数学杂志  2017, Vol. 37 Issue (5): 1029-1039   PDF    
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陈盈盈
蒋辉
带复合泊松跳扩散模型的点波动率门限估计量的渐近性质
陈盈盈, 蒋辉    
南京航空航天大学数学系, 江苏南京 210016
摘要:本文研究了带复合泊松跳扩散模型的点波动率门限估计量的渐近性质.利用门限方法和核函数技术,构造并证明了此模型点波动率估计量的渐近正态性.同时,应用Gärtner-Ellis定理及大偏差中的Delta方法,得到了估计量的中偏差原理.
关键词复合泊松过程    点波动率    渐近正态性    门限方法    中偏差原理    
ASYMPTOTIC PROPERTIES FOR SPOT VOLATILITY ESTIMATION OF DIFFUSIONS WITH COMPOUND POISSON JUMPS
CHEN Ying-ying, JIANG Hui    
Department of Mathematics, Nanjing University of Aeronautics and Astronautics, Nanjing 210016, China
Abstract: In this paper, we study the asymptotic behaviors for the threshold spot volatility estimator of the diffusion process with compound Poisson jumps. By the method of threshold criterion, we construct a kernel estimator for the volatility and study its asymptotic normality. Applying Gärtner-Ellis theorem, we obtain the moderate deviations.
Key words: compound Poisson process     spot volatility     asymptotic normality     threshold criterion     moderate deviations    
1 引言

波动率是度量金融市场风险的常用指标, 对波动率的估计和预测是近几十年来金融研究领域的重要课题之一.一个时刻点处的波动率常被称为点波动率(spot volatility), 其是套头交易, 期权定价, 风险分析和资产组合管理等金融活动中需要考虑的重要因素.随着电子化交易的普及和信息存储技术的发展, 以高精度时间“分”, “秒”为刻度来存储信息的高频环境逐步建立.高频数据可以迅速有效地捕捉市场信息, 比低频数据更能反映金融市场的真实状况, 为准确估计点波动率提供了途径.

关于点波动率的研究, Foster和Nelson[8]首次证明了卷样点波动率估计量的渐近正态性.但文中出现的条件和结果都十分抽象, 故Andreou和Ghysels[1]对文中出现的估计量进行了进一步研究.之后, Fan和Wang[7]在资产过程轨道连续情况下, 构建了点波动率的核密度估计量并得到了其渐近正态性.关于点波动率估计量的研究, 亦可参见Zu和Boswijk[12].

近年来, 大量金融理论和实证表明, 资产价格中常包含跳, 且跳的存在和类型对波动率估计量具有显著影响(夏登峰等[14]).在资产价格过程有复合泊松跳的情形下, 利用Mancini[10]中门限方法, 我们将Fan和Wang[7]提出的估计量进行推广和改进, 即剔除带跳部分对估计量的影响.同时证明了所构造估计量的渐近正态性与中偏差原理, 并给出了速率函数的精确表达式.此外, 关于积分波动率估计量的大偏差与中偏差原理, 可以参见Djellout等[5, 6], Hui[9]及Mancini[11].

本文的结构如下:在第二章中, 对模型进行介绍, 构造了点波动率的门限估计量并阐述了本文的主要结论.第三章给出了主要结论的证明.

2 点波动率门限估计量及主要结论

给定概率空间$(\Omega, \mathcal{F}, (\mathcal{F})_t, P)$, 令资产价格过程$X$服从跳扩散过程(见文献[3])

$\begin{equation}\label{df-jump-process} dX_t=\sigma_tdW_t+\mu_t dt+dL_t, \quad X_0=x_0, \quad t\in[0,], \end{equation}$ (2.1)

其中$\sigma_{t}$为关于$\mathcal{F}_t$适应的可料过程; $W_{t}$为标准布朗运动; $L_{t}$是一个与$\{W_{t}, t\geq0\}$独立的复合泊松过程, 即$L_t=\sum\limits_{i=1}^{N_t}Y_i$, 其中$\{Y_i, i\in\mathbb{N}\}$是一列独立同分布的随机变量, $\{N_{t}, t\geq0\}$是强度为$\lambda$的泊松过程, 且与$\{Y_i, i\in\mathbb{N}\}$独立.

定义点波动率为$\Gamma _{t}:=\sigma_{t}^{2}$.若$L\equiv0$, 过程$X$的二次变差为$[X, X]_t=\displaystyle\int_0^t\Gamma _{s}ds$.由于

$ \Gamma _{t}=\frac{d[X, X]_t}{dt}, $

通过在时间点$\{t_{i}=i/n, i=1, 2, \cdots, n\}$处对$X$进行的等距观测, Fan和Wang[7]构造了$\Gamma _{t}$的核密度估计量

$ \tilde{\Gamma}_{t}=\frac{1}{b_n}\sum\limits_{i\in I(t, b_n)}K(\frac{t_{i}-t}{b_n})(X_{t_{i}}-X_{t_{i-1}})^{2}, $

其中$I(t, b_n)=\{i: t_i\in[t-b_n, t+b_n]\}$, $K(x)$是支集为$[-1,1]$的核函数, $b_n$是带宽.

$L\neq0$, 可以得到过程$X$的二次变差为

$ [X, X]_t=\int_0^t\Gamma _{s}ds+\sum\limits_{0\leq s\leq t}(\Delta X_s)^2, \quad t\in[0,1], $

其中$\Delta X_s=X_s-X_{s-}$为过程$X$$s$点的振幅.为了估计$\Gamma_{t}$, 需要获得过程$X$的连续部分$X^c$:

$ X^c_t=\sigma_tdW_t+\mu_tdt, \quad t\in[0,1]. $

故而最主要的问题是如何将过程$X$的跳与连续部分区分开来.利用Mancini[10]及Fan和Wang[7]中的思想, 定义如下核估计量

$ \hat{\Gamma}_{t}=\frac{1}{b_n}\sum\limits_{i\in I(t, b_n)}K(\frac{t_{i}-t}{b_n})(X_{t_{i}}-X_{t_{i-1}})^{2} I_{\{{(X_{t_{i}}-X_{t_{i-1}})^{2}\leq r(1/n)}\}}, $

其中$r(1/n)$满足

$\begin{equation}\label{condition-threshold} \lim\limits_{n\to\infty}r(1/n)=0, \quad\lim\limits_{n\to\infty}\frac{\log n}{nr(1/n)}=0. \end{equation}$ (2.2)

为了研究$\hat{\Gamma}_{t}$的渐近性质, 给出如下假设(Fan和Wang[7])

(A1) $\sup\{|\sigma_t-\sigma_s|, t, s\in[0,1], |t-s|\leq a\}=O_p(a^{1/2}|\log a|^{1/2})$, $\mathop{\sup}\limits_{0\leq t\leq1}\sigma_{t}^{2}=O_{p}(1)$;

(A2) $\sup\{(\displaystyle\int_{t_{i-1}}^{t_i}({\sigma_s-\sigma_{t_{i-1}}) }dW_{s})^2, i=1, \cdots, n\}=O_{p}(n^{-2+\eta})$, 其中$\eta>0$, 且任意小;

(A3) $\mathop{\sup}\limits_{0\leq t\leq1}|\mu_{t}|=O_{p}(1)$;

(A4) $b_n\sim (n^{1/2}\log n)^{-1}$, 核函数$K(\cdot)$为取正值的二次可微函数, 支撑为$[-1,1]$, 满足

$ \int_{-1}^{1}K(x)dx=1, \quad\int_{-1}^{1}K^{2}(x)dx:=\lambda(K). $

首先, 给出估计量的渐近正态性.

定理1    令过程$X$满足(2.1) 式, 且条件(A1)-(A4) 成立.则对于任意的$t\in[0,1]$, 有

$ \sqrt{nb_n}({\frac{\hat{\Gamma}_{t}}{\Gamma_{t}}-1})\stackrel{L}{\longrightarrow} N(0, 2\lambda(K)), $

其中$\stackrel{L}{\longrightarrow} $代表依分布收敛.

推论1    在定理1的条件下, 可以推出$ \sqrt{nb_n}({\frac{\Gamma_{t}}{\hat{\Gamma}_{t}}-1})\stackrel{L}{\longrightarrow} N(0, 2\lambda(K)). $

假定$\{\lambda_n, n\geq0\}$为一列正实数且满足

$\begin{equation}\label{speed} \lambda_n\rightarrow\infty, \quad\frac{\lambda_n}{\sqrt{nb_n}}\to0. \end{equation}$ (2.3)

下面给出门限估计量$\hat{\Gamma}_t$的中偏差原理.

定理2    令过程$X$满足(2.1) 式, $\sigma_t$非随机且$\mu_t\equiv\mu\in\mathbb{R}$.若

$\begin{equation}\label{thr} r(1/n)=o(\sqrt{\frac{b_n}{n}}\lambda_n^{-1}), \end{equation}$ (2.4)

假定条件(A1) 及(A4) 成立, 则$\{\frac{\sqrt{nb_n}}{\lambda_n}(\frac{\hat{\Gamma}_{t}}{\Gamma_{t}}-1), \quad n\geq1\}$满足中偏差原理, 且速度为$\lambda_n^2$, 速率函数为$I_t(x)=\frac{x^2}{4\sigma^4_{t}\lambda(K)}$.特别地, 对任意的$x>0$, 有

$ \log P(\frac{\sqrt{nb_n}}{\lambda_n}\left|\frac{\hat{\Gamma}_t}{\Gamma_{t}}-1\right|\geq x) \sim -\frac{\lambda_n^2x^2}{4\lambda(K)}. $

应用大偏差中的Delta方法(Gao和Zhao[13]), 可以得到如下推论.

推论2    在定理2的条件下, 可以推出$\{\frac{\sqrt{nb_n}}{\lambda_n}(\frac{\Gamma_{t}}{\hat{\Gamma}_{t}}-1), \quad n\geq1\}$满足中偏差原理, 且速度为$\lambda_n^2$, 速率函数为$I_t(x)=\frac{x^2}{4\sigma^4_{t}\lambda(K)}$.

3 主要定理的证明

在这一节中, 将给出本文主要结论的证明.

3.1 ${\hat \Gamma _t}$的渐近分布

定理1的证明    对任意右连左极过程$Z$, 令$\Delta_{i}Z=Z_{t_{i}}-Z_{t_{i-1}}$.由Mancini[10]的定理3.1, 当$n$充分大时, 对每一个$i=1, 2, \cdots, n$, 有$I_{\{(\Delta_{i}X)^2\leq r(1/n)\}}=I_{\{\Delta_{i}N=0\}}$.从而可以得到

$\begin{array}{l} \sqrt {n{b_n}} ({{\hat \Gamma }_t} - {\Gamma _t}) = \sqrt {n{b_n}} (\frac{1}{{{b_n}}}\sum\limits_{i \in I(t,{b_n})} K (\frac{{{t_i} - t}}{{{b_n}}}){({\Delta _i}X)^2}{I_{\{ {\Delta _i}N = 0\} }} - {\Gamma _t})\\ \quad \quad \quad \quad \quad \quad = \sqrt {n{b_n}} (\frac{1}{{{b_n}}}\sum\limits_{i \in I(t,{b_n})} K (\frac{{{t_i} - t}}{{{b_n}}}){({\Delta _i}{X^c})^2} - {\Gamma _t})\\ \quad \quad \quad \quad \quad \quad - \sqrt {n{b_n}} (\frac{1}{{{b_n}}}\sum\limits_{i \in I(t,{b_n})} K (\frac{{{t_i} - t}}{{{b_n}}}){({\Delta _i}{X^c})^2}{I_{\{ {\Delta _i}N \ne 0\} }}). \end{array}$ (3.1)

由于$\Delta_{i}X^c=X_{t_{i}}^c-X_{t_{i-1}}^c=\displaystyle\int_{t_{i-1}}^{t_i}\sigma_s dW_s+\int_{t_{i-1}}^{t_i}\mu_s ds$, 易得到

$\begin{array}{l} \sqrt {n{b_n}} (\frac{1}{{{b_n}}}\sum\limits_{i \in I(t,{b_n})} K (\frac{{{t_i} - t}}{{{b_n}}}){({\Delta _i}{X^c})^2}{I_{\{ {\Delta _i}N \ne 0\} }})\\ \le 2\sqrt {n{b_n}} (\frac{1}{{{b_n}}}\sum\limits_{i \in I(t,{b_n})} K (\frac{{{t_i} - t}}{{{b_n}}}){(\int_{{t_{i - 1}}}^{{t_i}} {{\sigma _s}} d{W_s})^2}{I_{\{ {\Delta _i}N \ne 0\} }})\\ + 2\sqrt {n{b_n}} (\frac{1}{{{b_n}}}\sum\limits_{i \in I(t,{b_n})} K (\frac{{{t_i} - t}}{{{b_n}}}){(\int_{{t_{i - 1}}}^{{t_i}} {{\mu _s}} ds)^2}{I_{\{ {\Delta _i}N \ne 0\} }}). \end{array}$ (3.2)

首先, 由Lévy连续模定理知$\sup\{|W_{t_{i}}-W_{t_{i-1}}|, i=1, \cdots, n\}=O_{p}(n^{-1/2}\log^{1/2}n)$, 根据条件(A1) 及(A2) 即得

$\begin{array}{l} \quad \sup \left\{ {{{\left( {\int_{{t_{i - 1}}}^{{t_i}} {{\sigma _s}} d{W_s}} \right)}^2},i = 1, \cdots ,n} \right\}\\ \le 2\sup \{ (\int_{{t_{i - 1}}}^{{t_i}} {({\sigma _s} - {\sigma _{{t_{i - 1}}}})d{W_s}{)^2},i = 1, \cdots ,n\} + 2\sup \{ \sigma _{{t_{i - 1}}}^2{{({W_{{t_i}}} - {W_{{t_{i - 1}}}})}^2},i = 1, \cdots ,n\} } \\ = {O_p}({n^{ - 2 + \eta }}) + {O_p}({n^{ - 1}}\log n) = {O_p}({n^{ - 1}}\log n). \end{array}$

同时, 由泊松过程性质可知$N_{t+b_n}-N_{t-b_n}=O_{p}(b_n)$.从而结合条件(A4) 可得

$\begin{array}{l} \sqrt {n{b_n}} (\frac{1}{{{b_n}}}\sum\limits_{i \in I(t,{b_n})} K (\frac{{{t_i} - t}}{{{b_n}}}){(\int_{{t_{i - 1}}}^{{t_i}} {{\sigma _s}} d{W_s})^2}{I_{\{ {\Delta _i}N \ne 0\} }})\\ \le C({N_{t + {b_n}}} - {N_{t - {b_n}}})\sqrt {\frac{n}{{{b_n}}}} \sup \{ {(\int_{{t_{i - 1}}}^{{t_i}} {{\sigma _s}} d{W_s})^2},i = 1, \cdots ,n\} \\ = {O_p}({n^{ - \frac{3}{4}}}{\log ^{\frac{1}{2}}}n) = {o_p}(1), \end{array}$ (3.3)

其中$C$为只与核有关的常数.

再者, 条件(A3) 可推得$\displaystyle\int_{t_{i-1}}^{t_i}\mu_s ds=O_p(1/n)$.类似(3.3)式的证明知

$\begin{equation}\label{CLT-eq4} \sqrt{nb_n}(\frac{1}{b_n}\sum\limits_{i\in I(t, b_n)} K(\frac{t_{i}-t}{b_n})(\int_{t_{i-1}}^{t_i}\mu_s ds)^2 I_{\{\Delta_iN\neq0\}})=O_{p}(n^{-\frac{7}{4}}\log^{\frac{1}{2}}n)=o_p(1). \end{equation}$ (3.4)

因此结合(3.1), (3.3) 及(3.4) 式, 得到

$\begin{align*} \sqrt{nb_n}(\hat{\Gamma}_{t}-\Gamma_{t}) =\sqrt{nb_n}(\frac{1}{b_n}\sum\limits_{i\in I(t, b_n)}K(\frac{t_{i}-t}{b_n})((\Delta_{i}X^c))^2 -\Gamma_t)+o_p(1). \end{align*} $

由Fan和Wang[7]中定理1, 本文中定理1得证.

3.2 ${\hat \Gamma _t}$的中偏差原理

首先, 来计算$\hat{\Gamma}_t$的对数矩生成函数, 即对任意的$\theta\in\mathbb{R}$, 令

$ \Lambda_{n, t}(\theta)=\log E\exp(\sqrt{nb_n}\lambda_n\theta(\hat{\Gamma}_t-\Gamma_t)). $

引理1    在定理2的条件下, 有$\lim\limits_{n\to\infty}\frac{1}{\lambda_n^2}\Lambda_{n, t}(\theta)=\theta^2\sigma_t^4\lambda(K)$.

    由于过程$X$增量相互独立, 故

$\begin{array}{l} \quad \log E\exp (\sqrt {n{b_n}} {\lambda _n}\theta ({{\hat \Gamma }_t} - {\Gamma _t}))\\ = \sum\limits_{i \in I(t,{b_n})} {\log } E\exp (\sqrt {\frac{n}{{{b_n}}}} {\lambda _n}\theta K(\frac{{{t_i} - t}}{{{b_n}}}){({X_{{t_i}}} - {X_{{t_{i - 1}}}})^2}{I_{\{ {{({X_{{t_i}}} - {X_{{t_{i - 1}}}})}^2} \le r(1/n)\} }})\\ - \sqrt {n{b_n}} {\lambda _n}\theta \sigma _t^2. \end{array}$

由示性函数的定义可得

$\begin{array}{l} \quad E\exp (\sqrt {\frac{n}{{{b_n}}}} {\lambda _n}\theta K(\frac{{{t_i} - t}}{{{b_n}}}){({X_{{t_i}}} - {X_{{t_{i - 1}}}})^2}{I_{\{ {{({X_{{t_i}}} - {X_{{t_{i - 1}}}})}^2} \le r(1/n)\} }})\\ = E(\exp (\sqrt {\frac{n}{{{b_n}}}} {\lambda _n}\theta K(\frac{{{t_i} - t}}{{{b_n}}}){({X_{{t_i}}} - {X_{{t_{i - 1}}}})^2}){I_{\{ {{({X_{{t_i}}} - {X_{{t_{i - 1}}}})}^2} \le r(1/n)\} }})\\ \quad + P({({X_{{t_i}}} - {X_{{t_{i - 1}}}})^2} > r(1/n))\\ : = {J_1}\left( i \right) + {J_2}\left( i \right). \end{array}$

对于$J_1(ⅰ), J_2(ⅰ)$, 由$N$$W$的相互独立性可得

$\begin{array}{l} {J_1}\left( i \right) = E(\exp (\sqrt {\frac{n}{{{b_n}}}} {\lambda _n}\theta K(\frac{{{t_i} - t}}{{{b_n}}}){(X_{{t_i}}^c - X_{{t_{i - 1}}}^c)^2})|{N_{{t_i}}} - {N_{{t_{i - 1}}}} = 0)P({N_{{t_i}}} - {N_{{t_{i - 1}}}} = 0)\\ \quad \quad \quad - E(\exp (\sqrt {\frac{n}{{{b_n}}}} {\lambda _n}\theta K(\frac{{{t_i} - t}}{{{b_n}}}){(X_{{t_i}}^c - X_{{t_{i - 1}}}^c)^2}){I_{\{ {{(X_{{t_i}}^c - X_{{t_{i - 1}}}^c)}^2} > r(1/n)\} }}|{N_{{t_i}}} - {N_{{t_{i - 1}}}} = 0)\\ \quad \quad \quad \cdot P({N_{{t_i}}} - {N_{{t_{i - 1}}}} = 0)\\ \quad \quad \quad + E(\exp (\sqrt {\frac{n}{{{b_n}}}} {\lambda _n}\theta K(\frac{{{t_i} - t}}{{{b_n}}}){({X_{{t_i}}} - {X_{{t_{i - 1}}}})^2}){I_{\{ {{({X_{{t_i}}} - {X_{{t_{i - 1}}}})}^2} \le r(1/n),{N_{{t_i}}} - {N_{{t_{i - 1}}}} \ne 0\} }})\\ \quad \quad \quad : = {M_1}(i,\frac{1}{n}) - {M_2}(i,\frac{1}{n}) + {M_3}(i,\frac{1}{n}). \end{array}$

同理可得

$\begin{array}{l} {J_2}\left( i \right) = P({(X_{{t_i}}^c - X_{{t_{i - 1}}}^c)^2} > r(1/n)|{N_{{t_i}}} - {N_{{t_{i - 1}}}} = 0)P({N_{{t_i}}} - {N_{{t_{i - 1}}}} = 0)\\ \quad \quad \quad + P({({X_{{t_i}}} - {X_{{t_{i - 1}}}})^2} > r(1/n)|{N_{{t_i}}} - {N_{{t_{i - 1}}}} \ne 0)P({N_{{t_i}}} - {N_{{t_{i - 1}}}} \ne 0)\\ \quad \quad \quad : = {M_4}(i,\frac{1}{n}) + {M_5}(i,\frac{1}{n}). \end{array}$

为了完成引理1的证明, 对于$M_j(i, \frac{1}{n}), j=1, 2, 3, 4, 5$, 需要如下结论.

引理2    当$n\rightarrow\infty$, 对于$j=2, 3, 4, 5$, 有

$\begin{equation}\label{yinli2} \sup\limits_{1\leq i\leq n}\frac{nb_n}{\lambda_n^2}\cdot\frac{M_j(i, \frac{1}{n})}{M_1(i, \frac{1}{n})}\to0. \end{equation}$ (3.5)

    由于$\sigma_t$非随机, 结合条件(A1) 知$\sup\limits_{1\leq i\leq n}\displaystyle\int_{t_{i-1}}^{t_{i}}\sigma_s^2ds=O(n^{-1})$.根据(2.3) 式有

$\begin{equation}\label{MDP2} \sqrt{\frac{n}{b_n}}\lambda_n\sup\limits_{1\leq i\leq n}\int_{t_{i-1}}^{t_{i}}\sigma_s^2ds\rightarrow0, \quad n\to\infty. \end{equation}$ (3.6)

故当$n\rightarrow\infty$时,

$\begin{array}{l} {M_1}(i,\frac{1}{n}) = {e^{ - \lambda /n}}{(1 - 2\sqrt {\frac{n}{{{b_n}}}} {\lambda _n}\theta K(\frac{{{t_i} - t}}{{{b_n}}})\int_{{t_{i - 1}}}^{{t_i}} {\sigma _s^2} ds)^{ - \frac{1}{2}}}\\ \quad \quad \quad \quad \cdot \exp (\frac{{\sqrt {\frac{n}{{{b_n}}}} {\lambda _n}\theta K(\frac{{{t_i} - t}}{{{b_n}}}){\mu ^2}/{n^2}}}{{1 - 2\sqrt {\frac{n}{{{b_n}}}} {\lambda _n}\theta K(\frac{{{t_i} - t}}{{{b_n}}})\int\limits_{{t_{i - 1}}}^{{t_i}} {\sigma _s^2} ds}}). \end{array}$ (3.7)

根据高斯随机变量的指数不等式, 知

$\begin{array}{*{20}{l}} {{M_4}(i,\frac{1}{n}) = {e^{ - \lambda /n}}P({{(\int_{{t_{i - 1}}}^{{t_i}} {{\sigma _s}} d{W_s} + \frac{\mu }{n})}^2} > r(1/n))}\\ {\quad \quad \quad \quad \le 2{e^{ - \lambda /n}}\exp ( - \frac{{r(1/n) - \frac{{2{\mu ^2}}}{{{n^2}}}}}{{4\int_{{t_{i - 1}}}^{{t_i}} {\sigma _s^2ds} }})} \end{array}$ (3.8)

以及

$\begin{aligned} M_5(i, \frac{1}{n})\leq P(N_{t_i}-N_{t_{i-1}}\neq0)=1-e^{-\lambda/n}. \end{aligned}$ (3.9)

再来估计$M_2(i, \frac{1}{n})$.首先, 当$\theta\leq0$时,

$\begin{array}{*{20}{l}} {{M_2}(i,\frac{1}{n}) \le {e^{ - \lambda /n}}P({{(\int_{{t_{i - 1}}}^{{t_i}} {{\sigma _s}} d{W_s} + \frac{\mu }{n})}^2} > r(1/n))}\\ {\quad \quad \quad \quad \le 2{e^{ - \lambda /n}}\exp ( - \frac{{r(1/n) - \frac{{2{\mu ^2}}}{{{n^2}}}}}{{4\int_{{t_{i - 1}}}^{{t_i}} {\sigma _s^2ds} }}).} \end{array}$ (3.10)

其次, 当$\theta>0$时, 根据Hölder不等式, 对任意的$p>1, q>1$$\frac{1}{p}+\frac{1}{q}=1$, 当$n$充分大时有

$\begin{array}{*{20}{l}} {{M_2}(i,\frac{1}{n}) \le {e^{ - \lambda /n}}{E^{\frac{1}{p}}}\exp (\sqrt {\frac{n}{{{b_n}}}} {\lambda _n}\theta pK(\frac{{{t_i} - t}}{{{b_n}}}){{(\int_{{t_{i - 1}}}^{{t_i}} {{\sigma _s}} d{W_s} + \frac{\mu }{n})}^2})}\\ {\quad \quad \quad \quad \quad \cdot {P^{\frac{1}{q}}}((\int_{{t_{i - 1}}}^{{t_i}} {{\sigma _s}} d{W_s} + \frac{\mu }{n}) > r(1/n))}\\ {\quad \quad \quad \quad \le 2{e^{ - \lambda /n}}{{(1 - 2\sqrt {\frac{n}{{{b_n}}}} {\lambda _n}\theta pK(\frac{{{t_i} - t}}{{{b_n}}})\int_{{t_{i - 1}}}^{{t_i}} {\sigma _s^2} ds)}^{ - \frac{1}{{2p}}}}}\\ {\quad \quad \quad \quad \quad \cdot \exp ( - \frac{{r(1/n) - \frac{{2{\mu ^2}}}{{{n^2}}}}}{{4q\int_{{t_{i - 1}}}^{{t_i}} {\sigma _s^2} ds}} + \frac{{\sqrt {\frac{n}{{{b_n}}}} {\lambda _n}\theta K(\frac{{{t_i} - t}}{{{b_n}}}){\mu ^2}/{n^2}}}{{1 - 2\sqrt {\frac{n}{{{b_n}}}} {\lambda _n}\theta pK(\frac{{{t_i} - t}}{{{b_n}}})\int_{{t_{i - 1}}}^{{t_i}} {\sigma _s^2ds} }}).} \end{array}$ (3.11)

现在只需要估计$M_3(i, \frac{1}{n})$,

$ M_3(i, \frac{1}{n})\\ =E(\exp(\sqrt{\frac{n}{b_n}}\lambda_n\theta K(\frac{t_i-t}{b_n})(X^c_{t_i}-X^c_{t_{i-1}})^2) I_{\{(X^c_{t_i}-X^c_{t_{i-1}})^2\leq r(1/n), N_{t_i}-N_{t_{i-1}}\neq0, L_{t_i}-L_{t_{i-1}}=0\}})\\ \quad +E(\exp(\sqrt{\frac{n}{b_n}}\lambda_n\theta K(\frac{t_i-t}{b_n})(X_{t_i}-X_{t_{i-1}})^2)I_{\{(X_{t_i}-X_{t_{i-1}})^2\leq r(1/n), N_{t_i}-N_{t_{i-1}}\neq0, L_{t_i}-L_{t_{i-1}}\neq0\}})\\ :=M_{31}(i, \frac{1}{n})+M_{32}(i, \frac{1}{n}).$

一方面,

$M_{31}(i, \frac{1}{n})\leq E(\exp(\sqrt{\frac{n}{b_n}}\lambda_n\theta K(\frac{t_i-t}{b_n})(\int_{t_{i-1}}^{t_i}\sigma_sdW_s+\frac{\mu}{n})^2) I_{\{(\int_{t_{i-1}}^{t_i}\sigma_sdW_s+\frac{\mu}{n})^2\leq r(1/n)\}})\\ \quad \quad \quad \quad \quad \cdot P(N_{t_i}-N_{t_{i-1}}\neq0)\\ \quad \quad \quad \quad =(M_1(i, \frac{1}{n})-M_2(i, \frac{1}{n}))(e^{\lambda/n}-1).$

另一方面, 对于$M_{32}(i, \frac{1}{n})$:首先, 当$\theta\leq0$时,

$ \exp(\sqrt{\frac{n}{b_n}}\lambda_n\theta K(\frac{t_i-t}{b_n})(X_{t_i}-X_{t_{i-1}})^2)\leq1. $

其次, 当$\theta>0$$I_{\{(X_{t_i}-X_{t_{i-1}})^2\leq r(1/n)\}}=1$时,

$ \exp(\sqrt{\frac{n}{b_n}}\lambda_n\theta K(\frac{t_i-t}{b_n})(X_{t_i}-X_{t_{i-1}})^2) \leq\exp\{{\sqrt{\frac{n}{b_n}}\lambda_n\theta K(\frac{t_i-t}{b_n}) r(1/n)}\}. $

同时, 当$n$充分大时,

$\quad P((X_{t_i}-X_{t_{i-1}})^2\leq r(1/n), N_{t_i}-N_{t_{i-1}}\neq0, L_{t_i}-L_{t_{i-1}}\neq0)\\ =\lim\limits_{\varepsilon\to0}P((X_{t_i}-X_{t_{i-1}})^2\leq r(1/n), |L_{t_i}-L_{t_{i-1}}|>\varepsilon)\\ \leq \lim\limits_{\varepsilon\to0}P(\left|\int_{t_{i-1}}^{t_i}\sigma_sdW_s+\frac{\mu}{n}\right|\geq \varepsilon-\sqrt{r(1/n)})\leq2\exp({-\frac{r(1/n)-\frac{2\mu^2}{n^2}}{4\displaystyle\int_{t_{i-1}}^{t_{i}}\sigma_s^2ds}}).$

由此, 可以得到

$ M_3(i, \frac{1}{n}) \leq(M_1(i, \frac{1}{n})-M_2(i, \frac{1}{n}))(e^{\lambda/n}-1)\\ \quad \quad \quad \quad +\left\{\begin{array}{ll} 2\exp({-\frac{r(1/n)-\frac{2\mu^2}{n^2}}{4\displaystyle\int_{t_{i-1}}^{t_{i}}\sigma_s^2ds}}), \quad \theta \leq0, \\ 2\exp\{{\sqrt{\frac{n}{b_n}}\lambda_n\theta K(\frac{t_i-t}{b_n}) r(1/n)}\}\exp({-\frac{r(1/n)-\frac{2\mu^2}{n^2}}{4\displaystyle\int_{t_{i-1}}^{t_{i}}\sigma_s^2ds}}), \quad\theta >0.\end{array}\right.$ (3.12)

根据上述的分析, 若要证明(3.5) 式, 只需证

$\begin{equation}\label{24} \begin{aligned} Q:=\lim\limits_{n\rightarrow\infty}\frac{nb_n}{\lambda_n^2}\exp({-\frac{r(1/n)-\frac{2\mu^2}{n^2}}{4\sup\limits_{1\leq i\leq n}\displaystyle\int_{t_{i-1}}^{t_{i}}\sigma_s^2ds}})=0. \end{aligned} \end{equation}$ (3.13)

计算可得$\exp({\frac{2\mu^2}{4n^2\cdot\sup\limits_{1\leq i\leq n}\displaystyle\int_{t_{i-1}}^{t_{i}}\sigma_s^2ds}})\rightarrow1$, 故

$ Q=\lim\limits_{n\rightarrow\infty}\exp(\log\frac{nb_n}{\lambda_n^2}\cdot (1-\frac{r(1/n)}{4\sup\limits_{1\leq i\leq n}\displaystyle\int_{t_{i-1}}^{t_{i}}\sigma_s^2ds\log\frac{nb_n}{\lambda_n^2}})). $

一方面, 由(2.3) 式知$\log\frac{nb_n}{\lambda_n^2}\rightarrow+\infty.$

另一方面,

$ \frac{r(1/n)}{\log\frac{nb_n}{\lambda_n^2}\sup\limits_{1\leq i\leq n}\displaystyle\int_{t_{i-1}}^{t_{i}}\sigma_s^2ds} \geq\frac{nr(1/n)}{\log n+\log b_n-\log\lambda_n^2}. $

再由条件(A4) 及(2.3) 式知$\log b_n\sim -\frac{1}{2}\log n, \log\lambda_n^2\ll\frac{1}{2}\log n.$结合(2.2) 式即可得

$ \frac{r(1/n)}{\log\frac{nb_n}{\lambda_n^2}\sup\limits_{1\leq i\leq n}\displaystyle\int_{t_{i-1}}^{t_{i}}\sigma_s^2ds} \geq\frac{nr(1/n)}{\log n}\rightarrow+\infty. $

由此, (3.13) 式得证.从而引理2得证.

引理1的证明    利用引理2以及泰勒公式, 可以得到

$ \lim\limits_{n\to\infty}\frac{1}{\lambda_n^2}\Lambda_{n, t}(\theta)=\lim\limits_{n\to\infty}\frac{1}{\lambda_n^2}(-\sqrt{nb_n}\lambda_n\theta \sigma_t^2+\sum\limits_{i\in I(t, b_n)} \log M_1(i, \frac{1}{n}))\\ \quad \quad \quad \quad =\lim\limits_{n\to\infty}\frac{1}{\lambda_n^2}(-\sqrt{nb_n}\lambda_n\theta \sigma_t^2- \frac{1}{2}\sum\limits_{i\in I(t, b_n)}\log (1-2\sqrt{\frac{n}{b_n}}\lambda_n\theta K(\frac{t_i-t}{b_n})\int_{t_{i-1}}^{t_i}\sigma_s^2ds))\\ \quad \quad \quad \quad =\lim\limits_{n\to\infty}\frac{1}{\lambda_n^2}(-\sqrt{nb_n}\lambda_n\theta \sigma_t^2 +\sum\limits_{i\in I(t, b_n)}(\sqrt{\frac{n}{b_n}}\lambda_n\theta K(\frac{t_i-t}{b_n})\int_{t_{i-1}}^{t_i}\sigma_s^2ds))\\ \quad \quad \quad \quad \quad +\lim\limits_{n\to\infty}\frac{1}{\lambda_n^2}(1+\varepsilon_n) \sum\limits_{i\in I(t, b_n)}(n\lambda_n^2\theta^2\frac{1}{b_n}K^2(\frac{t_i-t}{b_n})(\int_{t_{i-1}}^{t_i}\sigma_s^2ds)^2)\\ \quad \quad \quad \quad \quad :=H_1+H_2, $

其中$|\varepsilon_n|\leq C|\theta|\sqrt{\frac{n}{b_n}}\lambda_n\sup\limits_{1\leq i\leq n}\displaystyle\int_{t_{i-1}}^{t_i}\sigma_s^2ds=o(1)$.

接下来, 分别处理$H_1, H_2$.

$ H_1=\lim\limits_{n\to\infty}\frac{\sqrt{nb_n}\theta}{\lambda_n} ( \frac{1}{b_n}\sum\limits_{i\in I(t, b_n)} ( K(\frac{t_i-t}{b_n})\int_{t_{i-1}}^{t_i}\sigma_s^2 ds )-\sigma_t^2 )\\ \quad \quad =\lim\limits_{n\to\infty}\frac{\sqrt{nb_n}\theta}{\lambda_n} ( \frac{1}{b_n}\sum\limits_{i\in I(t, b_n)} ( K(\frac{t_i-t}{b_n})\int_{t_{i-1}}^{t_i}\sigma_s^2 ds ) -\frac{1}{b_n}\sum\limits_{i\in I(t, b_n)} ( K(\frac{t_i-t}{b_n})\int_{t_{i-1}}^{t_i}\sigma_t^2 ds ) )\\ \quad \quad \quad +\lim\limits_{n\to\infty}\frac{\sqrt{nb_n}\theta}{\lambda_n}\sigma_t^2 ( \frac{1}{b_n}\sum\limits_{i\in I(t, b_n)} ( \int_{t_{i-1}}^{t_i}K(\frac{t_i-t}{b_n}) ds )-1 )\\ \quad \quad \quad :=H_{11}+H_{12}.$

一方面, 根据条件(A1) 知

$ \sup\{|\sigma_t-\sigma_s|, t, s\in[t-b_n, t+b_n]\}=O(b_n^{1/2}|\log b_n|^{1/2}). $

故有

$ H_{11}=\lim\limits_{n\to\infty}\frac{\sqrt{nb_n}\theta}{\lambda_n} ( \frac{1}{b_n}\sum\limits_{i\in I(t, b_n)} ( K(\frac{t_i-t}{b_n})\int_{t_{i-1}}^{t_i}(\sigma_s^2-\sigma_t^2 )ds ) )\\ \quad \quad \quad \leq \lim\limits_{n\to\infty}C\frac{\sqrt{nb_n}\theta}{\lambda_n}O(b_n^{1/2}|\log b_n|^{1/2})=0. $

另一方面, 由条件(A4) 中核函数$K(x)$的性质得

$\begin{equation}\label{huanyuan} \frac{1}{b_n}\sum\limits_{i\in I(t, b_n)}\int_{t_{i-1}}^{t_i}K(\frac{s-t}{b_n})ds =\frac{1}{b_n}\int_{t-b_n}^{t+b_n}K(\frac{s-t}{b_n})ds =\int_{-1}^{1}K(x)dx=1, \end{equation}$ (3.14)

$\begin{equation}\label{H12} H_{12}=\lim\limits_{n\to\infty}\frac{\sqrt{nb_n}\theta}{\lambda_n}\sigma_t^2 ( \frac{1}{b_n}\sum\limits_{i\in I(t, b_n)} ( \int_{t_{i-1}}^{t_i}(K(\frac {t_i-t}{b_n})-K(\frac{s-t}{b_n})) ds ) ). \end{equation}$ (3.15)

又由于$K(x)$是二次可微函数, 则由中值定理知, 对任意的$i$, 存在$\xi_i\in[\frac{s-t}{b_n}, \frac{t_i-t}{b_n}]$, 使得

$ K(\frac{t_i-t}{b_n})-K(\frac{s-t}{b_n})=K^{'}(\xi_i)(\frac{t_i-s}{b_n}). $

特别地, 当$s\in[t_{i-1}, t_i]$时, $K(\frac{t_i-t}{b_n})-K(\frac{s-t}{b_n})=O(\frac{1}{nb_n}).$结合(3.15) 式得

$H_{12}=\lim\limits_{n\to\infty}\frac{1}{\sqrt{nb_n}\lambda_n}=0. $

对于$H_2$而言,

$\begin{equation}\label{H2} H_2=\theta^2\lim\limits_{n\to\infty}\sum\limits_{i\in I(t, b_n)}\frac{n}{b_n}K^2(\frac{t_i-t}{b_n}) (\int_{t_{i-1}}^{t_i}\sigma_s^2ds)^2. \end{equation}$ (3.16)

一方面,

$ \int_{t_{i-1}}^{t_i}\sigma_s^2ds=\int_{t_{i-1}}^{t_i}(\sigma_t^2+\sigma_s^2-\sigma_t^2)ds =\frac{1}{n}(\sigma_t^2+O(b_n^{1/2}|\log b_n|^{1/2})). $

另一方面, 由于$\displaystyle\int_{t_{i-1}}^{t_i}ds=\frac{1}{n}$, 类似(3.14) 式的证明知

$ \frac{1}{b_n}\sum\limits_{i\in I(t, b_n)}\frac{1}{n}K^2(\frac{t_i-t}{b_n}) =\frac{1}{b_n}\sum\limits_{i\in I(t, b_n)}\int_{t_{i-1}}^{t_i}K^2(\frac{s-t}{b_n})ds\\ \quad \quad \quad \quad \quad \quad \quad \quad +\frac{1}{b_n}\sum\limits_{i\in I(t, b_n)}\int_{t_{i-1}}^{t_i} (K^2(\frac{t_i-t}{b_n})-K^2(\frac{s-t}{b_n}))ds\\ \quad \quad \quad \quad \quad \quad \quad \quad =\int_{-1}^{1}K^2(u)du+O(n^{-1/2}\log n)=\lambda(K)+o(1). $

因此结合(3.16) 式即得$H_2=\theta^2\lambda(K)\sigma_t^4$.由此引理1得证.

定理2的证明    根据Gärtner-Ellis定理, 定理2可以由引理1直接得到.

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