﻿ 由α-稳定过程驱动的线性自排斥扩散过程的参数估计

# 由α-稳定过程驱动的线性自排斥扩散过程的参数估计Parameter Estimation for the Linear Self-Repelling Diffusion Driven by α-Stable Motions

Abstract: In this paper, we consider parameter estimations of the linear self-repelling diffusion , where Mα is a symmetrical α-stable motion (1α＜2). The process is an analogue of the self-repelling diffusion. By using least squares method, we study estimators of unknown parameters, give their asymptotic distributions under the continuous observation and study the accuracy of the estimator.

1. 引言

1992年，Durrett和Rogers [1] 研究了一类增长物模型。在某种条件下，他们建立了如下随机微分方程解的渐近性质：

${X}_{t}={B}_{t}+{\int }_{0}^{t}{\int }_{0}^{s}f\left({X}_{s}-{X}_{u}\right)\text{d}u\text{d}s$

${X}_{t}={X}_{0}+{B}_{t}+{\int }_{0}^{t}\text{d}s{\int }_{ℝ}f\left(-x\right){\mathcal{L}}^{X}\left(s,{X}_{s}+x\right)\text{d}x$

${X}_{t}={B}_{t}-\theta {\int }_{0}^{t}{\int }_{0}^{s}\left({X}_{s}-{X}_{u}\right)\text{d}u\text{d}s+\nu t,\text{ }t\ge 0$

${X}_{t}^{\alpha }={M}_{t}^{\alpha }-\theta {\int }_{0}^{t}{\int }_{0}^{s}\left({X}_{s}^{\alpha }-{X}_{r}^{\alpha }\right)\text{d}r\text{d}s+\nu t$

${Y}_{s}^{\alpha }={\int }_{0}^{t}\left({X}_{s}^{\alpha }-{X}_{r}^{\alpha }\right)\text{d}s,\text{ }t\ge 0$

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

${\rho }_{n}\left(\theta ,\nu \right)={\int }_{0}^{T}{|{X}_{t}{}^{\alpha }-\left(\nu -\theta {Y}_{t}{}^{\alpha }\right)|}^{2}\text{d}t$

${\stackrel{^}{\theta }}_{T}=\frac{{X}_{T}^{\alpha }{\int }_{0}^{T}{Y}_{t}^{\alpha }\text{d}t-T{\int }_{0}^{T}{Y}_{t}^{\alpha }\text{d}{X}_{t}^{\alpha }}{T{\int }_{0}^{T}{\left({Y}_{t}^{\alpha }\right)}^{2}\text{d}t-{\left({\int }_{0}^{T}{Y}_{t}^{\alpha }\text{d}t\right)}^{2}}$

${\stackrel{^}{\nu }}_{T}=\frac{1}{T}\left({X}_{T}^{\alpha }+{\stackrel{^}{\theta }}_{T}{\int }_{0}^{T}{Y}_{t}^{\alpha }\text{d}t\right)$

${\stackrel{^}{\theta }}_{T}\to \theta$

$\stackrel{^}{\nu }\to \nu$

${T}^{1/\alpha -1}{\text{e}}^{-\frac{1}{2}\theta {T}^{2}}\left({\stackrel{^}{\theta }}_{T}-\theta \right)\to -\frac{2{\theta }^{1-1/\alpha }}{{\alpha }^{1/\alpha }}{\stackrel{˜}{M}}_{1}{|{\xi }_{\infty }-\frac{\nu }{\theta }|}^{-1}$ ,

${T}^{1/\alpha -1}\left(\stackrel{^}{\nu }-\nu -\frac{1}{T}{M}_{t}\right)\to \frac{2{\theta }^{-1/\alpha }}{{\alpha }^{1/\alpha }}{\stackrel{˜}{M}}_{1}\mathrm{sgn}\left({\xi }_{\infty }-\frac{\nu }{\theta }\right)$ .

2. 预备知识

$E\left[{\text{e}}^{iu\left({M}_{t}^{\alpha }-{M}_{s}^{\alpha }\right)}\right]={\text{e}}^{-\left(t-s\right)\lambda {|u|}^{\alpha }}$

$\text{d}{X}_{t}^{\alpha }=\text{d}{M}_{t}^{\alpha }-\theta \left({\int }_{0}^{t}\left({X}_{t}^{\alpha }-{X}_{r}^{\alpha }\right)\text{d}r\right)\text{d}t+\nu \text{d}t$ (1)

${h}_{\theta }\left(t,s\right)=\left\{\begin{array}{ll}1-\theta s{\text{e}}^{\frac{1}{2}\theta {s}^{2}}{\int }_{s}^{t}{\text{e}}^{-\frac{1}{2}\theta {u}^{2}}\text{d}u\hfill & t\ge s\hfill \\ 0\hfill & t

${X}_{t}^{\alpha }={\int }_{0}^{t}{h}_{\theta }\left(t,s\right)\text{d}{M}_{s}^{\alpha }+\nu {\int }_{0}^{t}{h}_{\theta }\left(t,s\right)\text{d}s$

3. 强相合性

${\Phi }_{T}:=T{\int }_{0}^{T}{\left({Y}_{t}\right)}^{2}\text{d}t-{\left({\int }_{0}^{T}{Y}_{t}\text{d}t\right)}^{2}$

${\stackrel{^}{\theta }}_{T}=\theta +\frac{1}{{\Phi }_{T}}\left({X}_{T}{\int }_{0}^{T}{Y}_{t}\text{d}t-T{\int }_{0}^{T}{Y}_{t}\text{d}{X}_{t}\right)$ (2)

$\stackrel{^}{\nu }=\nu +\frac{1}{T}{M}_{t}+\left({\stackrel{^}{\theta }}_{T}-\theta \right)\frac{1}{T}{\int }_{0}^{T}{Y}_{t}\text{d}t$ (3)

${Y}_{t}={\int }_{0}^{t}\left({x}_{s}-{x}_{r}\right)\text{d}s={\int }_{0}^{t}s\text{d}{x}_{s}={\int }_{0}^{t}s\text{d}{M}_{s}-\theta {\int }_{0}^{t}s{Y}_{s}\text{d}s+\nu {\int }_{0}^{t}s\text{d}s$

${Y}_{t}={\text{e}}^{-\frac{1}{2}\theta {t}^{2}}{\int }_{0}^{t}s{\text{e}}^{\frac{1}{2}\theta {s}^{2}}\text{d}{M}_{s}+\frac{\nu }{\theta }\left(1-{\text{e}}^{-\frac{1}{2}\theta {t}^{2}}\right)={\text{e}}^{-\frac{1}{2}\theta {t}^{2}}{\xi }_{t}+\frac{\nu }{\theta }\left(1-{\text{e}}^{-\frac{1}{2}\theta {t}^{2}}\right)$ (4)

$T{\text{e}}^{\frac{1}{2}\theta {T}^{2}}{\int }_{0}^{T}{Y}_{t}\text{d}t\to -\frac{1}{\theta }\left({\xi }_{\infty }-\frac{\nu }{\theta }\right)$

$T{\text{e}}^{\frac{1}{2}\theta {T}^{2}}{\int }_{0}^{T}{Y}_{t}\text{d}t=T{\text{e}}^{\frac{1}{2}\theta {T}^{2}}{\int }_{0}^{t}{\text{e}}^{-\frac{1}{2}\theta {t}^{2}}{\xi }_{t}\text{d}t+\frac{\nu }{\theta }{T}^{2}{\text{e}}^{\frac{1}{2}\theta {T}^{2}}-T{\text{e}}^{\frac{1}{2}\theta {T}^{2}}{\int }_{0}^{T}\frac{\nu }{\theta }{\text{e}}^{-\frac{1}{2}\theta {t}^{2}}\text{d}t\to -\frac{1}{\theta }\left({\xi }_{\infty }-\frac{\nu }{\theta }\right)$

${\xi }_{\infty }-{\xi }_{t}={\int }_{t}^{\infty }s{\text{e}}^{\frac{1}{2}\theta {s}^{2}}\text{d}Ms=-t{\text{e}}^{\frac{1}{2}\theta {t}^{2}}{M}_{t}^{2}-{\int }_{0}^{\infty }{M}_{s}^{\alpha }\left(1+\theta {s}^{2}\right){\text{e}}^{\frac{1}{2}\theta {s}^{2}}\text{d}s$

${\text{e}}^{\theta {T}^{2}}{\Phi }_{T}\to -\frac{1}{2\theta }{\left({\xi }_{\infty }-\frac{\nu }{\theta }\right)}^{2}$

${T}^{-2}{\text{e}}^{\frac{1}{2}\theta {T}^{2}}\left({M}_{T}{\int }_{0}^{T}{Y}_{t}\text{d}t-T{\int }_{0}^{T}{Y}_{t}\text{d}{M}_{t}\right)\to 0\text{ }\left(T\to \infty \right)\text{ }a.s.$

${T}^{-2}{\text{e}}^{\frac{1}{2}\theta {T}^{2}}\left(T{Y}_{t}{M}_{t}\right)=\frac{{M}_{t}}{T}\cdot \left({\text{e}}^{\frac{1}{2}\theta {T}^{2}}{Y}_{t}\right)\to 0$

${T}^{-2}{\text{e}}^{\frac{1}{2}\theta {T}^{2}}\left(T{\int }_{0}^{T}t{M}_{t}{\text{e}}^{-\frac{1}{2}\theta {t}^{2}}{\xi }_{t}\text{d}t\right)={T}^{-1}{\text{e}}^{\frac{1}{2}\theta {T}^{2}}{\int }_{0}^{T}t{M}_{t}{\text{e}}^{-\frac{1}{2}\theta {t}^{2}}{\xi }_{t}\text{d}t\to 0$

${T}^{-2}{\text{e}}^{\frac{1}{2}\theta {T}^{2}}\left(T{\int }_{0}^{T}t{\text{e}}^{-\frac{1}{2}\theta {t}^{2}}{M}_{t}\text{d}t\right)={T}^{-1}{\text{e}}^{\frac{1}{2}\theta {T}^{2}}{\int }_{0}^{T}t{\text{e}}^{-\frac{1}{2}\theta {t}^{2}}{M}_{t}\text{d}t\to 0$

${\stackrel{^}{\theta }}_{T}-\theta =\frac{1}{{\text{e}}^{\theta {T}^{2}}{\Phi }_{T}}{\text{e}}^{\theta {T}^{2}}\left({M}_{t}{\int }_{0}^{T}{Y}_{t}\text{d}t-T{\int }_{0}^{t}{Y}_{t}\text{d}{M}_{t}\right)\to 0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}a.s.$

$\begin{array}{c}{\stackrel{^}{\nu }}_{T}-\nu =\frac{1}{T}{M}_{t}+\left({\stackrel{^}{\theta }}_{T}-\theta \right)\frac{1}{T}{\int }_{0}^{T}{Y}_{t}\text{d}t\\ =\frac{1}{T}{M}_{t}+\frac{1}{{\text{e}}^{\theta {T}^{2}}{\Phi }_{T}}{T}^{-2}{\text{e}}^{\frac{1}{2}\theta {T}^{2}}\left({M}_{T}{\int }_{0}^{T}{Y}_{t}\text{d}t-T{\int }_{0}^{T}{Y}_{t}\text{d}{M}_{T}\right)T{\text{e}}^{\frac{1}{2}\theta {T}^{2}}\to 0\text{ }a.s.\end{array}$

4. 渐近分布

${T}^{1/\alpha }{\text{e}}^{\frac{1}{2}\theta {T}^{2}}{\left({\int }_{0}^{T}{|{Y}_{t}|}^{\alpha }\text{d}t\right)}^{1/\alpha }={\left(T{\text{e}}^{\frac{\alpha }{2}\theta {T}^{2}}{\int }_{0}^{T}{|{Y}_{t}|}^{\alpha }\text{d}t\right)}^{1/\alpha }\to -\frac{1}{{\left(\alpha \theta \right)}^{1/\alpha }}|{\xi }_{\infty }-\frac{\nu }{\theta }|$

$\begin{array}{c}{T}^{1/\alpha -1}{\text{e}}^{-\frac{1}{2}\theta {T}^{2}}\left({\stackrel{^}{\theta }}_{T}-\theta \right)=\frac{1}{{\text{e}}^{\theta {T}^{2}}{\Phi }_{T}}\left({T}^{1/\alpha -1}{\text{e}}^{\frac{1}{2}\theta {T}^{2}}{M}_{t}{\int }_{0}^{T}{Y}_{t}\text{d}t-{T}^{1/\alpha }{\text{e}}^{\frac{1}{2}\theta {T}^{2}}{\int }_{0}^{T}{Y}_{t}\text{d}{M}_{t}\right)\\ =\frac{1}{{\text{e}}^{\theta {T}^{2}}{\Phi }_{T}}\left[{Υ}_{1}\left(T\right)-{Υ}_{2}\left(T\right)\right]\end{array}$

${Υ}_{1}\left(T\right)={T}^{1/\alpha -1}{\text{e}}^{\frac{1}{2}\theta {T}^{2}}\frac{{M}_{t}}{T}T{\int }_{0}^{T}{Y}_{t}\text{d}t\to 0\text{ }a.s.$

${Υ}_{2}\left(T\right)={T}^{1/\alpha }{\text{e}}^{\frac{1}{2}\theta {T}^{2}}{\stackrel{˜}{M}}_{{\tau }_{\alpha }\left(T\right)}^{\alpha }=\frac{{\stackrel{˜}{M}}_{{\tau }_{\alpha }\left(T\right)}^{\alpha }}{{\tau }_{\alpha }^{1/\alpha }\left(T\right)}{\tau }_{\alpha }^{1/\alpha }\left(T\right){T}^{1/\alpha }{\text{e}}^{\frac{1}{2}\theta {T}^{2}}$

${Υ}_{2}\left(T\right)~-\frac{1}{{\left(\alpha \theta \right)}^{1/\alpha }}{\stackrel{˜}{M}}_{1}|{\xi }_{\infty }-\frac{\nu }{\theta }|$

${T}^{1/\alpha -1}{\text{e}}^{-\frac{1}{2}\theta {T}^{2}}\left({\stackrel{^}{\theta }}_{T}-\theta \right)\to -\frac{\frac{1}{{\left(\alpha \theta \right)}^{1/\alpha }}{\stackrel{˜}{M}}_{1}|{\xi }_{\infty }-\frac{\nu }{\theta }|}{\frac{1}{2\theta }{\left({\xi }_{\infty }-\frac{\nu }{\theta }\right)}^{2}}=-\frac{2{\theta }^{1-1/\alpha }}{{\alpha }^{1/\alpha }}{\stackrel{˜}{M}}_{1}{|{\xi }_{\infty }-\frac{\nu }{\theta }|}^{-1}$

${T}^{1/\alpha -1}\left(\stackrel{^}{\nu }-\nu -\frac{1}{T}{M}_{t}\right)={T}^{1/\alpha -1}{\text{e}}^{-\frac{1}{2}\theta {T}^{2}}\left(\stackrel{^}{\theta }-\theta \right){\text{e}}^{\frac{1}{2}\theta {T}^{2}}{\int }_{0}^{T}{Y}_{t}\text{d}t\to \frac{2{\theta }^{-1/\alpha }}{{\alpha }^{1/\alpha }}{\stackrel{˜}{M}}_{1}\mathrm{sgn}\left({\xi }_{\infty }-\frac{\nu }{\theta }\right)$

5. 统计模拟

${X}_{t}^{\alpha }={M}_{t}^{\alpha }-\theta {\int }_{0}^{t}{\int }_{0}^{s}\left({X}_{s}^{\alpha }-{X}_{r}^{\alpha }\right)\text{d}r\text{d}s+\nu t$

Figure 1. Curve: $\stackrel{^}{\theta }$ when h = 0.01

Figure 2. Curve: $\stackrel{^}{\theta }$ when h = 0.05

Figure 3. Curve: $\stackrel{^}{\nu }$ when h = 0.1

Figure 4. Curve: $\stackrel{^}{\nu }$ when h = 0.1

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