{\alpha }\text{d}t}\right)}^{1/\alpha }\\ ={\left({\left(1+T\right)}^2{\text{e}}^{-3^{-1}\theta \alpha \left({\left(1+T\right)}^3-1\right)}{\displaystyle {\int }_0^T{\left|Y_t^{\alpha }\right|}^{\alpha }\text{d}t}\right)}^{1/\alpha }~{\left(\theta \alpha \right)}^{-1/\alpha }\left|{\xi }_{\infty }^{\alpha }+\frac{\nu }{\theta }\right|\left(a.s.\right).\end{array}$

通过两个简单计算

$\begin{array}{l}{\left(1+T\right)}^{2/\alpha }{\text{e}}^{-3^{-1}\theta \left({\left(1+T\right)}^3-1\right)}{\tilde{M}}_{{\tau }_{\alpha }\left(T\right)}^{\alpha }~{\left(\theta \alpha \right)}^{-1/\alpha }\left|{\xi }_{\infty }^{\alpha }+\frac{\nu }{\theta }\right|{\tilde{M}}_1\\ T^{-1}{\left(1+T\right)}^{2/\alpha }{\text{e}}^{-3^{-1}\theta \left({\left(1+T\right)}^3-1\right)}{M}_T^{\alpha }{\displaystyle {\int }_0^T{Y}_t^{\alpha }\text{d}t}\to 0\text{,}\end{array}$

可以获得 ${\left(1+T\right)}^{2\left(1/\alpha -1\right)}{\text{e}}^{3^{-1}\theta \left({\left(1+T\right)}^3-1\right)}\left({\hat{\theta }}_T-\theta \right)~2{\theta }^{1-1/\alpha }{\alpha }^{-1/\alpha }{\left|{\xi }_{\infty }^{\alpha }+\frac{\nu }{\theta }\right|}^{-1}{\tilde{M}}_1$ 。同理，当T趋向于无穷时有

$\begin{array}{c}T{\left(1+T\right)}^{2/\alpha }\left({\hat{\nu }}_T-\nu -\frac{M_T^{\alpha }}{T}\right)=-\left({\left(1+T\right)}^2{\text{e}}^{-3^{-1}\theta \left({\left(1+T\right)}^3-1\right)}{\displaystyle {\int }_0^T{Y}_t^{\alpha }\text{d}t}\right)\left({\left(1+T\right)}^{2\left(1/\alpha -1\right)}{\text{e}}^{3^{-1}\theta \left({\left(1+T\right)}^3-1\right)}\left({\hat{\theta }}_T-\theta \right)\right)\\ ~-2{\left(\theta \alpha \right)}^{-1/\alpha }\mathrm{sgn}\left({\xi }_{\infty }^{\alpha }+\frac{\nu }{\theta }\right){\tilde{M}}_1\left(a.s.\right).\end{array}$

由此，该定理得到证明。

5. 总结

依据目前已研究的成果发现，很多学者主要研究关于布朗运动的自交互过程。然而随着研究的深入，这一类过程已不能满足现实的需求，需要进一步扩展。针对这一现状，本文研究一个自交互过程的特例，由 $\alpha$ 稳定过程驱动的线性自排斥过程的相关问题。文中我们利用收敛速度和分部积分得到一个递归收敛，并通过最小二乘法估计参数计算出渐近分布，有创新点，但也存在一定的局限性。事实上，本文研究的线性自排斥过程也只是自交互过程中的一类，是一种特殊的形式，还可以从其他方面入手对自交互过程进行更深入研究，包括改变噪声、更换函数、限定漂移项等。