# 基于离散观测的带有小的α稳定噪声的Vasicek利率模型的最小二乘估计Least Squares Estimators for Discretely Observed Vasicek Interest Rate Model with Small α Stable Noises

• 全文下载: PDF(686KB)    PP.539-549   DOI: 10.12677/SA.2017.65061
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dXt=(a-bXt)dt+δdZt

In this paper, we consider the problem of parameter estimation for Vasicek interest rate model with small α-stable noises, observed at n regularly spaced time points ti=(i/n,i=1,L,n) on [0,1]

dXt=(a-bXt)dt+δdZt

Least squares method is used to obtain a and b. The consistencies and asymptotic distributions of the LSE are established when δ→0 and δ→∞ simultaneously.

1. 引言

$\left(\Omega ,\mathcal{F},\text{P}\right)$ 为基本概率空间，具备右连续和增的σ-代数族 $\left({\mathcal{F}}_{t},t\ge 0\right)$ 。设 $Z=\left\{{Z}_{t},t\ge 0\right\}$ 为概率空间 $\left(\Omega ,\mathcal{F},\text{P}\right)$ 上的标准α-稳定Lévy运动， ${Z}_{1}~{S}_{\alpha }\left(1,\beta ,0\right)$$\beta \in \left[-1,1\right]$ 为偏度参数。本文我们考虑由标准稳定过程 驱动的Vasicek利率模型的参数估计问题。Vasicek利率模型是一种均值回复Ornstein-Uhlenbeck过程，由Vasicek在1977年首次提出(当 $\alpha =2$ ) [1] ，可由下列线性随机微分方程刻画：

$\text{d}{X}_{t}=\left(a-b{X}_{t}\right)\text{d}t+\sigma \text{d}{Z}_{t}$ (1)

${\rho }_{n,\sigma }\left(a,b\right)={\rho }_{n,\sigma }\left(a,b;{\left({X}_{{t}_{i}}\right)}_{i=1}^{n}\right)=\underset{i=1}{\overset{n}{\sum }}{|{X}_{{t}_{i}}-{X}_{{t}_{i-1}}+b{X}_{{t}_{i-1}}\Delta {t}_{i-1}-a\Delta {t}_{i-1}|}^{2}$

${\stackrel{^}{b}}_{n,\sigma }$${\stackrel{^}{a}}_{n,\sigma }$ 的最小二乘估计量即求 ${\stackrel{^}{b}}_{n,\sigma }=\mathrm{arg}{\mathrm{min}}_{b}{\rho }_{n,\sigma }\left(a,b\right)$${\stackrel{^}{a}}_{n,\sigma }=\mathrm{arg}{\mathrm{min}}_{a}{\rho }_{n,\sigma }\left(a,b\right)$ ，解得

${\stackrel{^}{b}}_{n,\sigma }=-\frac{n\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{X}_{{t}_{i-1}}\left({X}_{{t}_{i}}-{X}_{{t}_{i-1}}\right)}{\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{X}_{{t}_{i-1}}^{2}}+\frac{{\stackrel{^}{a}}_{n,\sigma }\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{X}_{{t}_{i-1}}}{\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{X}_{{t}_{i-1}}^{2}}$ (2)

${\stackrel{^}{a}}_{n,\sigma }=\underset{i=1}{\overset{n}{\sum }}\left({X}_{{t}_{i}}-{X}_{{t}_{i-1}}\right)+\frac{1}{n}\cdot {\stackrel{^}{b}}_{n,\sigma }\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{X}_{{t}_{i-1}}$ (3)

Vasicek利率模型的解为：

${X}_{{t}_{i}}={\text{e}}^{-\frac{b}{n}}{X}_{{t}_{i-1}}+\frac{a}{b}\left(1-{\text{e}}^{-\frac{b}{n}}\right)+\sigma {Z}_{i,\frac{1}{n}}$ (4)

${Z}_{i,\frac{1}{n}}=\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}\text{ }{\text{e}}^{-b\left({t}_{i}-s\right)}\text{d}{z}_{s}$

$\Upsilon \left(n,\sigma \right)=\sigma {n}^{2}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{X}_{{t}_{i}}{Z}_{i,\frac{1}{n}}-\sigma n\underset{i=1}{\overset{n}{\sum }}\text{ }{Z}_{i,\frac{1}{n}}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{X}_{{t}_{i-1}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Theta \left(n,\sigma \right)$ $=n\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{X}_{{t}_{i-1}}^{2}-{\left(\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{X}_{{t}_{i-1}}\right)}^{2}$

${\stackrel{^}{b}}_{n,\sigma }=\left(1-{\text{e}}^{-\frac{b}{n}}\right)n-\frac{\Upsilon \left(n,\sigma \right)}{\Theta \left(n,\sigma \right)}$

${\sigma }^{-1}\left({\stackrel{^}{b}}_{n,\sigma }-b\right){\to }_{d}\frac{\left({X}_{0}+\frac{a}{b}\right){\left(\frac{1-{\text{e}}^{-\alpha b}}{\alpha b}\right)}^{\frac{1}{\alpha }}}{\left({C}_{1}+{C}_{2}+{C}_{3}\right)-\left({C}_{4}+{C}_{5}+{C}_{6}\right)}U$

${\sigma }^{-1}\left({\stackrel{^}{a}}_{n,\sigma }-a-\sigma \underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{Z}_{i,\frac{1}{n}}\right){\to }_{d}D\frac{\left({X}_{0}+\frac{a}{b}\right){\left(\frac{1-{\text{e}}^{-\alpha b}}{\alpha b}\right)}^{\frac{1}{\alpha }}}{\left({C}_{1}+{C}_{2}+{C}_{3}\right)-\left({C}_{4}+{C}_{5}+{C}_{6}\right)}U$

2. 准备知识

2.1. Lévy过程

1) ${Z}_{0}=0$ a.s；

2) Z具有独立增量，i.e. ${Z}_{t}-{Z}_{s}$${\mathcal{F}}_{t}$ 独立， $0\le s

3) Z具有平稳增量，i.e. ${Z}_{t}-{Z}_{s}$${Z}_{t-s}$ 具有相同的分布， $0\le s

4) Z随机连续(依概率连续)，i.e.对所有的 $\epsilon >0$$s\ge 0$${\mathrm{lim}}_{t\to s}P\left(|{Z}_{t}-{Z}_{s}|>\epsilon \right)=0$

2.2. a-稳定分布

${\varphi }_{\alpha }\left(x\right)=E\mathrm{exp}\left\{ix\eta \right\}=\left\{\begin{array}{l}\mathrm{exp}\left\{-{\gamma }^{\alpha }{|x|}^{\alpha }\left(1-i\beta \mathrm{sgn}\left(x\right)\mathrm{tan}\frac{\alpha \text{π}}{2}\right)+i\mu x\right\},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\alpha \ne 1\\ \mathrm{exp}\left\{-\gamma |x|\left(1+i\frac{2}{\text{π}}\beta \mathrm{sgn}\left(x\right)\mathrm{log}|x|\right)+i\mu x\right\},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\alpha =1\end{array}$

${\left(E{|\underset{0}{\overset{t}{\int }}f\left(s\right)\text{d}{Z}_{s}|}^{q}\right)}^{\frac{1}{q}}=C\left(q,\alpha ,\beta \right){\left(\underset{0}{\overset{t}{\int }}{f}^{\alpha }\left(s\right)\text{d}s\right)}^{\frac{1}{\alpha }}$

3. 最小二乘估计量 ${\stackrel{^}{b}}_{n,\sigma }$${\stackrel{^}{a}}_{n,\sigma }$ 的相合性

$\begin{array}{c}{\stackrel{^}{b}}_{n,\sigma }-b=\left[\left(1-{\text{e}}^{-b/n}\right)n-b\right]-\frac{\sigma \underset{i=1}{\overset{n}{\sum }}\text{ }{X}_{{t}_{i-1}}{Z}_{i,\frac{1}{n}}-{n}^{-1}\sigma \underset{i=1}{\overset{n}{\sum }}\text{ }{Z}_{i,\frac{1}{n}}\underset{i=1}{\overset{n}{\sum }}\text{ }{X}_{{t}_{i-1}}}{{n}^{-1}\underset{i=1}{\overset{n}{\sum }}\text{ }{X}_{{t}_{i-1}}^{2}-{\left({n}^{-1}\underset{i=1}{\overset{n}{\sum }}\text{ }{X}_{{t}_{i-1}}\right)}^{2}}\\ :={\Lambda }_{n}-\frac{{\Phi }_{1}\left(n,\sigma \right)-{\Phi }_{2}\left(n,\sigma \right)}{{\Phi }_{3}\left(n,\sigma \right)-{\Phi }_{4}\left(n,\sigma \right)}\end{array}$

${X}_{{t}_{i-1}}={\text{e}}^{-b{t}_{i-1}}{X}_{0}+\frac{a}{b}\left(1-{\text{e}}^{-b{t}_{i-1}}\right)+\sigma \underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-b\left({t}_{i-1}-s\right)}\text{d}{Z}_{s}$

$\begin{array}{c}{\Phi }_{1}\left(n,\sigma \right)=\sigma {X}_{0}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-b{t}_{i-1}}\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}\text{ }{\text{e}}^{-b\left({t}_{i}-s\right)}\text{d}{Z}_{s}+\frac{\sigma a}{b}\underset{i=1}{\overset{n}{\sum }}\text{ }\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}\text{ }{\text{e}}^{-b\left({t}_{i}-s\right)}\text{d}{Z}_{s}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-\frac{\sigma a}{b}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-b{t}_{i-1}}\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}{\text{e}}^{-b\left({t}_{i}-s\right)}\text{d}{Z}_{s}+{\sigma }^{2}\underset{i=1}{\overset{n}{\sum }}\text{ }\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-b\left({t}_{i-1}-s\right)}\text{d}{Z}_{s}\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}\text{ }{\text{e}}^{-b\left({t}_{i}-s\right)}\text{d}{Z}_{s}\\ :=\underset{i=1}{\overset{4}{\sum }}\text{ }\text{ }{\Phi }_{1,i}\left(n,\sigma \right)\end{array}$

$\begin{array}{c}P\left(|{\Phi }_{1,1}\left(n,\sigma \right)|>\epsilon \right)\le {\epsilon }^{-1}E|\sigma {X}_{0}\underset{i=1}{\overset{n}{\sum }}\text{ }{\text{e}}^{-b{t}_{i-1}}\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}{\text{e}}^{-b\left({t}_{i}-s\right)}\text{d}{Z}_{s}|\\ \le {\epsilon }^{-1}\sigma E|{X}_{0}|E|\underset{0}{\overset{1}{\int }}\underset{i=1}{\overset{n}{\sum }}\text{ }{\text{e}}^{-b\left({t}_{i-1}+{t}_{i}\right)}{\text{e}}^{bs}{1}_{\left({t}_{i-1},{t}_{i}\right]}\text{d}{Z}_{s}|\\ \le C{\epsilon }^{-1}\sigma {\left(\underset{0}{\overset{1}{\int }}{|\underset{i=1}{\overset{n}{\sum }}\text{ }{\text{e}}^{-b\left({t}_{i-1}+{t}_{i}\right)}{\text{e}}^{bs}{1}_{\left({t}_{i-1},{t}_{i}\right]}|}^{\alpha }\text{d}s\right)}^{\frac{1}{\alpha }}\\ =C{\epsilon }^{-1}\sigma {\left(\frac{{\text{e}}^{\alpha b/n}-1}{\alpha b}\underset{i=1}{\overset{n}{\sum }}\text{ }{\text{e}}^{-\alpha b{t}_{i}}\right)}^{\frac{1}{\alpha }}\end{array}$

$n\to \infty$$\sigma \to 0$ 时，上式趋向于0。 ${\Phi }_{1,3}\left(n,\sigma \right)$${\Phi }_{1,1}\left(n,\sigma \right)$ 的证明类似，当 $n\to \infty$$\sigma \to 0$ 时， ${\Phi }_{1,3}\left(n,\sigma \right)$ 依概率收敛到0。

$\begin{array}{c}P\left(|{\Phi }_{1,2}\left(n,\sigma \right)|>\epsilon \right)\le {\epsilon }^{-1}E|\frac{\sigma a}{b}\underset{i=1}{\overset{n}{\sum }}\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}{\text{e}}^{-b\left({t}_{i}-s\right)}\text{d}{Z}_{s}|\\ \le C{\epsilon }^{-1}\sigma \underset{i=1}{\overset{n}{\sum }}{\left(\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}{\text{e}}^{-\alpha b\left({t}_{i}-s\right)}\text{d}s\right)}^{\frac{1}{\alpha }}\\ =C{\epsilon }^{-1}\sigma n{\left(\frac{1-{\text{e}}^{-\alpha b/n}}{\alpha b}\right)}^{\frac{1}{\alpha }}\\ =O\left(\sigma {n}^{1-\frac{1}{\alpha }}\right)\end{array}$ (5)

$\sigma {n}^{1-\frac{1}{\alpha }}\to 0$ 时上式趋向于0。

$\begin{array}{c}P\left(|{\Phi }_{1,4}\left(n,\sigma \right)|>\epsilon \right)\le {\epsilon }^{-1}E|{\sigma }^{2}\underset{i=1}{\overset{n}{\sum }}\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-b\left({t}_{i-1}-s\right)}\text{d}{Z}_{s}\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}{\text{e}}^{-b\left({t}_{i}-s\right)}\text{d}{Z}_{s}|\\ \le C{\epsilon }^{-1}{\sigma }^{2}\underset{i=1}{\overset{n}{\sum }}{\left(\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-\alpha b\left({t}_{i-1}-s\right)}\text{d}s\right)}^{\frac{1}{\alpha }}{\left(\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}{\text{e}}^{-\alpha b\left({t}_{i}-s\right)}\text{d}s\right)}^{\frac{1}{\alpha }}\\ =C{\epsilon }^{-1}{\sigma }^{2}{\left(\frac{1-{\text{e}}^{-\alpha b/n}}{\alpha b}\right)}^{\frac{1}{\alpha }}\underset{i=1}{\overset{n}{\sum }}{\left(\frac{1-{\text{e}}^{-\alpha b{t}_{i-1}}}{\alpha b}\right)}^{\frac{1}{\alpha }}=O\left({\sigma }^{2}{n}^{1-\frac{1}{\alpha }}\right)\end{array}$

$\begin{array}{c}{\Phi }_{2}\left(n,\sigma \right)={n}^{-1}\sigma \underset{i=1}{\overset{n}{\sum }}\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}{\text{e}}^{-b\left({t}_{i}-s\right)}\text{d}{Z}_{s}\underset{i=1}{\overset{n}{\sum }}\text{ }{\text{e}}^{-b{t}_{i-1}}{X}_{0}+\frac{\sigma a}{b}\underset{i=1}{\overset{n}{\sum }}\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}{\text{e}}^{-b\left({t}_{i}-s\right)}\text{d}{Z}_{s}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-{n}^{-1}\frac{\sigma a}{b}\underset{i=1}{\overset{n}{\sum }}\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}{\text{e}}^{-b\left({t}_{i}-s\right)}\text{d}{Z}_{s}\underset{i=1}{\overset{n}{\sum }}\text{ }{\text{e}}^{-b{t}_{i-1}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{n}^{-1}{\sigma }^{2}\underset{i=1}{\overset{n}{\sum }}\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}{\text{e}}^{-b\left({t}_{i}-s\right)}\text{d}{Z}_{s}\underset{i=1}{\overset{n}{\sum }}\text{ }\underset{0}{\overset{{t}_{i}}{\int }}\text{ }{\text{e}}^{-b\left({t}_{i}-s\right)}\text{d}{Z}_{s}\\ :=\underset{i=1}{\overset{4}{\sum }}\text{ }\text{ }{\Phi }_{2,i}\left(n,\sigma \right)\end{array}$

$\begin{array}{c}P\left(|{\Phi }_{2,1}\left(n,\sigma \right)|>\epsilon \right)\le {\epsilon }^{-1}E|{n}^{-1}\sigma \underset{i=1}{\overset{n}{\sum }}\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}{\text{e}}^{-b\left({t}_{i}-s\right)}\text{d}{Z}_{s}\underset{i=1}{\overset{n}{\sum }}\text{ }{\text{e}}^{-b{t}_{i-1}}{X}_{0}|\\ \le {\epsilon }^{-1}{n}^{-1}\sigma \underset{i=1}{\overset{n}{\sum }}\text{ }{\text{e}}^{-b{t}_{i-1}}E|{X}_{0}|\underset{i=1}{\overset{n}{\sum }}E|\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}{\text{e}}^{-b\left({t}_{i}-s\right)}\text{d}{Z}_{s}|\\ \le C{\epsilon }^{-1}\sigma \underset{i=1}{\overset{n}{\sum }}{\left(\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}{\text{e}}^{-\alpha b\left({t}_{i}-s\right)}\text{d}s\right)}^{\frac{1}{\alpha }}\\ =C{\epsilon }^{-1}\sigma {\left(\frac{1-{\text{e}}^{-\alpha b/n}}{\alpha b}\right)}^{\frac{1}{\alpha }}\to 0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}当n\to \infty ,\text{\hspace{0.17em}}\sigma \to 0\end{array}$

$\begin{array}{c}P\left(|{\Phi }_{2,4}\left(n,\sigma \right)|>\epsilon \right)\le {\epsilon }^{-1}E|{n}^{-1}{\sigma }^{2}\underset{i=1}{\overset{n}{\sum }}\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}{\text{e}}^{-b\left({t}_{i}-s\right)}\text{d}{Z}_{s}\underset{i=1}{\overset{n}{\sum }}\underset{0}{\overset{{t}_{i}}{\int }}\text{ }{\text{e}}^{-b\left({t}_{i}-s\right)}\text{d}{Z}_{s}|\\ \le C{\epsilon }^{-1}{n}^{-1}{\sigma }^{2}\underset{i=1}{\overset{n}{\sum }}{\left(\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}{\text{e}}^{-\alpha b\left({t}_{i}-s\right)}\text{d}s\right)}^{\frac{1}{\alpha }}\underset{i=1}{\overset{n}{\sum }}{\left(\underset{0}{\overset{{t}_{i}}{\int }}\text{ }{\text{e}}^{-\alpha b\left({t}_{i}-s\right)}\text{d}s\right)}^{\frac{1}{\alpha }}\\ =C{\epsilon }^{-1}{\sigma }^{2}{\left(\frac{1-{\text{e}}^{-\alpha b/n}}{\alpha b}\right)}^{\frac{1}{\alpha }}\underset{i=1}{\overset{n}{\sum }}{\left(\frac{1-{\text{e}}^{-\alpha b{t}_{i-1}}}{\alpha b}\right)}^{\frac{1}{\alpha }}\\ =O\left({\sigma }^{2}{n}^{1-\frac{1}{\alpha }}\right)\end{array}$

${\Phi }_{3}\left(n,\sigma \right)-{\Phi }_{4}\left(n,\sigma \right){\to }_{p}\left({C}_{1}+{C}_{2}+{C}_{3}\right)-\left({C}_{4}+{C}_{5}+{C}_{6}\right)$

$\begin{array}{c}{\Phi }_{3}\left(n,\sigma \right)=\frac{1}{n}\underset{i=1}{\overset{n}{\sum }}{\left({\text{e}}^{-b{t}_{i-1}}{X}_{0}+\frac{a}{b}\left(1-{\text{e}}^{-b{t}_{i-1}}\right)+\sigma \underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-b\left({t}_{i-1}-s\right)}\text{d}{Z}_{s}\right)}^{2}\\ =\frac{{X}_{0}^{2}}{n}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-2b{t}_{i-1}}+\frac{{a}^{2}}{{b}^{2}n}\underset{i=1}{\overset{n}{\sum }}{\left(1-{\text{e}}^{-b{t}_{i-1}}\right)}^{2}+\frac{{\sigma }^{2}}{n}\underset{i=1}{\overset{n}{\sum }}{\left(\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-b\left({t}_{i-1}-s\right)}\text{d}{Z}_{s}\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{2a}{bn}{X}_{0}\underset{i=1}{\overset{n}{\sum }}\left({\text{e}}^{-b{t}_{i-1}}-{\text{e}}^{-2b{t}_{i-1}}\right)+\frac{2\sigma }{n}{X}_{0}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-b{t}_{i-1}}\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-b\left({t}_{i-1}-s\right)}\text{d}{Z}_{s}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{2a\sigma }{bn}\underset{i=1}{\overset{n}{\sum }}\left(1-{\text{e}}^{-b{t}_{i-1}}\right)\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-b\left({t}_{i-1}-s\right)}\text{d}{Z}_{s}\\ :={\Phi }_{3,1}\left(n\right)+{\Phi }_{3,2}\left(n\right)+{\Phi }_{3,3}\left(n,\sigma \right)+{\Phi }_{3,4}\left(n\right)+{\Phi }_{3,5}\left(n,\sigma \right)+{\Phi }_{3,6}\left(n,\sigma \right)\end{array}$

${\Phi }_{3,1}\left(n\right)\to {X}_{0}^{2}\underset{0}{\overset{1}{\int }}\text{ }\text{ }{\text{e}}^{-2bs}\text{d}s=\frac{{X}_{0}^{2}}{2b}\left(1-{\text{e}}^{-2b}\right)$

${\Phi }_{3,2}\left(n\right)\to \frac{{a}^{2}}{{b}^{2}}-\frac{{a}^{2}}{2{b}^{3}}\left({\text{e}}^{-b}-1\right)\left({\text{e}}^{-b}-3\right)$

${\Phi }_{3,4}\left(n\right)\to \frac{a}{{b}^{2}}{X}_{0}{\left({\text{e}}^{-b}-1\right)}^{2}$

$\begin{array}{c}P\left(|{\Phi }_{3,3}\left(n,\sigma \right)|>\epsilon \right)\le {\epsilon }^{-\frac{p}{2}}E{|\frac{{\sigma }^{2}}{n}\underset{i=1}{\overset{n}{\sum }}{\left(\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-b\left({t}_{i-1}-s\right)}\text{d}{Z}_{s}\right)}^{2}|}^{\frac{p}{2}}\\ \le {\epsilon }^{-\frac{p}{2}}{\sigma }^{p}{n}^{-\frac{p}{2}}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-pb{t}_{i-1}}E{|\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-b\left({t}_{i-1}-s\right)}\text{d}{Z}_{s}|}^{p}\\ \le C{\epsilon }^{-\frac{p}{2}}{\sigma }^{p}{n}^{-\frac{p}{2}}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-pb{t}_{i-1}}{\left(\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-\alpha b\left({t}_{i-1}-s\right)}\text{d}s\right)}^{\frac{p}{\alpha }}\\ =O\left({\sigma }^{p}{n}^{1-\frac{p}{2}}\right)\end{array}$

$n\to \infty ,\text{\hspace{0.17em}}\sigma \to 0,\text{\hspace{0.17em}}\sigma {n}^{\frac{1}{p}-\frac{1}{2}}\to 0$ 时，上式趋向于0。

$\begin{array}{c}P\left(|{\Phi }_{3,5}\left(n,\sigma \right)|>\epsilon \right)\le {\epsilon }^{-1}E|\frac{2\sigma }{n}{X}_{0}\underset{i=1}{\overset{n}{\sum }}\text{ }{\text{e}}^{-2b{t}_{i-1}}\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{bs}\text{d}{Z}_{s}|\\ \le C{\epsilon }^{-1}\frac{\sigma }{n}\underset{i=1}{\overset{n}{\sum }}\text{ }{\text{e}}^{-2b{t}_{i-1}}{\left(\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{\alpha bs}\text{d}s\right)}^{\frac{1}{\alpha }}\end{array}$

$n\to \infty ,\sigma \to 0$ 时，上式收敛到0。与 ${\Phi }_{3,5}\left(n,\sigma \right)$ 证明类似， ${\Phi }_{3,6}\left(n,\sigma \right)$ 依概率收敛到0。下面分解 ${\Phi }_{4}\left(n,\sigma \right)$ 如下：

$\begin{array}{c}{\Phi }_{4}\left(n,\sigma \right)={n}^{-2}{\left(\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-b{t}_{i-1}}{X}_{0}+\frac{na}{b}-\frac{a}{b}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-b{t}_{i-1}}+\sigma \underset{i=1}{\overset{n}{\sum }}\text{ }\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-b\left({t}_{i-1}-s\right)}\text{d}{Z}_{s}\right)}^{2}\\ =\frac{{X}_{0}^{2}}{{n}^{2}}{\left(\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-b{t}_{i-1}}\right)}^{2}+\frac{{a}^{2}}{{n}^{2}{b}^{2}}{\left(n-\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-b{t}_{i-1}}\right)}^{2}+\frac{{\sigma }^{2}}{{n}^{2}}{\left(\underset{i=1}{\overset{n}{\sum }}\text{ }\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-b\left({t}_{i-1}-s\right)}\text{d}{Z}_{s}\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{2a{X}_{0}}{b{n}^{2}}\left[n\underset{i=1}{\overset{n}{\sum }}\text{ }{\text{e}}^{-b{t}_{i-1}}-{\left(\underset{i=1}{\overset{n}{\sum }}\text{ }{\text{e}}^{-b{t}_{i-1}}\right)}^{2}\right]+\frac{2\sigma {X}_{0}}{{n}^{2}}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-b{t}_{i-1}}\underset{i=1}{\overset{n}{\sum }}\text{ }\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-b\left({t}_{i-1}-s\right)}\text{d}{Z}_{s}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{2a\sigma }{b{n}^{2}}\left(n-\underset{i=1}{\overset{n}{\sum }}\text{ }{\text{e}}^{-b{t}_{i-1}}\right)\underset{i=1}{\overset{n}{\sum }}\text{ }\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-b\left({t}_{i-1}-s\right)}\text{d}{Z}_{s}\\ :={\Phi }_{4,1}\left(n\right)+{\Phi }_{4,2}\left(n\right)+{\Phi }_{4,3}\left(n,\sigma \right)+{\Phi }_{4,4}\left(n\right)+{\Phi }_{4,5}\left(n,\sigma \right)+{\Phi }_{4,6}\left(n,\sigma \right)\end{array}$

${\Phi }_{4,1}\left(n\right)\to \frac{{X}_{0}^{2}}{{b}^{2}}{\left({\text{e}}^{-b}-1\right)}^{2}$

${\Phi }_{4,2}\left(n\right)\to \frac{{a}^{2}}{{b}^{2}}+\frac{2{a}^{2}}{{b}^{3}}\left({\text{e}}^{-b}-1\right)+\frac{{a}^{2}}{{b}^{4}}{\left({\text{e}}^{-b}-1\right)}^{2}$

${\Phi }_{4,4}\left(n\right)\to -\frac{2a{X}_{0}}{{b}^{2}}\left({\text{e}}^{-b}-1\right)+\frac{2a{X}_{0}}{{b}^{3}}{\left({\text{e}}^{-b}-1\right)}^{2}$

$\begin{array}{c}P\left(|{\Phi }_{4,3}\left(n,\sigma \right)|>\epsilon \right)\le {\epsilon }^{-1}E|\frac{{\sigma }^{2}}{{n}^{2}}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-b\left({t}_{i-1}-s\right)}\text{d}{Z}_{s}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-b\left({t}_{i-1}-s\right)}\text{d}{Z}_{s}|\\ \le C{\epsilon }^{-1}\frac{{\sigma }^{2}}{{n}^{2}}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-\alpha b{t}_{i-1}}{\left(\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{\alpha bs}\text{d}s\right)}^{\frac{1}{\alpha }}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-\alpha b{t}_{i-1}}{\left(\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{\alpha bs}\text{d}s\right)}^{\frac{1}{\alpha }}\end{array}$

$n\to \infty ,\sigma \to 0$ 时，上式收敛到0。

$\begin{array}{c}P\left(|{\Phi }_{4,5}\left(n,\sigma \right)|>\epsilon \right)\le {\epsilon }^{-1}E|\frac{2\sigma {X}_{0}}{{n}^{2}}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-b{t}_{i-1}}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-b\left({t}_{i-1}-s\right)}\text{d}{Z}_{s}|\\ \le C{\epsilon }^{-1}\frac{\sigma }{{n}^{2}}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-b{t}_{i-1}}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-b{t}_{i-1}}{\left(\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{\alpha bs}\text{d}s\right)}^{\frac{1}{\alpha }}\to 0\end{array}$

$n\to \infty$ 时，由 ${\Lambda }_{n}\to 0$ 和引理3.1、引理3.2，我们直接得到 ${\stackrel{^}{b}}_{n,\sigma }-b={\Lambda }_{n}-\frac{{\Phi }_{1}\left(n,\sigma \right)-{\Phi }_{2}\left(n,\sigma \right)}{{\Phi }_{3}\left(n,\sigma \right)-{\Phi }_{4}\left(n,\sigma \right)}{\to }_{p}0$

$\begin{array}{l}{\stackrel{^}{a}}_{n,\sigma }-a=\underset{i=1}{\overset{n}{\sum }}\left({X}_{{t}_{i}}-{X}_{{t}_{i-1}}\right)+\frac{1}{n}\cdot {\stackrel{^}{b}}_{n,\sigma }\underset{i=1}{\overset{n}{\sum }}\text{ }{X}_{{t}_{i-1}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\underset{i=1}{\overset{n}{\sum }}\left[{\text{e}}^{-\frac{b}{n}}{X}_{{t}_{i-1}}+\frac{a}{b}\left(1-{\text{e}}^{-\frac{b}{n}}\right)+\sigma {Z}_{i,\frac{1}{n}}-{X}_{{t}_{i-1}}\right]+\frac{1}{n}{\stackrel{^}{b}}_{n,\sigma }\underset{i=1}{\overset{n}{\sum }}\text{ }{X}_{{t}_{i-1}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=n\left({\text{e}}^{-\frac{b}{n}}-1\right)\cdot \frac{1}{n}\underset{i=1}{\overset{n}{\sum }}\text{ }{X}_{{t}_{i-1}}+b\frac{1}{n}\underset{i=1}{\overset{n}{\sum }}\text{ }{X}_{{t}_{i-1}}+\frac{na}{b}\left(1-{\text{e}}^{-\frac{b}{n}}\right)-a\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\sigma \underset{i=1}{\overset{n}{\sum }}\text{ }{Z}_{i,\frac{1}{n}}+\left({\stackrel{^}{b}}_{n,\sigma }-b\right)\frac{1}{n}\underset{i=1}{\overset{n}{\sum }}\text{ }{X}_{{t}_{i-1}}\to 0\end{array}$

4. 最小二乘估计量 ${\stackrel{^}{b}}_{n,\sigma }$${\stackrel{^}{a}}_{n,\sigma }$ 的渐近性

$\begin{array}{c}{\sigma }^{-1}\left({\stackrel{^}{b}}_{n,\sigma }-b\right)={\sigma }^{-1}\left[\left(1-{\text{e}}^{-b/n}\right)n-b\right]-\frac{\underset{i=1}{\overset{n}{\sum }}\text{ }{X}_{{t}_{i-1}}{Z}_{i,\frac{1}{n}}-{n}^{-1}\underset{i=1}{\overset{n}{\sum }}\text{ }{Z}_{i,\frac{1}{n}}\underset{i=1}{\overset{n}{\sum }}\text{ }{X}_{{t}_{i-1}}}{{n}^{-1}\underset{i=1}{\overset{n}{\sum }}\text{ }{X}_{{t}_{i-1}}^{2}-{\left({n}^{-1}\text{ }{X}_{{t}_{i-1}}\right)}^{2}}\\ :={\sigma }^{-1}{\Lambda }_{n}-\frac{{\sigma }^{-1}{\Phi }_{1}\left(n,\sigma \right)-{\sigma }^{-1}{\Phi }_{2}\left(n,\sigma \right)}{{\Phi }_{3}\left(n,\sigma \right)-{\Phi }_{4}\left(n,\sigma \right)}\end{array}$

$\begin{array}{c}{\Psi }_{1,1}\left(n\right)={X}_{0}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-b\left({t}_{i-1}+{t}_{i}\right)}{U}_{i}\\ ~{X}_{0}{\left(\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-\alpha b\left({t}_{i-1}+{t}_{i}\right)}\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}{\text{e}}^{\alpha bs}\text{d}s\right)}^{\frac{1}{\alpha }}{S}_{\alpha }\left(1,\beta ,0\right)\\ \triangleq {X}_{0}{\left(\frac{1-{\text{e}}^{-\alpha b}}{\alpha b}\right)}^{\frac{1}{\alpha }}U\end{array}$

${\Psi }_{1,2}\left(n\right)=\frac{a}{b}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-b{t}_{i}}{U}_{i}~\frac{a}{b}U$

${\Psi }_{1,3}\left(n\right)=\frac{a}{b}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-b\left({t}_{i-1}+{t}_{i}\right)}{U}_{i}~\frac{a}{b}{\left(\frac{1-{\text{e}}^{-\alpha b}}{\alpha b}\right)}^{\frac{1}{\alpha }}U$

$\begin{array}{c}P\left(|{\Psi }_{1,4}\left(n,\sigma \right)|>\epsilon \right)\le {\epsilon }^{-1}E|\sigma \underset{i=1}{\overset{n}{\sum }}\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-b\left({t}_{i-1}-s\right)}\text{d}{Z}_{s}\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}{\text{e}}^{-b\left({t}_{i-1}-s\right)}\text{d}{Z}_{s}|\\ \le {\epsilon }^{-1}\sigma \underset{i=1}{\overset{n}{\sum }}\text{ }C{\left(\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-\alpha b\left({t}_{i-1}-s\right)}\text{d}s\right)}^{\frac{1}{\alpha }}C{\left(\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}{\text{e}}^{-\alpha b\left({t}_{i-1}-s\right)}\text{d}s\right)}^{\frac{1}{\alpha }}\\ =C{\epsilon }^{-1}\sigma {\left(\frac{1-{\text{e}}^{-\alpha b/n}}{\alpha b}\right)}^{\frac{1}{\alpha }}\underset{i=1}{\overset{n}{\sum }}{\left(\frac{1-{\text{e}}^{-\alpha b{t}_{i-1}}}{\alpha b}\right)}^{\frac{1}{\alpha }}=O\left(\sigma {n}^{1-\frac{1}{\alpha }}\right)\end{array}$

$\sigma {n}^{1-\frac{1}{\alpha }}\to 0$ 时上式趋向于0。由类似的证明方法， ${\Psi }_{2}\left(n,\sigma \right)$ 依概率收敛于 $\frac{a}{b}U$

${\sigma }^{-1}\left({\stackrel{^}{b}}_{n,\sigma }-b\right){\to }_{d}\frac{\left({X}_{0}+\frac{a}{b}\right){\left(\frac{1-{\text{e}}^{-\alpha b}}{\alpha b}\right)}^{\frac{1}{\alpha }}}{\left({C}_{1}+{C}_{2}+{C}_{3}\right)-\left({C}_{4}+{C}_{5}+{C}_{6}\right)}U$

$\begin{array}{l}\frac{1}{n}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{X}_{{t}_{i-1}}=\frac{1}{n}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-b{t}_{i-1}}{X}_{0}+\frac{a}{b}-\frac{a}{nb}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }{\text{e}}^{-b{t}_{i-1}}+\frac{\sigma }{n}\underset{i=1}{\overset{n}{\sum }}\text{ }\text{ }\underset{0}{\overset{{t}_{i-1}}{\int }}{\text{e}}^{-b\left({t}_{i-1}-s\right)}\text{d}{Z}_{s}\\ \to \frac{a-{X}_{0}}{b}\left({\text{e}}^{-b}-1\right)+\frac{a}{b}\triangleq D\end{array}$

${\sigma }^{-1}\sigma {\sum }_{i=1}^{n}{Z}_{i,\frac{1}{n}}=O\left({n}^{1-\frac{1}{\alpha }}\right)$ 发散，所以

${\sigma }^{-1}\left({\stackrel{^}{a}}_{n,\sigma }-a-\sigma {\sum }_{i=1}^{n}{Z}_{i,\frac{1}{n}}\right){\to }_{d}D\frac{\left({X}_{0}+\frac{a}{b}\right){\left(\frac{1-{\text{e}}^{-\alpha b}}{\alpha b}\right)}^{\frac{1}{\alpha }}}{\left({C}_{1}+{C}_{2}+{C}_{3}\right)-\left({C}_{4}+{C}_{5}+{C}_{6}\right)}U$

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