数学物理学报  2016, Vol. 36 Issue (1): 168-175   PDF (297 KB)    
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王志
荆卉婷
闫理坦
Skorokhod空间中多参数分数布朗单的一个逼近结果
王志1, 荆卉婷2, 闫理坦3    
1 宁波工程学院理学院 浙江 宁波 315211;
2 宁波工程学院经济管理学院 浙江 宁波 315211;
3 东华大学数学系 上海 201620
摘要: 证明了一类由独立同分布随机变量构成的随机游走在Skorokhod空间中依分布收敛于多参数分数布朗单.
关键词: 多参数分数布朗单     随机游走     弱收敛    
Approximation of Multidimensional Parameter Fractional Brownian Sheet in Skorokhod Space
Wang Zhi1, Jing Huiting2, Yan Litan3    
1 School of Sciences, Ningbo University of Technology, Zhejiang Ningbo 315211;
2 School of Economics and Management, Ningbo University of Technology, Zhejiang Ningbo 315211;
3 Department of Mathematics, Donghua University, Shanghai 201620
Abstract: In this paper, we prove a convergence in law of a random walk constructed using a family of independent identically distributed random variables to the multidimensional parameter fractional Brownian sheet in the Skorokhod space.
Key words: Multidimensional parameter fractional Brownian sheet     Random walk     Weak convergence    
1 引言

近年来,分数布朗运动作为一个特定的高斯过程引起了人们的广泛关注和研究, 具有长(或短)记忆性,自相似性, Hölder 连续性,平稳增量性等众多良好的性质, 且在很多领域如数理金融、救护、水文、交通分析、成像加工、电子通讯等有着相当广泛的应用.可参见文献[3, 10, 11, 14, 17] 关于分数布朗运动的详细论述. 分数布朗运动 $B^H=\{B_t^H,t\geq0\}$ 是一个定义在 $(\Omega,{\cal F},P)$ 上的样本轨道连续的中心化高斯过程并且具有协方差函数 $$ R^H(s,t):=E[B^{H}_tB^{H}_s]=\frac{1}{2}[s^{2H}+t^{2H}-|t-s|^{2H}].$$ 称 $H\in(0,1)$ 为Hurst参数. 特别地, 当 $H=\frac12$ 时,$ B^H$ 退化为标准的布朗运动 $B$.$B^H$ 具有如下形式关于布朗运动的积分表现 \begin{equation}\label{ir} B^H_t=\int_0^{t}K_H(t,s){\rm d}B_s,t\geq 0, \tag{1.1} \end{equation} 其中 $K_H$ 是核函数 \begin{equation}\label{representation-2} K_H(t,s)=c_H(t-s)^{H-\frac{1}{2}}+ \Big(\frac{1}{2}-H\Big)c_H\int_s^{t}(u-s)^{H-\frac{3}{2}}\left(1- \Big(\frac su\Big)\right)^{\frac{1}{2}-H}{\rm d}u, \tag{1.2} \end{equation} $c_H>0$ 是常数 $$ c_H=\sqrt{\frac{2H\Gamma(\frac{3}{2}-H)}{ \Gamma(H+\frac{1}{2})\Gamma(2-2H)}}.$$

一般而言,分数布朗运动有两类常见的多参数扩张. 其一是Lévy 意义下的分数布朗域(fractional Brownian random field), 参见文献[5].另一类推广是由文献[12] 引入的各向异性分数布朗域 (anisotropic fractional Brownian random field),它是一个中心化的高斯过程 $B^{\alpha}=\{B_t^{\alpha},t\in {\Bbb R}_+^d\}$ 具有如下形式的协方差函数 $$ E[B^{\alpha}_tB^{\alpha}_s]=\prod_{k=1}^d\frac{1}{2}[s_k^{2\alpha_k}+t_k^{2\alpha_k}-|t_k-s_k|^{2\alpha_k}], $$ 其中 $\alpha=(\alpha_1,\alpha_2,\cdots,\alpha_d)\in(0,1)^d$,称之为 $d$ -参数分数布朗单.当 $\alpha_1=\alpha_2=\cdots=\alpha_d=\frac12$ 时,$B^{\alpha}$ 退化为标准的 $d$ -参数布朗单 $$W=\{W_t,t\in {\Bbb R}_+^d\}. $$ 这类过程在坐标轴上取值为零且具有连续的等价样本.

类似于分数布朗运动的积分表现(1.1),$d$ -参数分数布朗单 $B^{\alpha}$ 具有如下的积分表现 \begin{equation}\label{irs} B^{\alpha}_t=\int_0^{t_d}\cdots \int_0^{t_1}K_{\alpha_1}(t_1,u_1)K_{\alpha_2}(t_2,u_2)\cdots K_{\alpha_d}(t_d,u_d){\rm d}W_u.\tag{1.3} \end{equation} 基于此积分表现和经典的Donsker 定理,Bardina-Florit[1] 建立了 ${B^\alpha }$ 的一个逼近结果, 构造了一个多维随机游走在连续函数空间中依分布收敛于分数布朗单. 本文的主要工作是下面的定理,在Skorokhod 空间中给出了多参数分数布朗单的一个弱逼近结果.

定理 1.1  令 $\alpha_k>\frac12$,$k=1,2,\cdots,d$. 设 $\left\{\xi_{i_1,i_2,\cdots,i_d}^{n},i_k=1,2,\cdots\right\}$ 是一族独立同分布的随机变量满足均值为零, $$E(\xi_{i_1,i_2,\cdots,i_d}^{n})^2=1~~ \mbox{ 且}~~ E(\xi_{i_1,i_2,\cdots,i_d}^{n})^m<+\infty. $$ 对任意的 $n\geq 1$,$t=(t_1,t_2,\cdots,t_d)\in [0, 1]^d$,令 \begin{equation}\label{bs} B^n_t:=\frac{1}{n^{\frac d2}}\sum_{k=1}^{d}\sum_{i_k=1}^{\lfloor nt_k\rfloor}\xi_{i_1,i_2,\cdots,i_d}^{n}, \tag{1.4} \end{equation} \begin{eqnarray}\label{zn} Z^n_t:&=&\int_{0}^{t_d}\cdots \int_{0}^{t_1} K^{n}_{\alpha_1}(t_1,u_1)\cdots K^{n}_{\alpha_d}(t_d,u_d){\rm d}B^n_u onumber\\&=&n^{\frac d2}\sum_{k=1}^{d}\sum_{i_k=1}^{\lfloor nt_k\rfloor} \xi_{i_1,i_2,\cdots,i_d}^{n} \int_{\frac{i_d-1}{n}}^{\frac{i_d}{n}}\cdots\int_{\frac{i_1-1}{n}}^{\frac{i_1}{n}} K_{\alpha_1} \bigg(\frac{\lfloor nt_1\rfloor}{n},u_1\bigg)\cdots onumber\\ && \times K_{\alpha_d} \bigg(\frac{\lfloor nt_d\rfloor}{n},u_d\bigg){\rm d}u_1\cdots {\rm d}u_d, \tag{1.5}\end{eqnarray} 其中 $K_{\alpha_k}$ 由后面(2.1)式给出,由(2.2)式定义的序列 $\{K^{n}_{\alpha_k},\; n=1,2,\cdots\}$ 是核函数 $K_{\alpha_k}$ 的一个估计.则 $\{Z^n\}$ 在~Skorokhod 空间 $D([0, 1]^d)$ 中依分布收敛于 $d$ -参数分数布朗单 $B^{\alpha}$.

当 $d=1$ 时,关于分数布朗运动的弱逼近结果可参见文献 [6, 8, 9, 13, 15, 16, 19, 20]等.若 $d=2$,关于双参数分数布朗单的弱逼近结果可参见[2, 18, 21, 22, 23] 等相关文献.

2 预备知识

令 $(\Omega,{\cal F},P)$ 是一个完备概率空间, $\{{\cal F}_{t}; t\in [0, 1]^d\}$ 是 ${\cal F}$ 的一族子 $\sigma$ -域,满足对任意的 $s\leq t$, ${\cal F}_{s}\subseteq {\cal F}_{t}$.给定 $s\leq t$, $\Delta_{s}X_t$ 表示随机过程 $X$ 在区间 $(s,t]=\Pi_{i=1}^d(s_i,t_i]\subset {\Bbb R}^d$ 上的全增量.

假设 $\Lambda$ 是形如 $\lambda(t)=(\lambda_1(t_1),\cdots,\lambda_d(t_d))$ 的映射族 $$\lambda: [0, 1]^d\rightarrow [0, 1]^d, $$ 其中 $\lambda_i: [0, 1]\rightarrow [0, 1]$ 是一个连续、严格单调递增、边界值固定的函数.Skorokhod 空间 $D=D([0, 1]^d)$ 是定义在 $[0, 1]^d$ 上的从上连续且下极限存在的函数空间装备了度量 \begin{eqnarray*} d(x,y):=\inf \{\min(\|x-y\lambda\|,\|\lambda\|): \lambda \in \Lambda\}, \end{eqnarray*} 其中 $\|x-y\lambda\|=\sup \{|x(t)-y(\lambda(t))| : t\in [0, 1]^d\}$, $\|\lambda\|=\sup \{|\lambda(t)-t| : t\in [0, 1]^d\}.$ 在此度量下,$D$ 是一个完备可分的度量空间.更多关于Skorokhod 空间的相关理论可参见文献[4].设 $X,X^n$ 为 $D$ 中的随机过程,称 $X^n$ 依分布收敛于 $X$ 如果对任意的有界连续函数 $f: D\rightarrow {\Bbb R}$,当 $n$ 趋于无穷时,有 $$Ef(X^n)\rightarrow Ef(X).$$

假定 $H>\frac{1}{2}$,由(1.2) 式给定的核函数 $K_H$ 可以简记为下式(参见文献 [7])\begin{equation}\label{representation} K_H(t,s)= (H-\frac{1}{2})c_H s^{\frac{1}{2}-H}\int_s^{t}u^{H-\frac{1}{2}}(u-s)^{H-\frac{3}{2}}{\rm d}u.\tag{2.1} \end{equation} 且对任意的 $0<t_0<t_0'$,$0<t<t'$,满足 $$ \int_{t_0}^{t'_0}(K_H(t',x)-K_H(t,x))^{2}{\rm d}x\leq C_H(t_0'-t_0)^{2-2H}.$$ 定义 \begin{equation}\label{sec1-eq1-3} K^{n}_H(t,s):=n\int_{s-\frac{1}{n}}^{s} K_H \Big(\frac{\lfloor nt\rfloor}{n},u\Big){\rm d}u,n=1,2,\cdots, \tag{2.2} \end{equation} $\lfloor x\rfloor$ 记作 $x$ 的整部.

3 定理1.1 的证明

本节主要证明定理1.1,我们需要证明序列 $Z_n$ 的胎紧性和有限维分布的弱收敛性.首先证明胎紧性. 考虑到 $Z_n$ 在坐标轴上取值为零,由文献[4],只需证明如下定理.

定理 3.1  假设 $Z_n(t,s)$ 是由(1.5)式给定的随机过程,则对任意的 $s,t\in [0, 1]^d$,$s<t$ 和偶数 $m\geq 2$,存在常数 $C_m>0$,使得 $$ \sup_n E(\Delta_{s}Z^n_t)^{m}\leq C_m \prod_{k=1}^d(t_k-s_k)^{m\alpha_k}.$$

  注意到 \begin{eqnarray*} \Delta_{s}Z^n_t &=& \int_{s_d}^{t_d}\cdots \int_{s_1}^{t_1} (K^{n}_{\alpha_1}(t_1,u_1)-K^{n}_{\alpha_1}(s_1,u_1)) \cdots (K^{n}_{\alpha_d}(t_d,u_d)-K^{n}_{\alpha_1}(s_d,u_d)){\rm d}B^n_u \\ & =&n^{\frac d2}\sum_{k=1}^{d}\sum_{i_k=1}^{\lfloor nt_k\rfloor} \xi_{i_1,i_2,\cdots,i_d}^{n} \int_{\frac{i_d-1}{n}}^{\frac{i_d}{n}}\cdots\int_{\frac{i_1-1}{n}}^{\frac{i_1}{n}} \bigg(K_{\alpha_1}\bigg(\frac{\lfloor nt_1\rfloor}{n},u_1\bigg)-K_{\alpha_1}\bigg(\frac{\lfloor ns_1\rfloor}{n},u_1\bigg)\bigg)\cdots \\ && \times \bigg(K_{\alpha_d}\bigg(\frac{\lfloor nt_d\rfloor}{n},u_d\bigg)-K_{\alpha_d} \bigg(\frac{\lfloor ns_d\rfloor}{n},u_d\bigg)\bigg){\rm d}u_1\cdots {\rm d}u_d \\ & =&n^{\frac d2}\sum_{k=1}^{d}\sum_{i_k=1}^{\lfloor nt_k\rfloor} \xi_{i_1,i_2,\cdots,i_d}^{n} \int_{\frac{i_1-1}{n}}^{\frac{i_1}{n}} \bigg(K_{\alpha_1}\bigg(\frac{\lfloor nt_1\rfloor}{n},u_1\bigg)-K_{\alpha_1} \bigg(\frac{\lfloor ns_1\rfloor}{n},u_1\bigg)\bigg) {\rm d}u_1 \cdots \\ &&\times \int_{\frac{i_d-1}{n}}^{\frac{i_d}{n}} \bigg(K_{\alpha_d}\bigg(\frac{\lfloor nt_d\rfloor}{n},u_d\bigg)-K_{\alpha_d} \bigg(\frac{\lfloor ns_d\rfloor}{n},u_d\bigg)\bigg){\rm d}u_d. \end{eqnarray*} 因此, \begin{eqnarray*} && E(\Delta_{s}Z^n_t)^{m} \\ &=&n^{\frac{dm}{2}} E \bigg[ \sum_{k=1}^{d}\sum_{i_k=1}^{\lfloor nt_k\rfloor} \xi_{i_1,i_2,\cdots,i_d}^{n} \int_{\frac{i_1-1}{n}}^{\frac{i_1}{n}} \bigg(K_{\alpha_1}\bigg(\frac{\lfloor nt_1\rfloor}{n},u_1\bigg)-K_{\alpha_1} \bigg(\frac{\lfloor ns_1\rfloor}{n},u_1\bigg)\bigg) {\rm d}u_1 \cdots \\ &&\times \int_{\frac{i_d-1}{n}}^{\frac{i_d}{n}} \bigg(K_{\alpha_d}\bigg(\frac{\lfloor nt_d\rfloor}{n},u_d\bigg)-K_{\alpha_d} \bigg(\frac{\lfloor ns_d\rfloor}{n},u_d\bigg)\bigg){\rm d}u_d \bigg]^m\\ & \leq& n^{\frac{dm}{2}}C \bigg[ \sum_{k=1}^{d}\sum_{i_k=1}^{\lfloor nt_k\rfloor} \int_{\frac{i_1-1}{n}}^{\frac{i_1}{n}} \bigg(K_{\alpha_1}\bigg(\frac{\lfloor nt_1\rfloor}{n},u_1\bigg)-K_{\alpha_1} \bigg(\frac{\lfloor ns_1\rfloor}{n},u_1\bigg)\bigg) {\rm d}u_1 \cdots \\ &&\times \int_{\frac{i_d-1}{n}}^{\frac{i_d}{n}}\cdots \bigg(K_{\alpha_d}\bigg(\frac{\lfloor nt_d\rfloor}{n},u_d\bigg)-K_{\alpha_d} \bigg(\frac{\lfloor ns_d\rfloor}{n},u_d\bigg)\bigg){\rm d}u_d \bigg]^m\\ & =&C \bigg[ \prod_{k=1}^{d} \bigg( \sqrt{n}\sum_{i_k=1}^{\lfloor nt_k\rfloor} \int_{\frac{i_{k-1}}{n}}^{\frac{i_k}{n}} \bigg(K_{\alpha_k}\bigg(\frac{\lfloor nt_k\rfloor}{n},u_k\bigg)-K_{\alpha_k} \bigg(\frac{\lfloor ns_k\rfloor}{n},u_k\bigg)\bigg){\rm d}u_k \bigg)^2 \bigg]^{\frac m2}.\end{eqnarray*} 由Cauchy-Schwarz 不等式可知,上式小于等于 \begin{eqnarray*} &&C \prod_{k=1}^{d} \bigg[ \sum_{i_k=1}^{\lfloor nt_k\rfloor} \int_{\frac{i_{k-1}}{n}}^{\frac{i_k}{n}} \bigg(K_{\alpha_k}\bigg(\frac{\lfloor nt_k\rfloor}{n},u_k\bigg)-K_{\alpha_k} \bigg(\frac{\lfloor ns_k\rfloor}{n},u_k\bigg)\bigg)^2 {\rm d}u_k \bigg]^{\frac m2}\\ & \leq& C \prod_{k=1}^{d} \bigg[ \int_{0}^{t_k} \bigg(K_{\alpha_k}\bigg(\frac{\lfloor nt_k\rfloor}{n},u_k\bigg)-K_{\alpha_k} \bigg(\frac{\lfloor ns_k\rfloor}{n},u_k\bigg)\bigg)^2 {\rm d}u_k \bigg]^{\frac m2}\\ & =& C \prod_{k=1}^{d} \bigg|\frac{\lfloor nt_k\rfloor-\lfloor ns_k\rfloor}{n}\bigg|^{m\alpha_k}.\end{eqnarray*} 对任意的 $0<s_k<t_k$,$\frac12<\alpha_k<1$.若 $nt_k-ns_k\geq 1$,可以得到 $$\left|\frac{\lfloor nt_k\rfloor-\lfloor ns_k\rfloor}{n}\right|^{2\alpha_k}\leq |2(t_k-s_k)|^{2\alpha_k}. $$ 此外,若 $nt_k-ns_k<1$,则存在整数 $m$,使得 $t_k$,$s_k$ 必定属于区间 $[\frac{m}{n},\frac{m+1}{n})$,蕴含着 $$\left|\frac{\lfloor nt_k\rfloor-\lfloor ns_k\rfloor}{n}\right|^{2\alpha_k}=0.$$ 因此,对任意的 $n\geq 1$, 可以得到 $$ \left|\frac{\lfloor nt_k\rfloor-\lfloor ns_k\rfloor}{n}\right|^{2\alpha_k}\leq |2(t_k-s_k)|^{2\alpha_k}.$$ 证毕.

定理 3.2  由 (1.5) 式给定的随机过程 $\{Z^n_t\}$ 依分布收敛于多参数分数布朗单 $B^{\alpha}$.

  对任意的 $N\in{\Bbb N}$,$a_1,\cdots,a_N \in {\Bbb R}$ 和 $t^1,\cdots,t^N \in [0, 1]^d$.往证 $ Y^n:=\sum\limits_{j=1}^{N}a_jZ^n_{t^j} $ 依分布收敛于一个高斯随机变量满足均值为零,方差为~ $ E\Big(\sum\limits_{j=1}^{N}a_jB^{\alpha}_{t^j}\Big)^{2}.$ 事实上,均值为零是平凡的. 注意到 \begin{eqnarray*} (\sigma^{n})^{2}:&=&E(Y^n)^2\\ &=& \sum_{j,l=1}^Na_ja_l EZ^n_{t^j}Z^n_{t^l}\\ &=& \sum_{j,l=1}^Na_ja_l n^d \sum_{k=1}^{d}\sum_{i_k=1}^{n} \\ && \int_{\frac{i_d-1}{n}}^{\frac{i_d}{n}}\cdots\int_{\frac{i_1-1}{n}}^{\frac{i_1}{n}} K_{\alpha_1}\bigg(\frac{\lfloor nt^j_1\rfloor}{n},u_1\bigg)\cdots K_{\alpha_d} \bigg(\frac{\lfloor nt^j_d\rfloor}{n},u_d\bigg){\rm d}u_1\cdots {\rm d}u_d\\ && \times \int_{\frac{i_d-1}{n}}^{\frac{i_d}{n}}\cdots\int_{\frac{i_1-1}{n}}^{\frac{i_1}{n}} K_{\alpha_1}\bigg(\frac{\lfloor nt^l_1\rfloor}{n},u_1\bigg)\cdots K_{\alpha_d} \bigg(\frac{\lfloor nt^l_d\rfloor}{n},u_d\bigg){\rm d}u_1\cdots {\rm d}u_d\\ &=& \sum_{j,l=1}^Na_ja_l n^d \prod_{k=1}^{d}\sum_{i_k=1}^{n} \int_{\frac{i_k-1}{n}}^{\frac{i_k}{n}} K_{\alpha_k}\bigg(\frac{\lfloor nt^j_k\rfloor}{n},u\bigg){\rm d}u \int_{\frac{i_k-1}{n}}^{\frac{i_k}{n}} K_{\alpha_k}\bigg(\frac{\lfloor nt^l_k\rfloor}{n},u\bigg){\rm d}u.\end{eqnarray*} 由中值定理可知,存在 $u_{i,k}^{n},u_{i,l}^{n}\in(\frac{i_k-1}{n},\frac{i_k}{n}]$,使得 \begin{eqnarray}\label{ri} &&n\sum_{i_k=1}^{n} \int_{\frac{i_k-1}{n}}^{\frac{i_k}{n}} K_{\alpha_k}\bigg(\frac{\lfloor nt^j_k\rfloor}{n},u\bigg){\rm d}u \int_{\frac{i_k-1}{n}}^{\frac{i_k}{n}} K_{\alpha_k}\bigg(\frac{\lfloor nt^l_k\rfloor}{n},u\bigg){\rm d}u onumber\\ &=& \frac{1}{n} \sum_{i_k=1}^{n} K_{\alpha_k}\bigg(\frac{\lfloor nt^j_k\rfloor}{n},u_{i,k}^{n}\bigg)K_{\alpha_k}\bigg(\frac{\lfloor nt^l_k\rfloor}{n},u_{i,l}^{n}\bigg).\tag{3.1}\end{eqnarray} 考虑到核函数 $K_\cdot(t,\cdot)$ 是连续且是单调递减的,从而存在 $$ u_{i}^{n}\in \left[\min(u_{i,k}^{n},u_{i,l}^{n}), \max(u_{i,k}^{n},u_{i,l}^{n})\right]\subseteq \left(\frac{i_k-1}{n},\frac{i_k}{n}\right], $$ 使得 (3.1)式等于 \begin{eqnarray}\label{ri2} \frac{1}{n} \sum_{i_k=1}^{n} K_{\alpha_k}\bigg(\frac{\lfloor nt^j_k\rfloor}{n},u_{i}^{n}\bigg)K_{\alpha_k}\bigg(\frac{\lfloor nt^l_k\rfloor}{n},u_{i}^{n}\bigg).\tag{3.2}\end{eqnarray}

另一方面,当 $\frac12<H<1$ 时,核函数 $K_H$ 关于两个变量是连续的; 映射 $t\mapsto \frac{\lfloor nt\rfloor}{n}$ 在 $[0,T]$ 上一致收敛于恒等映射.因此(3.2)式是Riemann 和. 从而(3.1)式收敛于 $$ \int_{0}^{1}K_{\alpha_k}(t^j_k,u)K_{\alpha_k}(t^l_k,u){\rm d}u.$$ 进而,$(\sigma^{n})^{2}$ 收敛于 $$ \sum_{j,l=1}^Na_ja_l \prod_{k=1}^{d} \int_{0}^{1}K_{\alpha_k}(t^j_k,u)K_{\alpha_k}(t^l_k,u){\rm d}u = E\bigg(\sum\limits_{j=1}^{N}a_jB^{\alpha}_{t^j}\bigg)^{2}.$$

改写 $Y^n$ 如下 \begin{eqnarray*} Y^n&=& \sum\limits_{j=1}^{N}a_jZ^n_{t^j}\\ &=& \sum_{k=1}^{d}\sum_{i_k=1}^{n} n^{\frac d2} \xi_{i_1,i_2,\cdots,i_d}^{n} \sum\limits_{j=1}^{N}a_j\\ &&\times \int_{\frac{i_d-1}{n}}^{\frac{i_d}{n}}\cdots\int_{\frac{i_1-1}{n}}^{\frac{i_1}{n}} K_{\alpha_1}\bigg(\frac{\lfloor nt^j_1\rfloor}{n},u_1\bigg)\cdots K_{\alpha_d} \bigg(\frac{\lfloor nt^j_d\rfloor}{n},u_d\bigg){\rm d}u_1\cdots {\rm d}u_d\\ &:=&\sum_{k=1}^{d}\sum_{i_k=1}^{n} Y^n_{i_1,i_2,\cdots,i_d}.\end{eqnarray*} 至此,只需证明如下形式的Lindeberg 条件成立: 对任意的 $\varepsilon>0$ \begin{eqnarray}\label{lind} \lim_{n\rightarrow \infty}\frac{1}{(\sigma^{n})^2} \sum_{k=1}^{d}\sum_{i_k=1}^{n} E\left[(Y^n_{i_1,i_2,\cdots,i_d})^2 1_{\{|Y^n_{i_1,i_2,\cdots,i_d}|>\varepsilon \sigma^{n}\}}\right]=0.\tag{3.3}\end{eqnarray} 事实上,当 $\frac12<H<1$时,$K_H(t,s)$ 关于 $t$ 递增,关于 $s$ 递减, 由Cauchy-Schwarz 不等式, 可知 \begin{eqnarray*} &&(Y^n_{i_1,i_2,\cdots,i_d})^2 \\ &=&n^d(\xi_{i_1,i_2,\cdots,i_d}^{n})^2 \bigg(\sum\limits_{j=1}^{N}a_j \int_{\frac{i_d-1}{n}}^{\frac{i_d}{n}}\cdots\int_{\frac{i_1-1}{n}}^{\frac{i_1}{n}} K_{\alpha_1}\bigg(\frac{\lfloor nt^j_1\rfloor}{n},u_1\bigg)\cdots K_{\alpha_d} \bigg(\frac{\lfloor nt^j_d\rfloor}{n},u_d\bigg){\rm d}u_1\cdots {\rm d}u_d\bigg)^2\\ &\leq& n^d(\xi_{i_1,i_2,\cdots,i_d}^{n})^2 A \bigg( \int_{\frac{i_1-1}{n}}^{\frac{i_1}{n}} K_{\alpha_1}(1,u_1){\rm d}u_1 \times \cdots \times \int_{\frac{i_d-1}{n}}^{\frac{i_d}{n}} K_{\alpha_d}(1,u_d) {\rm d}u_d\bigg)^2 \\ &\leq &(\xi_{i_1,i_2,\cdots,i_d}^{n})^2 A \int_{\frac{i_1-1}{n}}^{\frac{i_1}{n}} K^2_{\alpha_1}(1,u_1){\rm d}u_1 \times \cdots \times \int_{\frac{i_d-1}{n}}^{\frac{i_d}{n}} K^2_{\alpha_d}(1,u_d) {\rm d}u_d \\&\leq& (\xi_{i_1,i_2,\cdots,i_d}^{n})^2A \prod_{k=1}^d \int_{0}^{\frac{1}{n}} K^2_{\alpha_k}(1,u){\rm d}u\\ & =&(\xi_{i_1,i_2,\cdots,i_d}^{n})^2A\delta^{n}, \end{eqnarray*} 其中 $A:=\Big(\sum\limits_{j=1}^{N}a_j\Big)^2$, $\delta^{n}:=\prod\limits_{k=1}^d \int_{0}^{\frac{1}{n}} K^2_{\alpha_k}(1,u){\rm d}u$.因此 $$ \left\{|Y_{i_1,i_2,\cdots,i_d}^{n}|>\varepsilon \sigma^{n}\right\} = \left\{(Y^n_{i_1,i_2,\cdots,i_d})^2>\varepsilon^2(\sigma^{n})^2 \right\} \subseteq \left\{(\xi_{i_1,i_2,\cdots,i_d}^{n})^2A\delta^{n}> \varepsilon^2(\sigma^{n})^2\right\}.$$ 从而对任意的 $i_k=1,2,\cdots,n$,$k=1,2,\cdots,d$, $$ E\left[(Y_{i_1,i_2,\cdots,i_d}^{n})^2 1_{\{|Y_{i_1,i_2,\cdots,i_d}^{n}|>\varepsilon \sigma^{n}\}}\right]\leq E\left[(\xi_{i_1,i_2,\cdots,i_d}^{n})^2A\delta^{n}1_{\{(\xi_{i_1,i_2,\cdots,i_d}^{n})^2A\delta^{n} >\varepsilon^2(\sigma^{n})^2\}}\right] $$ 且因为 $n\rightarrow \infty $时有 $\delta^{n}\rightarrow0$,故 \begin{eqnarray*} &&\frac{1}{(\sigma^{n})^2} \sum_{k=1}^{d}\sum_{i_k=1}^{n} E\left[(Y_{i_1,i_2,\cdots,i_d}^{n})^2 1_{\{|Y_{i_1,i_2,\cdots,i_d}^{n}|>\varepsilon \sigma^{n}\}}\right]\\ & \leq &\frac{1}{(\sigma^{n})^2} \sum_{k=1}^{d}\sum_{i_k=1}^{n} E\left[(\xi_{i_1,i_2,\cdots,i_d}^{n})^2A\delta^{n}1_{\{(\xi_{i_1,i_2,\cdots,i_d}^{n})^2A\delta^{n}>\varepsilon^2(\sigma^{n})^2\}}\right] \\ &\leq& E\left[\xi^2 1_{\{(\xi_{i_1,i_2,\cdots,i_d}^{n})^2A\delta^{n}>\varepsilon^2(\sigma^{n})^2\}}\right] \rightarrow 0, \end{eqnarray*} 其中 $\xi=\xi_{1,1,\cdots,1}^{1}$.

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