扩散模型在金融,工程和军事等领域有着广泛的应用. 近年来, 扩散过程的统计推断已成为概率与统计学家关注的热点问题,其中, 非时齐扩散模型统计推断的研究越来越受到关注. 比如,在金融领域中, 经济条件随时间变动已是基本的事实, 因此有必要设想金融资产的瞬时期望收益和瞬时波动率不但与指 定的状态变量有关,而且也应该依赖于时间. 这意味着基础状态变量应 该是一个非时齐的扩散过程. 对于非时齐扩散模型的研究, 迄今已做了大量的工作. 例如文献 Ho和Lee^[1], Hull和White^[2],Black和Karasinski^[3]和Fan 等^[4,5] 等考虑了时变参数的连续扩散模型. 以上文献主要研究的是非时齐扩散模型的参数估计问题. 对于非时齐扩散模型的非参数估计问题, 陈萍等^[6]和马雷等^[7] 分别考虑了时变扩散系数的小 波估计和核估计,他们都证明了估计量的强收敛性.

本文研究如下非时齐扩散模型中时变扩散系数的局部非参数估计问题. 考虑定义在概率空间$(\Omega,F,P)$上的随机过程 $\{X_t,0\leq t\leq T\}$,满足如下的非齐次扩散模型 $${\rm d}X_t=\mu(t,X_t){\rm d}t+\sigma(t,X_t){\rm d}B_t,\ \ X_0=x_0 (1.1)$$ 其中$\{B_t,0\leq t\leq T\}$是标准布朗运动, 二元函数$\mu(t,X_t)$和$\sigma(t,X_t)$分别称为随机过程$X_t$ 的漂移系数和扩散系数,初值$x_0\in L^2$,且与$\{B_t,0\leq t\leq T\}$ 独立. 令$F_t$ 为$\{X_s,0\leq s\leq t\}$生成的$\sigma$ -代数, 本文中,$C_i,i=1,2,3$皆为不依赖与任何变量的常数. 本文将基于非时齐扩散模型的离散观测样本,利用局部近似的方法, 构造扩散系数的局部估计量,并证明估计量的强相合性和渐近正态性. 下面给出模型(1.1)需要满足的假设条件.

(A1) (线性增长条件)~ 函数$\mu$和$\sigma$是可测函数,且满足 $$|\mu(t,x)|+|\sigma(t,x)|\leq C_1(1+|x|),x\in R,t\in [0,T].$$

(A2) (Lipschitz条件)~ $$|\mu(t,x)-\mu(t,y)|+|\sigma(t,x)-\sigma(t,y)|\leq C_2|x-y|,x,y\in R,t\in [0,T].$$

(A3)~ $|\sigma(t_1,x)-\sigma(t_2,x)|\leq C_3|t_1-t_2|,$且$\sigma(t,x)>0,x\in R,t_1,t_2,t\in [0,T]$.

假设条件(A1)和(A2)保证了扩散模型(1.1)有唯一的强解, 条件(A3)保证了我们可以对$\sigma(t,x)$关于$t$进行局部近似.

为便于讨论估计量的渐近性质,给出如下预备知识.

定义1.1若一个连续随机过程$X_t$可以分解为 $$X_t=LX_t+A_t, $$ 其中$LX_t$是一个连续局部鞅, $A_t$是具有有限方差的连续适应过程,则称$X_t$是一个连续半鞅.

定义1.2若$X_t$是一个连续半鞅, 则存在一个关于$t$非减的随机过程$L_X(t,a)$,使得 $$L_X(t,a)=\lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon}\int_0^tI_{[a,a+\epsilon]}(X_s){\rm d}[X]_s$$ 几乎处处成立,并称$L_X(t,a)$为$X_t$在$a$点处的局部时, 其中$[X]_t$表示$X_t$的二次互偏差过程,$I_A$表示集合$A$的示性函数.

局部时$L_X(t,a)$是一个表示过程$X_t$在点$a$的邻域内发生的次数的量. 若扩散过程$X_t$有极小的二阶矩$\sigma^2(X_t)$, 则其在点$a$处的局部时可以表示为 $$L_X(t,a)=\lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon}\int_0^tI_{[a,a+\epsilon]}(X_s)\sigma^2(X_s){\rm d}s.$$

引理1.1令$X_t$是一个连续半鞅, 且有二次互偏差过程$[X]_t$,若$f$是一个正的二元波雷尔可测函数,则有 $$\int_0^tf(X_s,s){\rm d}[X]_s=\int_{-\infty}^{+\infty}{\rm d}a\int_0^tf(a,s){\rm d}L_X(s,a),$$ 特别地,若$f$是一个正的一元波雷尔可测函数,则上式可以为 $$\int_0^tf(X_s){\rm d}[X]_s=\int_{-\infty}^{+\infty}f(a)L_X(t,a){\rm d}a.$$

引理1.2若$X_t$满足假设条件(A1)--(A3), $r$和$a$是任意固定的实数,则当$\lambda\rightarrow\infty$时,有 $$\frac{1}{2}\sqrt{\lambda}\left\{L_X(t,r+\frac{a}{\lambda})-L_X(t,r)\right\}\rightarrow_d \left \{ \begin{array} {ll} W(L_X(t,r),a),& a>0 ,\\ W(L_X(t,r),-a),& a<0, \end{array} \right .$$ 其中符号$``\rightarrow_d"$表示``依分布收敛", $W(t,a)$是标准Brownian Sheet,且与$X_t$独立.

引理1.3设$\{X_t,0\leq t\leq T\}$是满足假设条件(A1)--(A3)随机过程,令$s({\rm d}x)=\frac{2{\rm d}x}{S'(x)\sigma^2(x)}$是过程$X_t$的速率测度,这里$S'(x)$是尺度函数的一阶导数,$\sigma^2(x)$ 是波动率函数. $f(\cdot)$ 和 $g(\cdot)$是关于速率测度$s({\rm d}x)$可积的任意两个波雷尔可测函数,则有 $$P\left\{\lim_{T\rightarrow\infty}\frac{\int_{0}^Tf(X_s){\rm d}s}{\int_{0}^Tg(X_s){\rm d}s}=\frac{\int_{-\infty}^{+\infty}f(x)s({\rm d}x)} {\int_{-\infty}^{+\infty}g(x)s({\rm d}x)}\right\}=1.$$

注引理1.1的证明参见文献^[8]的第6章的 推论1.6和练习题1.15,引理1.2和引理1.3的证明分别参见文献^[9]中的引理5和引理6.

令$\{X_{t_i},i=0,1,2,\cdots,n\}$是扩散模型(1.1)的等距离散观测样本, 其中$t_i=iT/n$,$i=0,1,2,\cdots,n$. 记$\Lambda_n=T/n$, 则$t_i=i\Lambda_n,i=0,1,2,\cdots,n$. 下面我们基于该离散观测样本来构造扩散系数$\sigma^2(t,x)$的局部估计量, 并讨论该估计量的渐近性质.

由假设条件(A3)知,我们可以对$\sigma(t,x)$关于$t$进行局部近似, 即对于任意给定的时间点$t^*$,取$\gamma_n>0$, 使得当$n\rightarrow\infty$时,$\gamma_n\rightarrow 0$ 且$n\gamma_n\rightarrow\infty$, 当$t\in[t^*-\gamma_n,t^*+\gamma_n]$时,令 $$\sigma^2(t,x)=\sigma^2(t^*,x),(2.1)$$ 即在$t^*$的$\gamma_n$邻域内,扩散系数$\sigma^2(t,x)$仅是$x$的函数, 而与$t$无关.

对于观测样本$\{X_{t_i},i=0,1,2,\cdots,n\}$,令 $$i_0^{(t^*)}=\inf\left\{i:t_i\in[t^*-\gamma_n,t^*+\gamma_n]\right\},\,\,i_1^{(t^*)}=\sup\left\{i:t_i\in[t^*-\gamma_n,t^*+\gamma_n]\right\}.$$ 令 $$n_{\gamma_n}=\left[\big(i_1^{(t^*)}-i_0^{(t^*)}\big)/ \Lambda_n\right], $$ 其中$[x]$表示不大于数$x$的最大整数. 记 $$t_i^{(\gamma_n)}=t_{i_0^{(t^*)}}+i\Lambda_n,~~i=0,1,2, \cdots,n_{\gamma_n}, $$ 这样就得到了时间区间$[t^*-\gamma_n,t^*+ \gamma_n]$上的观测值. 选取适当的$\varepsilon_{\gamma_n}>0$, 对每个观测值$X_{t_i^{(\gamma_n)}}$定义停时序列如下 $$\tau_{i,0}^{(\gamma_n)}=\inf\left\{t_l\geq0:\left|X_{t_l}-X_{t_i^{(\gamma_n)}}\right|\leq\varepsilon_{\gamma_n}, l\in\left[i_0^{(t^*)},i_1^{(t^*)}\right]\right\}, $$ $$ \tau_{i,j+1}^{(\gamma_n)}=\inf\left\{t_l\geq\tau_{i,j}^{(\gamma_n)}+\Delta_{\gamma_n}: \left|X_{t_l}-X_{t_i^{(\gamma_n)}}\right|\leq\varepsilon_{\gamma_n}, l\in\left[i_0^{(t^*)},i_1^{(t^*)}\right]\right\},$$ $$j=0,1,2,\cdots,m_{\gamma_n}(t_i^{(\gamma_n)}), \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ 其中$\Delta_{\gamma_n}=(2\gamma_n)/n_{\gamma_n}$, $m_{\gamma_n}(t_i^{(\gamma_n)}) =\sum\limits_{l=1}^{n_{\gamma_n}}I_{\{|X_{t_l}-X_{t_i^{(\gamma_n)}}|\leq\varepsilon_{\gamma_n}\}}$表示落在$X_{t_i^{(\gamma_n)}}$的$\varepsilon_{\gamma_n}$邻域内的观测值的个数. 对每一个$i=0,1,2,\cdots,n_{\gamma_n}$,令 $$ \tilde{\sigma}^2(t^*,X_{t_i^{(\gamma_n)}})=\frac{1}{m_{\gamma_n}(t_i^{(\gamma_n)})\Delta_{\gamma_n}} \sum\limits_{j=0}^{m_{\gamma_n}(t_i^{(\gamma_n)})-1}\left[X_{\tau_{i,j}^{(\gamma_n)}+\Delta_{\gamma_n}}-X_{\tau_{i,j}^{(\gamma_n)}}\right]^2. (2.2) $$ 基于式(2.2)构造扩散系数的局部估计量为: 对任意给定的时间的$t^*$,令 $$ \hat{\sigma}^2(t^*,x)=\frac{\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)\tilde{\sigma}^2(t^*,X_{t_i^{(\gamma_n)}})} {\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)}, (2.3) $$ 其中$K_h(\cdot)=K(\cdot/h)/h$, $K(\cdot)$是一个非负的权重函数,称其为核函数. 假设核函数$K(\cdot)$满足如下条件

(1)~ $K(\cdot)$是一个连续可微的对称函数,且其导函数$K'(\cdot)$绝对可积.

(2)~ $\int_{-\infty}^{+\infty}K(x){\rm d}x=1,\int_{-\infty}^{+\infty} K^2(x){\rm d}x<\infty,\int_{-\infty}^{+\infty}x^2K(x){\rm d}x <\infty$和$\sup\limits_{x\in R}K(x)<C$,其中$C$是一个正的常数.

下面的定理2.1和定理2.2分别给出了局部估计量(2.3) 式的强相合性和渐近正态性.

定理2.1假设条件(A1)--(A3)成立, 如果有：$h_{\gamma_n}\rightarrow0$,使得 $$\frac{1}{h_{\gamma_n}}\left[\Delta_{\gamma_n}\log(1/\Delta_{\gamma_n})\right]^{1/2}\rightarrow0,$$ 且$\gamma_n\rightarrow0$,使得 $$\frac{\gamma_n^{1/2}}{n_{\gamma_n}}\left[\Delta_{\gamma_n}\log(1/\Delta_{\gamma_n})\right]^{1/2}\rightarrow0,$$ 则对任意给定的时间点$t^*$,有 $$\hat{\sigma}^2(t^*,x)\rightarrow_{\rm a.s.} \sigma^2(t^*,x),$$ 其中符号$``\rightarrow_{\rm a.s.}"$表示``依概率1收敛"或``几乎处处收敛".

定理2.2假设条件(A1)--(A3)成立, 如果有$h_{\gamma_n}=o(\varepsilon_{\gamma_n})$且$n_{\gamma_n}\varepsilon_{\gamma_n}^4\rightarrow 0$,则对任意给定的时间点$t^*$,有 $$\bar{L}_X(2\gamma_n,x)\sqrt{n_{\gamma_n}\varepsilon_{\gamma_n}}/(2\gamma_n)\left\{\hat{\sigma}^2(t^*,x)-\sigma^2(t^*,x)\right\} \rightarrow_dN\left(0,2\sigma^4(t^*,x)\right),$$ 如果有$h_{\gamma_n}=o(\varepsilon_{\gamma_n})$且$n_{\gamma_n}\varepsilon_{\gamma_n}^4\rightarrow \infty$,则对任意给定的时间点$t^*$,有 $$\bar{L}_X(2\gamma_n,x)\varepsilon_{\gamma_n}^{-\frac{3}{2}}\left\{\hat{\sigma}^2(t^*,x)-\sigma^2(t^*,x)\right\} \rightarrow_dN\left(0,16\varphi^{ind}\left[\frac{\partial\sigma(t^*,x)}{\partial x}\right]^2\right),$$ 其中$\bar{L}(2\gamma_n,x)=\frac{1}{\sigma^2(t^*,x)}L_X(2\gamma_n,x) =\frac{1}{\sigma^2(t^*,x)}\lim\limits_{\epsilon\rightarrow 0}\frac{1}{\epsilon}\int_0^{2\gamma_n}I_{[x,x+\epsilon]} (X_s)\sigma^2(t^*,X_s){\rm d}s$, $\varphi^{ind}=2\int_0^\infty\int_0^\infty ts\left(\frac{1}{2}I_{|t|\leq 1}\right)\left(\frac{1}{2}I_{|s|\leq 1}\right)\min(t,s){\rm d}t{\rm d}s$.

证明局部估计量的渐近性质需要用到如下引理.

引理3.1如果有$h_{\gamma_n}\rightarrow0$,使得 $$\frac{1}{h_{\gamma_n}}\left[\Delta_{\gamma_n}\log(1/\Delta_{\gamma_n})\right]^{1/2}\rightarrow0,$$ 则有 $$\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}} \left(X_{t_i^{(\gamma_n)}}-x\right)\rightarrow_{\rm a.s.}\bar{L}(2\gamma_n,x).$$ 引理3.1的证明参见文献[9,定理1].

定理2.1的证明考虑(2.3)式,其可以表示为 \begin{eqnarray} \hat{\sigma}^2(t^*,x)&=&\frac{\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)\tilde{\sigma}^2(t^*,X_{t_i^{(\gamma_n)}})} {\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)} \nonumber\\ &=&\frac{\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)\left[\tilde{\sigma}^2(t^*,X_{t_i^{(\gamma_n)}})- \sigma^2(t^*,X_{t_i^{(\gamma_n)}})\right]} {\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)} \\ %(3.1)$$ &&+\frac{\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right) \sigma^2(t^*,X_{t_i^{(\gamma_n)}})} {\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)} . %(3.2)$$ \end{eqnarray} 首先证明(3.2)式为 \begin{eqnarray*} &&\frac{\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right) \sigma^2(t^*,X_{t_i^{(\gamma_n)}})} {\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)} \\ &=&\frac{\int_{t^*-\gamma_n}^{t^*+\gamma_n}K_{h_{\gamma_n}}\left(X_{s}-x\right)\sigma^2(t^*,X_{s}){\rm d}s+O_{\rm a.s.}\left(\frac{1}{h_{\gamma_n}} \left[\Delta_{\gamma_n}\log(1/\Delta_{\gamma_n})\right]^{1/2}\right)} {\int_{t^*-\gamma_n}^{t^*+\gamma_n}K_{h_{\gamma_n}}\left(X_{s}-x\right){\rm d}s+O_{\rm a.s.}\left(\frac{1}{h_{\gamma_n}} \left[\Delta_{\gamma_n}\log(1/\Delta_{\gamma_n})\right]^{1/2}\right)}, \end{eqnarray*} 其中符号$``O_{\rm a.s.}"$表示``几乎处处有界或依概率1有界". 考虑下式 \begin{eqnarray} \Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right) \sigma^2(t^*,X_{t_i^{(\gamma_n)}})-\int_{t^*-\gamma_n}^{t^*+\gamma_n}K_{h_{\gamma_n}}\left(X_{s}-x\right)\sigma^2(t^*,X_{s}){\rm d}s %(3.3)$$ \end{eqnarray} 由假设(A1)--(A3)和核函数$K(\cdot)$的性质,对于(3.3)式, 有 \begin{eqnarray*} &&\left|\sum\limits_{i=0}^{n_{\gamma_n}-1}\int_{b_i}^{b_{i+1}}\left[K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right) \sigma^2(t^*,X_{t_i^{(\gamma_n)}})-K_{h_{\gamma_n}}\left(X_{s}-x\right)\sigma^2(t^*,X_{s})\right]{\rm d}s\right| \\ &&+\left|\Delta_{\gamma_n}K_{h_{\gamma_n}}\left(X_{t_0^{(\gamma_n)}}-x\right) \sigma^2(t^*,X_{t_0^{(\gamma_n)}})\right|+\left|\Delta_{\gamma_n}K_{h_{\gamma_n}}\left(X_{t_{n_{\gamma_n}}^{(\gamma_n)}}-x\right) \sigma^2(t^*,X_{t_{n_{\gamma_n}}^{(\gamma_n)}})\right| \\ &\leq& \left|\sum\limits_{i=0}^{n_{\gamma_n}-1}\int_{b_i}^{b_{i+1}}\left[K_{h_{\gamma_n}}\left(X_{s}-x\right) \sigma^2(t^*,X_{t_i^{(\gamma_n)}})-K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)\sigma^2(t^*,X_{t_i^{(\gamma_n)}})\right]{\rm d}s\right| \\ &&+\left|\sum\limits_{i=0}^{n_{\gamma_n}-1}\int_{b_i}^{b_{i+1}}\left[K_{h_{\gamma_n}}\left(X_{s}-x\right) \sigma^2(t^*,X_{s})-K_{h_{\gamma_n}}\left(X_{s}-x\right)\sigma^2(t^*,X_{t_i^{(\gamma_n)}})\right]{\rm d}s\right| \\ &&+C_1O_{\rm a.s.}\left(\frac{\Delta_{\gamma_n}}{h_{\gamma_n}}\right) \\ &\leq& \frac{1}{h_{\gamma_n}}\sum\limits_{i=0}^{n_{\gamma_n}-1}\int_{b_i}^{b_{i+1}}\left|K'\left(\frac{\tilde{X}_{is}-x}{h_{\gamma_n}}\right)\right| \left|\frac{X_{s}-X_{t_i^{(\gamma_n)}}}{h_{\gamma_n}}\right|\left|\sigma^2(t^*,X_{t_i^{(\gamma_n)}})\right|{\rm d}s \\ &&+\left|\sum\limits_{i=0}^{n_{\gamma_n}-1}\int_{b_i}^{b_{i+1}}K_{h_{\gamma_n}}\left(X_{s}-x\right) \left[\sigma^2(t^*,X_{s})-\sigma^2(t^*,X_{t_i^{(\gamma_n)}})\right]{\rm d}s\right|+C_1O_{\rm a.s.}\left(\frac{\Delta_{\gamma_n}}{h_{\gamma_n}}\right), \end{eqnarray*} 其中$b_i=t^*-\gamma_n+i\Delta_{\gamma_n},i=0,1,2,\cdots,n_{\gamma_n}-1$, $C_1$是常数, $\tilde{X}_{is}$是$X_s$和$X_{t_i^{(\gamma_n)}}$之间的一个值. 定义 \begin{eqnarray} \kappa_{\gamma_n}=\max_{0\leq i\leq n_{\gamma_n}}\sup_{b_i\leq s\leq b_{i+1}}\left|X_{s}-X_{t_i^{(\gamma_n)}}\right|. %(3.4)$$ \end{eqnarray} 由文献^[10]中第2章的定理9.25,得 \begin{eqnarray} P\left(\left[\limsup_{\Delta_{\gamma_n}\rightarrow0}\frac{\kappa_{\gamma_n}} {\left(\Delta_{\gamma_n}\log(1/\Delta_{\gamma_n})\right)^{1/2}}=C_2\right]\right)=1, %(3.5)$$ \end{eqnarray} 其中$C_2$是某个常数. 因此 $$\kappa_{\gamma_n}=O_{\rm a.s.}\left(\Delta_{\gamma_n}\log(1/\Delta_{\gamma_n})\right)^{1/2},$$ 若$h_{\gamma_n}\rightarrow0$, 使得 $$\frac{1}{h_{\gamma_n}}\left[\Delta_{\gamma_n}\log(1/\Delta_{\gamma_n})\right]^{1/2}\rightarrow0, $$ 则有 \begin{eqnarray} \frac{\kappa_{\gamma_n}}{h_{\gamma_n}}\rightarrow_{\rm a.s.}0. %(3.6)$$ \end{eqnarray} 由(3.6)式,得 \begin{eqnarray} K'\left(\frac{\tilde{X}_{is}-x}{h_{\gamma_n}}\right)=K'\left(\frac{X_{s}-x}{h_{\gamma_n}}+o_{\rm a.s.}(1)\right),i=0,1,2,\cdots,n_{\gamma_n}. %(3.7) \end{eqnarray} 由(3.4)式和(3.7)式, $K'$的绝对可积性以及$\bar{L}_X(t,\cdot)$和$\sigma^2(\cdot,\cdot)$ 的连续性, 得到 \begin{eqnarray*} &&\frac{\kappa_{\gamma_n}}{h_{\gamma_n}}\frac{1}{h_{\gamma_n}}\int_{t^*-\gamma_n}^{t^*+\gamma_n}\left|K'\left(\frac{X_{s}-x}{h_{\gamma_n}} +o_{\rm a.s.}(1)\right)\right|\left|\sigma^2(t^*,X_s+o_{\rm a.s.}(1))\right|{\rm d}s \\ &=&\frac{\kappa_{\gamma_n}}{h_{\gamma_n}}\frac{1}{h_{\gamma_n}}\int_{-\infty}^{+\infty}\left|K'\left(\frac{p-x}{h_{\gamma_n}} +o_{\rm a.s.}(1)\right)\right|\left|\sigma^2(t^*,p)\right|\bar{L}_X(2\gamma_n,p){\rm d}p \\ &=&\frac{\kappa_{\gamma_n}}{h_{\gamma_n}}\frac{1}{h_{\gamma_n}}\int_{-\infty}^{+\infty}\left|K'(q +o_{\rm a.s.}(1))\right|\left|\sigma^2(t^*,qh_{\gamma_n}+x)\right|\bar{L}_X(2\gamma_n,qh_{\gamma_n}+x){\rm d}q \\ &\leq & C_3\left(\frac{\kappa_{\gamma_n}}{h_{\gamma_n}}\right)O_{\rm a.s.}\left(\bar{L}_X(2\gamma_n,x)\right) \\ &\leq & C_3\left(\frac{\kappa_{\gamma_n}}{h_{\gamma_n}}\right)O_{\rm a.s.}\left(\gamma_n^{1/2}\right) , \end{eqnarray*} 其中最后一个不等式成立是由于 $$\bar{L}_X(2\gamma_n,x)=\gamma_n^{1/2}\frac{1}{\sigma}L_w\left(1,\frac{x}{\sigma\gamma_n^{1/2}}\right)=O_{\rm a.s.}(\gamma_n^{1/2}). $$ 同理可证 $$\left|\sum\limits_{i=0}^{n_{\gamma_n}-1}\int_{b_i}^{b_{i+1}}K_{h_{\gamma_n}}\left(X_{s}-x\right) \left[\sigma^2(t^*,X_{s})-\sigma^2(t^*,X_{t_i^{(\gamma_n)}})\right]{\rm d}s\right|\leq C_4 \kappa_{\gamma_n}O_{\rm a.s.}(\gamma_n^{1/2}).$$ 由上述可得 \begin{eqnarray*} &&\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right) \sigma^2(t^*,X_{t_i^{(\gamma_n)}}) \\ &=&\int_{t^*-\gamma_n}^{t^*+\gamma_n}K_{h_{\gamma_n}}\left(X_{s}-x\right)\sigma^2(t^*,X_{s}){\rm d}s+O_{\rm a.s.}\left(\frac{\gamma_n^{1/2}}{h_{\gamma_n}} \left[\Delta_{\gamma_n}\log(1/\Delta_{\gamma_n})\right]^{1/2}\right). \end{eqnarray*} 由类似的方法可得 \begin{eqnarray*} &&\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right) \\ &=&\int_{t^*-\gamma_n}^{t^*+\gamma_n}K_{h_{\gamma_n}}\left(X_{s}-x\right){\rm d}s+O_{\rm a.s.}\left(\frac{\gamma_n^{1/2}}{h_{\gamma_n}} \left[\Delta_{\gamma_n}\log(1/\Delta_{\gamma_n})\right]^{1/2}\right). \end{eqnarray*} 故对于(3.2)式,如果有$\gamma_n\rightarrow0, h_{\gamma_n}\rightarrow0$,使得 $$\frac{\gamma_n^{1/2}}{n_{\gamma_n}}\left[\Delta_{\gamma_n}\log(1/\Delta_{\gamma_n})\right]^{1/2}\rightarrow0,$$ 则由引理1.3得 \begin{eqnarray*} &&\frac{\int_{t^*-\gamma_n}^{t^*+\gamma_n}K_{h_{\gamma_n}} \left(X_{s}-x\right)\sigma^2(t^*,X_{s}){\rm d}s+ O_{\rm a.s.}\left(\frac{1}{h_{\gamma_n}} \left[\Delta_{\gamma_n}\log(1/\Delta_{\gamma_n})\right]^{1/2}\right)} {\int_{t^*-\gamma_n}^{t^*+\gamma_n}K_{h_{\gamma_n}} \left(X_{s}-x\right){\rm d}s+O_{\rm a.s.}\left(\frac{1}{h_{\gamma_n}} \left[\Delta_{\gamma_n}\log(1/\Delta_{\gamma_n})\right]^{1/2}\right)} \\ &=&\frac{\sigma^2(t^*,x)s(x)+o_{\rm a.s.}(1)}{s(x)+ o_{\rm a.s.}(1)}\rightarrow_{\rm a.s.}\sigma^2(t^*,x), \end{eqnarray*} 其中$s(x)$是过程的速率函数.

现在考虑(3.1)式,下面证明: 对固定的$X_{t_i^{(\gamma_n)}}$,有下式成立 $$\tilde{\sigma}^2(t^*,X_{t_i^{(\gamma_n)}})=\sigma^2(t^*,X_{t_i^{(\gamma_n)}})+o_{\rm a.s.}(1).$$ 由$\sigma^2(\cdot,\cdot)$的Lipschitz性质,有 \begin{eqnarray*} &&\frac{1}{m_{\gamma_n}(t_i^{(\gamma_n)})\Delta_{\gamma_n}} \sum\limits_{j=0}^{m_{\gamma_n}(t_i^{(\gamma_n)})-1}\left[X_{\tau_{i,j}^{(\gamma_n)}+\Delta_{\gamma_n}}-X_{\tau_{i,j}^{(\gamma_n)}}\right]^2 -\sigma^2(t^*,X_{t_i^{(\gamma_n)}}) \\ &=&\frac{1}{2m_{\gamma_n}(t_i^{(\gamma_n)})\Delta_{\gamma_n}} \sum\limits_{j=0}^{m_{\gamma_n}(t_i^{(\gamma_n)})-1}\int_{\tau_{i,j}^{(\gamma_n)}}^{\tau_{i,j}^{(\gamma_n)}+\Delta_{\gamma_n}}2\left(X_s-X_{t_i^{(\gamma_n)}} \right)\sigma(t^*,X_s){\rm d}B_s \\ &&+\frac{1}{2m_{\gamma_n}(t_i^{(\gamma_n)})\Delta_{\gamma_n}} \sum\limits_{j=0}^{m_{\gamma_n}(t_i^{(\gamma_n)})-1}\int_{\tau_{i,j}^{(\gamma_n)}}^{\tau_{i,j}^{(\gamma_n)}+\Delta_{\gamma_n}}2\left(X_s-X_{t_i^{(\gamma_n)}} \right) \mu(t^*,X_s){\rm d}s +C_5O_{\rm a.s.}(\kappa_{\gamma_n}), \end{eqnarray*} 其中 $$\int_{\tau_{i,j}^{(\gamma_n)}}^{\tau_{i,j}^{(\gamma_n)}+\Delta_{\gamma_n}}2\left(X_s-X_{t_i^{(\gamma_n)}}\right) \mu(t^*,X_s){\rm d}s=o_{\rm a.s.} \bigg(\int_{\tau_{i,j}^{(\gamma_n)}}^{\tau_{i,j}^{(\gamma_n)} +\Delta_{\gamma_n}}2\left(X_s-X_{t_i^{(\gamma_n)}}\right) \sigma(t^*,X_s){\rm d}B_s\bigg).$$ 定义积分为 $$y_{\tau_{i,j}^{(\gamma_n)}+\Delta_{\gamma_n}}=\int_{\tau_{i,j}^{(\gamma_n)}}^{\tau_{i,j}^{(\gamma_n)}+\Delta_{\gamma_n}}2 \left(X_s-X_{t_i^{(\gamma_n)}}\right) \sigma(t^*,X_s){\rm d}B_s, $$ 上式定义的积分关于$F_{\tau_{i,j}^{(\gamma_n)}+\Delta_{\gamma_n}}$ 可测, 由It\^{o}积分的性质, 得$E\big(y_{\tau_{i,j}^{(\gamma_n)}+\Delta_{\gamma_n}}\big)=0$, 又由It\^{o}积分的等距性,则对所有的$j\leq m_{\gamma_n}$,有 $$\theta_{\tau_{i,j}^{(\gamma_n)}+\Delta_{\gamma_n}}= {\rm Var}\left(y_{\tau_{i,j}^{(\gamma_n)}+\Delta_{\gamma_n}}\right) <\infty. $$ 因此$\big(y_{\tau_{i,j}^{(\gamma_n)}+\Delta_{\gamma_n}}, F_{\tau_{i,j}^{(\gamma_n)}+\Delta_{\gamma_n}}\big)$是一个均值为零, 方差有限的鞅差序列,由强大数定律知, 当$m_{\gamma_n}(t_{i}^{(\gamma_n)})\rightarrow\infty$时,有 $$\frac{1}{2m_{\gamma_n}(t_i^{(\gamma_n)})\Delta_{\gamma_n}} \sum\limits_{j=0}^{m_{\gamma_n}(t_i^{(\gamma_n)})-1}y_{\tau_{i,j}^{(\gamma_n)}+\Delta_{\gamma_n}} =o_{\rm a.s.}(1).$$ 故对每个$0\leq i\leq n_{\gamma_n}$,有 $$\tilde{\sigma}^2(t^*,X_{t_i^{(\gamma_n)}})\rightarrow_{\rm a.s.}\sigma^2(t^*,X_{t_i^{(\gamma_n)}}). $$ 下面考虑收敛速度问题 \begin{eqnarray} &&\frac{1}{2m_{\gamma_n}(t_i^{(\gamma_n)})\Delta_{\gamma_n}} \sum\limits_{j=0}^{m_{\gamma_n}(t_i^{(\gamma_n)})-1}\int_{\tau_{i,j}^{(\gamma_n)}}^{\tau_{i,j}^{(\gamma_n)}+\Delta_{\gamma_n}}2\left(X_s-X_{t_i^{(\gamma_n)}} \right)\sigma(t^*,X_s){\rm d}B_s \nonumber\\ &=&\frac{\frac{1}{2\varepsilon_{\gamma_n}}\sum\limits_{j=0}^{n_{\gamma_n}}I_{\{|X_{t_j}^{(\gamma_n)}-X_{t_i}^{(\gamma_n)}|\leq\varepsilon_{\gamma_n}\}} \int_{t_j^{(\gamma_n)}}^{t_{j+1}^{(\gamma_n)}}2\left(X_s-X_{t_i^{(\gamma_n)}} \right)\sigma(t^*,X_s){\rm d}B_s}{\frac{\Delta_{\gamma_n}}{2\varepsilon_{\gamma_n}} \sum\limits_{j=0}^{n_{\gamma_n}}I_{\{|X_{t_j}^{(\gamma_n)}-X_{t_i}^{(\gamma_n)}|\leq\varepsilon_{\gamma_n}\}}}. %(3.8)$$ \end{eqnarray} 类似于文献[6,定理2]的证明,得到 \begin{eqnarray*} &&\sqrt{\bar{L}_X(2\gamma_n,X_{t_i^{(\gamma_n)}})\varepsilon_{\gamma_n}} \frac{\frac{1}{2\varepsilon_{\gamma_n}}\sum\limits_{j=0}^{n_{\gamma_n}}I_{\{|X_{t_j}^{(\gamma_n)}-X_{t_i}^{(\gamma_n)}|\leq\varepsilon_{\gamma_n}\}} \int_{t_j^{(\gamma_n)}}^{t_{j+1}^{(\gamma_n)}}2\left(X_s-X_{t_i^{(\gamma_n)}} \right)\sigma(t^*,X_s){\rm d}B_s}{\frac{\Delta_{\gamma_n}}{2\varepsilon_{\gamma_n}} \sum\limits_{j=0}^{n_{\gamma_n}}I_{\{|X_{t_j}^{(\gamma_n)}-X_{t_i}^{(\gamma_n)}|\leq\varepsilon_{\gamma_n}\}}} \\ &=&O_{\rm a.s.}(1). \end{eqnarray*} 故 \begin{eqnarray*} &&\tilde{\sigma}^2(t^*,X_{t_i^{(\gamma_n)}})-\sigma^2(t^*,X_{t_i^{(\gamma_n)}}) \\ &=&\frac{\frac{1}{2\varepsilon_{\gamma_n}}\sum\limits_{j=0}^{n_{\gamma_n}}I_{\{|X_{t_j}^{(\gamma_n)}-X_{t_i}^{(\gamma_n)}|\leq\varepsilon_{\gamma_n}\}} \int_{t_j^{(\gamma_n)}}^{t_{j+1}^{(\gamma_n)}}2\left(X_s-X_{t_i^{(\gamma_n)}} \right)\sigma(t^*,X_s){\rm d}B_s}{\frac{\Delta_{\gamma_n}}{2\varepsilon_{\gamma_n}} \sum\limits_{j=0}^{n_{\gamma_n}}I_{\{|X_{t_j}^{(\gamma_n)}-X_{t_i}^{(\gamma_n)}|\leq\varepsilon_{\gamma_n}\}}} +C_5O_{\rm a.s.}(\kappa_{\gamma_n}) \\ &=&O_{\rm a.s.}\left(\frac{1}{\sqrt{\bar{L}_X(2\gamma_n,X_{t_i^{(\gamma_n)}})\varepsilon_{\gamma_n}}}\right)+C_5O_{\rm a.s.}(\kappa_{\gamma_n}). \end{eqnarray*} 选择适当的带宽参数$\varepsilon_{\gamma_n}$, 则有 $$O_{\rm a.s.}\left(\frac{1}{\sqrt{\bar{L}_X(2\gamma_n,X_{t_i^{(\gamma_n)}})\varepsilon_{\gamma_n}}}\right)=o_{\rm a.s.}(1). $$ 因此,定理得证.

定理2.2的证明考虑下式 \begin{eqnarray} &&\hat{\sigma}^2(t^*,x)-\sigma^2(t^*,x) \nonumber\\ &=&\frac{\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)\tilde{\sigma}^2(t^*,X_{t_i^{(\gamma_n)}})} {\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)}-\sigma^2(t^*,x) \nonumber\\ &=&\frac{\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)\tilde{\sigma}^2(t^*,X_{t_i^{(\gamma_n)}})} {\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)}-\frac{\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}} -x\right)\sigma^2(t^*,X_{t_i^{(\gamma_n)}})}{\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)} %(3.9)$$ \\ &&+\frac{\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)\sigma^2(t^*,X_{t_i^{(\gamma_n)}})} {\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)}-\frac{\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}} -x\right)\sigma^2(t^*,x)}{\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)}. %(3.10)$$ \end{eqnarray} 首先考虑(3.10)式,由引理1.1和核函数$K(\cdot)$的对称性, (3.10)式的分子部分可以表示为 \begin{eqnarray*} &&\int_{-\infty}^{+\infty}K(c)\frac{\partial\sigma^2(t^*,x)}{\partial x}ch_{\gamma_n}\left(\frac{L_X(2\gamma_n,x+ch_{\gamma_n})-L_X(2\gamma_n,x)} {\sigma^2(t^*,x+ch_{\gamma_n})}\right){\rm d}c \\ &&+\int_{-\infty}^{+\infty}K(c)\frac{\partial\sigma^2(t^*,x)}{\partial x}ch_{\gamma_n}\left(\frac{\sigma^2(t^*,x)-\sigma^2(t^*,x+ch_{\gamma_n})} {\sigma^2(t^*,x+ch_{\gamma_n})\sigma^2(t^*,x)}\right)L_X(2\gamma_n,x){\rm d}c \\ &&+\int_{-\infty}^{+\infty}K(c)\frac{1}{2}\frac{\partial^2\sigma^2(t^*,x)}{\partial x^2}c^2h_{\gamma_n}^2\left(\frac{L_X(2\gamma_n,x+ch_{\gamma_n})} {\sigma^2(t^*,x+ch_{\gamma_n})}\right){\rm d}c \\ &\triangleq & A_1+A_2+A_3. \end{eqnarray*} 由引理1.2,$A1$有以下极限形式 \begin{eqnarray*} A1&=&2h_{\gamma_n}^{3/2}\frac{\partial\sigma^2(t^*,x)}{\partial x}\frac{1}{\sigma^2(t^*,x)}\int_{-\infty}^{+\infty}cK(c)\frac{1}{2\sqrt{h_{\gamma_n}}}\left[L_X(2\gamma_n,x+ch_{\gamma_n}) -L_X(2\gamma_n,x)\right]{\rm d}c \\ &=&2h_{\gamma_n}^{3/2}\frac{\partial\sigma^2(t^*,x)}{\partial x}\frac{1}{\sigma^2(t^*,x)}\int_{0}^{+\infty}cK(c)\frac{1}{2\sqrt{h_{\gamma_n}}}\left[L_X(2\gamma_n,x+ch_{\gamma_n})-L_X(2\gamma_n,x)\right]{\rm d}c \\ &&+2h_{\gamma_n}^{3/2}\frac{\partial\sigma^2(t^*,x)}{\partial x}\frac{1}{\sigma^2(t^*,x)}\int_{-\infty}^{0}cK(c)\frac{1}{2\sqrt{h_{\gamma_n}}}\left[L_X(2\gamma_n,x+ch_{\gamma_n})-L_X(2\gamma_n,x)\right]{\rm d}c \\ &\rightarrow_d&2h_{\gamma_n}^{3/2}\frac{\partial\sigma^2(t^*,x)}{\partial x}\frac{1}{\sigma^2(t^*,x)}\int_{-\infty}^{0}cK(c)W^{\otimes}(L_X(2\gamma_n,x),-c){\rm d}c \\ &&+2h_{\gamma_n}^{3/2}\frac{\partial\sigma^2(t^*,x)}{\partial x}\frac{1}{\sigma^2(t^*,x)}\int_{0}^{+\infty}cK(c)W^{\oplus}(L_X(2\gamma_n,x),c){\rm d}c \\ &=_d&2h_{\gamma_n}^{3/2}\frac{\partial\sigma^2(t^*,x)}{\partial x}\frac{1}{\sigma(t^*,x)}\sqrt{\bar{L}_X(2\gamma_n,x)} \bigg(\int_{0}^{+\infty}cK(c)W^{\oplus}(1,c){\rm d}c \\ &&+\int_{-\infty}^{0}cK(c)W^{\otimes}(1,x),-c){\rm d}c \bigg), \end{eqnarray*} 其中$W^{\oplus}$和$W^{\otimes}$是相互独立的Brownian sheet (参见文献^[8]中第13章定理2.3),符号$``=_d"$表示``依分布等价". 又知 \begin{eqnarray*} \int_{0}^{+\infty}cK(c)W^{\oplus}(1,c){\rm d}c &=_d&\int_{0}^{+\infty}cK(c)W^{\oplus}(c){\rm d}c \\ &=_d&N\left(0,\int_{0}^{+\infty}\int_{0}^{+\infty} usK(u)K(s)\min(u,s){\rm d}u{\rm d}s\right), \end{eqnarray*} 类似地 \begin{eqnarray*} \int_{-\infty}^{0}cK(c)W^{\otimes}(1,-c){\rm d}c &=_d&\int_{-\infty}^{0}cK(c)W^{\otimes}(-c){\rm d}c \\ &=_d&N\left(0,-\int_{-\infty}^{0}\int_{-\infty}^{0} usK(u)K(s)\max(u,s){\rm d}u{\rm d}s\right). \end{eqnarray*} 故 \begin{eqnarray*} &&2h_{\gamma_n}^{3/2}\frac{\partial\sigma^2(t^*,x)}{\partial x}\frac{1}{\sigma(t^*,x)}\sqrt{\bar{L}_X(2\gamma_n,x)} \bigg(\int_{0}^{+\infty}cK(c)W^{\oplus}(1,c){\rm d}c \\ &&+ \int_{-\infty}^{0}cK(c)W^{\otimes}(1,x),-c){\rm d}c \bigg) \\ &=_d&2h_{\gamma_n}^{3/2}\frac{\partial\sigma^2(t^*,x)}{\partial x}\frac{1}{\sigma(t^*,x)}\sqrt{\bar{L}_X(2\gamma_n,x)}N \left(0,2\int_{0}^{+\infty}\int_{0}^{+\infty} usK(u)K(s)\min(u,s){\rm d}u{\rm d}s\right) \\ &=_d&h_{\gamma_n}^{3/2}MN\left(0,16\varphi \left(\frac{\partial\sigma(t^*,x)}{\partial x}\right)^2\bar{L}_X(2\gamma_n,x)\right), \end{eqnarray*} 其中$\varphi=2\int_{0}^{+\infty}\int_{0}^{+\infty}usK(u)K(s) \min(u,s){\rm d}u{\rm d}s$. 因此 \begin{eqnarray*} && \frac{1}{h_{\gamma_n}^{3/2}}\left(\frac{\int_{-\infty}^{+\infty}K(c)\frac{\partial\sigma^2(t^*,x)}{\partial x}ch_{\gamma_n}\left(\frac{L_X(2\gamma_n,x+ch_{\gamma_n})-L_X(2\gamma_n,x)} {\sigma^2(t^*,x+ch_{\gamma_n})}\right){\rm d}c}{\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)}\right) \\ &&\rightarrow_dMN\left(0,16\varphi\left(\frac{\partial\sigma(t^*,x)}{\partial x}\right)^2/\bar{L}_X(2\gamma_n,x)\right). \end{eqnarray*} 对于$A_2,A_3$,由$\sigma^2(\cdot,\cdot)$的Lipschitz性质,得 $$\frac{A_2}{\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)}+ \frac{A_3}{\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)}=O_{\rm a.s.}(h_{\gamma_n}^2). $$ 故对(3.10)式,有 \begin{eqnarray*} &&\frac{1}{h_{\gamma_n}^{3/2}}\left[\frac{\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)\sigma^2(t^*,X_{t_i^{(\gamma_n)}})} {\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)}-\frac{\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}} -x\right)\sigma^2(t^*,x)}{\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)}\right] \\ &=_d&\frac{1}{h_{\gamma_n}^{3/2}}\left(\frac{\int_{-\infty}^{+\infty}K(c)\frac{\partial\sigma^2(t^*,x)}{\partial x}ch_{\gamma_n}\left(\frac{L_X(2\gamma_n,x+ch_{\gamma_n})-L_X(2\gamma_n,x)} {\sigma^2(t^*,x+ch_{\gamma_n})}\right){\rm d}c}{\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)} +O_{\rm a.s.}(h_{\gamma_n}^2)\right) \\ &=_d&MN\left(0,16\varphi\left(\frac{\partial\sigma(t^*,x)}{\partial x}\right)^2/\bar{L}_X(2\gamma_n,x)\right). \end{eqnarray*} 下面考虑(3.9)式, 由It\^{o}引理,(3.9)式的分子部分可以表示为 \begin{eqnarray*} &&\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)\frac{\sum\limits_{j=0}^{n_{\gamma_n}} I_{\{|X_{t_j}^{(\gamma_n)}-X_{t_i}^{(\gamma_n)}|\leq\varepsilon_{\gamma_n}\}}\frac{1}{2}\left[\int_{t_j^{(\gamma_n)}}^{t_{j+1}^{(\gamma_n)}} (\sigma^2(t^*,x)-\sigma^2(t^*,X_{t_i^{(\gamma_n)}})){\rm d}s\right]} {\sum\limits_{j=0}^{n_{\gamma_n}}I_{\{|X_{t_j}^{(\gamma_n)}-X_{t_i}^{(\gamma_n)}|\leq\varepsilon_{\gamma_n}\}}} \\ &&+\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)\frac{\sum\limits_{j=0}^{n_{\gamma_n}} I_{\{|X_{t_j}^{(\gamma_n)}-X_{t_i}^{(\gamma_n)}|\leq\varepsilon_{\gamma_n}\}}\frac{1}{2}\left[\int_{t_j^{(\gamma_n)}}^{t_{j+1}^{(\gamma_n)}} 2(X_s-X_{t_i^{(\gamma_n)}})\sigma(t^*,X_s){\rm d}B_s\right]} {\sum\limits_{j=0}^{n_{\gamma_n}}I_{\{|X_{t_j}^{(\gamma_n)}-X_{t_i}^{(\gamma_n)}|\leq\varepsilon_{\gamma_n}\}}} \\ &&+\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)\frac{\sum\limits_{j=0}^{n_{\gamma_n}} I_{\{|X_{t_j}^{(\gamma_n)}-X_{t_i}^{(\gamma_n)}|\leq\varepsilon_{\gamma_n}\}}\frac{1}{2}\left[\int_{t_j^{(\gamma_n)}}^{t_{j+1}^{(\gamma_n)}} 2(X_s-X_{t_i^{(\gamma_n)}})\mu(t^*,X_s){\rm d}s\right]} {\sum\limits_{j=0}^{n_{\gamma_n}}I_{\{|X_{t_j}^{(\gamma_n)}-X_{t_i}^{(\gamma_n)}|\leq\varepsilon_{\gamma_n}\}}} \\ &&+O_{\rm a.s.}\left(\frac{\Delta_{\gamma_n}}{\varepsilon_{\gamma_n}}\right) \\ &\triangleq & B_1+B_2+B_3+O_{\rm a.s.}\left(\frac{\Delta_{\gamma_n}}{\varepsilon_{\gamma_n}}\right). \end{eqnarray*} 类似于文献[9,定理3]的证明,我们有 若$h_{\gamma_n}=o(\varepsilon_{\gamma_n})$,则 \begin{eqnarray} \sqrt{\frac{\varepsilon_{\gamma_n}}{\Delta_{\gamma_n}}}\left(\frac{B_2} {\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)}\right) \rightarrow_dMN\left(0,\frac{2\sigma^4(t^*,x)}{\bar{L}_X(2\gamma_n,x)}\right) %(3.11)$$ \end{eqnarray} 且 \begin{eqnarray*} B_1& =_d&\int_{-\infty}^{+\infty}K(c)\frac{\frac{1}{2}\int_{-\infty}^{+\infty}I_{\{|a|\leq1\}}\left(\frac{\partial\sigma^2(t^*,x)}{\partial x}\varepsilon_{\gamma_n}a\right)\left(\frac{L_X(2\gamma_n,x+a\varepsilon_{\gamma_n})-L_X(2\gamma_n,x)} {\sigma^2(t^*,x+a\varepsilon_{\gamma_n})}\right){\rm d}a}{\frac{1}{2}\int_{-\infty}^{+\infty}I_{\{|a|\leq1\}} \bar{L}_X(2\gamma_n,x+a\varepsilon_{\gamma_n}){\rm d}a} \\ &&\cdot \bar{L}_X(2\gamma_n,x+ch_{\gamma_n}){\rm d}c \\ &&+\int_{-\infty}^{+\infty}K(c)\frac{\frac{1}{2}\int_{-\infty}^{+\infty}I_{\{|a|\leq1\}}\left(\frac{\partial\sigma^2(t^*,x)}{\partial x}\varepsilon_{\gamma_n}a\right)\left(\frac{\sigma^2(t^*,x)-\sigma^2(t^*,x+a\varepsilon_{\gamma_n})} {\sigma^2(t^*,x+a\varepsilon_{\gamma_n})\sigma^2(t^*,x)}\right)L_X(2\gamma_n,x){\rm d}a}{\frac{1}{2}\int_{-\infty}^{+\infty} I_{\{|a|\leq1\}} \bar{L}_X(2\gamma_n,x+a\varepsilon_{\gamma_n}){\rm d}a} \\ &&\cdot\bar{L}_X(2\gamma_n,x+ch_{\gamma_n}){\rm d}c \\ &&+\int_{-\infty}^{+\infty}K(c)\frac{\frac{1}{2}\int_{-\infty}^{+\infty}I_{\{|a|\leq1\}}\left(\frac{1}{2}\frac{\partial^2\sigma^2(t^*,x)}{\partial x^2}\varepsilon_{\gamma_n}^2a^2\right)\bar{L}_X(2\gamma_n,x+a\varepsilon_{\gamma_n}){\rm d}a}{\frac{1}{2}\int_{-\infty}^{+\infty}I_{\{|a|\leq1\}} \bar{L}_X(2\gamma_n,x+a\varepsilon_{\gamma_n}){\rm d}a} \\ &&\cdot\bar{L}_X(2\gamma_n,x+ch_{\gamma_n}){\rm d}c \\ &&-\int_{-\infty}^{+\infty}K(c)\left(\frac{\partial\sigma^2(t^*,x)}{\partial x}h_{\gamma_n}c+\frac{1}{2}\frac{\partial^2\sigma^2(t^*,x)}{\partial x^2}h_{\gamma_n}^2c^2\right)\bar{L}_X(2\gamma_n,x+ch_{\gamma_n}){\rm d}c. \end{eqnarray*} 由引理1.2,得 \begin{equation} \frac{1}{\varepsilon_{\gamma_n}^{3/2}}\left(\frac{B_1} {\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)}\right) \rightarrow_dMN\left(0,16\varphi^{ind}\left(\frac{\partial\sigma(t^*,x)}{\partial x}\right)^2\frac{1}{\bar{L}_X(2\gamma_n,x)}\right), % (3.12)$$ \end{equation} 其中 $$\varphi^{ind}=2\int_0^\infty\int_0^\infty ts\left(\frac{1}{2}I_{|t|\leq 1}\right)\left(\frac{1}{2}I_{|s|\leq 1}\right)\min(t,s){\rm d}t{\rm d}s. $$

注意到 $$\frac{B_3}{\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}} K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)}=O_{\rm a.s.}(\kappa_{\gamma_n}). $$ 综合(3.10)式和(3.9)式的分析,得 \begin{eqnarray*} \hat{\sigma}^2(t^*,x)-\sigma^2(t^*,x) &=&\frac{A_1}{\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)} +\frac{A_2}{\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)} \\ &&+\frac{A_3}{\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)} +\frac{B_1}{\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)} \\ &&+\frac{B_2}{\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)} +\frac{B_3}{\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)}. \end{eqnarray*} 故当$h_{\gamma_n}=o(\varepsilon_{\gamma_n})$和 $\varepsilon_{\gamma_n}^4/\Delta_{\gamma_n}\rightarrow0$时, 由(3.11)式,得 \begin{eqnarray*} &&\sqrt{\frac{\varepsilon_{\gamma_n}}{\Delta_{\gamma_n}}}\left\{\frac{\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}} \left(X_{t_i^{(\gamma_n)}}-x\right)\tilde{\sigma}^2(t^*,X_{t_i^{(\gamma_n)}})} {\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)}-\sigma^2(t^*,x)\right\} \\ &=_d& \sqrt{\frac{\varepsilon_{\gamma_n}}{\Delta_{\gamma_n}}}\left\{\frac{B_2} {\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)}+O_p\left(\varepsilon_{\gamma_n}^{3/2}\right)\right\} \\ &\rightarrow_d& MN\left(0,\frac{2\sigma^4(t^*,x)}{\bar{L}_X(2\gamma_n,x)}\right). \end{eqnarray*} 若当$h_{\gamma_n}=o(\varepsilon_{\gamma_n})$和$\varepsilon_{\gamma_n}^4/\Delta_{\gamma_n}\rightarrow\infty$时, 由(3.12)式,得 \begin{eqnarray*} &&\frac{1}{\varepsilon_{\gamma_n}^{3/2}}\left\{\frac{\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}} \left(X_{t_i^{(\gamma_n)}}-x\right)\tilde{\sigma}^2(t^*,X_{t_i^{(\gamma_n)}})} {\Delta_{\gamma_n}\sum\limits_{i=0}^{n_{\gamma_n}}K_{h_{\gamma_n}}\left(X_{t_i^{(\gamma_n)}}-x\right)}-\sigma^2(t^*,x)\right\} \\ &&\rightarrow_dMN\left(0,16\varphi^{ind}\left(\frac{\partial\sigma(t^*,x)}{\partial x}\right)^2\frac{1}{\bar{L}_X(2\gamma_n,x)}\right). \end{eqnarray*} 证毕.