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.

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.

若$L\neq0$, 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$X$在$s$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$I_t(x)=\frac{x^2}{4\sigma^4_{t}\lambda(K)}$.

在这一节中, 将给出本文主要结论的证明.

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.

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$\lim\limits_{n\to\infty}\frac{1}{\lambda_n^2}\Lambda_{n, t}(\theta)=\theta^2\sigma_t^4\lambda(K)$.

证    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$N$与$W$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)

证    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$p＞1, q＞1$且$\frac{1}{p}+\frac{1}{q}=1$, 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$\theta＞0$且$I_{\{(X_{t_i}-X_{t_{i-1}})^2\leq r(1/n)\}}=1$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)

则

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

(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.