## $\boldsymbol{\alpha}$ -稳定过程驱动的非线性随机微分方程的参数估计: 非遍历情形

1安徽工程大学数理与金融学院 安徽芜湖241000

2东华大学理学院 上海201620

## Parameter Estimation for Nonlinear Stochastic Differential Equations Driven by $\boldsymbol\alpha$-Stable Processes: Non-ergodic Case

Zhang Xuekang,1,*, Wan Shanlin,1, Shu Huisheng,2

1School of Mathematics-Physics and Finance, Anhui Polytechnic University, Anhui Wuhu 241000

2College of Science, Donghua University, Shanghai 201620

 基金资助: 国家自然科学基金.  12101004国家自然科学基金.  62073071国家自然科学基金.  12271003安徽工程大学引进人才科研启动基金.  2020YQQ064安徽省高端装备智能控制国际联合研究中心开放基金.  IRICHE-05

 Fund supported: The NSFC.  12101004The NSFC.  62073071The NSFC.  12271003Startup Foundation for Introducing Talent of Anhui Polytechnic University.  2020YQQ064Open Project of Anhui Province Center for International Reasearch of Intelligent Control of High-end Equipment.  IRICHE-05

Abstract

The present paper deals with the parameter estimation problem for nonlinear stochastic differential equations driven by $\alpha$-stable processes based on continuous-time observation. We first discuss the consistency and the rate of convergence of the weighted trajectory fitting estimator. Then, we have established the asymptotic distribution of the estimator.

Keywords： Non-ergodic case ; $\alpha$-stable processes ; Nonlinear stochastic differential equations ; Consistency ; Asymptotic distribution

Zhang Xuekang, Wan Shanlin, Shu Huisheng. Parameter Estimation for Nonlinear Stochastic Differential Equations Driven by $\boldsymbol\alpha$-Stable Processes: Non-ergodic Case. Acta Mathematica Scientia[J], 2023, 43(1): 249-260 doi:

## 1 引言

$\begin{matrix}\label{1.1} {\rm d}X_t=\theta a(X_t){\rm d}t+{\rm d}B_t,\quad X_0=x_0,\quad t\geq 0, \end{matrix}$

$\begin{matrix}\label{model} {\rm d}X_t=\theta a(X_t){\rm d}t+{\rm d}Z_t,\quad X_0=x_0, \end{matrix}$

$A_t=\int_0^ta(X_s){\rm d}s,\quad t\geq 0.$

$$$\label{1.40} X_t=x_0+\theta A_t+Z_t.$$$

$\omega_t$ 为非负确定函数, 上述等式两边同乘以 $\omega_t$ 可得

$\omega_tX_t=\omega_tx_0+\omega_t\theta A_t+\omega_tZ_t.$

$\int_0^T|\omega_tX_t-(\omega_tx_0+\theta\omega_t\theta A_t)|^2{\rm d}t$

$\begin{matrix}\label{1.41} \widehat{\theta}_T=\frac{\int_0^T\omega_t^2(X_t-x_0)A_t{\rm d}t}{\int_0^T\omega_t^2A_t^2{\rm d}t}, \quad T> 0. \end{matrix}$

$\begin{matrix}\label{1.4} \widehat{\theta}_T-\theta=\frac{\int_0^T\omega_t^2A_t Z_t{\rm d}t}{\int_0^T\omega_t^2A_t^2{\rm d}t}. \end{matrix}$

## 2 预备知识

$\begin{eqnarray*} \phi_\eta(u)={\Bbb E}\exp\{{\rm i}u\eta\}=\left\{ \begin{array}{ll} \exp\left\{-\sigma^\alpha|u|^\alpha\left(1-{\rm i}\beta {\rm sgn}(u)\tan\frac{\alpha\pi}{2}\right)+ {\rm i}\mu u\right\},\quad \alpha\neq 1\\ \exp\left\{-\sigma |u|\left(1+{\rm i}\beta\frac{2}{\pi} {\rm sgn}(u)\log|u|\right)+{\rm i}\mu u\right\},\quad \alpha= 1, \end{array} \right. \end{eqnarray*}$

$\mu=0$, 则称 $\eta$ 为严 $\alpha$ -稳定分布. 若 $\beta=0$, 则称 $\eta$ 为对称 $\alpha$ -稳定分布 (见文献[1,12,20]及其参考文献).

(i) $Z_0=0$, a.s.;

(ii) $Z_t-Z_s\sim S_{\alpha}((t-s)^{\frac{1}{\alpha}},\beta,0), t>s\geq 0$;

(iii) 对任意有限点 $0\leq s_0<s_1<\cdots<s_m<\infty$, 随机变量 $Z_{s_0}, Z_{s_1}-Z_{s_0},\cdots,Z_{s_m}-Z_{s_{m-1}}$ 相互独立, 则称随机过程 $\{Z_t\}_{t\geq 0}$ 为标准 $\alpha$ -稳定过程.

$\lim\limits_{T\rightarrow\infty}\int_0^Tf_t\varphi_T({\rm d}t)=f_\infty,$

$h_1(T)=\int_0^T\omega_t^2{\rm e}^{2c\theta t}{\rm d}t,\quad h_2(T)=\int_0^T\omega_t^2{\rm e}^{c\theta t}{\rm d}t.$

(A1) 假设存在正常数 $c$ 使得

$a(x)=cx+r(x),\,\,x\in\Bbb R,$

$r(x)$ 满足

$|r(x)-r(y)|\leq L|x-y|,\,\, (x,y)\in\Bbb R^2$

$|r(x)|\leq K(1+|x|^\gamma),\,\, x\in\Bbb R,$

$a(x)=cx+|x|^\gamma,\,\, x\in\Bbb R.$

(A3) 当 $T\rightarrow\infty$ 时, $h_i(T)\rightarrow\infty$, $i=1,2$, 并对每一个 $M>0$, 当 $T\rightarrow\infty$ 时,$\frac{h_i(M)}{h_i(T)}\rightarrow 0$.

(A4) 非负加权函数 $\omega_t$ 满足下列条件

$\omega_{t_1t_2}=\omega_{t_1}\omega_{t_2},\,\, t_1,t_2\geq 0.$

(A5) 存在常数 $C'>0$$b<0 使得当 T 足够大时, 有 \frac{\omega_t^2{\rm e}^{c\theta t}}{h_2(T)}\leq C'{\rm e}^{b(T-t)},\,\, t\in[T]. ## 3 加权拟合估计量 \widehat{\theta_T} 的相合性 定理3.1 (i) 若条件 (A1) 和 (A3) 成立且 \theta>0, 则有 \begin{matrix}\label{2.1} \lim\limits_{T\rightarrow\infty}\widehat{\theta}_T=\theta,\quad\mbox{a.s.,} \end{matrix} 即, 加权拟合估计量 \widehat{\theta}_T 强相合的; (ii) 若条件 (A2) 和 (A4) 成立且 \theta=0, 则当 T\rightarrow\infty 时, \widehat{\theta}_T\rightarrow_p\theta. (i) 若条件 (A1) 成立且\theta>0, 则方程(1.2)的解 X 是非常返的. 对函数 {\rm e}^{-c\theta t}X_t 利用 It\hat{o}公式可得 \begin{eqnarray*} d{\rm e}^{-\theta ct}X_t&=&-c\theta {\rm e}^{-c\theta t}X_t{\rm d}t+{\rm e}^{-\theta c t}(\theta cX_t{\rm d}t+\theta r(X_t){\rm d}t+{\rm d}Z_t)\\ &=&\theta r(X_t) {\rm e}^{-c\theta t}{\rm d}t+ {\rm e}^{-c\theta t}{\rm d}Z_t. \end{eqnarray*} 两边从 0$$t$ 积分, 可得

$\begin{matrix}\label{e} {\rm e}^{-c\theta t}X_t=x_0+\int_0^t \theta r(X_s) {\rm e}^{-c\theta s}{\rm d}s+\int_0^t{\rm e}^{-c\theta s}{\rm d}Z_s:=\eta_t. \end{matrix}$

$\xi_t=\int_0^t{\rm e}^{c\theta u}{\rm d}Z_u$. 易知 $\{\xi_t\}_{t\geq 0}$ 是一个 $L^p\,(1<p<\alpha)$ 有界的右连左极鞅, 且 $\xi_t$ 是一个分布为 $S_\alpha(\tau_t^{\frac{1}{\alpha}},\beta,0)$$\alpha -稳定随机变量, 其中 \tau_t=\int_0^t|{\rm e}^{-c\theta s}|^\alpha {\rm d}s=\frac{1-{\rm e}^{-\alpha c \theta t}}{\alpha c \theta}.t\rightarrow\infty 时, \xi_t 收敛到一个分布为 S_\alpha((\alpha c\theta)^{-1/\alpha},\beta,0)$$\alpha$ -稳定随机变量. 由鞅收敛定理可知

$\begin{matrix}\label{xi} \lim\limits_{t\rightarrow\infty} \xi_t=\int_0^\infty {\rm e}^{-c\theta s}{\rm d}Z_s:=\xi_\infty,\quad \mbox{a.s.} \end{matrix}$

$\begin{matrix}\label{non2.0} |\rho_t|&\leq&\theta K\int_0^t{\rm e}^{-c\theta s}(1+|X_s|^\gamma){\rm d}s\nonumber\\ &=&\theta K\int_0^t{\rm e}^{-c\theta s}{\rm d}s+\theta K\int_0^t{\rm e}^{-c\theta (1-\gamma)s}|\eta_s|^\gamma {\rm d}s\nonumber\\ &\leq& 2\theta K\int_0^t{\rm e}^{-c\theta(1-\gamma) s}{\rm d}s+\theta K\int_0^t{\rm e}^{-c\theta (1-\gamma)s}|\eta_s|{\rm d}s\nonumber\\ &\leq&\frac{2K}{c(1-\gamma)}+\theta K\int_0^t{\rm e}^{-c\theta (1-\gamma)s}|\eta_s|{\rm d}s. \end{matrix}$

$\begin{matrix} |\eta_t|\leq K_1+\theta K\int_0^t{\rm e}^{-c\theta (1-\gamma)s}|\eta_s|{\rm d}s, \end{matrix}$

$\begin{matrix} |\eta_t|\leq K_1\exp\left\{\theta K\int_0^t{\rm e}^{-c\theta (1-\gamma)s}{\rm d}s\right\},\quad t\geq 0. \end{matrix}$

$\begin{matrix}\label{eta*} \eta_*:=\sup\limits_{t\geq 0}|\eta_t|\leq K_1{\rm e}^{\frac{K}{c(1-\gamma)}}<\infty, \end{matrix}$

$|\rho_t|\leq \frac{K}{c(1-\gamma)}(2+\eta_*).$

$|\rho_t-\rho_s|\leq 2\theta K\int_s^t{\rm e}^{-c\theta(1-\gamma) u}{\rm d}u+\theta K\eta_*\int_s^t{\rm e}^{-c\theta (1-\gamma)u}{\rm d}u,\quad 0\leq s\leq t.$

$\begin{matrix}\label{rho} \lim\limits_{t\rightarrow\infty} \rho_t=\int_0^\infty {\rm e}^{-c\theta s}\theta r(X_s){\rm d}s,\quad \mbox{a.s.} \end{matrix}$

$\begin{matrix}\label{eta0} \lim\limits_{t\rightarrow\infty}{\rm e}^{-c\theta t}X_t=x_0+\int_0^\infty \theta r(X_s) {\rm e}^{-c\theta s}{\rm d}s+\int_0^\infty {\rm e}^{-c\theta s}{\rm d}Z_s:=\eta_\infty, \quad \mbox{a.s.} \end{matrix}$

$\begin{eqnarray*} 0 & \leq & \lim\limits_{t\rightarrow\infty}\frac{\int_0^t|r(X_s)|}{{\rm e}^{c\theta t}}\leq\lim\limits_{t\rightarrow\infty}\frac{\int_0^tK(1+|X_s|^\gamma){\rm d}s}{{\rm e}^{c\theta t}}\nonumber\\ & = & \lim\limits_{t\rightarrow\infty}\frac{K\int_0^t(|{\rm e}^{-c\theta s}X_s|^\gamma-\eta_\infty^\gamma){\rm e}^{c\theta\gamma s}{\rm d}s+K\eta_\infty^\gamma\int_0^t{\rm e}^{c\theta\gamma s}{\rm d}s}{{\rm e}^{c\theta t}}\nonumber\\ & = & \lim\limits_{t\rightarrow\infty}\frac{K}{{\rm e}^{c\theta(1-\gamma) t}}\left(\frac{\lambda_t^1}{{\rm e}^{c\theta\gamma t}}\int_0^t(|{\rm e}^{-c\theta s}X_s|^\gamma-\eta_\infty^\gamma)\frac{{\rm e}^{c\theta\gamma s}}{\lambda_t^1}{\rm d}s+\eta_\infty^\gamma\frac{\int_0^t{\rm e}^{c\theta\gamma s}{\rm d}s} {{\rm e}^{c\theta\gamma t}}\right)\nonumber\\ & = & 0,\quad \mbox{a.s.}, \end{eqnarray*}$

$\begin{matrix}\label{r(x)} \lim\limits_{t\rightarrow\infty}\frac{\int_0^tr(X_s)}{{\rm e}^{c\theta t}}=0,\quad \mbox{a.s.} \end{matrix}$

$\begin{matrix}\label{non2.8} \lim\limits_{t\rightarrow\infty}\frac{A_t}{{\rm e}^{c\theta t}}&=&\lim\limits_{t\rightarrow\infty}\frac{c\int_0^tX_s{\rm d}s}{{\rm e}^{c\theta t}}+\lim\limits_{t\rightarrow\infty}\frac{\int_0^tr(X_s){\rm d}s}{{\rm e}^{c\theta t}}\nonumber\\ &=&\lim\limits_{t\rightarrow\infty}\frac{c \int_0^t({\rm e}^{-c\theta s}X_s-\eta_\infty){\rm e}^{c\theta s}{\rm d}s+c\eta_\infty\int_0^t{\rm e}^{c\theta s}{\rm d}s}{{\rm e}^{c\theta t}}\nonumber\\ &=&\lim\limits_{t\rightarrow\infty}\left(\frac{\lambda_t^2}{{\rm e}^{c\theta t}}\int_0^t({\rm e}^{c\theta s}X_s-\eta_\infty)\frac{{\rm e}^{c\theta s}}{\lambda_t^2}{\rm d}s+c\eta_\infty\frac{\int_0^t{\rm e}^{c\theta s}{\rm d}s} {{\rm e}^{c\theta t}}\right)\nonumber\\ &=&\frac{\eta_\infty}{\theta},\quad \mbox{a.s.}, \end{matrix}$

$\begin{matrix}\label{law} \lim\limits_{t\rightarrow\infty}\frac{Z_t}{{\rm e}^{c\theta t}}=0,\quad \mbox{a.s.} \end{matrix}$

$\begin{matrix} \lim\limits_{T\rightarrow\infty}(\widehat{\theta}_T-\theta)&=&\lim\limits_{T\rightarrow\infty}\frac{\int_0^T\omega_t^2A_tZ_t{\rm d}t}{\int_0^T\omega_t^2A_t^2} =\lim\limits_{T\rightarrow\infty}\frac{\int_0^T\frac{A_t}{{\rm e}^{c\theta t}}\frac{Z_t}{{\rm e}^{c\theta t}}\frac{\omega_t^2{\rm e}^{2c\theta t}}{h_1(T)}{\rm d}t}{\int_0^T\left(\frac{A_t}{{\rm e}^{c\theta t}}\right)^2\frac{\omega_t^2{\rm e}^{2c\theta t}}{h_1(T)}{\rm d}t} =0,\quad \mbox{a.s.} \end{matrix}$

(ii) 若 $\theta=0$ 和条件 (A2) 成立, 则

$X_t=x_0+Z_t.$

$\begin{eqnarray*} \int_0^T\omega_t^2A_tZ_t{\rm d}t&=&T^{1+\frac{1}{\alpha}}\int_0^1\omega_{sT}^2\frac{Z_{sT}}{T^{\frac{1}{\alpha}}}\left(\int_0^{sT}\left(c(Z_{\nu}+x_0)+|Z_{\nu}+x_0|^\gamma\right) {\rm d}\nu\right){\rm d}s\\ &=&T^{2+\frac{2}{\alpha}}\omega_T^2\int_0^1\omega_s^2\frac{Z_{sT}}{T^{\frac{1}{\alpha}}}\left(c\int_0^{s}\frac{Z_{uT}+x_0}{T^{\frac{1}{\alpha}}}{\rm d}u\right){\rm d}s\\ &&+T^{2+\frac{\gamma+1}{\alpha}}\omega_T^2\int_0^1\omega_s^2\frac{Z_{sT}}{T^{\frac{1}{\alpha}}}\left(\int_0^{s}\left|\frac{Z_{uT}+x_0}{T^{\frac{1}{\alpha}}} \right|^\gamma {\rm d}u\right){\rm d}s \end{eqnarray*}$

$\begin{eqnarray*} \int_0^T\omega_t^2A_t^2{\rm d}t&=&T\int_0^1\omega_{sT}^2\left(\int_0^{sT}\left(c(Z_{\nu}+x_0)+|Z_{\nu}+x_0|^\gamma\right) {\rm d}\nu\right)^2{\rm d}s\\ &=&T^{3+\frac{2}{\alpha}}\omega_T^2c^2\int_0^1\omega_s^2\left(\int_0^s\frac{Z_{uT}+x_0}{T^{\frac{1}{\alpha}}}{\rm d}u\right)^2{\rm d}s\\ &&+T^{3+\frac{2\gamma}{\alpha}}\omega_T^2\int_0^1\omega_s^2\left(\int_0^s\left|\frac{Z_{uT}+x_0}{T^{\frac{1}{2}}}\right|^\gamma {\rm d}u\right)^2{\rm d}s\\ &&+2T^{3+\frac{1+\gamma}{\alpha}}\omega_T^2c\int_0^1\omega_s^2\left(\int_0^s\frac{Z_{uT}+x_0}{T^{\frac{1}{\alpha}}}{\rm d}u\right)\left(\int_0^s\left|\frac{Z_{uT}+x_0}{T^{\frac{1}{\alpha}}}\right|^\gamma {\rm d}u\right){\rm d}s. \end{eqnarray*}$

$\alpha$ -稳定过程的自相似性质, 可得对任意 $T>0$, 有

$\{Z_t,t\geq0\}\stackrel{d}{=}\left\{T^{\frac{1}{\alpha}}\widetilde{Z}_{\frac{t}{T}},t\geq 0\right\},$

$\{Z_s,s\geq 0\}\stackrel{d}{=}\{T^{\frac{1}{\alpha}}\widetilde{Z}_{\frac{s}{T}},s\geq 0\}$, 其中 $\{\widetilde{Z}_t,t\geq 0\}$$\alpha -稳定过程. 又由Slutsky定理可知当 T\rightarrow\infty $$\label{dis} \frac{\int_0^T\omega_t^2A_tZ_t{\rm d}t}{\int_0^T\omega_t^2A_t^2{\rm d}t} \stackrel{d}{=}\frac{1}{T}\frac{f_1+f_2}{g_1+g_2+g_3} \rightarrow_p 0,$$ 其中 \begin{eqnarray*} &&f_1=c^2\int_0^1\omega_s^2\widetilde{Z}_s\left(\int_0^{s}\left(\widetilde{Z}_u+\frac{x_0}{T^{\frac{1}{\alpha}}}\right){\rm d}u\right){\rm d}s,\\ &&f_2=T^{-\frac{1-\gamma}{\alpha}}\int_0^1\omega_s^2\widetilde{Z}_s\left(\int_0^{s}\left|\widetilde{Z}_u+\frac{x_0}{T^{\frac{1}{\alpha}}}\right|^\gamma {\rm d}u\right){\rm d}s,\\ &&g_1=c^2\int_0^1\omega_s^2\left(\int_0^s\left(\widetilde{Z}_u+\frac{x_0}{T^{\frac{1}{\alpha}}}\right){\rm d}u\right)^2{\rm d}s,\\ &&g_2=T^{-\frac{2(1-\gamma)}{\alpha}}\int_0^1\left(\int_0^s\left|\widetilde{Z}_u+\frac{x_0}{T^{\frac{1}{\alpha}}}\right|^\gamma {\rm d}u\right)^2{\rm d}s,\\ &&g_3=2T^{-\frac{1-\gamma}{\alpha}}\int_0^1\omega_s^2\int_0^s\left(\widetilde{Z}_u+\frac{x_0}{T^{\frac{1}{\alpha}}}\right){\rm d}u\int_0^s\left|\widetilde{Z}_u+\frac{x_0}{T^{\frac{1}{\alpha}}}\right|^\gamma {\rm d}u{\rm d}s. \end{eqnarray*} 所以, 由(1.5)与(3.14)式可知当 T\rightarrow\infty 时, \widehat{\theta}_T\rightarrow_p\theta. 证毕. 注3.1 (i) 若 \theta>0$$a(x)=cx$, 则定理 3.1(i) 退化为文献[9,定理 4.1].

(ii) 若 $\theta=0$$a(x)=|x|^\gamma, 则定理 3.1(ii) 退化为文献[28,定理 2.1(ii)]. ## 4 加权拟合估计量 \widehat{\theta_T} 的渐近分布 定理4.1 若条件 (A1), (A3), (A5) 成立及 \theta>0, 则当 T\rightarrow\infty \begin{matrix}\label{th3.1} \frac{h_1(T)}{h_2(T)T^{\frac{1}{\alpha}}}(\widehat{\theta}_T-\theta)\Rightarrow\theta \frac{U}{\eta_\infty}, \end{matrix} 其中 \eta_\infty 满足(3.9)式和 U 是一个服从 \alpha -稳定分布 S_\alpha(1,\beta,0) 的随机变量, 且与随机变量 \eta_\infty 相互独立. 根据条件 (A1) 和(1.5)式, 可得 \begin{matrix}\label{3.1} \frac{h_1(T)}{h_2(T)T^{\frac{1}{\alpha}}}(\widehat{\theta}_T-\theta) &=&\frac{h_2^{-1}(T)T^{-\frac{1}{\alpha}}\int_0^T\omega_t^2A_tZ_t{\rm d}t}{h_1^{-1}(T)\int_0^T\omega_t^2A_t^2{\rm d}t}\nonumber\\ &=&\frac{\eta_T^2}{h_1^{-1}(T)\int_0^T\omega_t^2A_t^2{\rm d}t}\left(\frac{h_2^{-1}(T)T^{-\frac{1}{\alpha}}Z_T\int_0^T\omega_t^2A_t{\rm d}t}{\eta_T^2}\right.\nonumber\\ &&\left.+\frac{h_2^{-1}(T)T^{-\frac{1}{\alpha}}\int_0^T\omega_t^2A_t(Z_T-Z_t){\rm d}t}{\eta_T^2}\right)\nonumber\\ :&=&H_1(T)(H_2(T)+H_3(T)). \end{matrix} 接下来, 我们将分别研究 H_i(T),\,i=1,2,3 的渐近行为. 由条件 (A3) 和引理 2.2 可知 \lim\limits_{T\rightarrow\infty}h_1^{-1}(T)\int_0^T\omega_t^2A_t^2{\rm d}t=\lim\limits_{T\rightarrow\infty}\int_0^T\left(\frac{A_t}{{\rm e}^{c\theta t}}\right)^2\frac{\omega_t^2{\rm e}^{2c\theta t}}{h_1(T)}{\rm d}t=\frac{\eta_\infty^2}{\theta^2},\quad \mbox{a.s.} 显然, $$\label{3.2} \lim\limits_{T\rightarrow\infty}H_1(T)=\theta^2,\quad \mbox{a.s.}$$ 事实上 $$\label{3.3} H_2(T)=\frac{h_2^{-1}(T)\int_0^T\omega_t^2A_t{\rm d}t}{\eta_T}\frac{T^{-\frac{1}{\alpha}}Z_T}{\eta_T}.$$ 又由引理 2.2 可知 \lim\limits_{T\rightarrow\infty}h_2^{-1}(T)\int_0^T\omega_t^2 A_t{\rm d}t=\lim\limits_{T\rightarrow\infty}\int_0^T\left(\frac{A_t}{{\rm e}^{c\theta t}}\right)\frac{\omega_t^2{\rm e}^{c\theta t}}{h_2(T)}{\rm d}t=\frac{\eta_\infty}{\theta},\quad \mbox{a.s.} 这意味着, $$\label{3.4} \lim\limits_{T\rightarrow\infty}\frac{h_2^{-1}(T)\int_0^T\omega_t^2A_t{\rm d}t}{\eta_T}=\frac{1}{\theta},\quad\mbox{a.s.}$$ H_2(T) 中的第二项 \frac{T^{-\frac{1}{\alpha}}Z_T}{\eta_T} 可表示为 \frac{T^{-\frac{1}{\alpha}}Z_T}{\eta_T}=\frac{T^{-\frac{1}{\alpha}}(Z_T-Z_{T^{\frac{1}{\alpha}}})+T^{-\frac{1}{\alpha}}Z_{T^{\frac{1}{\alpha}}}}{\eta_{T^{\frac{1}{\alpha}}}+(\eta_T-\eta_{T^{\frac{1}{\alpha}}})}. 我们可以得到以下几个结论 (1) 随机变量 \frac{1}{T^\frac{1}{\alpha}}(Z_T-Z_{T^\frac{1}{\alpha}}) 服从 \alpha -稳定分布S_\alpha((1-T^{\frac{1}{\alpha}-1})^\frac{1}{\alpha},\beta,0), 且当 T\rightarrow\infty 时, 随机变量 \frac{1}{T^\frac{1}{\alpha}}(Z_T-Z_{T^\frac{1}{\alpha}}) 弱收敛于一个分布为 S_\alpha(1,\beta,0)$$\alpha$ -稳定随机变量 $U$.

(2) 由鞅强大数定律可得

$\lim\limits_{T\rightarrow\infty}\frac{Z_{T^{\frac{1}{\alpha}}}}{T^{\frac{1}{\alpha}}}=0,\quad \mbox{a.s.}$

(3) 显然,

$\lim\limits_{T\rightarrow\infty}\eta_{T^{\frac{1}{\alpha}}}=\eta_\infty,\quad \mbox{a.s.}$

(4) 随机变量 $T^{-\frac{1}{\alpha}}(Z_T-Z_{T^{\frac{1}{\alpha}}})$$\eta_{T^{\frac{1}{\alpha}}} 相互独立. (5) 当 T\rightarrow\infty 时, 有 \eta_T-\eta_{T^\frac{1}{\alpha}}\rightarrow_p 0. 证明结论 (5). 由 \eta_t 的定义可得 \begin{matrix}\label{3.5} \eta_T-\eta_{T^{\frac{1}{\alpha}}}=\theta\int_{T^{\frac{1}{\alpha}}}^T{\rm e}^{-c\theta s}r(X_s){\rm d}s+\int_{T^{\frac{1}{\alpha}}}^T{\rm e}^{-c\theta s}{\rm d}Z_s. \end{matrix} 这意味着, \begin{matrix}\label{non3.6} |\eta_T-\eta_{T^{\frac{1}{\alpha}}}| &\leq& \theta K\int_{T^{\frac{1}{\alpha}}}^T{\rm e}^{-c\theta s}(1+|X_s|^\gamma ){\rm d}s+\left|\int_{T^{\frac{1}{\alpha}}}^T{\rm e}^{-c\theta s}{\rm d}Z_s\right|\nonumber\\ &\leq& \theta K \frac{{\rm e}^{-c\theta T^{\frac{1}{\alpha}}}-{\rm e}^{-c\theta T}}{c\theta}+\theta K\int_{T^{\frac{1}{\alpha}}}^T|{\rm e}^{-c\theta s}X_s|^\gamma {\rm e}^{-c\theta s(1-\gamma)}{\rm d}s+\left|\int_{T^{\frac{1}{\alpha}}}^T{\rm e}^{-c\theta s}{\rm d}Z_s\right|\nonumber\\ &\leq& \theta K \frac{{\rm e}^{-c\theta T^{\frac{1}{\alpha}}}-{\rm e}^{-c\theta T}}{c\theta}+K\eta_*^\gamma \frac{{\rm e}^{-c\theta (1-\gamma) T^{\frac{1}{\alpha}}}-{\rm e}^{-c\theta (1-\gamma) T}}{c(1-\gamma)}+\left|\int_{T^{\frac{1}{\alpha}}}^T{\rm e}^{-c\theta s}{\rm d}Z_s\right|, \end{matrix} 其中 \eta_* 满足(3.7)式. 不难发现, 当 T\rightarrow\infty 时, 不等式(4.7)右边的第一项与第二项收敛于 0. 对于不等式(4.7)右边的第三项 \left|\int_{T^{\frac{1}{\alpha}}}^T{\rm e}^{-\theta s}{\rm d}Z_s\right|, 根据 Markov 不等式和引理, 可得对任意给定的 \delta>0, 当 T\rightarrow\infty \begin{matrix}\label{3.7} {\Bbb P}\left\{\left|\int_{T^{\frac{1}{\alpha}}}^T{\rm e}^{-c\theta s}{\rm d}Z_s\right|>\delta\right\}&\leq& \delta^{-1}{\Bbb E}\left|\int_{T^{\frac{1}{\alpha}}}^T{\rm e}^{-c\theta s}{\rm d}Z_s\right|\nonumber\\ &\leq & \delta^{-1}C_2\left(\int_{T^{\frac{1}{\alpha}}}^T{\rm e}^{-\alpha c\theta s}{\rm d}s\right)^{\frac{1}{\alpha}}\nonumber\\ &\leq & \delta^{-1}C_2\left(\frac{{\rm e}^{-\alpha c \theta T}-{\rm e}^{-\alpha c\theta T^{\frac{1}{\alpha}}}}{\alpha c\theta}\right)^{\frac{1}{\alpha}} \end{matrix} 趋近于 0.(4.6)-(4.8)式联立可得结论(5) 成立. 由结论 (1)-(5), 可知当 T\rightarrow\infty \begin{matrix}\label{3.8} \frac{T^{-\frac{1}{\alpha}}Z_T}{\eta_T}\Rightarrow\frac{U}{\eta_\infty}, \end{matrix} 其中 U$$\eta_\infty$ 相互独立. 根据(4.4),(4.5), 及(4.9)式, 可得当 $T\rightarrow\infty$

$\begin{matrix}\label{3.9} H_2(T)\Rightarrow\frac{U}{\theta\eta_\infty}. \end{matrix}$

$\begin{matrix}\label{4.0} &&\left|T^{-\frac{1}{\alpha}}h_2^{-1}(T)\int_0^T\omega_t^2A_t(Z_T-Z_t){\rm d}t\right|\nonumber\\ &\leq & T^{-\frac{1}{\alpha}}h_2^{-1}(T)\int_0^T\omega_t^2\left|\int_0^t(cX_s+r(X_s)){\rm d}s\right||Z_T-Z_t|{\rm d}t\nonumber\\ &\leq & T^{-\frac{1}{\alpha}}h_2^{-1}(T)\int_0^T\left(\int_0^t\left(c{\rm e}^{c\theta s}|{\rm e}^{-c\theta s}X_s|+K\left(1+|X_s{\rm e}^{-c\theta s}|^\gamma {\rm e}^{c\theta\gamma s}\right)\right){\rm d}s\right)|Z_T-Z_t|\omega_t^2{\rm d}t\nonumber\\ &\leq & \frac{\eta_*}{\theta}T^{-\frac{1}{\alpha}}h_2^{-1}(T)\int_0^T|Z_T-Z_t|\omega_t^2{\rm e}^{c\theta t}{\rm d}t+KT^{-\frac{1}{\alpha}}h_2^{-1}(T)\int_0^T|Z_T-Z_t|\omega_t^2t{\rm d}t\nonumber\\ &&+\frac{K\eta_*^\gamma}{c\theta\gamma}T^{-\frac{1}{\alpha}}h_2^{-1}(T)\int_0^T|Z_T-Z_t|\omega_t^2{\rm e}^{c\theta\gamma t}{\rm d}t\nonumber\\ :&=&G_1(T)+G_2(T)+G_3(T), \end{matrix}$

$\begin{matrix}\label{4.1} {\Bbb E}\left[T^{-\frac{1}{\alpha}}h_2^{-1}(T)\int_0^T|Z_T-Z_t|\omega_t^2{\rm e}^{c\theta t}{\rm d}t\right]&\leq& T^{-\frac{1}{\alpha}}\int_0^T{\Bbb E}[|Z_T-Z_t|]\frac{\omega_t^2{\rm e}^{c\theta t}}{h_2(T)}{\rm d}t\nonumber\\ &\leq & C'T^{-\frac{1}{\alpha}}\int_0^TC(1,\alpha)(T-t)^{\frac{1}{\alpha}}{\rm e}^{b(T-t)}{\rm d}t\nonumber\\ &=&C'T^{-\frac{1}{\alpha}}\int_0^TC(1,\alpha)u^{\frac{1}{\alpha}}{\rm e}^{bu}{\rm d}u\nonumber\\ &\leq& C(1,\alpha)C'T^{-\frac{1}{\alpha}}\int_0^\infty u^{\frac{1}{\alpha}}{\rm e}^{bu}{\rm d}u\nonumber\\ &\leq & C(1,\alpha)C'T^{-\frac{1}{\alpha}}\Gamma\left(1+\frac{1}{\alpha}\right)|b|^{-(1+\frac{1}{\alpha})}, \end{matrix}$

$\begin{eqnarray*} {\Bbb E}\left[T^{-\frac{1}{\alpha}}h_2^{-1}(T)\int_0^T|Z_T-Z_t|\omega_t^2{\rm e}^{c\theta t}{\rm d}t\right]\rightarrow 0. \end{eqnarray*}$

$$$\label{4.2} G_1(T)\rightarrow_p0.$$$

$\begin{matrix}\label{4.3} {\Bbb E}[G_2(T)]&\leq & KT^{-\frac{1}{\alpha}}\int_0^T{\Bbb E}[|Z_T-Z_t|]\frac{\omega_t^2{\rm e}^{c\theta t}}{h_2(T)}\frac{t}{{\rm e}^{c\theta t}}{\rm d}t\nonumber\\ &\leq & \frac{1}{c\theta} KT^{-\frac{1}{\alpha}}\int_0^T{\Bbb E}[|Z_T-Z_t|]\frac{\omega_t^2{\rm e}^{c\theta t}}{h_2(T)}{\rm d}t \end{matrix}$

$\begin{matrix}\label{4.4} {\Bbb E}[G_3(T)]&\leq&\frac{K\eta_*^\gamma}{c\theta\gamma}T^{-\frac{1}{\alpha}}\int_0^T{\Bbb E}[|Z_T-Z_t|]\frac{\omega_t^2{\rm e}^{c\theta t}}{h_2(T)}\frac{1}{{\rm e}^{c\theta(1-\gamma) t}}{\rm d}t\nonumber\\ &\leq&\frac{K\eta_*^\gamma}{c\theta\gamma}T^{-\frac{1}{\alpha}}\int_0^T{\Bbb E}[|Z_T-Z_t|]\frac{\omega_t^2{\rm e}^{c\theta t}}{h_2(T)}{\rm d}t. \end{matrix}$

$\begin{matrix}\label{4.5} G_2(T)\rightarrow_p0,\quad G_3(T)\rightarrow_p0. \end{matrix}$

$\begin{matrix}\label{3.11} H_3(T)\rightarrow_p0. \end{matrix}$

$h_1(T)=\int_0^T {\rm e}^{rt}{\rm e}^{2c\theta t}{\rm d}t=\frac{{\rm e}^{(2c\theta+r) T}-1}{2c\theta+r},$
$h_2(T)=\int_0^T {\rm e}^{rt}{\rm e}^{c\theta t}{\rm d}t=\frac{{\rm e}^{(c\theta+r) T}-1}{c\theta+r}.$

$\frac{\omega_t^2{\rm e}^{c\theta t}}{h_2(T)}\leq C'{\rm e}^{-(r+c\theta)(T-t)}.$

$\begin{matrix}\label{th3.2} T\widehat{\theta}_T\Rightarrow\frac{\int_0^1\omega_s^2\tilde{Z}_s\left(\int_0^{s}\tilde{Z}_u {\rm d}u\right){\rm d}s}{\int_0^1\omega_{s}^2\left(\int_0^1\tilde{Z}_u {\rm d}u\right)^2{\rm d}s}, \end{matrix}$

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