数学物理学报, 2023, 43(5): 1519-1528

基于LENQD序列生成的线性过程误差回归函数小波估计Berry-Esseen界

李永明,1,*, 庞伟才,2, 李乃医,3

1上饶师范学院数学与计算机科学学院 江西上饶 334001

2南宁师范大学数学与统计学院 南宁 530001

3广东海洋大学数学与计算机学院 广东湛江 524088

Berry-Esseen Bound of Wavelet Estimator for Regression Model with Linear Process Errors Generated by LENQD Sequence

Li Yongming,1,*, Pang Weicai,2, Li Naiyi,3

1School of Mathematics and Computer Science, Jiangxi Shangrao Normal University, Jiangxi Shangrao 334001

2School of Mathematics and Statistics, Nanning Normal University, Nanning 530001

3School of Mathematics and Computer, Guangdong Ocean University, Guangdong Zhanjiang 524088

通讯作者: * 李永明,Email: lym1019@163.com

收稿日期: 2022-06-6   修回日期: 2023-01-12  

基金资助: 国家自然科学基金项目(12161074)
江西省自然科学基金重点项目(20212ACB201006)
广东省自然科学基金项目(2022A1515010978)
上饶市科技计划(2020K006)

Received: 2022-06-6   Revised: 2023-01-12  

Fund supported: NSFC(12161074)
NSFJ(20212ACB201006)
NSFG(2022A1515010978)
Shangrao Science and Technology Talent Plan(2020K006)

作者简介 About authors

庞伟才,Email:pangweicai2021@163.com;

李乃医,Email:linaiyi1979@163.com

摘要

在LENQD相依序列生产的线性过程误差下, 研究了固定设计非参数回归模型小波估计. 利用LENQD序列的矩不等式和特征函数不等式, 建立了未知回归函数小波估计的Berry-Esseen界. 通过选取适当的参数, 其界可达$O(n^{-1/6})$. 所得结果推广了相关文献的研究结果.

关键词: 固定设计回归模型; 线性过程; LENQD序列; 小波估计; Berry-Esseen界

Abstract

Based on linear process random errors generated by LENQD random sequence, the wavelet estimator of nonparametric fixed design regression model is considered. By the characteristic function inequality and moment inequality of LENQD random sequence, the Berry-Esseen bounds of the wavelet estimator for unknown regression function are obtained. And by choosing some suitable constants, their bounds can reach $O(n^{-\frac{1}{6}})$. The obtained results generalize the corresponding results in recent literature.

Keywords: Fixed design regression; Linear process; LENQD random sequence; Wavelet estimator; Berry-Esseen bound

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本文引用格式

李永明, 庞伟才, 李乃医. 基于LENQD序列生成的线性过程误差回归函数小波估计Berry-Esseen界[J]. 数学物理学报, 2023, 43(5): 1519-1528

Li Yongming, Pang Weicai, Li Naiyi. Berry-Esseen Bound of Wavelet Estimator for Regression Model with Linear Process Errors Generated by LENQD Sequence[J]. Acta Mathematica Scientia, 2023, 43(5): 1519-1528

1 引言

考虑非参数固定设计回归模型:

$ \begin{equation} Y_{ni}=g(t_{ni})+\varepsilon_{ni}, \ 1\leq i\leq n, \end{equation} $

其中$g(\cdot)$是定义在$[0,1]$区间上的未知回归函数, $\{t_{ni}, 1\leq i\leq n\}$是固定设计点列, 满足$0=t_{n0}\leq t_{n1} \leq t_{nn-1}\leq x_{nn}=1$, $\{\varepsilon_{ni}, 1\leq i\leq n\}$是随机误差. 由于回归模型(1.1)在实际问题中有非常广泛的应用, 许多学者对未知回归函数$g(\cdot)$提出了不同的估计方法, 并进行了相关的研究工作.

譬如: Georgiev[1]提出了如下形式的回归函数加权核估计

$ \begin{equation} \widetilde{g}_n(t)=\sum\limits_{i=1}^{n}\omega_{ni}(t)Y_{ni}, \end{equation} $

其中, 权函数$\omega_{ni}(t)$依赖于固定设计点$t_{n1}, \cdots, t_{nn}$和样本观察数$n$. 之后, 一些学者对之进行了研究. 在独立误差情形[2], 在混合相依误差情形, Roussas 等[3], 杨善朝和李永明[4]等分别研究了加权估计的渐近正态和一致渐近正态性等; 在负相依误差情形, Yang[5]研究了加权估计的一致渐近正态性; 在线性过程误差情形, Liang和Li[6]基于NA序列生成的线性过程误差下获得回归函数加权估计量的Berry-Esseen界, Ding等[7]基于LNQD序列生成的线性过程误差下, 研究了回归函数加权估计量的Berry-Esseen界; 等等, 这里就不逐一列举相关文献.

在二十世纪九十年代, 小波估计方法被成功运用到估计未知密度函数、回归函数等统计推断中. 鉴于小波估计在统计应用中, 对待估函数要求较低的情形下也能得到令人满意的一些统计性质, 从而得到学者们广泛研究. 如Antoniadis等[8]对模型(1.1)给出了未知回归函数$g(\cdot)$的小波估计形如

$ \begin{equation}\widehat{g}_n(t)=\sum\limits_{i=1}^{n}Y_{ni}\int_{A_i}E_m(t,x){\rm d}x, \end{equation} $

这里$t_{ni}\in A_i$, $A_i=[d_{i-1}, d_i],\ 1\leq i \leq n$是区间$[0,1]$上的一个分割, $d_0=0$, $d_i=(t_{ni}+t_{n(i+1)})/2$, $1\leq i\leq n-1, d_n=1;$ 小波再生核$E_m(t, x)$是由刻度函数$\varphi(t)$产生的, 即

$E_m(t,x)=2^m E_0(2^m t,2^m x),\ \ E_0(t, x)=\sum\limits_{j\in Z}\varphi(t-j)\varphi(x-j),$

此处$m=m(n)>0$是光滑参数仅与$n$有关.

对小波估计(1.3)进行研究的文献, 可参见文献[9,10,11,12,13,14,15,16,17]等等, 其中文献[14,15,16]在由混合序列生成的线性过程误差下, 研究了非参数回归函数小波估计的Berry-Esseen界等渐近性质.

由于LNQD相依序列在多元统计分析理论和可靠性理论等方面有着广泛应用, 受到学者们的广泛关注和研究. 2022年Li等[18]提出了LENQD(Linearly extended negative quadrant dependent)序列, 是LNQD相依序列的拓广. 鉴于此, 本文将在Ding等[7]基于LNQD序列生成的线性误差下研究回归函数加权估计的相关结果基础上, 推广研究在LENQD序列生成的线性过程误差下回归函数小波估计的Berry-Esseen界. 下面先回顾LENQD序列的定义.

定义1 称随机序列$\{X_n, n\geq 1 \}$是LENQD的, 如果对任意不交子集$A, B\subset \{1, 2, \cdots,\}$和正实数$\{l_1, l_2, \cdots, \}$, 或者负实数$\{l_1, l_2, \cdots, \}$, 有

$\sum\limits_{i\in A}l_iX_i \ {和} \ \sum\limits_{j\in B}l_jX_j\ \hbox{是 ENQD}.$

称随机变量$X$$Y$是ENQD的, 如果存在一个控制常数$M\geq 1$, 对每个实数$x, y$, 有

${\rm P}(X\leq x, Y\leq y)\leq M {\rm P} (X\leq x) {\rm P}(Y\leq y)$

${\rm P}(X> x, Y> y)\leq M {\rm P} (X> x) {\rm P}(Y> y).$

称随机序列$\{X_n, n\geq 1\}$是两两 ENQD 的, 如果对所有的$i, j \geq 1$, $i\neq j$, $X_i$$X_j$是ENQD的.

由定义1.1可知, 当$M=1$时, LENQD序列就是LNQD的, 从而LEQND是一列更广泛的相依结构, 且弱于NQD、NOD、NA、LNQD和END等相依序列. 由此, 本文利用LEQND序列的性质, 矩不等式和特征函数不等式, 基于LENQD相依序列生成的线性过程误差下, 研究非参数回归函数小波估计的Berry-Esseen界是有意义的.

下面给出本文需要的LENQD序列的性质, 特征函数不等式和矩不等式, 具体参见文献[18].

引理1.1 [18] 设随机序列$\{X_n, n\geq 1 \}$是LENQD的, $f_n$是一列同为非增(或非降)的函数序列, 则随机函数序列$\{f_n(X_n), n\geq 1 \}$也是LENQD的, 且其控制系数不变.

引理1.2 [18] 设随机序列$\{X_1,\cdots, X_m\}$是LENQD的, 有有限的二阶矩. 则对任意的实数$t_1, \cdots,t_m\in R$, 有

$\begin{eqnarray*} \Big |{\rm E}\exp \Big({\rm i}\sum\limits_{j=1}^{m}t_jX_j\Big)-\prod_{j=1}^{m}{\rm E}\exp ({\rm i} t_jX_j)\Big|& \leq& 2\sum\limits_{1\leq j<l<\leq m }|t_jt_l| |{\rm Cov}(X_j, X_l)|. \end{eqnarray*}$

引理1.3 [18] 设随机序列$\{X_1, \cdots, X_n\}$是LENQD的. 如果对$p\geq 2$, 有${\rm E}X_j=0$${\rm E}|X_j|^p< \infty$, $j=1,\cdots, n$. 则存在$M>0$, 有

${\rm E}\big|\sum\limits_{j=1}^{n} X_j\big|^p\leq C_p\Big(\sum\limits_{j=1}^n{\rm E}|X_j|^p+ M \Big(\sum\limits_{j=1}^n {\rm E}(X_j^2)\Big)^{p/2} \Big), $
${\rm E}|\sum\limits_{j=1}^{n} X_j|^p\leq C_p n^{p/2-1} M \sum\limits_{j=1}^n{\rm E}|X_j|^p, $

此处$C_p$是仅依赖于$p$的某一正常数. 进一步, 如果$\{b_j, j\geq 1\}$是实数列, 则有

${\rm E}\Big|\sum\limits_{j=1}^{n}b_j X_j\Big|^p\leq C_p\Big(\sum\limits_{j=1}^n{\rm E}|b_j X_j|^p+ M \Big(\sum\limits_{j=1}^n {\rm E}\Big(b_j X_j\Big )^2\Big)^{p/2} \Big),$
${\rm E}\Big|\sum\limits_{j=1}^{n} b_j X_j \Big|^p\leq C_p n^{p/2-1} M \sum\limits_{j=1}^n{\rm E}\Big|b_j X_j\Big|^p. $

2 主要结果

为了定理叙述简洁, 下面给出一些假设.

(A1) 设$\{\varepsilon_{ni}, 1\leq i\leq n\}$线性表示为$\varepsilon_{ni}=\sum\limits_{j=-\infty}^{\infty}a_{j}e_{i-j}$, $\{a_j\}$为实数列且满足$\sum\limits_{j=-\infty}^{\infty}|a_j|<\infty$, $\{e_j, -\infty< j< \infty \}$是同分布LENQD序列, ${\rm E}e_0=0, {\rm E}|e_0|^{2+\rho}<\infty$, $\rho>0$.

(A2) $\{\varepsilon_ {ni}\}$的谱函数$f(\omega)$满足$0<c_1\leq f(\omega)\leq c_2<\infty$, $\omega\in (-\pi,\pi].$

(A3) $({\rm i})$ 刻度函数$\varphi(\cdot)$$\gamma$-正则, $\gamma$是正整数, 即对任意的 $l\leq \gamma$和整数$z$, $\left|d^l \varphi/dx^l\right| \leq c_z(1+|x|)^{-z},$其中$c_z$是一仅依赖于$z$的常数;

$({\rm ii})$ 刻度函数$\varphi(\cdot)$满足一阶Lipschitz条件, 且有紧支撑. 又当$\xi\to \infty$时, 有$|\hat{\varphi}(\xi)-1|=O(\xi), $ 其中$\hat{\varphi}$$\varphi$的Fourier变换. $\max\limits_{1\leq i \leq n}|d_i-d_{i-1}-n^{-1}|=o(n^{-1})$.

(A4) $({\rm i})$ 回归函数$g(\cdot)$满足一阶Lipschitz条件;

$({\rm ii})$ 回归函数$g(\cdot)$属于$\nu$-阶Sobolev空间${\rm H}^{\nu}$, 即$g\in {\rm H}^\nu$${\rm H}^\nu=\big\{g: \int|\overline{g}(w)|^2(1+w^2)^\nu {\rm d}w<\infty\big\}, $ 此处$\overline{g}$$g$的Fourier变换.

(A5) 令$p=p_n$, $q=q_n$, $k=k_n=[3n/(p+q)]$为正整数, 使得$p+q\leq 3n$, $qp^{-1}\to 0$, $\gamma_{in}\to 0$, $i=1,2,3,4$, 其中$\gamma_{1n}=qp^{-1} 2^m, $$\gamma_{2n}=p\frac{2^m}{n},$$\gamma_{3n}=n\big(\sum\limits_{|j|>n}|a_{j}|\big)^2, $$\gamma_{4n}=u(q)(2^m/n)^2$, 这里$u(q)=\sup\limits_{j\geq 1}\sum\limits_{j:|j-i|\geq q}|{\rm Cov}(e_i, e_j)|.$

$\sigma_n^2=\sigma_n^2(t)={\rm Var}(\widehat{g}_n(t)),\ \ S_n=S_n(t)=\sigma_n^{-1} \{\widehat{g}_n(t)-{\rm E}\widehat{g}_n(t)\}.$

下面给出本文的主要结果.

定理2.1 假设(A1)-(A5)成立. 则对任意的$t\in [0,1]$,

$\sup\limits_{u}\big|{\rm P}(S_n(t)\leq u) -\Phi (u)\big| \leq C \left\{\gamma _{1n}^{1/3}+\gamma _{2n}^{1/3}+\gamma _{2n}^{\rho/2}+ \gamma _{3n}^{1/3}+ \gamma _{4n}^{1/3} + u(q) \right\}. $

推论 2.1 假设(A1)-(A5)成立, 且$u(1)<\infty$, 则对任意的$t\in [0,1]$,

$\sup\limits_{u}\big|{\rm P}(S_n(t)\leq u) -\Phi (u)\big|=o(1). $

推论 2.2 假设(A1)-(A5)成立, 取$\rho=\frac{2}{3}$. 如果$\frac{2^m}{n}=O(n^{-\theta})$, $u(n)=O\big(n^{-\frac{\theta-\tau}{3(2\tau-1)}}\big)$, $\frac{1}{2}<\tau\leq \theta<1$, 以及 $\sup\limits_{n\geq 1} \big(n^{\frac{\theta-\tau+1}{2}}\big)\sum\limits_{|j|>n}|a_j| <\infty, $ 则对任意的$t\in [0,1]$, $\sup\limits_{u}\big|{\rm P}(S_n(t)\leq u) -\Phi (u)\big|= O\big(n^{-\frac{ \theta-\tau }{3}}\big).$

注2.1 关于假设条件的注记

$({\rm i})$ 条件(A2)-(A4)是回归函数小波估计的常规性假设. 见文献[8,9,10,11,12,13,14,15].

$({\rm ii})$ 条件(A5)中, 适当选取$p, q, m$可使得$\gamma _{in}\to 0, i=1, 2, 4$成立. 如$p, q$$2^m$选取如下: $p\sim n^{\tau_1}$, $q\sim n^{\tau_2}$, $2^m\sim n^{\tau_3}$, 其中$\tau_1>\tau_2>0,\ \tau_3>0$, $\tau_2+\tau_3<\tau_1<1-\tau_3$, 此处$x_n\sim y_n$表示$x_n/y_n\to C$. 故取$\tau_1=0.58, \tau_2=0.30, \tau_3=0.20$, 则有$\gamma _{in}\to 0, i=1, 2, 4$.

(${\rm iii}$) 在一些正规条件情况下, 常用的AR、MA和ARMA过程广泛用于序列相关数据的建模, $\gamma_{3n}\to 0$是容易得到的, 见文献[6,注2.1(c)].

注2.2 由推论2.2知, 当$\theta\approx1$,$\tau>\frac{1}{2}$且接近于$\frac{1}{2}$时, Berry-Esseen界逼近$O(n^{-\frac{ 1}{6}})$.

注2.3 本文所得结果把文献[14]基于混合序列生成的线性过程误差的小波估计的Berry-Esseen界, 推广到基于LENQD序列生成的线性过程误差情形. 同时, 也把文献[7]基于LNQD序列生成的线性过程误差的回归函数加权核估计的Berry-Esseen界, 推广到基于LENQD序列生成的线性过程误差回归函数小波估计.

3 辅助结果

为了记号和叙述方便, 把$S_n$分解,

$\begin{eqnarray*} S_n& = & \sigma_n^{-1}\sum\limits_{i=1}^n \Big(\int_{A_i} E_m(t,s){\rm d}s\Big) \varepsilon_{ni} \nonumber\\ &=&\sigma_n^{-1}\sum\limits_{i=1}^n \Big(\int_{A_i} E_m(t,s){\rm d}s\Big) \Big(\sum\limits_{j=-n}^na_je_{i-j}\Big)+\sigma_n^{-1}\sum\limits_{i=1}^n \Big(\int_{A_i} E_m(t,s){\rm d}s\Big) \Big(\sum\limits_{|j|>n}a_je_{i-j}\Big) \\ &\stackrel{\Delta}{=}& S_{1n}+S_{2n}. \end{eqnarray*}$

$S_{1n}$可表示为

$\begin{eqnarray*} S_{1n}=\sum\limits_{l=1-n}^{2n}\sigma_n^{-1}\Big (\sum\limits_{i=\max\{1,l-n\}}^{\min\{n,l+n\}}a_{i-l}\int_{A_i} E_m(t,s){\rm d}s\Big)e_l \stackrel{\Delta}{=}\sum\limits_{l=1-n}^{2n}Z_{nl}, \end{eqnarray*}$

且可分解为

$ S_{1n}=S_{1n}^{\prime}+S_{1n}^{\prime\prime}+S_{1n}^{\prime\prime\prime}, $

这里

$S_{1n}^{\prime}=\sum\limits_{w=1}^ky_{nw}, \ S_{1n}^{\prime\prime}=\sum\limits_{w=1}^ky_{nw}^{\prime}, \ S_{1n}^{\prime\prime\prime}=y^{\prime}_{n k+1},$
$y_{nw}=\sum\limits_{i=m_w}^{m_w+p-1}Z_{ni}, \ y_{nw}^{\prime}=\sum\limits_{i=l_w}^{l_w+q-1}Z_{ni}, \ y_{nk+1}^{\prime}=\sum\limits_{i=k(p+q)-n+1}^{2n}Z_{ni},$
$ m_w=(w-1)(p+q)+1-n, \ l_w=(w-1)(p+q)+p+1-n,\ w=1, \cdots, k.$

从而$S_n$可表示为

$S_n=S_{1n}^{\prime}+S_{1n}^{\prime\prime}+S_{1n}^{\prime\prime\prime}+S_{2n}.$

为了证明主要结论, 先给出辅助结果.

引理3.1 [13,引理3.1] 假设(A3)-(A4)成立, 则有

${\rm (i)}$$|\int_{A_i} E_m(t,s){\rm d}s|=O(\frac{2^m}{n}),i=1,2,\cdots,n;$

${\rm (ii)}$$\sum\limits_{i=1}^{n}(\int_{A_i} E_m(t,s) {\rm d}s)^{2}=O(\frac{2^m}{n});$

${\rm (iii)} $$\sup\limits_{m}\int_{0}^{1} |E_m(t,s){\rm d}s|\leq C;$

$\rm{(iv)}$$\sum\limits_{i=1}^{n}|\int_{A_i} E_m(t,s){\rm d}s|\leq C.$

引理3.2 [14,引理3.5] 假设(A1)-(A5)成立, 则有

$\sigma_n^2(t)\geq C 2^mn^{-1}, \ \ \sigma_n^{-2}(t)|\int_{A_i} E_m(t,s){\rm d}s|\leq C.$

引理3.3 假设(A1)-(A5)成立, 则有

$({\rm i})$${\rm E}(S_{1n}^{\prime\prime})^2\leq C\gamma _{1n}, \ {\rm E}(S_{1n}^{\prime\prime\prime})^2\leq C\gamma _{2n},\ {\rm E}(S_{2n})^2\leq C\gamma _{3n}$;

$({\rm ii})$${\rm P}(|S''_{1n}|\geq \gamma _{1n}^{1/3}) \leq C \gamma _{1n}^{1/3}, \ {\rm P}(|S'''_{1n}|\geq \gamma _{2n}^{1/3}) \leq C \gamma _{2n}^{1/3}, \ {\rm P}(|S_{2n}|\geq \gamma _{3n}^{1/3}) \leq C \gamma _{3n}^{1/3}.$

在条件(A1)-(A5)下, 由引理3.2和引理1.3, 计算可得

$ \begin{matrix} {\rm E}(S_{1n}^{\prime\prime})^2 &\leq & C \sum\limits_{w=1}^k \sum\limits_{i=l_w}^{l_w+q-1} \sigma _n^{-2} \Big(\sum\limits_{j=\max\{1,i-n\}}^{\min\{n,i+n\}} \ a_{j-i} \int_{A_j} E_m(t,s){\rm d}s\Big)^2 {\rm E} e_i^{2} \nonumber\\ &\leq & C \sum\limits_{w=1}^k \sum\limits_{i=l_w}^{l_w+q-1}\frac{2^m}{n} \Big(\sum\limits_{j=\max\{1,i-n\}}^{\min\{n,i+n\}}\big|a_{j-i}\big|\Big)^2 \leq C kq\frac{2^m}{n}\Big(\sum\limits_{j=-\infty}^{\infty}\big|a_{j}\big|\Big)^2 \nonumber\\ &\leq & C qp^{-1}2^m = C\gamma _{1n},\end{matrix} $
$ \begin{matrix} {\rm E}(S_{1n}^{\prime\prime\prime})^2 &\leq & C \sum\limits_{i=k(p+q)-n+1}^{2n}\sigma _n^{-2} \Big(\sum\limits_{j=\max\{1,i-n\}}^{\min\{n,i+n\}}a_{j-i}\int_{A_j} E_m(t,s){\rm d}s\Big )^2 {\rm E}e_i^{2} \nonumber\\ &\leq & C \sum\limits_{i=k(p+q)-n+1}^{2n}\frac{2^m}{n} \Big(\sum\limits_{j=\max\{1,i-n\}}^{\min\{n,i+n\}}\big|a_{j-i}\big|\Big )^2 \nonumber\\ &\leq & C \big(3n-k(p+q)\big)\frac{2^m}{n} \Big(\sum\limits_{j=-\infty}^{\infty}\big|a_{j}\big|\Big )^2 \leq C p\cdot \frac{2^m}{n} = C\gamma _{2n}. \end{matrix} $

利用引理1.3, 引理1.2(iv)和引理3.2, 注意到${\rm E}e_0^2<\infty$, 计算可得

$ \begin{matrix} {\rm E}(S_{2n}^2)&=& {\rm E} \Big( \sigma_n^{-1}\sum\limits_{i=1}^n \Big(\int_{A_i} E_m(t,s){\rm d}s\Big) \Big(\sum\limits_{|j|>n}a_je_{i-j}\Big) \Big)^2 \nonumber\\ &=& {\rm E} \Big| \sigma_n^{-1}\sum\limits_{i_1=1}^n \Big(\int_{A_{i1}} E_m(t,s){\rm d}s\Big) \Big(\sum\limits_{|j_1|>n}a_{j_1}e_{i_1-j_1}\Big) \Big| \\ &&\times \Big| \sigma_n^{-1}\sum\limits_{i_2=1}^n \Big(\int_{A_{i2}} E_m(t,s){\rm d}s\Big) \Big(\sum\limits_{|j_2|>n}a_{j_2}e_{i_2-j_2}\Big) \Big| \nonumber\\ && \leq {\rm E} \Big\{ \sum\limits_{i_1=1}^n \Big|\int_{A_{i1}} E_m(t,s){\rm d}s\Big| \cdot \Big(\sum\limits_{i_2=1}^n \Big|\sum\limits_{|j_1|>n}a_{j_1}e_{i_1-j_1}\Big| \Big|\sum\limits_{|j_2|>n}a_{j_2}e_{i_2-j_2}\Big| \Big) \Big\} \nonumber \\ && \leq C n \Big(\sum\limits_{|j|>n}a_{j} \Big)^2= C\gamma _{3n}. \end{matrix} $

结合(3.1)-(3.3)式, 证得引理3.1(i). 再利用Markov不等式和引理3.3(i), 引理3.3(ii)立即得证.

引理3.4 假设(A1)-(A5)成立. 记$s_n^2=\sum\limits_{w=1}^{k}{{\rm Var}(y_{nw})}$, 则有

$|s_n^2-1|\leq C \Big(\gamma _{1n}^{1/2}+\gamma _{2n}^{1/2}+\gamma _{3n}^{1/2}+ u(q)\Big).$

$\Gamma_n=\sum\limits_{1\leq i<j\leq k} {{\rm Cov}(y_{ni},y_{nj})}$, 则$s_{n}^2={\rm E}(S_{1n}^{\prime})^2-2\Gamma _n$. 注意到${\rm E}(S_n)^2=1$, 由引理3.3(i)可得

$ \begin{matrix} |{\rm E}(S_{1n}^{\prime})^2-1| &=& |{\rm E}(S_{1n}^{\prime\prime}+S_{1n}^{\prime\prime\prime}+S_{2n})^2 - 2{\rm E}[S_n(S_{1n}^{\prime\prime}+S_{1n}^{\prime\prime\prime}+S_{2n})]| \nonumber \\&\leq & C(\gamma _{1n}^{1/2}+\gamma _{2n}^{1/2}+\gamma _{3n}^{1/2}). \end{matrix} $

又由引理3.2和引理3.1(iv), 以及假设条件(A1), 可得

$ \begin{matrix} |\Gamma _n| &&\leq \sum\limits_{1\leq i<j\leq k}\sum\limits_{s_1=k_i}^{k_i+p-1} \sum\limits_{t_1=k_j}^{k_j+p-1}{\big|{\rm Cov}(Z_{ns_1},Z_{nt_1})\big|} \nonumber\\&& \leq \sum\limits_{1\leq i<j\leq k}\sum\limits_{s_1=k_i}^{k_i+p-1}\sum\limits_{t_1=k_j}^{k_j+p-1} \sum\limits_{u=\max\{1,s_1-n\}}^{\min\{n,s_1+n\}} \sum\limits_{v=\max\{1,t_1-n\}}^{\min\{n,t_1+n\}}\sigma _n^{-2} \nonumber\\&& \times \Big|\int\limits_{A_{u}} E_m(t,s){\rm d}s\int\limits_{A_{v}} E_m(t,s){\rm d}s\Big| \cdot\big|a_{u-s_1}a_{v-t_1}\big|\cdot \big|{\rm Cov}(e_{s_1},e_{t_1})\big| \nonumber\\&& \leq C \sum\limits_{i=1}^{k-1} \sum\limits_{s_1=k_i}^{k_i+p-1} \sum\limits_{u=\max\{1,s_1-n\}}^{\min\{n,s_1+n\}} \Big|\int_{A_{u}} E_m(t,s){\rm d}s\Big| \cdot |a_{u-s_1}| \nonumber \\&& \times \sum\limits_{j=i+1}^k \sum\limits_{t_1=k_j}^{k_j+p-1} \sum\limits_{v=\max\{1,t_1-n\}}^{\min\{n,t_1+n\}} |a_{v-t_1}|\cdot \big|{\rm Cov}(e_{s_1},e_{t_1})\big| \nonumber\\ && \leq C \sum\limits_{i=1}^{k-1}\sum\limits_{s_1=k_i}^{k_i+p-1} \sum\limits_{u=\max\{1,s_1-n\}}^{\min\{n,s_1+n\}} \Big|\int_{A_{u}} E_m(t,s){\rm d}s\Big| \cdot |a_{u-s_1}| \cdot \sup\limits_{t_1\geq 1} \sum\limits_{t_1:|t_1-s_1|\geq q} \big|{\rm Cov}(e_{s_1},e_{t_1})\big| \nonumber \\ && \leq C u(q)\sum\limits_{i=1}^{k-1}\sum\limits_{s_1=k_i}^{k_i+p-1} \sum\limits_{u=1}^{n} \Big |\int_{A_{u}} E_m(t,s){\rm d}s \Big| |a_{u-s_1}| \nonumber\\&& \leq C u(q)\sum\limits_{u=1}^{n}\Big|\int_{A_{u}} E_m(t,s){\rm d}s\Big|\Big(\sum\limits_{i=1}^{k-1}\sum\limits_{s_1=k_i}^{k_i+p-1} |a_{u-s_1}|\Big) \leq C u(q). \end{matrix} $

因此, 由(3.4)式和(3.5)式可得

$ \big|s_n^2-1\big|\leq \big| {\rm E} (S_{1n}^{\prime})^2-1\big|+2|\Gamma _n|\leq C\big(\gamma _{1n}^{1/2}+\gamma _{2n}^{1/2}+\gamma _{3n}^{1/2}+ u(q)\big). $

引理证明完毕.

假设随机变量$\{\eta _{nw}:w=1, \cdots, k\}$是独立, 且对任意$w=1, \cdots, k$,$\eta_{nw}$$y_{nw}$是同分布的随机变量. 令$T_n=\sum\limits_{w=1}^{k}{\eta _{nw}}$, 则有

引理3.5 假设(A1)-(A5)成立, 则有

$\sup\limits_{u}\big|{\rm P}\big(T_n/s_n \leq u\big)-\Phi (u)\big|\leq C \gamma_{2n}^{\rho/2}.$

由文献[19,定理5.7] (Berry-Esseen不等式), 当$r\geq 2$时, 计算可得

$ \begin{equation} \sup\limits_{u}\big|{\rm P}\big(T_n/s_n\leq u\big)-\Phi (u)\big| \leq C \frac{\sum\limits _{w=1}^k {\rm E}|y_{nw}|^{r}}{s_n^r}. \end{equation} $

根据引理3.2和引理1.3, 利用$C_r$-不等式, 取$r=2+\rho$, 则有

$ \begin{matrix} \sum\limits_{w=1}^k {\rm E}|y_{nw}|^{2+\rho} &\leq& C\bigg\{\sum\limits_{w=1}^k \sum\limits_{j=m_w}^{m_w+p-1}\Big|\sum\limits_{i=\max\{1,j-n\}}^{\min\{n,j+n\}} \sigma_{n}^{-1}a_{i-j}\int_{A_{i}}E_m(t,s){\rm d}s \Big|^{2+\rho}{\rm E}|e_j|^{2+\rho} \nonumber\\ && \ \ +M\sum\limits_{w=1}^k \Big(\sum\limits_{j=m_w}^{m_w+p-1} \Big( \sum\limits_{i=\max\{1,j-n\}}^{\min\{n,j+n\}} \sigma_{n}^{-1}a_{i-j}\int_{A_{i}}E_m(t,s){\rm d}s \Big)^2 {\rm E}e_j^2 \Big)^{1+\rho/2}\bigg\} \nonumber\\ &\leq& C\sigma_{n}^{-(2+\rho)}\sum\limits_{w=1}^k \sum\limits_{j=m_w}^{m_w+p-1} \sum\limits_{i=\max\{1,j-n\}}^{\min\{n,j+n\}} \Big|a_{i-j}\int\limits_{A_{i}}E_m(t,s){\rm d}s\Big| \\ &&\times \Big|\sum\limits_{i=\max\{1,j-n\}}^{\min\{n,j+n\}} a_{i-j}\int\limits_{A_{i}}E_m(t,s){\rm d}s \Big|^{1+\rho} \nonumber\\ && +C M\sum\limits_{w=1}^k \Big( \sum\limits_{j=m_w}^{m_w+p-1} \Big( \sum\limits_{i=\max\{1,j-n\}}^{\min\{n,j+n\}} \sigma_{n}^{-1}a_{i-j}\int_{A_{i}}E_m(t,s){\rm d}s \Big)^2 \Big)^{1+\rho/2} \nonumber\\ & \leq& C(\frac{2^m}{n})^{\rho/2} \sum\limits_{i=1}^{n}\Big|\int_{A_{i}}E_m(t,s){\rm d}s\Big| \Big(\sum\limits_{w=1}^k\sum\limits_{j=m_w}^{m_w+p-1} |a_{i-j}|\Big) \nonumber\\ && +C M p^{\rho/2} \sum\limits_{w=1}^k \sum\limits_{j=m_w}^{m_w+p-1} \Big( \sum\limits_{i=\max\{1,j-n\}}^{\min\{n,j+n\}} \sigma_{n}^{-1}a_{i-j}\int_{A_{i}}E_m(t,s){\rm d}s \Big)^{2+\rho} \nonumber \\ & \leq& C(\frac{2^m}{n})^{\rho/2} +C p^{\rho/2}\sigma_{n}^{-(2+\delta)} \sum\limits_{w=1}^k \sum\limits_{j=m_w}^{m_w+p-1} \sum\limits_{i=\max\{1,j-n\}}^{\min\{n,j+n\}} \big|a_{i-j}\big| \\ \nonumber\\&& \times \big|\int_{A_{i}}E_m(t,s){\rm d}s \big| \Big|\sum\limits_{i=\max\{1,j-n\}}^{\min\{n,j+n\}} a_{i-j}\int_{A_{i}}E_m(t,s){\rm d}s \Big|^{(1+\rho)} \nonumber\\& \leq& C(\frac{2^m}{n})^{\rho/2} +CM (\frac{2^m}{n})^{\rho/2} p^{\rho/2} \sum\limits_{i=1}^{n}\Big |\int_{A_{i}}E_m(t,s){\rm d}s\Big |\Big(\sum\limits_{w=1}^k \sum\limits_{j=m_w}^{m_w+p-1} |a_{i-j}|\Big) \nonumber\\&\leq & C(\frac{2^m}{n})^{\rho/2}p^{\rho/2}=C \gamma_{2n}^{\rho/2}. \end{matrix} $

因此, 根据引理3.4, 结合(3.6)式和(3.7)式, 引理得证.

引理3.6 假设(A1)-(A5)成立, 则有 $\sup\limits_{u}\big|{\rm P}(S_{1n}^{\prime}\leq u)-{\rm P}(T_n\leq u)\big| \leq C \big\{ \gamma _{2n}^{\rho/2} + \gamma _{4n}^{1/3} \big\}. $

假设$S_{1n}^{\prime}$$T_n$的特征函数分别为$\phi_1 (t)$$\psi_1 (t)$. 则由文献[19,定理5.3]得

$ \begin{matrix} && \sup\limits_{u}\big|{\rm P}(S_{1n}^{\prime}\leq u)-{\rm P}(T_n\leq u)\big| \nonumber \\ &\leq & \int_{-T}^{T}{\Big|\frac{\phi_1(t)-\psi_1(t)}{t}\Big|{\rm d}t} + T \sup\limits_{u}\int_{|t|\leq c/T} \big|{\rm P}(T_n\leq u+t)-{\rm P}(T_n\leq u)\big|{\rm d}t. \end{matrix} $

又由引理1.2和引理3.1, 以及假设条件(A1), 计算可得

$\begin{eqnarray*} \big|\phi_1 (t)-\psi_1 (t)\big| &=& \Big|{\rm E}\exp( {\rm \text{i}}t \sum\limits_{w=1}^{k} y_{nw})-\prod\limits_{w=1}^{k} {\rm E}\exp({\rm \text{i}}t y_{nw})\Big| \\& \leq& 4 t^2 \sum\limits_{1\leq i <j\leq k} \sum\limits_{s_0=m_i}^{m_i+p-1} \sum\limits_{t_0=m_j}^{m_j+p-1} \big|Cov(Z_{ns_0},\ Z_{nt_0})\big| \\& \leq& 4 t^2 \sigma_{n}^{-2} \sum\limits_{1\leq i <j\leq k} \sum\limits_{s_0=m_i}^{m_i+p-1} \sum\limits_{t_0=m_j}^{m_j+p-1} \sum\limits_{u=\max\{1,s_0-n\}}^{\min\{n, s_0+n\}} \sum\limits_{v=\max\{1,t_0-n\}}^{\min\{n, t_0+n\}} |a_{u-s_0} a_{v-t_0}| \\&& \ \ \cdot \big| \int_{A_{u}}E_m(t,s){\rm d}s\big| \cdot \big|\int_{A_{v}}E_m(t,s){\rm d}s\big|\cdot \big|Cov(e_{s_0},\ e_{t_0})\big| \\ & \leq& C t^2 \sigma_{n}^{-2} (2^m/n)^2 u(q) \leq C t^2 \gamma_{4n}. \end{eqnarray*}$

于是我们有

$ \begin{equation} \int_{-T}^{T} \Big|\frac{\phi_1 (t)-\psi_1 (t)}{t}\Big|{\rm d}t \leq C \gamma _{4n}T^2. \end{equation} $

又由引理3.5, 通过计算可得

$\begin{eqnarray*} && \sup\limits_{u} \big|{\rm P}(T_n\leq u+t) -{\rm P}(T_n\leq u)\big| \\ &\leq& \sup\limits_{u} \Big|{\rm P}\big(\frac{T_n}{s_n}\leq \frac{u+t}{s_n}\big) -{\rm \Phi}\big(\frac{u+t}{s_n}\big )\Big| +\sup\limits_{u} \Big|{\rm P}\big(\frac{T_n}{s_n}\leq \frac{u}{s_n}\big) -{\rm \Phi}\big(\frac{u}{s_n}\big)\Big| +\sup\limits_{u} \Big|{\rm \Phi}\big(\frac{u+t}{s_n}\big) -{\rm \Phi}\big(\frac{u}{s_n}\big)\Big| \\ &\leq & C \Big( \gamma_{2n}^{\rho/2}+ \big|\frac{t}{s_n}\big|\Big) \leq C \Big( \gamma_{2n}^{\rho/2}+ \big|t|\Big). \end{eqnarray*}$

从而有

$ \begin{equation} T \sup\limits_{u}\int_{|t|\leq c/T} \big|{\rm P}(T_n \leq u+t)-{\rm P}(T_n \leq u )\big|{\rm d}t \leq C \big\{\gamma _{2n}^{\rho/2}+1/T\big\}. \end{equation} $

再结合(3.8)式, (3.9)式和(3.10)式, 取$T=\gamma _{4n}^{-1/3}$, 即可得

$\begin{eqnarray*} && \sup\limits_{u}\big|{\rm P}(S_{1n}^{\prime}\leq u)-{\rm P}(T_n\leq u)\big| \leq C\Big\{ \gamma _{4n} T^2 + \gamma _{2n}^{\rho/2}+\frac{1}{T}\Big\} = C\big\{ \gamma _{2n}^{\rho/2} + \gamma _{4n}^{1/3} \big\}. \end{eqnarray*}$

这样引理证明完毕.

引理3.7 [引理A3] 假设$\{\zeta _n: n\geq 1\}$, $\{\eta _n: n\geq 1\}$, $\{\xi _n: n\geq 1\}$$\{\varsigma_n: n\geq 1\}$是随机变量序列, $\{\gamma _n:n\geq 1\}$是一列正常数序列, $\gamma _n\to 0$. 如果 $\sup\limits_{u}\big|F_{\zeta _n}(u)-\Phi (u)\big| \leq C\gamma _n,$ 则对任意的 $\varepsilon_1> 0$, $\varepsilon_2> 0$$\varepsilon_3> 0$

$\begin{eqnarray*} && \sup\limits_{u}\Big|F_{\zeta _n+\eta _n +\xi_n+\varsigma_n}(u)-\Phi (u)\Big| \leq C\Big\{\gamma _n+ \sum\limits_{i=1}^3\varepsilon_i\! +\! P(|\eta _n|\geq \varepsilon_1)+ P(|\xi _n|\geq \varepsilon_2)\!+\! P(|\varsigma_n|\geq \varepsilon_3) \Big\}. \end{eqnarray*}$

4 主要结论的证明

定理2.1的证明 由引理3.7可得

$ \begin{matrix} \sup\limits_{u}\Big|{\rm P}(S_n\leq u) -\Phi (u)\Big| &\leq & C\Big\{\sup\limits_{u}\Big|{\rm P}(S_{1n}^{\prime}\leq u)-\Phi(u)\Big| +\sum\limits_{i=1}^{3}\gamma _{in}^{1/3} + {\rm P}\Big(\big|S_{1n}^{\prime\prime}\big|\geq \gamma _{1n}^{1/3}\Big) \nonumber \\&& +{\rm P}\Big(\big|S_{1n}^{\prime\prime\prime}\big|\geq \gamma _{2n}^{1/3}\Big) +{\rm P}\Big(\big|S_{2n}\big|\geq \gamma _{3n}^{1/3}\Big)\Big\}. \end{matrix} $

而由引理3.6, 引理3.5和引理3.4, 计算可得

$ \begin{matrix} & & \sup\limits_{u}\Big|{\rm P}(S_{1n}^{\prime}\leq u)-\Phi (u)\Big| \nonumber\\ &\leq & \sup\limits_{u}\Big|{\rm P}(S_{1n}^{\prime}\leq u)-{\rm P}(T_n\leq u)\Big| + \sup\limits_{u}\Big|{\rm P}(T_n\leq u)-{\rm \Phi}(u/s_n)\Big| + \sup\limits_{u}\big|{\rm \Phi}(u/s_n)-{\rm \Phi}(u)\big| \nonumber\\ &\leq & C \Big\{\sup\limits_{u}\big|{\rm P}(S_{1n}^{\prime}\leq u)-{\rm P}(T_n\leq u)\big| + \sup\limits_{u}\big|{\rm P}(T_n\leq u)-{\rm \Phi}(u/s_n)\big| + \big| s_n^2-1\big|\Big\} \nonumber\\ &\leq & C \Big\{\big(\gamma_{2n}^{\rho/2} + \gamma _{4n}^{1/3}\big) +\gamma_{2n}^{\rho/2} + \big(\gamma _{1n}^{1/2}+\gamma _{2n}^{1/2}+\gamma _{3n}^{1/2}+ u(q)\big)\Big\} \nonumber \\ &\leq & C \left\{\gamma _{1n}^{1/2}+\gamma _{2n}^{1/2}+\gamma_{2n}^{\rho/2}+\gamma_{3n}^{1/2}+ \gamma _{4n}^{1/3} + u(q)\right\}. \end{matrix} $

这样, 结合(4.1)式和(4.2)式, 利用引理3.3(ii), 计算可得

$\begin{eqnarray*} \sup\limits_{u}\Big|{\rm P}(S_n\leq u) -\Phi (u)\Big| &\leq & C\Big\{\gamma _{1n}^{1/2}+\gamma _{2n}^{1/2}+\gamma_{2n}^{\rho/2}+\gamma_{3n}^{1/2}+ \gamma _{4n}^{1/3} + u(q) +\sum\limits_{i=1}^{3}\gamma _{in}^{1/3} \Big\} \\ &\leq & C \Big\{\gamma _{1n}^{1/3}+\gamma _{2n}^{1/3}+\gamma_{2n}^{\rho/2}+ \gamma _{3n}^{1/3}+ \gamma _{4n}^{1/3} + u(q) \Big\}. \end{eqnarray*}$

定理2.1证明完毕.

推论2.1的证明 由于$u(1)<\infty$, 则有$u(q)\rightarrow 0$$\gamma_{4n}=u(q) (2^m/n)^2\rightarrow0$. 再由假设条件(A5)和定理2.1, 推论2.1立即得证.

推论2.2的证明$p=[n^\tau]$,$q=[n^{2\tau-1}]$. 由于$ 2^m/n=O(n^{-\theta})$, 取$\tau<\theta$, 计算得到

$\begin{eqnarray*} \gamma _{1n}^{1/3}&=& \gamma_{2n}^{1/3}=O\big(n^{-\frac{ \theta-\tau }{3}}\big), \\ \gamma_{3n}^{1/3}&=&n^{1/3}\Big(\sum\limits_{|j|>n}|a_{j}|\Big)^{2/3}=n^{1/3}\cdot n^{-\frac{\theta-\tau+1}{3}}\Big(n^{\frac{\theta-\tau+1}{2}}\sum\limits_{|j|>n}|a_{j}|\Big)^{2/3}=O\big(n^{-\frac{ \theta-\tau }{3}}\big), \\ \gamma _{4n}^{1/3}&=& \gamma_{4n}=\big [u(q) (2^m/n)^2\big]^{1/3}= O\big(n^{-\frac{ \theta-\tau }{3}}\cdot n^{-2\theta}\big)^{1/3}=O\big(n^{-\frac{ 7\theta-\tau }{9}}\big)\leq O\big(n^{-\frac{ \theta-\tau }{3}}\big), \\ u(q)&=& O\big( q^{-\frac{\theta-\tau}{3(2\tau-1)}}\big)=O\big(n^{-\frac{ \theta-\tau }{3}}\big). \end{eqnarray*}$

从而, 根据定理2.1, 推论2.2立即得证.

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