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(Ω,F,P)上的随机过程 {Xt,0≤t≤T},满足如下的非齐次扩散模型 dXt=μ(t,Xt)dt+σ(t,Xt)dBt, X0=x0(1.1) 其中{Bt,0≤t≤T}是标准布朗运动, 二元函数μ(t,Xt)和σ(t,Xt)分别称为随机过程Xt 的漂移系数和扩散系数,初值x0∈L2,且与{Bt,0≤t≤T} 独立. 令Ft 为{Xs,0≤s≤t}生成的σ -代数, 本文中,Ci,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μ和σ是可测函数,且满足 |μ(t,x)|+|σ(t,x)|≤C1(1+|x|),x∈R,t∈[0,T].
(A2) (Lipschitz条件)~ |μ(t,x)−μ(t,y)|+|σ(t,x)−σ(t,y)|≤C2|x−y|,x,y∈R,t∈[0,T].
(A3)~ |σ(t1,x)−σ(t2,x)|≤C3|t1−t2|,且σ(t,x)>0,x∈R,t1,t2,t∈[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σ(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Xt可以分解为 Xt=LXt+At, 其中LXt是一个连续局部鞅, At是具有有限方差的连续适应过程,则称Xt是一个连续半鞅.
定义1.2若Xt是一个连续半鞅, 则存在一个关于t非减的随机过程LX(t,a),使得 LX(t,a)=limϵ→01ϵ∫t0I[a,a+ϵ](Xs)d[X]s 几乎处处成立,并称LX(t,a)为Xt在a点处的局部时, 其中[X]t表示Xt的二次互偏差过程,IA表示集合A的示性函数.
局部时LX(t,a)是一个表示过程Xt在点a的邻域内发生的次数的量. 若扩散过程Xt有极小的二阶矩σ2(Xt), 则其在点a处的局部时可以表示为 LX(t,a)=limϵ→01ϵ∫t0I[a,a+ϵ](Xs)σ2(Xs)ds.
引理1.1令Xt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 ∫t0f(Xs,s)d[X]s=∫+∞−∞da∫t0f(a,s)dLX(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 ∫t0f(Xs)d[X]s=∫+∞−∞f(a)LX(t,a)da.
引理1.2若Xt满足假设条件(A1)--(A3), r和a是任意固定的实数,则当λ→∞时,有 12√λ{LX(t,r+aλ)−LX(t,r)}→d{W(LX(t,r),a),a>0,W(LX(t,r),−a),a<0, 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.