设$ \mathbf{N} $为自然数集, $ \{X_{i};i\in \mathbf{N}\} $是概率空间$ (\Omega, \mathcal{A}, P) $上的随机变量序列.

定义1.1   对随机变量序列$ \{X_{n};n\geq1\} $, 如果当$ n\rightarrow\infty $时,

$ \alpha(n) = \sup\limits_{k\geq 1}\sup\limits_{\mathcal{A}\in\mathcal{F}_{1}^{k}, \mathcal{B}\in\mathcal{F}_{k+n}^{\infty}} \big|P(\mathcal{A}\mathcal{B})-P(\mathcal{A})P(\mathcal{B})\big|\rightarrow 0, $

其中$ \sigma $域$ \mathcal{F}_{1}^{k} = \sigma(X_{j};1\leq j\leq k), \mathcal{F}_{k+n}^{\infty} = \sigma(X_{j};j\geq k+n) $, 则称$ \{X_{n};n\geq 1\} $是$ \alpha $混合的.

设$ Y_{1}, Y_{2}, \cdots, Y_{n} $是固定点$ x_{1}, x_{2}, \cdots, x_{n} $的$ n $个观察值, 适合回归模型

$ \begin{equation} Y_{i} = g(x_{i})+\varepsilon_{i}, i = 1, 2, \cdots, n, \end{equation} $

(1.1)

其中回归函数$ g(x) $是[0, 1]上的未知函数, 且当$ x\notin[0, 1] $时, 令$ g(x) = 0 $, $ \{\varepsilon_{i}\} $是随机误差序列.

假定$ 0 = x_{0}\leq x_{1}\leq\cdots\leq x_{n-1}\leq x_{n} = 1 $.考虑Gasser和Müller^[1]对回归函数$ g(x) $提出的一种积分权核估计

$ \begin{equation} g_{n}(x) = {h_{n}^{-1}}\sum\limits_{i = 1}^{n}Y_{i}\int_{x_{i-1}}^{x_{i}}K(\frac{x-s}{h_{n}})\mathrm{d} s, \end{equation} $

(1.2)

其中核函数$ K(u) $是$ \mathcal{R}^{1} $上可测函数, 且当$ n\rightarrow \infty $时, $ 0<h_{n}\rightarrow 0 $.

$ \alpha $混合序列是一类重要的非独立序列, 引起了诸多学者的关注.在$ \alpha $混合序列下, 文献[2]研究了回归加权核估计的强大数律; 文献[3]研究了一般回归加权函数估计的一致渐近正态性; 文献[4]研究了核型分位数估计的渐近正态性; 文献[5]讨论了风险度量ES非参数估计部分和的渐近正态性; 文献[6]讨论了回归加权核估计的强相合性.讨论研究该混合序列背景下的非参数估计的大样本极限性质可更好地为进一步的实证研究提供理论基础.不少文献(如[7-15])讨论了在独立情形与混合相依情形下Priestley-Chao加权核回归估计的完全收敛性和强相合性, 获得了很好的结果.早期, 对于文献[1]中Gasser和Müller提出的一种积分权核估计(1.2), 文献[16-18]研究了独立序列情形下估计(1.2)的完全收敛性、渐近正态性等大样本性质.另外, 文献[19-21]给出了研究$ \alpha $混合随机变量序列大样本性质所需的部分和、加权和的不等式工具.较少文献在$ \alpha $混合序列背景下研究估计(1.2)的大样本极限性质.本文则在$ \alpha $混合样本下给出回归函数核估计(1.2)的强相合性与一致强相合性结论.利用文献[21]的指数不等式, 在较弱的条件下, 本文得到较理想的收敛结论.由于在$ \alpha $混合样本下研究具体的统计估计问题还是比较少, 因此本文的研究是有意义的.

记$ \tilde{\delta_{i}} = x_{i}-x_{i-1}, \delta_{n} = \max\limits_{1\leq i\leq n}(x_{i}-x_{i-1}) $, 本文用$ C $表示与$ n $无关的大于0常数, $ \|x\|_{r} = (E|x|^{r})^{1/r} $, 且$ [x] $表示不超过$ x $的最大整数, 并使用如下基本条件

(a) $ \int_{-\infty}^{+\infty}K(u)\mathrm{d} u = 1, \int_{-\infty}^{+\infty}|K(u)|\mathrm{d} u<\infty $, $ K(\cdot) $在$ \mathcal{R}^{1} $上有界, $ K(\cdot) $在$ \mathcal{R}^{1} $上满足$ \beta(\beta>0) $阶Lipschitz条件, 即$ \forall x, y\in\mathcal{R}^{1} $, 有$ |K(x)-K(y)|\leq C|x-y|^{\beta} $;

(b) $ g(\cdot) $在[0, 1]上一致连续;

(c) 当$ n\rightarrow\infty $时, $ h_{n}\rightarrow 0, \delta_{n}\rightarrow 0 $.

定理1.2   设基本条件(a)–(c)成立.又假设(ⅰ) $ E\varepsilon_{i} = 0, i = 1, 2, \cdots, n $, 且$ |\varepsilon_{i}|\leq b $ a.s.; (ⅱ) $ \{\varepsilon_{i}\} $为$ \alpha $混合相依, 且$ \alpha(n) = O(n^{-\lambda}) $, 其中$ \lambda>1 $; (ⅲ)当$ n\rightarrow\infty $时,

$ \begin{equation} \frac{\delta_{n}}{h_{n}} = O(n^{-s}), s>0. \end{equation} $

(1.3)

(1) 若

$ \begin{equation} \lambda>\frac{2}{s}-1, \end{equation} $

(1.4)

则有$ \lim\limits_{n\rightarrow\infty}g_{n}(x) = g(x) $ a.s., $ x\in(0, 1); $

(2) 若存在$ d>0 $使得$ n^{d}h_{n}\rightarrow\infty $以及

$ \begin{equation} \lambda>\frac{d(1+\beta)/\beta+2}{s}-1, \end{equation} $

(1.5)

则对任给的$ 0<\tau<1/2 $, 都有$ \lim\limits_{n\rightarrow\infty}\sup\limits_{x\in[\tau, 1-\tau]}|g_{n}(x)-g(x)| = 0 $ a.s..

注  定理1.2给出了$ \alpha $混合相依序列有界情形下, 估计(1.2)的强相合与一致强相合性.

下面的定理1.3是在该序列无界条件下得到该估计的强相合与一致强相合性.

定理1.3   设基本条件(a)–(c)成立.又假设(ⅰ) $ E\varepsilon_{i} = 0, i = 1, 2, \cdots, n $, 且$ \sup \limits_{i} E|\varepsilon_{i}|^{r}<\infty $, 其中$ r>2 $; (ⅱ) $ \{\varepsilon_{i}\} $为$ \alpha $混合相依, 且$ \alpha(n) = O(n^{-\lambda}) $, 其中$ \lambda>r/(r-2) $; (ⅲ)当$ n\rightarrow\infty $时,

$ \begin{equation} \frac{\delta_{n}}{h_{n}} = O(n^{-s}), \frac{1}{r}<s\leq 1. \end{equation} $

(1.6)

(1) 若

$ \begin{equation} \lambda>\frac{2}{s-1/r}-1, \end{equation} $

(1.7)

则有$ \lim\limits_{n\rightarrow\infty}g_{n}(x) = g(x) $ a.s. $ \; x\in(0, 1); $

(2) 若存在$ d>0 $使得$ n^{d}h_{n}\rightarrow\infty $以及

$ \begin{equation} \lambda>\frac{2+1/(r\beta)+d(1+\beta)/\beta}{s-1/r}-1, \end{equation} $

(1.8)

则对任给的$ 0<\tau<1/2 $, 有$ \lim\limits_{n\rightarrow\infty}\sup\limits_{x\in[\tau, 1-\tau]}|g_{n}(x)-g(x)| = 0 $ a.s..

为了证明定理的结论, 本节给出一些引理.

引理2.1^[19]  若$ \{X_{i};i\geq1\} $是零均值的$ \alpha $混合随机变量序列, 且$ \{a_{i};i\geq1\} $是实数列.

(ⅰ)如果$ E|X_{i}|^{2+\delta}<\infty $, 这里$ \delta> 0 $, 则

$ \begin{equation} E(|\sum\limits_{i = 1}^{n}a_{i}X_{i}|^{2}) \leq (1+20\sum\limits_{m = 1}^{n}\alpha^{\frac{\delta}{2+\delta}}(m))\sum\limits_{i = 1}^{n}a_{i}^{2}\|X_{i}\|_{2+\delta}^{2}; \end{equation} $

(2.1)

(ⅱ)如果$ |X_{i}|<b_{i} $ a.s., 则

$ \begin{equation} E ( |\sum\limits_{i = 1}^{n}a_{i}X_{i} |^{2} ) \leq (1+8\sum\limits_{m = 1}^{n}\alpha(m) )\sum\limits_{i = 1}^{n}a_{i}^{2}b_{i}^{2}. \end{equation} $

(2.2)

引理2.2^[21]  令$ \{X_{i};i\geq1\} $是零均值的$ \alpha $混合随机变量序列, 且$ |X_{i}|\leq b<\infty $ a.s., 进一步假定正整数$ k_{n} $满足$ 1\leq k_{n}\leq n/2 $.则对任意的$ \epsilon>0 $, 有

$ \begin{equation} P ( |\sum\limits_{i = 1}^{n}X_{i} |>n\epsilon ) \leq 4\exp (-\frac{n\epsilon^{2}}{12(3\sigma_{n}^{2}+bk_{n}\epsilon)} )+36b\epsilon^{-1}\alpha(k_{n}), \end{equation} $

(2.3)

其中$ \sigma_{n}^{2} = n^{-1}\sum\limits_{j = 1}^{2m_{n}+1}E|U_{j}|^{2}, \; m_{n} = [\frac{n}{2k_{n}}], U_{j} = \sum\limits_{(j-1)k_{n}<i\leq jk_{n}}\; \; X_{i}\; (j = 1, 2, \cdots, 2m_{n}) $, 和$ U_{2m_{n}+1} = \sum\limits_{i = 2m_{n}k_{n}+1}^{n}X_{i} $.

引理2.3   设$ g(x) $在[0, 1]上一致连续, $ \int_{-\infty}^{+\infty}K(u)\mathrm{d}u = 1, \int_{-\infty}^{+\infty}|K(u)\mathrm{d}u<\infty $.如果$ h_n\to 0 $和$ \delta_n\to 0 $, 则

$ \begin{equation} \lim\limits_{n\rightarrow\infty}Eg_{n}(x) = g(x), \quad x\in(0, 1). \end{equation} $

(2.4)

对任给的$ 0<\tau<1/2 $, 有

$ \begin{equation} \lim\limits_{n\rightarrow\infty}\sup\limits_{x\in[\tau, 1-\tau]}|Eg_{n}(x)-g(x)| = 0. \end{equation} $

(2.5)

证  由(1.1)和(1.2)式, 有

$ \begin{align} Eg_{n}(x)-g(x) = E [h_{n}^{-1}\sum\limits_{i = 1}^{n}Y_{i}\int_{x_{i-1}}^{x_{i}}K (\frac{x-s}{h_{n}} )\mathrm{d}s ]-g(x)\\ = E [h_{n}^{-1}\sum\limits_{i = 1}^{n}(g(x_{i})+\varepsilon_{i})\int_{x_{i-1}}^{x_{i}}K (\frac{x-s}{h_{n}} )\mathrm{d}s ]-g(x)\\ = h_{n}^{-1}\sum\limits_{i = 1}^{n}g(x_{i})\int_{x_{i-1}}^{x_{i}}K (\frac{x-s}{h_{n}} )\mathrm{d}s-g(x). \end{align} $

(2.6)

从而

$ \begin{align} Eg_{n}(x)-g(x) = & [h_{n}^{-1}\sum\limits_{i = 1}^{n}\int_{x_{i-1}}^{x_{i}}g(x_{i})K (\frac{x-s}{h_{n}} )\mathrm{d}s -h_{n}^{-1}\sum\limits_{i = 1}^{n}\int_{x_{i-1}}^{x_{i}}g(s)K (\frac{x-s}{h_{n}} )\mathrm{d}s ]\\ &+ [h_{n}^{-1}\sum\limits_{i = 1}^{n}\int_{x_{i-1}}^{x_{i}}g(s)K (\frac{x-s}{h_{n}} )\mathrm{d}s-g(x) ]\\ : = &I_{n1}(x)+I_{n2}(x). \end{align} $

(2.7)

要证明(2.4)式, 只需分别证明:对$ x\in(0, 1) $, $ I_{n1}(x)\to 0 $和$ I_{n2}(x)\to 0 $.记$ B_1 = \int_{-\infty}^{+\infty}|K(u)|\mathrm{d}u<\infty $, 令$ u = \frac{x-s}{h_{n}} $.由$ g(\cdot) $在[0, 1]上一致连续知, $ \forall\varepsilon>0 $, 当$ n $充分大时, 有

$ \begin{align} |I_{n1}(x)|&\leq h_{n}^{-1}\sum\limits_{i = 1}^{n}\int_{x_{i-1}}^{x_{i}}|g(x_{i})-g(s)| |K(\frac{x-s}{h_{n}}) |\mathrm{d} s\\ &\leq \frac{\varepsilon}{2B_1} h_{n}^{-1}\sum\limits_{i = 1}^{n}\int_{x_{i-1}}^{x_{i}} |K(\frac{x-s}{h_{n}}) |\mathrm{d} s\\ & = \frac{\varepsilon}{2B_1} h_{n}^{-1}\int_{0}^{1} |K(\frac{x-s}{h_{n}}) |\mathrm{d} s = \frac{\varepsilon}{2B_1}\int^{x/h_n}_{(x-1)/h_n}|K(u)|\mathrm{d}u\\ &\leq \frac{\varepsilon}{2}. \end{align} $

(2.8)

所以$ I_{n1}(x)\to 0 $.另一方面, 由条件$ \int_{-\infty}^{+\infty}K(u)\mathrm{d}u = 1 $, 有

$ \begin{align} I_{n2}(x) = &h_{n}^{-1}\int_{0}^{1}g(s)K(\frac{x-s}{h_{n}})\mathrm{d}s-g(x) = \int^{x/h_n}_{(x-1)/h_n}g(x-uh_n)K(u)\mathrm{d}u-g(x)\int_{-\infty}^{+\infty}K(u)\mathrm{d}u\\ = &\int^{x/h_n}_{(x-1)/h_n}[g(x-uh_n)-g(x)]K(u)\mathrm{d}u -g(x)\int_{-\infty}^{(x-1)/h_n}K(u)\mathrm{d}u-g(x)\int_{x/h_n}^{\infty}K(u)\mathrm{d}u. \end{align} $

(2.9)

当$ n $充分大时, 由条件$ \int_{-\infty}^{+\infty}|K(u)|\mathrm{d}u<\infty, h_n\to 0 $, 知

$ \begin{equation} \int^{(x-1)/h_n}_{-\infty}|K(u)|\mathrm{d}u\to 0, \ \ \int^{\infty}_{x/h_n}|K(u)|\mathrm{d}u\to 0. \end{equation} $

(2.10)

注意到$ g(x) $在[0, 1]上有界, 记$ B_2 = \sup\limits_{x\in[0, 1]}|g(x)| $, 取$ \delta>0 $, 可有

$ \begin{align} \label{eq:19} &\left|\int^{x/h_n}_{(x-1)/h_n}[g(x-uh_n)-g(x)]K(u)\mathrm{d}u \right| \leq\int_{(x-1)/h_n}^{x/h_n}|K(u)|\cdot|g(x-uh_{n})-g(x)|\mathrm{d}u\\ \leq&\int_{|u|<\delta/h_n}|K(u)|\cdot|g(x-uh_{n})-g(x)|\mathrm{d}u +\int_{|u|\geq\delta/h_n}|K(u)|\cdot|g(x-uh_{n})-g(x)|\mathrm{d}u\\ \leq& B_1\sup\limits_{|u|<\delta/h_n} |g(x-uh_{n})-g(x)|+2B_2\int_{|u|\geq\delta/h_n}|K(u)|\mathrm{d}u. \end{align} $

(2.11)

由$ g(x) $在$ x $处连续以及$ \int_{-\infty}^{+\infty}|K(u)|\mathrm{d}u<\infty $知, $ \forall\varepsilon>0 $, 当$ n $充分大时, 有

$ \begin{equation} |\int^{x/h_n}_{(x-1)/h_n}[g(x-uh_n)-g(x)]K(u)\mathrm{d}u |<\varepsilon. \end{equation} $

(2.12)

即

$ \begin{equation} \int^{x/h_n}_{(x-1)/h_n}[g(x-uh_n)-g(x)]K(u)\mathrm{d}u\to 0. \end{equation} $

(2.13)

由(2.9), (2.10)和(2.13)式知, $ I_{n2}(x)\to 0 $.从而(2.4)式成立.由于$ g(x) $有界, 所以上述证明过程对(2.5)式也成立.证毕.

引理2.4   设$ \int_{-\infty}^{+\infty}|K(u)|\mathrm{d}u<\infty $且$ h_n\to 0 $, 则对任意的$ x\in(0, 1) $, 有

$ \begin{equation} \lim\limits_{n\rightarrow\infty} h_{n}^{-1}\sum\limits_{i = 1}^{n}\int_{x_{i-1}}^{x_{i}} |K (\frac{x-s}{h_{n}} ) |\mathrm{d} s = \int_{-\infty}^{+\infty}|K(u)|\mathrm{d}u. \end{equation} $

(2.14)

而对任给的$ 0<\tau<1/2 $, 有

$ \begin{equation} \lim\limits_{n\rightarrow\infty}\sup\limits_{x\in[\tau, 1-\tau]} h_{n}^{-1}\sum\limits_{i = 1}^{n}\int_{x_{i-1}}^{x_{i}} |K (\frac{x-s}{h_{n}} ) |\mathrm{d} s = \int_{-\infty}^{+\infty}|K(u)|\mathrm{d} u. \end{equation} $

(2.15)

证  对任意的$ x\in(0, 1) $, 令$ u = \frac{x-s}{h_{n}} $, 显然

$ \begin{align} &h_{n}^{-1}\sum\limits_{i = 1}^{n}\int_{x_{i-1}}^{x_{i}} |K(\frac{x-s}{h_{n}}) |\mathrm{d} s-\int_{-\infty}^{+\infty}|K(u)|\mathrm{d}u\\ = &h_{n}^{-1}\int_{0}^{1} |K(\frac{x-s}{h_{n}}) |\mathrm{d} s-h_{n}^{-1}\int_{-\infty}^{+\infty} |K(\frac{x-s}{h_{n}}) |\mathrm{d} s\\ = &-h_{n}^{-1}\int_{-\infty}^{0}|K(\frac{x-s}{h_{n}})|\mathrm{d}s-h_{n}^{-1}\int_{1}^{+\infty}|K(\frac{x-s}{h_{n}})| \mathrm{d} s. \end{align} $

(2.16)

由条件$ \int_{-\infty}^{+\infty}|K(u)|\mathrm{d}u<\infty $与$ h_n\to 0, $有

$ \begin{equation} h_{n}^{-1}\int_{-\infty}^{0}\|K(\frac{x-s}{h_{n}})|\mathrm{d} s = \int_{x/h_n}^{+\infty}|K(u)|\mathrm{d}u\to 0 \end{equation} $

(2.17)

和

$ \begin{equation} h_{n}^{-1}\int^{+\infty}_{1}\|K(\frac{x-s}{h_{n}})|\mathrm{d} s = \int^{(x-1)/h_n}_{-\infty}|K(u)|\mathrm{d}u\to 0. \end{equation} $

(2.18)

联合(2.16)–(2.18)式得结论(2.14).另外, 对任给的$ 0<\tau<1/2 $, 有

$ \begin{align} \sup\limits_{x\in[\tau, 1-\tau]}h_{n}^{-1}\int_{-\infty}^{0}|K(\frac{x-s}{h_{n}})|\mathrm{d} s & = \sup\limits_{x\in[\tau, 1-\tau]}\int_{x/h_n}^{+\infty}|K(u)|\mathrm{d}u \leq\int_{\tau/h_n}^{+\infty}|K(u)|\mathrm{d}u \to 0 \end{align} $

(2.19)

和

$ \begin{align} \sup\limits_{x\in[\tau, 1-\tau]}h_{n}^{-1}\int^{+\infty}_{1}|K\Big(\frac{x-s}{h_{n}}\Big)|\mathrm{d} s & = \sup\limits_{x\in[\tau, 1-\tau]}\int^{(x-1)/h_n}_{-\infty}|K(u)|\mathrm{d}u \leq \int^{-\tau/h_n}_{-\infty}|K(u)|\mathrm{d}u\to 0. \end{align} $

(2.20)

由(2.16), (2.19)和(2.20)式得结论(2.15).证毕.

引理2.5^[14]  设$ \{Z_{i};i\geq1\} $是随机变量序列, 若存在常数$ \rho>0 $, 使得$ E|Z_{i}| = O(i^{-(1+\rho)}) $, 则$ \Sigma_{i = 1}^{\infty}Z_{i} $ a.s.收敛.

记$ S_{n}(x) = \sum\limits_{i = 1}^{n}W_{ni}(x)\varepsilon_{i}, W_{ni}(x) = h_{n}^{-1}\int_{x_{i-1}}^{x_{i}}K\big(\frac{x-s}{h_{n}}\big)\mathrm{d} s $.可由积分中值定理, 存在$ \vartheta_{i}\in(0, 1) $, 使得

$ \begin{equation} W_{ni}(x) = h_{n}^{-1}K (\frac{x-x_{i}+\vartheta_{i}\tilde{\delta}_{i}}{h_{n}} )\tilde{\delta}_{i} \leq C\frac{\delta_{n}}{h_{n}}. \end{equation} $

(3.1)

由于

$ \begin{equation} |g_{n}(x)-g(x)|\leq |\sum\limits_{i = 1}^{n}W_{ni}(x)\varepsilon_{i} | +|Eg_{n}(x)-g(x)|. \end{equation} $

(3.2)

由(3.2)式、引理2.3以及Borel-Cantelli引理可知, 定理1.2与定理1.3的证明归结为证明如下式子

$ \begin{equation*} |\sum\limits_{i = 1}^{n}W_{ni}(x)\varepsilon_{i} |\stackrel{c}\rightarrow 0, \quad x\in(0, 1). \end{equation*} $

与对任给的$ 0<\tau<1/2 $, $ \sup\limits_{x\in[\tau, 1-\tau]} |\sum\limits_{i = 1}^{n}W_{ni}(x)\varepsilon_{i} |\stackrel{c}\rightarrow 0. $即$ \forall\epsilon>0 $, 有

$ \begin{equation} \sum\limits_{n = 1}^{\infty}P ( |\sum\limits_{i = 1}^{n}W_{ni}(x)\varepsilon_{i} |>\epsilon )<\infty, x\in(0, 1) \end{equation} $

(3.3)

与

$ \begin{equation} \sum\limits_{n = 1}^{\infty}P (\sup\limits_{x\in[\tau, 1-\tau]} |\sum\limits_{i = 1}^{n}W_{ni}(x)\varepsilon_{i} |>\epsilon )<\infty. \end{equation} $

(3.4)

又由引理2.4及条件(a), 当$ n\rightarrow\infty $时, 可有

$ \begin{equation} \sum\limits_{i = 1}^{n}|W_{ni}(x)|\rightarrow\int_{-\infty}^{+\infty}|K(u)|\mathrm{d} u<C, \sup\limits_{x\in[\tau, 1-\tau]}\sum\limits_{i = 1}^{n}|W_{ni}(x)|\rightarrow\int_{-\infty}^{+\infty}|K(u)|\mathrm{d} u<C. \end{equation} $

(3.5)

定理1.2的证明  首先证明(3.3)式.记$ X_{i} = W_{ni}(x)\varepsilon_{i} $, 由条件(ⅰ)及(3.1)式, 有$ EX_{i} = 0 $, 且有$ |X_{i}| = |W_{ni}(x)\varepsilon_{i}|\leq C\frac{\delta_{n}}{h_{n}}\leq Cn^{-s} $, 由于$ \alpha(n) = O(n^{-\lambda}) $且$ \lambda>1 $, 所以在引理2.1中的$ \sum\limits_{m = 1}^{\infty}\alpha(m)\leq C\sum\limits_{m = 1}^{\infty}m^{-\lambda}<\infty. $根据引理2.1的(2.2)式可得

$ \begin{equation*} \begin{split} E|U_{j}|^{2} = E |\sum\limits_{i = (j-1)k_{n}+1}^{jk_{n}}W_{ni}(x)\varepsilon_{i} |^{2} \leq C\sum\limits_{i = (j-1)k_{n}+1}^{jk_{n}}W_{ni}^{2}(x), \end{split} \end{equation*} $

又根据(3.1)、(3.5)以及(1.3)式, 可有

$ \begin{equation*} \begin{split} \sigma_{n}^{2} = \frac{1}{n}\sum\limits_{j = 1}^{2m_{n}+1}E|U_{j}|^{2} &\leq C\frac{1}{n}\sum\limits_{i = 1}^{n}W_{ni}^{2}(x) \leq C\frac{1}{n}\sup\limits_{x\in(0, 1)}|W_{ni}(x)|\sum\limits_{i = 1}^{n}|W_{ni}(x)|\\ &\leq Cn^{-1}\frac{\delta_{n}}{h_{n}}\leq Cn^{-(1+s)}. \end{split} \end{equation*} $

取$ k_{n} = n^{s}/(\log n\log\log n) $, 其中$ 0<s\leq 1 $, 当$ n $充分大时, 可保证$ k_{n}\leq \frac{n}{2} $, 从而由引理2.2, $ \forall\epsilon>0, x\in(0, 1) $, 可有

$ \begin{equation} \begin{split} &P ( |\sum\limits_{i = 1}^{n}W_{ni}(x)\varepsilon_{i} |>\epsilon ) = P ( |\sum\limits_{i = 1}^{n}X_{i} |>\epsilon )\\ \leq& C\exp (-\frac{n(\frac{\epsilon}{n})^{2}}{12(3\sigma_{n}^{2}+bk_{n}\epsilon/n)} )+36nb\epsilon^{-1}\alpha(k_{n})\\ \leq& C\exp (-\frac{C}{n^{-s}+n^{-s}k_{n}} )+Cn^{1-s}k_{n}^{-\lambda} \leq C\exp (-\frac{C}{n^{-s}k_{n}} )+Cn^{1-s}k_{n}^{-\lambda}\\ = &C\exp(-C\log n\log\log n)+Cn^{-[s(\lambda+1)-1]}(\log n\log\log n)^{\lambda} \leq Cn^{-[s(\lambda+1)-1]}(\log n\log\log n)^{\lambda}. \end{split} \end{equation} $

(3.6)

由(1.4)式知, (3.3)式成立.

其次证明(3.4)式.选取$ l_{n} $个中心在$ t_{1}, t_{2}, \cdots, t_{l_{n}} $, 半径为$ R_{n} = (h_n^{1+\beta}/(\log n\log\log n))^{1/\beta} $的邻域$ B_{1}, B_{2}, \cdots, B_{l_{n}} $覆盖$ [0, 1] $, 其中$ l_{n} = O\big((h_n^{-(1+\beta)}\log n\log\log n)^{1/\beta}\big) $.由条件(ⅰ)知$ |\varepsilon_{i}|<b $ a.s.和基本条件(a)知$ K(\cdot) $满足$ \beta $阶Lipschitz条件, 对正整数$ k, 1\leq k\leq l_{n} $, 可得

$ \begin{equation*} \begin{split} \sup\limits_{x\in B_{k}}|S_{n}(x)-S_{n}(t_{k})| = &\sup\limits_{x\in B_{k}} |h_{n}^{-1}\sum\limits_{i = 1}^{n}\varepsilon_{i}\int_{x_{i-1}}^{x_{i}} [K (\frac{x-s}{h_{n}} ) -K (\frac{t_{k}-s}{h_{n}} ) ]\mathrm{d} s |\\ \leq&C h_{n}^{-1}\sup\limits_{x\in B_{k}}\sum\limits_{i = 1}^{n}\tilde{\delta}_{i} |\frac{x-t_{k}}{h_{n}} |^{\beta} \ \ \text {a.s. } \leq C h_{n}^{-1-\beta}R_{n}^{\beta}\\ = &Ch_{n}^{-(1+\beta)}[(h_n^{1+\beta}/\log n\log\log n)^{1/\beta}]^{\beta} = (\log n\log\log n)^{-1}. \end{split} \end{equation*} $

即$ \forall\epsilon>0 $, 当$ n $充分大时,

$ \begin{equation} \begin{split} \sup\limits_{x\in B_{k}}|S_{n}(x)-S_{n}(t_{k})|<\epsilon/2 \ \ \text {a.s.}. \end{split} \end{equation} $

(3.7)

对上述$ \forall\epsilon>0 $, 当$ n $充分大时, 由(3.6)与(3.7)式, 以及条件(2)中$ n^{d}h_{n}\rightarrow\infty $可知$ h_{n}^{-1}\leq Cn^{d} $, 从而有

$ \begin{equation*} \begin{split} P (\sup\limits_{x\in[\tau, 1-\tau]}|S_{n}(x)|>\epsilon ) \leq& P (\bigcup\limits_{k = 1}^{l_{n}}(\sup\limits_{x\in B_k}|S_{n}(x)|>\epsilon) )\\ \leq&\sum\limits_{k = 1}^{l_{n}}P(\sup\limits_{x\in B_k}|S_{n}(x)-S_{n}(t_{k})|+|S_{n}(t_{k})|>\epsilon)\\ \leq& C\sum\limits_{k = 1}^{l_{n}} P(|S_{n}(t_{k})|>\epsilon/2)\\ \leq& Cl_{n}n^{-[s(\lambda+1)-1]}(\log n\log\log n)^{\lambda}\\ \leq& C(h_n^{-(1+\beta)}\log n\log\log n)^{1/\beta}n^{-s(\lambda+1)+1}(\log n\log\log n)^{\lambda}\\ \leq& Cn^{\frac{d(1+\beta)}{\beta}-s(\lambda+1)+1}(\log n\log\log n)^{\lambda+1/\beta}. \end{split} \end{equation*} $

由(1.5)式知, (3.4)式成立.

定理1.3的证明  令

$ \xi_{i} = \varepsilon_{i}I(|\varepsilon_{i}|\leq n^{1/r})-E\varepsilon_{i}I(|\varepsilon_{i}|\leq n^{1/r}), \tilde{\xi}_{i} = \varepsilon_{i}I(|\varepsilon_{i}|>n^{1/r})-E\varepsilon_{i}I(|\varepsilon_{i}|>n^{1/r}) $

则有

$ S_{n}(x) = \sum\limits_{i = 1}^{n}W_{ni}(x)\varepsilon_{i} = \sum\limits_{i = 1}^{n}W_{ni}(x)\xi_{i}+\sum\limits_{i = 1}^{n}W_{ni}(x)\tilde{\xi}_{i}: = S_{1n}(x)+S_{2n}(x). $

首先证明$ |S_{1n}(x)|\rightarrow 0 $ a.s., $ x\in(0, 1) $与$ \sup\limits_{x\in[\tau, 1-\tau]}|S_{1n}(x)|\rightarrow 0 $ a.s..为此只需证明:对$ \forall\epsilon>0 $, 有

$ \begin{equation} \sum\limits_{n = 1}^{\infty}P\big(|S_{1n}(x)|>\epsilon\big)<\infty, x\in(0, 1) \end{equation} $

(3.8)

与

$ \begin{equation} \sum\limits_{n = 1}^{\infty}P\Big(\sup\limits_{x\in[\tau, 1-\tau]}|S_{1n}(x)|>\epsilon\Big)<\infty. \end{equation} $

(3.9)

记$ V_{i} = W_{ni}(x)\xi_{i} $, 显然有$ EV_{i} = 0 $, 且由条件(3.1)式有$ |V_{i}| = |W_{ni}(x)\xi_{i}|\leq Cn^{1/r}\frac{\delta_{n}}{h_{n}}\leq Cn^{-s+1/r} $.现在对序列$ V_1, V_2, \cdots $, 利用引理2.2, 为此令$ U_{j} = \sum\limits_{i = (j-1)k_{n}+1}^{jk_{n}}V_{i} $.由于$ \alpha(n) = O(n^{-\lambda}) $且$ \lambda>r/(r-2) $, 所以在引理2.1中的

$ \sum\limits_{m = 1}^{\infty}\alpha^{\delta/(2+\delta)}(m) = \sum\limits_{m = 1}^{\infty}\alpha^{(r-2)/r}(m)\leq C\sum\limits_{m = 1}^{\infty}m^{-\lambda(r-2)/r}<\infty. $

根据引理2.1的(2.1)式可得

$ \begin{equation*} \begin{split} E|U_{j}|^{2} = E |\sum\limits_{i = (j-1)k_{n}+1}^{jk_{n}}W_{ni}(x)\xi_{i} |^{2} \leq C\sum\limits_{i = (j-1)k_{n}+1}^{jk_{n}}W_{ni}^{2}(x). \end{split} \end{equation*} $

又根据(3.1)、(3.5)以及(1.6)式, 可有

$ \begin{equation*} \begin{split} \sigma_{n}^{2} = \frac{1}{n}\sum\limits_{j = 1}^{2m_{n}+1}E|U_{j}|^{2}\leq C\frac{1}{n}\sum\limits_{i = 1}^{n}W_{ni}^{2}(x)\leq Cn^{-(1+s)}. \end{split} \end{equation*} $

取$ k_{n} = n^{s-1/r}/(\log n\log\log n) $, 其中$ 0<s\leq 1 $, 当$ n $充分大时, 可保证$ k_{n}\leq \frac{n}{2} $, 从而由引理2.2, $ \forall\epsilon>0, x\in(0, 1) $, 可有

$ \begin{equation} \begin{split} &P ( |S_{1n}(x) |>\varepsilon ) = P ( |\sum\limits_{i = 1}^{n}V_{i} |>\varepsilon )\\ \leq& C\exp (-\frac{n\big(\varepsilon/n\big)^{2}}{12(3\sigma_{n}^{2}+bk_{n}\varepsilon/n)} )+36nb\varepsilon^{-1}\alpha(k_{n})\\ \leq& C\exp (-\frac{C}{n^{-s}+n^{-s+1/r}k_{n}} )+Cn^{1-s+1/r}k_{n}^{-\lambda}\\ = &C\exp(-C\log n\log\log n)+Cn^{1-s+1/r-\lambda(s-1/r)}(\log n\log\log n)^{\lambda}\\ \leq& Cn^{-[(s-1/r)(\lambda+1)-1]}(\log n\log\log n)^{\lambda}. \end{split} \end{equation} $

(3.10)

由(1.7)式知, (3.8)式成立.

选取$ \tilde{l}_{n} $个中心在$ \tilde{t}_{1}, \tilde{t}_{2}, \cdots, \tilde{t}_{\tilde{l}_{n}} $, 半径为$ \tilde{R}_{n} = (n^{-1/r}h_n^{1+\beta}/(\log n\log\log n))^{1/\beta} $的邻域$ \tilde{B}_{1}, \tilde{B}_{2}, \cdots, \tilde{B}_{\tilde{l}_{n}} $覆盖$ [0, 1] $, 其中$ \tilde{l}_{n} = O\big((n^{1/r}h_n^{-(1+\beta)}\log n\log\log n)^{1/\beta}\big) $.注意到$ |\xi_{i}|\leq 2n^{1/r} $和基本条件(a)知$ K(\cdot) $满足$ \beta $阶Lipschitz条件, 对正整数$ k, 1\leq k\leq \tilde{l}_{n} $, 可得

$ \begin{equation*} \begin{split} \sup\limits_{x\in \tilde{B}_{k}}|S_{1n}(x)-S_{1n}(\tilde{t}_{k})| = &\sup\limits_{x\in \tilde{B}_{k}} |h_{n}^{-1}\sum\limits_{i = 1}^{n}\xi_{i}\int_{x_{i-1}}^{x_{i}} [K (\frac{x-s}{h_{n}} ) -K (\frac{\tilde{t}_{k}-s}{h_{n}} ) ]\mathrm{d} s |\\ \leq& Cn^{1/r}h_{n}^{-1}\sup\limits_{x\in \tilde{B}_{k}}\sum\limits_{i = 1}^{n}\tilde{\delta}_{i} |\frac{x-\tilde{t}_{k}}{h_{n}} |^{\beta} \ \ \text {a.s.}\\ \leq& C n^{1/r} h_{n}^{-(1+\beta)}\tilde{R}_{n}^{\beta} \leq C(\log n\log\log n)^{-1}. \end{split} \end{equation*} $

即$ \forall\epsilon>0 $, 当$ n $充分大时,

$ \begin{equation} \begin{split} \sup\limits_{x\in \tilde{B}_{k}}|S_{1n}(x)-S_{1n}(\tilde{t}_{k})|<\epsilon/2 \ \ \text {a.s.}. \end{split} \end{equation} $

(3.11)

对上述$ \forall\epsilon>0 $, 当$ n $充分大时, 由(3.10)与(3.11)式, 以及条件(2)中$ n^{d}h_{n}\rightarrow\infty $可知$ h_{n}^{-1}\leq Cn^{d} $, 从而有

$ \begin{equation*} \begin{split} &P (\sup\limits_{x\in[\tau, 1-\tau]}|S_{1n}(x)|>\epsilon ) \leq P (\bigcup\limits_{k = 1}^{\tilde{l}_{n}}\Big(\sup\limits_{x\in \tilde{B}_k}|S_{1n}(x)|>\epsilon\Big) )\\ \leq&\sum\limits_{k = 1}^{\tilde{l}_{n}}P\Big(\sup\limits_{x\in \tilde{B}_k}|S_{1n}(x)-\tilde{S}_{1n}(\tilde{t}_{k})|+|\tilde{S}_{1n}(\tilde{t}_{k})|>\epsilon\Big) \leq C\sum\limits_{k = 1}^{\tilde{l}_{n}} P(|S_{1n}(\tilde{t}_{k})|>\epsilon/2)\\ \leq &C\tilde{l}_{n}n^{-[(s-1/r)(\lambda+1)-1]}(\log n\log\log n)^{\lambda}\\ \leq &C(n^{1/r}h_n^{1+\beta}\log n\log\log n)^{1/\beta}n^{-(s-1/r)(\lambda+1)+1}(\log n\log\log n)^{\lambda}\\ \leq& Cn^{\frac{1}{r\beta}+\frac{d(1+\beta)}{\beta}-(s-1/r)(\lambda+1)+1}(\log n\log\log n)^{\lambda+1/\beta}. \end{split} \end{equation*} $

结合条件(1.8)知, (3.9)式成立.

其次, 证明$ \forall\epsilon>0 $, 有

$ \begin{equation} |S_{2n}(x)|\rightarrow 0\; \text {a.s.}\; \; x\in(0, 1) \end{equation} $

(3.12)

与

$ \begin{equation} \sup\limits_{x\in[\tau, 1-\tau]}|S_{2n}(x)|\rightarrow 0\; \text {a.s.}. \end{equation} $

(3.13)

记$ \widetilde{S}_{2n}(x) = \sum\limits_{i = 1}^{n}W_{ni}(x)\varepsilon_{i}I(|\varepsilon_{i}|>n^{1/r}) $.由引理2.4及条件(ⅰ)中的$ \sup\limits_{i}E|\varepsilon_{i}|^{r}<\infty $, 这里$ r>2 $, 当$ n $充分大时, 可有

$ \begin{equation} \begin{split} |E\widetilde{S}_{2n}(x)| &\leq\sum\limits_{i = 1}^{n}|W_{ni}(x)|E\big[|\varepsilon_{i}|I(|\varepsilon_{i}|>n^{1/r})\big]\\ &\leq n^{-(r-1)/r}\sum\limits_{i = 1}^{n}|W_{ni}(x)|E\big[|\varepsilon_{i}|^{r}I(|\varepsilon_{i}|>n^{1/r})\big]\\ &\leq Cn^{-(r-1)/r}\rightarrow 0. \end{split} \end{equation} $

(3.14)

由(1.6)与(3.1)式, 可得

$ \begin{equation*} \begin{split} |\widetilde{S}_{2n}(x)| \leq \sum\limits_{i = 1}^{n}|W_{ni}(x)||\varepsilon_{i}|I(|\varepsilon_{i}|>n^{1/r}) \leq Cn^{-s}\sum\limits_{i = 1}^{n}|\varepsilon_{i}|I(|\varepsilon_{i}|>i^{1/r}): = CJ_{n}. \end{split} \end{equation*} $

记$ \eta_{i} = i^{-s}|\varepsilon_{i}|I(|\varepsilon_{i}|>i^{1/r}) $, 可有$ E|\eta_{i}|\leq i^{-s}i^{\frac{1-r}{r}}E|\varepsilon_{i}|^{r}\leq Ci^{-1-(s-\frac{1}{r})} $, 得到$ E|\eta_{i}| = O(i^{-1-(s-\frac{1}{r})}) $, 又由引理2.5可知$ \sum\limits_{i = 1}^{\infty}\eta_{i} $ a.s.收敛, 根据Kronecker引理(见文献[22]), 当$ n\rightarrow\infty $时, 有$ J_{n}\rightarrow 0 $ a.s., 从而(3.12)式成立.在前面两式的左边加上$ \sup\limits_{x\in[\tau, 1-\tau]} $运算符, 不等式仍然成立, 所以(3.13)式也成立.

因此(3.3)与(3.4)式成立, 定理1.3得证.