我们只是证明(3.1)(ii)作为(3.1)(i) 是类似的。
步骤1。设置\(\波浪号{g}_{i} =g(t_{i})-\sum_{j=1}^{n} 克(t_{j})int_{A_{j}}E_{m}(t_{i},s)\,ds\),\(tilde{varepsilon}_{i}=\varepsilen_{我}-\sum_{j=1}^{n}\varepsilon_{j}\int_{A{j}}E_{m}(t_{i},s)\,ds\)很容易理解
$$\开始{aligned}[b]\hat{beta}_{西}-\beta&=T_{n}^{-2}\Biggl[\sum_{i=1}^{n} 一个_{i} \波浪线{x}_{i} \varepsilon_{我}-\sum_{i=1}^{n} 一个_{i} \波浪线{x}_{i} \Biggl(\sum_{j=1}^{n}\varepsilon_{j}\int _{A_{j}}E_{m}(t_{i},s)\,ds\Biggr)+\sum _{i=1}^{n} 一个_{i} \波浪线{x}_{i} \波浪线{g}_{i} \Biggr]\\&:=A_{1n}-A_{2n}+A_{3n}。\结束{对齐}$$
(4.1)
首先,我们证明\(A_{1n}\右箭头0\)a.s.注意\(A_{1n}=\sum_{i=1}^{n}(T_{n}^{-2}一个_{i} \波浪线{x}_{i} \sigma{i})e_{i}=:\sum{i=1}^{n}c_{ni}e(镍)_{i} \)它由(A1)(i)、(A2)(ii)、(2.6)、和\(2^{m}/n=O(n^{-1/2})\)那个
$$开始{聚集}\max_{1\leqi\leqn}\vertc_{ni}\vert\leq\max_}\frac{verta{i}\tilde{x}_{i} \vert}{T_{n}}\cdot\max_{1\leqi\leqn}\frac{\sigma_{i}}{T_}n}=O\bigl(n^{-1/2}\biger),\\sum_{i=1}^{n} c(c)^{2}_{ni}=T_{n}^{-4}\sum_{i=1}^{n} 一个_{i} \波浪线{x}^{2}_{i} \cdot a_{i}\西格玛^{2}_{i} =O\bigl(n^{-1}\bigr)。\结束{聚集}$$
因此,根据引理A.1款,我们有\(A_{1n}\右箭头0\)a.s.显然,
$$A_{2n}=\sum_{j=1}^{n}\Biggl(\sum_{i=1}^}n}T_{n}^{-2}a_{i} \波浪号{x}_{i} \sigma_{j}\int_{A{j}}E_{m}(t_{i},s)\,ds\Biggr)E_{j}=:\sum_{j=1}^{n} d日_{nj}e_{j} ●●●●$$
通过(A2)(ii),引理答6, (2.6)、和\(2^{m}/n=O(n^{-1/2})\)我们可以获得
$$\begin{聚集}\max_{1\leqj\leqn}\vert d_{nj}\vert\leq\Bigl(\max_}1\leq j\leq n}\sigma_{j}\Bigr)\biggl(\max_1\leq i,j\lequen}\int_{A{j}}\Bigl\vert E_{m}(t_{i},s)\Bigr\vert,ds\biggr)\biggl(t_{n}^{-2}\求和{i=1}^{n}\转换A{i}\波浪线{x}_{i} \vert\Biggr)=O\bigl(n^{-1/2}\bigr),\\begin{aligned}\sum_{j=1}^{n} d日^{2}_{nj}&\leq CT_{n}^{-2}\sum_{j=1}^{n}\sum_{i=1}^}\n}\biggl(int_{A_{j}}E_{m}(t_{i},s)^{n} 一个_{i} \波浪线{x}^{2}_{i} \Biggr)\\&\leq C T_{n}^{-2}\sum_{i=1}^{n}\sum_{j=1}^}n}\biggl(int_{A{j}}E_{m}(T_{i},s)\,ds\Biggr)^{2}=O\bigl(2^{m}/n\biger)=O\bigl(n^{-1/2}\bigr)。\结束{对齐}\结束{聚集}$$
因此\(A_{2n}\右箭头0\)a.s.由引理A.1款显然,来自(2.6)和(2.7)我们获得
$$\vert A_{3n}\vert\leq\Bigl(\max_{1\leqi\leqn}\vert_tilde{g}_{i} \vert\Bigr)\cdot\Biggl(T_{n}^{-2}\sum_{i=1}^{n}\verta{i}\波浪线{x}_{i} \vert\Biggr)=O\bigl(2^{-m}+n^{-1}\biger)\rightarrow0$$
第2步。我们证明了(3.2)(i) ,作为(3.2)(ii)类似。我们可以看到
$$\beart{collected}\max_{1\leq i\leq n}\bigl\vert\hat开始{g}_{五十} (t_{i})-g(t_{i})\bigr\vert\\\quad\leq\max_{1\leqi\leqn}\Biggl\{vert\beta-\hat{beta}_{L}\vert\cdot\Biggl \vert\sum_{j=1}^{n} x个_{j} \int_{A_{j}}E_{m}(t_{i},s)\,ds\Biggr\vert+\vert\tilde{g}_{i} \vert+\Biggl\vert\sum_{j=1}^{n}\sigma_{j} e(电子)_{j} \ int_{A_{j}}E_{m}(t_{i},s)\,ds\Biggr\vert\Biggr\}\\quad:=B_{1n}+B_{2n}+B_{3n}。\结束{聚集}$$
与一起(3.1)(i) 在附加假设下,可以得出以下结论\(B_{1n}\右箭头0\)a.s.我们从(A2)(ii)和引理得到答6那个\(B_{3n}\右箭头0\)a.s.通过应用引理A.2款.因此(3.2)(i) 由证明(2.7). □
步骤1。首先,我们证明(3.3)(i) ●●●●。我们有
$$\开始{aligned}[b]\vert\hat{beta}_{左}-\beta\vert&=S_{n}^{-2}\Biggl[\sum_{i=1}^{n}\波浪线{x}_{i} \varepsilon(瓦雷普西隆)_{我}-\sum_{i=1}^{n}\波浪线{x}_{i} \Biggl(sum_{j=1}^{n}\varepsilon_{j}\int_{A{j}}E_{m}(t_{i},s)\,ds\Biggr)+\sum_{i=1}^}{x}_{i} \波浪线{g}_{i} \Biggr]\\&:=C_{1n}-C_{2n}+C_{3n}。\结束{对齐}$$
(4.2)
请注意
$$C_{1n}=\sum_{i=1}^{n}\bigl(S_{n}^{-2}\波浪线{x}_{i} \sima_{i}\bigr)e_{i}:=\sum_{i=1}^{n} c(c)'_{ni}e(镍)_{我}$$
和
$$C_{2n}=\sum_{j=1}^{n}\Biggl(\sum_{i=1}^}n}S_{n}^{-2}\tilde{x}_{i} \sigma_{j}\int_{A{j}}E_{m}(t_{i},s)\,ds\Biggr)E_{j}:=\sum_{j=1}^{n}d'_{nj}e_{j} ●●●●$$
类似于(3.1)(ii),我们有
$$开始{聚集}\max_{1\leqi\leqn}\bigl\vertc'_{ni}\bigr\vert\leq\biggl{x}_{i} \vert}{S_{n}}\biggr)\cdot\frac{1}{S_}}=O\bigl(n^{-1/2}\bigr),\qquad\sum_{i=1}^{n} c(c)^{\prime2}_{ni}\leq C\sum_{i=1}^{n}\波浪线{x}_{i} ^{2}/S_{n}^{4}=O\bigl(n^{-1}\biger)(t_{i},S)\bigr\vert\,ds\biggr)\biggl(S_{n}^{-2}\sum_{i=1}^{n}\vert\tilde{x}_{i} \vert\Biggr)=O\bigl(2^{m}/n\bigr)=O\bigl(n^{-1/2}\bigr),\\begin{aligned}\sum_{j=1}^{n} d日^{\prime2}_{nj}&\leq CS_{n}^{-2}\sum_{j=1}^{n}\sum_{i=1}^}\n}\biggl(int_{A_{j}}E_{m}(t_{i},s)\,ds\biggr)^{2}_{i} \Biggr)\\&\leq C S_{n}^{-2}\sum_{i=1}^{n}\sum_{j=1}^}n}\biggl(int_{A{j}}E_{m}(t_{i},S)\,ds\Biggr)^{2}=O\bigl(2^{m}/n\biger)=O\bigl(n^{-1/2}\bigr)。\结束{对齐}\结束{聚集}$$
因此,应用引理A.1款和采取\(α=4\),我们得到\(C_{in}=o(n^{-1/4})\)美国。,\(i=1,2).至于\(C_{3n}\),由\(2^{m}/n=O(n^{-1/2})\), (2.6)、和(2.7)我们很容易看到
$$\vert C_{3n}\vert\leq\Bigl(\max_{1\leqi\leqn}\vert_tilde{g}_{i} \vert\Bigr)\cdot\Biggl(S_{n}^{-2}\sum_{i=1}^{n}\vert\tilde{x}_{i} \vert\Biggr)=O\bigl(2^{-m}+n^{-1}\bigr)=O\bigle(n^{-1-4}\biger)$$
第2步。我们证明了(3.3)(ii)。注意到\(\帽子{f}_{n} (u)=\sum_{i=1}^{n}[\波浪线{x}_{i} (\beta-\hat{\beta}_{L})+\波浪线{g}_{i} +\tilde{\varepsilon}_{i}]^{2}\int_{B_{i{}E_{m}(u,s)\,ds\),我们可以看到
$$\开始{aligned}&\max_{1\leq-j\leq-n}\bigl\vert\hat{f}_{n} (u{j})-f(u{j})\bigr\vert\\&\quad\leq\max_{1\leqj\leqn}\Biggl\vert\sum_{i=1}^{n}\varepsilon^{2}_{i} \nint_{B_{i}}E_{m}(u_{j},s)\,ds-f{n}\varepsilon_{j}\int_{A{j}}E_{m}(t_{i},s)\,ds\Biggr)\Biggr\vert\\&\qquad{}+\max_{1\leqj\leqn}\Biggl\vert\sum_{i=1}^{n},ds\Biggl(sum_{j=1}^{n}\varepsilon_{j}\int_{A{j}}E_{m}(t_{i},s),ds\Bigr){g}_{i} int_{B_{i}}E_{m}(u_{j},s),ds\Biggr\vert\&\qquad{}+2\vert\beta-\hat{beta}_{L}\vert\max_{1\leqj\leqn}\Biggl\vert\sum_{i=1}^{n}\ tilde{x}_{i} tilde{varepsilon}{i}\int_{B_{i}E_{m}(u_{j},s),ds\Biggr\vert+(\beta-\hat{beta}_{L})^{2}\max_{1\leqj\leqn}\Biggl\vert\sum_{i=1}^{n}\ tilde{x}_{i} ^{2}\int_{B_{i}}E_{m}(u_{j},s){x}_{i} \波浪线{g}_{i} \int_{B_{i}}E_{m}(u_{j},s)\,ds\Biggr\vert+\max_{1\leqj\leqn}\Biggl\vert\sum_{i=1}^{n}\tilde{g}^{2}_{i} \ int _{B_{i}}E_{m}(u_{j},s)\,ds\Biggr\vert\\&&\ quad:=\sum_{i=1}^{8} 天_{in}。\结束{对齐}$$
至于\(D_{1n}\),我们有
$$\开始{对齐}D_{1n}\leq{}&\max_{1\leqj\leqn}\Biggl\vert\sum_{i=1}^{n} (f)(u{i})\bigl(e_{i{^{2}-1\大)\int_{B_{i}}E_{m}(u_{j},s)\,ds\Biggr\vert\\&+\max_{1\leqj\leqn}\Biggl\vert\sum_{i=1}^{n} (f)(u{i})\int_{B_{i}}E_{m}(u{j},s)\,ds-f(u}j})\ Biggr\vert\=:{}&D{11n}+D{12n}。\结束{对齐}$$
请注意\(Ee_{i}^{2}=1\),所以\(e_{i}^{2}-1=[(e^{+}_{i})^{2} -E类(e^{+}{i})^{2}]+[(e)^{-}_{i} )^{2} -E类(e)^{-}_{i} )^{2}]:=\xi{i{1}}+\xi{2}}\)、和
$$D_{11n}\leq\max_{1\leqj\leqn}\Biggl\vert\sum_{i=1}^{n}\Biggl}\Biggl(f(u{i})\int_{B_{i}}E_{m}(u{j},s)\,ds\biggr)\xi_{i{2}}\biggr\vert$$
自\({\xi{i{1}},i\geq1\}\)和\({\xi{i{2}},i\geq1\}\)是平均值为零的NA随机变量,\(\sup_{i} 电子|\xi_{i_{j}}|^{p/2}\leq C\sup_{i} E类|e_{i}|^{p}<\infty\),\(j=1,2).通过(A2)(ii)和引理答6我们有
$$\max_{1\leqi,j\leqn}\biggl\vertf(u_{i})\int_{B_{i{}E_{m}(u{j},s)\,ds\biggr\vert=O\bigl(2^{m}/n\biger)=O\bigl(n^{-1/2}\bigr)$$
和
$$\max_{1\leq_j\leq-n}\sum_{i=1}^{n}\biggl\vert f(u_{i})\int_{B_{i{}E_{m}(u_j},s)\,ds\biggr\vert=O(1)$$
因此\(D_{11n}\右箭头0\)莱玛的a.sA.2款.签署人(2.7)我们有\(|D_{12n}|=O(2^{-m}+n^{-1})\),所以\(D_{1n}\rightarrow0\)美国。
请注意
$$\vert D_{2n}\vert\leq2\Biggl(\max_{1\leq i\leq n}\Biggl\vert\sum_{j=1}^{n}\sigma)_{j} e(电子)_{j} \int_{A_{j}}E_{m}(t_{i},s)\,ds\Biggr\vert\Biggr)\Biggl(\max_{1\leqj\leqn}\Biggl\vert\sum_{i=1}^{n}\sigma_{i} e(电子)_{i} \int_{B_{i}}E_{m}(u_{j},s)\,ds\Biggr\vert\Biggr)$$
类似于\(D_{11n}\),我们获得\(D_{2n}\rightarrow0\)莱玛的a.sA.2款.
应用假设,从(2.6), (2.7), (3.3)(i) ,引理A.2款、和引理答6由此可见
$$开始{aligned}&\vertD_{3n}\vert\leq\max_{1\leqi\leqn}\Biggl(\sum_{j=1}^{n}\sigma_{j} e(电子)_{j} \int_{A_{j}}E_{m}(t_{i},s)\,ds\Biggr)^{2}\max_{1\leqj\leqn}\sum_{i=1}^{n}\biggl\vert\int_{B_{i}}E_}leq{}&2\Bigl(\max_{1\leqi\leqn}\vert\tilde{g}_{i} \vert\Bigr)\cdot\Biggl(\max_{1\leq-j\leq-n}\Biggl\vert\sum_{i=1}^{n}{\varepsilon}_{i}\int_{B_{i{}}E_{m}(u_{j},s)\,ds\Biggr\vert\Biggr)+2\Bigl(\max_1\leq-i\leqn-}\vert\tilde{g}_{i} \vert\Bigr)\\&\cdot\Biggl(\max_{1\leqi\leqn}\Biggl\vert\sum_{k=1}^{n}\varepsilon_{k}\int_{A_{k}}E_{m}{i}}E_{m}(u_{j},s)\,ds\Biggr\vert\Biggr)\\rightarrow{}&0\quad\mbox{A.s.},\end{aligned}\\&\vertD_{6n}\vert\leq(\beta-\hat{\beta}_{L})^{2}\cdot\biggl(\max_{1\leqi,j\leqn}\int_{B_{i}}\bigl\vertE_{m}(u_{j},s)\bigr\vert\,ds\biggr)\biggl(\sum_{i=1}^{n}\tilde{x}_{i} ^{2}\Biggr)=o(1)\quad\mbox{a.s.},\\&\vert D_{7n}\vert\leq2\vert\beta-\hat{beta}_{L}\vert_cdot\Bigl(\max_{1\leqi\leqn}\vert\tilde{g}_{i} \vert\Bigr)\cdot\biggl(\max_{1\leqi,j\leqn}\int_{B_{i}}\bigl\vertE_{m}(u_{j},s)\Bigr\vert\,ds\biggr)\biggl(\sum_{i=1}^{n}\vert\tilde{x}_{i} \vert\Biggr)\rightarrow0\quad\mbox{a.s.},\\&\vert D_{8n}\vert\leq\Bigl(\max_{1\leqi\leqn}\vert_tilde{g}_{i} \vert\Bigr)^{2}\cdot\Biggl(\max_{1\leqj\leqn}\sum_{i=1}^{n}\Biggl\vert\int_{B_{i}}E_{m}(u_{j},s)\,ds\biggr\vert\biggr)\rightarrow0,\\&\vertD_{5n}\vert\leq2\vert\beta-\hat{beta}_{L}\vert\cdot\ Biggl{1\leqj\leqn}\sqrt{\Biggl(\sum_{i=1}^{n}\波浪线{x}^{2}_{i} \Biggr)\sum_{i=1}^{n}\tilde{\varepsilon}^{2}_{i} \biggl(\int_{B_{i}}E_{m}(u_{j},s)\,ds\biggr)^{2}}\biggr)。\结束{对齐}$$
为了证明这一点\(D_{5n}\右箭头0\)美国,这足以表明
$$\max_{1\leqj\leqn}\sum_{i=1}^{n}\tilde{\varepsilon}^{2}_{i} \biggl(\int_{B_{i}}E_{m}(u_{j},s)\,ds\biggr)^{2}=O\bigl(n^{-1/2}\biger)\quad\mbox{a.s.}$$
(4.3)
至于(4.3),我们可以分开
$$\begin{collected}\max_{1\leqj\leqn}\sum_{i=1}^{n}\tilde{\varepsilon}^{2}_{i} \biggl(\int_{B_{i}}E_{m}(u_{j},s)\,ds\biggr)^{2}\\quad\leq\max_{1\leq j\leq n}\sum_{i=1}^{n}{\varepsilon}^{2}_{i} \biggl(int_{B_{i}}E_{m}(u_{j},s)\,ds\biggr biggl(sum_{k=1}^{n}{\varepsilon}_{k}\int_{A{k}}E_{m}(t_{i},s),ds\biggr)\biggl(\int_{B_{i}}E_{m}(u_{j},s)\,ds\biggr)^{2}\biggr)\cdot\biggl(\sum_{k=1}^{n}{\varepsilon}_{k}\int_{A_{k}}E_{m}(t_{i},s)\,ds\biggr)^{2}\\quad:=D_{51n}+D_{52n}+D_{53n}。\结束{聚集}$$
通过引理A.2款和答6,自\(2^{m}/n=O(n^{-1/2})\),我们有
$$\begin{aligned}&\begin}aligned{}\vertD_{52n}\vert\leq{}&2\Biggl(\max_{1\leqi\leqn}\Biggl\vert\sum_{k=1}^{n}{\varepsilon}_{k}\int_{A{k}}E_{m}(t_{i},s)\,ds\Biggr\vert\Biggr)\cdot\Biggl(\ max_{1\leqi,j\ leqn}\int_{B_{i}}\bigl\vertE_{m}(u_{j},s)\bigr\vert\,ds\Biggr)\\&\cdot\Biggl(\max_{1\leqj\leqn}\Biggl\vert\sum_{i=1}^{n} {\varepsilon}_{i}\nint_{B_{i{}E_{m}(u_{j},s),ds\Biggr\vert\Biggr)\\={}&o\bigl(n^{-1/2}\biger)\quad\mbox{a.s.},\end{aligned}\\&\begin{alinged}\vertD_{53n}\vert\leq{}&\Biggl(max_{1\leqi\leqn}\Biggl\ vert\sum_{k=1}^{n}{\varepsilon}_{k}\int_{a{k}}E_{m}(t_{i},s)\,ds\Biggr\vert^{2}\Biggr)\cdot\Biggl(\max_{1\leqi,j\leqn}\int_{B_{i}}\bigl\vertE_{m}(u_{j},s)\bigr\vert\,ds\biggr)\&\cdot\Biggl(\max_{1\leqj\leqn}\sum_{i=1}^{n}\Biggl\vert\int_B_{i}}E_{m{(u{j},s)\,ds\biggr\vert\biggr)\\={}&o\bigl(n^{-1/2}\biger)\quad\mbox{a.s.}\end{aligned}\end{aligned}$$
请注意
$$开始{aligned}\vertD_{51n}\vert\leq{}&\max_{1\leqj\leqn}\Biggl\vert\sum_{i=1}^{n}\sigma^{2}_{i} \xi_{i_{1}}\biggl(\int_{B_{i}}E_{m}(u_{j},s)\,ds\biggr)^{2}\biggr\vert+\max_{1\leqj\leqn}\Bigl\vert\sum_{i=1}^{n}\sigma^{2}_{i} \xi_{i_{2}}\biggl(\int_{B_{i}}E_{m}(u_{j},s)\,ds\biggr)^{2}\biggr\vert\\&+\max_{1\leqj\leqn}\Bigl\vert\sum_{i=1}^{n}\sigma^{2}_{i} Ee公司^{2}_{i} \biggl(int_{B_{i}}E_{m}(u_{j},s)\,ds\biggr)$$
哪里\(\xi{i{1}}=[(e^{+}{i})^{2} -E类(e^{+}_{i})^{2}]\)和\(xi{i{2}}=[(e^{-}_{i} )^{2} -E类(e)^{-}_{i} )^{2}]\)是具有零均值的NA随机变量。类似于\(D_{11n}\),我们得到\(D'_{51n}=o(n^{-1/2})\)a.s.和\(D''_{51n}=o(n^{-1/2})\)a.s.由引理A.2款另一方面,
$$\bigl\vert D''{51n}\bigr\vert\leq\biggl(\max_{1\leqj\leqn}\sigma^{2}_{i} \int_{B_{i}}\bigl\vertE_{m}(u_{j},s)\bigr\vert\,ds\biggr)\Biggl$$
因此我们证明了\(D_{51n}=O(n^{-1/2})\)a.s.这就完成了(3.3)(ii)。
步骤3。接下来,我们证明(3.3)(iii)通过(A2)(ii)和(3.3)(ii)。何时n个足够大,很容易理解
$$0\leqm'_{0}\leq\min_{1\leqi\leqn}\hat{f}_{n} (u{i})\leq\max{1\leqi\leqn}\hat{f}_{n} (u_{i})\leq M'_{0}\leq\infty$$
(4.4)
\(C_{5}\leqW_{n}^{2}/n\leqC_{6}\)、和\(W^{-2}_{n} \sum_{i=1}^{n}|a_{ni}\波浪线{x}_{i} |\leq C\)。因此,我们有
$$\vert\tilde{\beta}_{无}-\β\vert\leq\frac{n}{W_{n}^{2}}\Biggl\vert\frac{1}{n}\sum_{i=1}^{n}a_{ni}\波浪线{x}_{i} \ tilde{\varepsilon}_{i}\Biggr\vert+\frac{1}{W_{n}^{2}}\Bigl\vert\sum_{i=1}^{n}a_{ni}\ tilde{x}_{i} \波浪线{g}_{i} \Biggr\vert=:\frac{n}{W_{n}^{2}}E_{1n}+E_{2n}$$
(4.5)
与一起(2.7)和(4.5),我们得到
$$\begin{collected}\vertE_{2n}\vert\leq\Bigl(\max_{1\leqi\leqn}\vert_tilde{g}_{i} \vert\Bigr)\cdot\Biggl(W_{n}^{-2}\sum_{i=1}^{n}\verta_{ni}\tilde{x}_{i} \vert\Biggr)\rightarrow0,\\开始{aligned}\vertE_{1n}\vert\leq{}&\Biggl\vert\frac{1}{n}\sum_{i=1}^{n}(a_{ni}-a_{i} )\波浪号{x}_{i} \tilde{\varepsilon}_{i}\Biggr\vert+\Biggl\vert\frac{1}{n}\sum_{i=1}^{n}a_{i}\tilde{x}_{i} \ tilde{\varepsilon}_{i}\Biggr\vert\\\leq{}&\Biggl\vert\frac{1}{n}\sum_{i=1}^{n}(a_{ni}-a_{i} )\波浪号{x}_{i} {\varepsilon}_{i}\Biggr\vert+\Biggl\vert\frac{1}{n}\sum_{i=1}^{n}(a_{ni}-a_{i} )\波浪号{x}_{i} Biggl(sum_{j=1}^{n}\varepsilon_{j}\int_{A{j}}E_{m}(t_{i},s)\,ds\Biggr){x}_{i} \ tilde{\varepsilon}_{i}\Biggr\vert\\=:{}&E_{11n}+E_{12n}+E2{13n}。\结束{对齐}\结束{聚集}$$
我们从引理中知道A.2款那个
$$\开始{对齐}[b]n^{-1}\sum_{i=1}^{n} 电子_{i} ^{2}&=n^{-1}\sum_{i=1}^{n}\bigl[\bigl(e^{+}_{i}\biger)^{2} -E类\bigl(e^{+}_{i}\bigr)^{2}\biger]+n^{-1}\sum_{i=1}^{n}\bigl[\bigl^{-}_{i} \更大)^{2} -E类\比格尔^{-}_{i} \bigr)^{2}\bigr]+n^{-1}\sum_{i=1}^{n} Ee公司^{2}_{i} \\&=O(1)\quad\mbox{a.s.}\end{aligned}$$
(4.6)
应用引理A.2款并结合(4.4)–(4.6)带有(3.3)(ii),我们获得
$$\begin{collected}\vertE_{11n}\vert\leq\biggl(\max_{1\leqi\leqn}\frac{\vert\hat{f}_{n} (u{i})-f(u{i})\vert}{\hat{f}_{n} (u{i})f(u{i})}\biggr)\sqrt{\Biggl(\frac{1}{n}\sum{i=1}^{n}\波浪线{x}^{2}_{i} \Biggr)\Biggl(\frac{1}{n}\sum_{i=1}^{n}{e}^{2}_{i} \Biggr)}\rightarrow0\quad\mbox{a.s.},\\开始{aligned}\vert E_{12n}\vert&\leq\biggl(\max_{1\leqi\leqn}\frac{\vert\hat{f}_{n} (u{i})-f(u{i})\vert}{\hat{f}_{n} (u{i})f(u{i})}\biggr)\cdot\Biggl(\frac{1}{n}\sum_{i=1}^{n}\vert\tilde{x}_{i} \vert\Biggr)\cdot\Biggl(\max_{1\leq i\leq n}\Biggl\vert\sum _{j=1}^{n}\varepsilon\{j}\int _{A_{j}}E_{m}(t_{i},s)\,ds\Biggr\vert\Biggr)\\&&\rightarrow0\quad\mbox{A.s.}\end{aligned}\end{collected}$$
至于\(E_{13n}\),我们有\(E_{13n}=\压裂{T_{n}^{2}}{n}(|A{1n}+A{2n}|)\右箭头0\)因此\(E_{1n}\右箭头0\)美国。 □
证明(3.4)与的相似(3.2)(i) ,因此我们省略了它。
我们只是证明(3.5)(i) ,作为(3.5)(ii)类似。从定义\({\beta}_{L}\)我们有
$$\开始{aligned}[b]S_{n}^{2}(\hat{\beta}_{左}-\β)&=\sum_{i=1}^{n}\波浪线{x}_{i} \西格玛_{i} e(电子)_{我}-\sum_{i=1}^{n}\波浪线{x}_{i} \Biggl(\sum_{j=1}^{n}\西格玛_{j} e(电子)_{j} \nint_{A_{j}}E_{m}(t_{i},s)\,ds\Biggr)+\sum_{i=1}^{n}\波浪线{x}_{i} \波浪线{g}_{i} \\&:=L_{1n}-L_{2n}+L_{3n}。\结束{对齐}$$
(4.7)
设置\(Z_{ni}=frac{tilde{x}_{i} \西格玛_{i} e(电子)_{i} }{\西格玛{1n}}\),我们采用伯恩斯坦的大块和小块程序。让\(y{nm}=sum{i=k{m}}^{k{m{+p-1}Z{ni}),\(y'{nm}=sum{i=l{m}}^{l{m{+q-1}Z{ni}),\(y'{nk+1}=sum{i=k(p+q)+1}^{n} Z_{ni}\),\(k{m}=(m-1)(p+q)+1),\(l{m}=(m-1)(p+q)+p+1),\(m=1,2,\ldot,k).然后
$$\西格玛^{-1}_{1n}左_{1n}:=\波浪{左}_{1n}=\sum_{i=1}^{n}Z_{ni}=\sum _{m=1}^{k} 年_{nm}+\总和{m=1}^{k} 年'{nm}+y'{nk+1}=:L_{11n}+L_{12n}+L_{13n}$$
我们观察到
$$\开始{对齐}L_{2n}&=\sum_{i=1}^{n}\Biggl[\tilde{h}_{i} +伏_{我}-\Biggl(\sum_{k=1}^{n} v(v)_{k} \int_{A_{k}}E_{m}(t_{i},s)\,ds\Biggr)\Biggr]\cdot\Biggl(\sum_{j=1}^{n}\sigma_{j} e(电子)_{j} \ int_{A_{j}}E_{m}(t_{i},s)\,ds\Biggr)\\&=\sum_{i=1}^{n}\波浪线{h}_{i} \Biggl(\sum_{j=1}^{n}\sigma_{j} e(电子)_{j} \nint_{A_{j}}E_{m}(t_{i},s)\,ds\Biggr)+\sum_{i=1}^{n} v(v)_{i} \Biggl(\sum_{j=1}^{n}\西格玛_{j} e(电子)_{j} \int_{A_{j}}E_{m}(t_{i},s)\,ds\Biggr)\\&\quad-\sum_{i=1}^{n}\Biggl(\sum_{k=1}^{n} v(v)_{k} \int_{A_{k}}E_{m}(t_{i},s)\,ds\Biggr)\Biggl(\sum_{j=1}^{n}\sigma_{j} e(电子)_{j} \ int_{A_{j}}E_{m}(t_{i},s)\,ds\Biggr)\\&:=L_{21n}+L_{22n}-L_{23n}。\结束{对齐}$$
所以我们可以写
$$\frac{S_{n}^{2}(\hat{\beta}_{左}-\β)}{\西格玛{1n}}=L_{11n}+L_{12n}+L_{13n}+\西格马{1n{^{-1}(L_{21n}+L_{22n}-L_{23n}+L_{3n})$$
通过应用引理A.4款我们有
$$\开始{对齐}[b]&\sup_{y}\biggl\vert P\biggl(\frac{S_{n}^{2})_{左}-\β)}{\sigma_{1n}}\leqy\biggr}^{1/3}\biger)\\&\qquad{}+P\bigl(\sigma_{1n}^{-1}\vertL_{21n}\vert>\lambda_{3n}^{1/3}\bigr)+P\bigl(\sigma_{1n}^{-1}\vert L_{22n}\vert>\lambda_{4n}^{1/3}\bigr)+P\bigl(\sigma _{1n}^{-1}\vert L_{23n}\vert>\lambda_{4n}^{1/3}\bigr)\\&\qquad{}+\frac{1}{\sqrt{2\pi}}\Biggl(\sum_{k=1}^{3}\lambda_{kn}^{1/3}+2\lambda_{4n}^{1/3}\Biggl gr)+\frac{1}{\sqrt{2 \pi}}\sigma\{1n}^{-1}\vert L_{3n}\vert\\&&quad=\sum_{k=1}^{8} 我_{kn}。\结束{对齐}$$
(4.8)
因此,为了证明(3.5)(i) ,这足以表明\(\sum_{k=2}^{8} 我_{kn}=O(\mu_{1n})和\(I{1n}=O(\upsilon_{1nneneneep+\lambda^{1/2}_{1n}+\lampda^{1/2}_{2n})这里我们需要以下Abel不等式(参见Härdle et al[5]). 让\(A_{1},\ldots,A_{n})和\(B_{1},\ldots,B_{n})(\(B_{1}\geq B_{2}\geq\cdots\geq B_)是两个实数序列,让\(S_{k}=\sum_{i=1}^{k} A类_{i} \),\(M_{1}=\min_{1\leqk\leqn}S_{k}\)、和\(M_{2}=max_{1\leqk\leqn}S_{k}).然后
十亿美元_{1} M(M)_{1} {k=1}^{n} A类_{k} B类_{k} \列克B_{1} M(M)_{2}. $$
(4.9)
请注意
$$\sigma_{1n}^{2}=E\Biggl(\sum_{i=1}^{n}\波浪线{x}_{i} \西格玛_{i} e(电子)_{i} \Biggr)^{2}=\ int _{-\pi}^{\pi}\psi(\omega)\Biggl\vert\sum _{k=1}^{n}\波浪{x}_{k} \西格玛_{k} e(电子)^{-ik\omega}\Biggr\vert^{2}\,d\omega$$
和
$$\varGamma_{n}^{2}(t)=E\Biggl(\sum_{i=1}^{n}\sigma)_{i} e(电子)_{i} \ int_{A_{i}}E_{m}(t,s)\,ds\Biggr)^{2}=\int_{-\pi}^{\pi}\psi(\omega)\Biggl\vert\sum_{k=1}^{n}\sigma_{k}\int_A_{k{}}E_ m}$$
根据(A2)(ii)、(A5)、(2.6)、和引理答6由此可见
$$\开始{对齐}和C_{7}n\leq C_{1}\sum_{i=1}^{n}\波浪线{x}_{i} ^{2}\leq\sigma_{1n}^{2{\leqC_{2}\sum_{i=1}^{n}\波浪线{x}_{i} ^{2}\leq C_{8}n,\end{对齐}$$
(4.10)
$$\开始{对齐}和C_{9}\sum_{i=1}^{n}\biggl(\int_{A_{i}}E_{m}(t,s)\,ds\biggr)^{2}\leq\varGamma_{n}^{2{(t)\leqC_10}\sum _{i=1}^{n}\bigl(int_{A_{i}E_{m}(t,s)\,ds\biggr)^{2}=O\bigl(2^{m}/n\bigr)。\结束{对齐}$$
(4.11)
步骤1。我们首先证明\(\sum_{k=2}^{8} 我_{kn}=O(μ_{1n})\).使用引理答3, (2.7)、和(2.8),来自(A0)(i),(A1)–(A6)(4.10)、和(4.11)由此可见
$$\开始{聚集}我_{2n}\leq\fracc{E\vertL_{12n}\vert^{2}}{\lambda_{1n}^{2/3}}\leq \frac{C}{n\lambda_{1n{2/3}}\sum_{m=1}^{k}\sum_{i=L_{m}}^{L_{m}+q-1}\ tilde{x}_{i} ^{2}\西格玛^{2}_{i} \leq\frac{Ckq}{n\lambda{1n}^{2/3}}\leq C\lambda{1n}^{1/3},\\i_{3n}\leq\frac{E\vert L_{13n}\vert ^{2}}}{\lambda{2n}^{2/3}}\leq\frac{C}{n\lambda{2n}^{2/3}}}\sum _{i=k(p+q)+1}^{n}\tilde{x}_{i} ^{2}\西格玛^{2}_{i} \leq\frac{Cp}{n\lambda_{2n}^{2/3}}\leq-C\lambda_{2n{1/3},\\begin{对齐}I_{4n}&\leq\frac{\sigma{1n}^{-2}东\vert L_{21n}\vert^{2}}{\lambda_{3n}^{2/3}}\leq\frac{C}{n\lambda_{3n{2/3}}\sum_{j=1}^{n}\Biggl(\sum_{i=1}^}n}\tilde{h}_{i} \int_{A_{j}}E_{m}(t_{i},s)\,ds\Biggr)^{2}\sigma^{2}_{j} \\&\leq\frac{C}{n\lambda_{3n}^{2/3}}\Bigl(\max_{1\leqi\leqn}\vert\tilde{h}_{i} \vert^{2}\Bigr)\cdot\sum_{j=1}^{n}\Biggl(\max_{1\leqj\leqn}\sum_{i=1}^}\n}\int_{A_{j}}\bigl\vertE_{m}(t_{i},s)\Bigr\vert\,ds\Biggr)^{2{\\leq\frac{C(2^{-m}+n^{-1})^{2\vertE_m}\lambda_{3n}^{2/3}}\leq C\lambda _{3n}^{1/3},\end{aligned}\\begin{对齐}I_{5n}&\leq\frac{\sigma{1n}^{-2}E\vert L_{22n}\vert^{2}}{\lambda_{4n}^{2/3}}\leq\frac{C}{n\lambda{4n{2/3}}\sum_{j=1}^{n}\Biggl(\sum_{i=1}^{n} v(v)_{i} \int_{A_{j}}E_{m}(t_{i},s)\,ds\Biggr)^{2}\sigma^{2}_{j} \\&\leq\frac{C}{n\lambda_{4n}^{2/3}}\max_{1\leqi,j\leqn}\biggl(int_{A{j}}\bigl\vertE_m}(t_{i},s)\bigr\vert\,ds\biggr)\cdot\max_1\leqi\leqn}\sum_{j=1}^{n}\biggl\vert\int_{A{j}}\bigl\vert E_{m}(t_{i},s)\bigr\vert\,ds\biggr\vert\cdot\biggl(\max_{1\leq-m\leq-n}\biggl\vert\sum_{i=1}^{m} v(v)_{j_{i}}\Biggr\vert\Biggr)^{2}\\&\leq\frac{C(2^{m} n个^{-1}\log^{2} n个)}{\lambda_{4n}^{2/3}}\leq C\lambda_{4n}^{1/3},\end{aligned}\\begin{alinged}I_{6n}&\leq\frac{\sigma_{1n}^{-2}东\vert L_{23n}\vert^{2}}{\lambda_{4n}^{2/3}}\leq\frac{C}{n\lambda_{4n{2/3}}\sum_{j=1}^{n}\Biggl(\sum_{L=1}^{n} v(v)_{l} \int_{A{l}}E_{m}(t_{i},s)^{2}_{j} \\&\leq\frac{C}{n\lambda_{4n}^{2/3}}\max_{1\leqi,j\leqn}\biggl(int_{A_{j}}\bigl\vertE_m}(t_{i},s)\bigr\vert\,ds\biggr)\cdot\max_1\leqi\leqn}\sum_{j=1}^{n}\biggl\vert\int_A_{j}}E_{m}(t_{i},s)\,ds\biggr\vert\&\quad\cdot\biggl(\max_{1\leq-l\leq-n}\sum_{i=1}^{n}\biggl\vert\int_{A{l}}E_{m{,ds\biggr\vert\max_{1\leqm\leqn}\Biggl\vert\sum_{i=1}^{m} v(v)_{j_{i}}\Biggr\vert\Biggr)^{2}\\&\leq\frac{C(2^{m} n个^{-1}\log^{2} n个)}{\lambda_{4n}^{2/3}}\leq C\lambda_{4n}^{1/3}。\结束{对齐}\结束{聚集}$$
至于\(I_{8n}\),我们有
$$\开始{对齐}I_{8n}&=\sigma_{1n}^{-1}\Biggl\vert\sum_{i=1}^{n}\波浪线{x}_{i} \波浪线{g}_{i} \Biggr\vert\leq\frac{C}{\sqrt{n}}\Biggl(\Biggl\vert\sum_{i=1}^{n} v(v)_{i} \波浪号{g}_{i} \Biggr\vert+\Biggl\vert\sum_{i=1}^{n}\波浪线{h}_{i} \波浪线{g}_{i} \Biggr\vert+\Biggl\vert\sum _{i=1}^{n}\波浪号{g}_{i} \sum_{j=1}^{n} v(v)_{j} \nint_{A_{j}}E_{m}(t_{i},s)\,ds\Biggr\vert\Biggr)\\&\leq\frac{C}{\sqrt{n}}\max_{1\leqi\leqn}\vert\tilde{g}_{i} \vert\max_{1\leq-m\leq-n}\Biggl\vert\sum_{i=1}^{m} v(v)_{j_{i}}\Biggr\vert+\frac{Cn}{\sqrt{n}}\max_{1\leqi\leqn}\vert\tilde{h}_{i} \vert\max_{1\leq-i\leq-n}\vert\tilde{g}_{i} \vert\\&\quad+\frac{C}{\sqrt{n}}\max_{1\leqi\leqn}\vert\tilde{g}_{i} \vert\cdot\max_{1\leqj\leqn}\sum_{i=1}^{n}\int_{A_{j}}\bigl\vertE_{m}(t_{i},s)\bigr\vert\,ds\cdot\max_{1\\leqm\leqn}\Biggl\vert\sum_{i=1}^{m} v(v)_{j_{i}}\Biggr\vert\\&\leq C\bigl(2^{-m}+n^{-1}\bigr)\bigle(\log n+\sqrt{n}\bigl.(2^}-m}+n ^{-1{\bigr.)\biger)=C\lambda_{5n}。\结束{对齐}$$
因此,根据之前的估计,我们得出\(\sum_{k=2}^{8} 我_{kn}=O(\mu_{1n}).
第2步。我们核实\(I{1n}=O(λ.让\({eta{nm}:m=1,2,\ldots,k\})是具有相同分布的独立随机变量\(y{nm}\),\(m=1,2,\ldot,k).设置\(H_{n}=\sum_{m=1}^{k}\eta_{nm}\)和\(s_{n}^{2}=\sum_{m=1}^{k}\operatorname{Var}(y_{nm})\).遵循Liang和Li定理2.1的证明方法[27]和Li等人[20],我们很容易看到
$$\boot{aligned}[b]I_{1n}&=\sup_{y}\bigl\vert P(L_{11n}\leq y)-\varPhi(y)\bigr\vert\\&&\leq\sup_{y}\bigl\vert P(L_{11n}\leq y)-P(H_{n}\leq y)\bigr\vert\\&&quad+\sup_{y}\bigl\vert P(H_{n}\leq y)-\varPhi(y/s_{n})\bigr \vert+\sup_{y}\bigl\vert\varPhi(y/s_{n})-\varPhi(y)\bigr\vert\\&:=I_{11n}+I_{12n}+I_{13n}。\结束{对齐}$$
(4.12)
(i) 我们评估\(s_{n}^{2}\).注意到\(s_{n}^{2}=E{L}^{2}_{11n}-2\sum_{1\leqi<j\leqk}\操作符名{Cov}(y_{ni},y_{nj})和\(波浪号{左}_{1n}^{2}=1\),我们可以
$$\bigl\vert E(L_{11n})^{2}-1\bigr\vert\leq C\bigl(\lambda_{1n}^{1/2}+\lambda _{2n}^}_1/2}\bigr)$$
(4.13)
另一方面,从(A1),(A2)(4.10)、和(2.8)由此可见
$$开始{对齐}[b]\biggl\vert\sum_{1\leqi<j\leqk}\operatorname{Cov}(y_{ni},y_{nj}^{k_{j}+p-1}\垂直\波浪线{x}_{s} \波浪线{x}_{t} \sigma_{s}\sigma _{t}\vert\cdot\bigl\vert\operatorname{Cov}(e_{s},e_{t})\bigr\vert\&\leq Ckpn^{-1}u(q) \leq铜(q)。\结束{对齐}$$
(4.14)
因此,从(4.13)和(4.14)由此可见\(|s{n}^{2}-1|\leq C(\lambda_{1n}^{1/2}+\lambda_{2n}^}{1/2{+u(q)).
(ii)应用Berry–Esséen不等式(见Petrov[28],定理5.7),用于\(增量>0),我们得到
$$\sup_{y}\bigl\vert P(H_{n}/s_{n{}\leqy)-\varPhi(y)\bigr\vert\leq C\sum_{m=1}^{k}\bigr(E\vert y_{nm}\vert^{2+\delta}/s_}n}^{2+/delta}\biger)$$
(4.15)
按引理A.3从(A0)、(A1)、(A2)、(4.10)、和(2.8)我们可以推断出
$$\开始{对齐}[b]\sum_{m=1}^{k} 电子\vert y_{nm}\vert ^{2+\delta}&=\sum_{m=1}^{k} E类\Biggl\vert\sum_{j=k_{m}^{k_{m}+p-1}Z_{ni}\Biggr\vert^{2+\delta}\&&\leqC\sigma_{1n}^{-(2+\delta)}\sum_{m=1}^{k}\Biggl\{sum_{i=k_{m}}^{k_{m}+p-1}E\vert\tilde{x}_{i} \西格玛_{i} e(电子)_{i} \vert^{2+\delta}+\Biggl[\sum_{i=k_{m}}^{k_{m}+p-1}E(\波浪线{x}_{i} \西格玛_{i} e(电子)_{i} )^{2}\Biggr]^{1+\delta/2}\bigr\}\\&\leq C\bigl。\结束{对齐}$$
(4.16)
自\(s_{n}\右箭头1\)由(4.13)和(4.14),来自(4.15)和(4.16)我们很容易看到\(I_{12n}\leq C\lambda^{delta/2}_{2n}\)。请注意\(I_{13n}=O(|s_{n}^{2}-1|)=O(\lambda_{1n}^{1/2}+\lambda_{2n}^}{1/2{+u(q)).
(iii)接下来,我们评估\(I_{11n}\).让\(\varphi_{1}(t)\)和\(瓦尔斐{2}(t))是的特征函数\(L_{11n}\)和\(H_{n}\)分别是。因此应用Esséen不等式(见Petrov[28],定理5.3),对于任何\(T>0\),我们有
$$开始{对齐}[b]&\sup_{t}\bigl\vert P(L_{11n}\leqt)-P t}\bigl\vert P(H_{n}\lequ+t)-P。\结束{对齐}$$
(4.17)
从引理答5和(4.14)由此可见
$$\开始{aligned}\bigl\vert\varphi_{1}(t)-\varphi_}(t)\bigr\vert&=\Biggl\vert E\exp\Biggl(\mathrm{i} t吨\sum_{m=1}^{k}y_{nm}\Biggr)-\prod_{m=1{^{k{E\exp{(\mathrm{i} t吨y_{nm})}\Biggr\vert\\&&\leq4t^{2}\sum_{1\leq i<j\leq k}\sum_{s_{1}=k_{i}^{k_{i}+p-1}\sum_{t_{1}=k_{j}^{k_{j}+p-1}\bigl\vert\operatorname{Cov}(Z_{ns_{1}},Z_{nt_{1}})\bigr\vert\\&&\leq4Ct^{2} 单位(q) ,\结束{对齐}$$
这意味着
$$I’_{11n}=\int_{-T}^{T}\biggl\vert\frac{\varphi_{1}(T)-\varphi_{2}(T)}{T}\biggr\vert\,dt\leq-Cu(q)T^{2}$$
(4.18)
因此(4.15)和(4.16)我们有
$$开始{对齐}[b]&\sup_{t}\bigl\vert P(H_{n}\leq t+u)-P u}{s{n}}\biggr)\biggr\vert\\&\qquad+\sup_{t}\bigl\vert P\biggl(\frac{H_{n}{s_{n{}}\leq\frac}{t}{s_n}}\ biggr}\biggr)\biggr\vert+\sup_{t}\biggl\vert\varPhi\biggl(\frac{t+u}{s_{n}}\bigr)-\varPhi\ biggl gr)-\varPhi(t)\biggr\vert+\sup_{t}\biggl\vert\varPhi\biggl(\frac{t+u}{s_{n}}\bigr)-\varPhi\ biggl\biggr\vert\\&\quad\leq C\biggl(\lambda_{2n}^{delta/2}+\biggl\vert\frac{u}{s_{n}}\biggr\vert\biggr)\leq C\ bigl。\结束{对齐}$$
(4.19)
发件人(4.19)由此可见
$$I''{11n}=T\sup_{T}\int_{vert u\vert\leq C/T}\bigl\vert P(H_{n}\leq T+u)-P(H_}n}\leq T)\bigr\vert,du\leq C\bigl(\lambda_{2n}^{delta/2}+1/T\bigr)$$
(4.20)
组合(4.17), (4.18)带有(4.20)和选择\(T=u^{-1/3}(q)\),我们很容易看到\(I_{11n}\leqC(u^{1/3}(q)+\lambda_{2n}^{delta/2}).所以\(I{1n}\leq C(λ.这就完成了定理的证明3.3从步骤1和步骤2开始。 □
在定理中3.3,选择\(p=\floor n^{\theta}\floor),\(q=地板n^{2\theta-1}\rfloor),\(增量=1),何时\(1/2<theta\leq7/10),我们有\(\mu_{1n}=O(n^{-(θ-1)/3})和\(upsilon_{1n}=O(n ^{-(θ-1)/3}).因此(3.6)直接遵循定理3.3. □
我们只证明了\({g}(t)={g}_{L} (t)\),作为\({g}(t)={g}_{W} (t)\)是类似的。
根据的定义\(\帽子{g}_{五十} (t)\)我们很容易看到
$$\开始{aligned}\varGamma_{n}^{-1}(t)\bigl(\hat{g}_{五十} (t)-E\hat{g}_{五十} (t)\bigr)={}&\varGamma_{n}^{-1}^{n} x个_{i} (E\hat{\beta}_{左}-\beta)\nint_{A_{i}}E_{m}(t,s)\,ds\Biggr)\\&+\varGamma_{n}^{-1}(t)\Biggl(\sum_{i=1}^{n} x个_{i} (\beta-\hat{\beta}_{L})\int_{A{i}}E_{m}(t,s)\,ds\Biggr)\\:={}&J{1n}+J{2n}+J{3n}。\结束{对齐}$$
设置\(Z'_{ni}=frac{sigma_{i} e(电子)_{i} \int_{A_{i}}E_{m}(t,s)\,ds}{\varGamma_{n}(t)}\)。类似于\(\波浪号{左}_{1n}\),我们可以分开\(J_{1n}\)作为\(J{1n}=sum{i=1}^{n}Z'{ni}:=J{11n}+J{12n}+J{13n}),其中\(J{11n}=sum{m=1}^{k}chi{nm}),\(J{12n}=sum{m=1}^{k}\chi'{nm}),\(J{13n}=chi'{nk+1}),\(chi{nm}=sum{i=k{m}}^{k{m{+p-1}Z'{ni}),\(chi'{nm}=sum{i=l{m}}^{l{m{+q-1}Z'{ni}),\(chi'{nk+1}=sum{i=k(p+q)+1}^{n} Z'_{ni}\),\(k{m}=(m-1)(p+q)+1),\(l{m}=(m-1)(p+q)+p+1),\(m=1,2,\ldot,k).
应用引理A.4款,我们有
$$\开始{对齐}[b]&\sup_{y}\biggl\vert P\biggl(\frac{\hat{g}_{五十} (t)-E\hat{g}_{五十} (t)}{\varGamma_{n}(t){\leqy\biggr)-\varPhi(y)\biggr\vert\\&\quad\leq\sup_{y}\bigl\vert P(J_{11n}\leqy)-\valPhi(y)\bigr\vert+P\bigl(\vert J_{12n}\vert>\gamma_{1n}^{1/3}\biger)+P\bigl{2n}^{1/3}\biger)\\&\qquad+\frac{vertJ{2n{vert}{sqrt{2\pi}}+P\bigl(vertJ}3n}\vert>\gamma{3n}^}(2+\delta)/(3+\delta)}\bigr)+\frac{1}{\sqrt{2\pi}}\Biggl(\sum_{k=1}^{2}\gamma_{kn}^{1/3}+\gamma_{3n}^{\frac{2+\delta}{3+\delta}}\Biggr)\\\quad=\sum_{k=1}^{6} 克_{kn}。\结束{对齐}$$
(4.21)
因此,这足以表明\(\sum_{k=2}^{6} G公司_{kn}=O(\mu_{2n})和\(G_{1n}=O(伽玛^{1/2}_{1n}+\gamma^{1/2}_{2n}+\upsilon_{2n}).
步骤1。我们首先证明\(\sum_{k=2}^{6} 克_{kn}=O(\mu_{2n}).类似于\(一)_{2n}-I_{8n}\)在定理中3.3,我们有
$$\开始{聚集}G_{2n}\leq\frac{E\vert J_{12n}\vert^{2}}{\gamma_{1n}^{2/3}}\leq \frac}{C}{\varGamma_{n}^}2}(t)\gamma_{1n{2/3}}\sum_{m=1}^{k}\sum_{i=l_{m}}^{l_{m}+q-1}\biggl(int_{A_{i}}E_{m}(t,s)\,ds\biggr)^{2}\sigma^{2}_{i} \leq\frac{Ckq2^{m}}{n\gamma{1n}^{2/3}}\leq C\gamma_{1n{1/3},\\G{3n}\leq\ frac{E\vert J{13n}\vert^{2}}{gamma{2n}^2,3}}\leq\frac{C}{varGamma_{n}^2}(t)\gamma_2n}^{2/3}}\sum_{i=k(p+q)+1}^{n}\biggl(int_{A{i}}E_{m}(t,s)\,ds\biggr)^{2}\sigma^{2}_{i} \leq\frac{C2^{m}p}{n\gamma_{2n}^{2/3}}\leq-C\gamma_2n}^}{1/3}。\结束{聚集}$$
注意,如果\(\xi_{n}\rightarrow\xi\sim n(0,1)\),然后\(E|\xi_{n}|\rightarrow E|\xi|=\sqrt{2/\pi}\)和\(E|\xi_{n}|^{2+\delta}\rightarrow E|\xi|^{2+\ delta}\).根据定理3.3(i) 和(2.6)由此可见
$$\begin{aligned}&\vert\beta-E\hat{beta}_{L}\vert\leq E\vert\beta-\hat{beta{{L}\ vert=O\bigl(\sigma{1n}/S_{n}^{2}\bigr)=O\bigl(n^{-1/2}\biger),\end{alinged}$$
(4.22)
$$\begin{aligned}&E\vert\beta-\hat{\beta}_{L}\vert^{2+\delta}\leq O\bigl(\bigle(\sigma_{1n}/S_{n}^2}\bigr)^{2+/delta}\ bigr。\结束{对齐}$$
(4.23)
因此,应用Abel不等式(4.9)并将(A1)(iii)和(A2)(i)与引理结合答6,来自(4.22)和(4.23)我们得到
$$\beart{aligned}[b]\vert G_{4n}\vert&=\frac{1}{\sqrt{2\pi}\varGamma_{n}(t)}\cdot\vert\beta-E\hat{\beta}_{L}\vert\cdot\Biggl\vert\sum_{i=1}^{n} x个_{i} \int_{A_{i}}E_{m}(t,s)\,ds\Biggr\vert\\&\leq C\varGamma_{n}^{-1}(t)n^{-1/2}\Biggl vert\,ds\cdot\max_{1\leql\leqn}\Biggl\vert\sum_{i=1}^{l} v(v)_{j_{i}}\Biggr\vert\Biggr)\\&\leq C\bigl(2^{-m/2}+\sqrt{2^{m}/n}\log n\bigr)=C\gamma_{3n}\end{aligned}$$
(4.24)
和
$$开始{对齐}[b]\vert J_{3n}\vert^{2+\delta}&=\varGamma^{-(2+\delta)}_{n}(t)E\vert\beta-\hat{beta}_{L}\vert ^{2+/delta}\Biggl\vert\sum_{i=1}^{n} x个_{i} \int_{A_{i}}E_{m}(t,s)\,ds\Biggr\vert^{2+delta}\\&\leq C\varGamma_{n}^{-(2+delta)}l\vert E_{m}(t,s)\bigr\vert\,ds\cdot\max_{1\leq l\leq n}\Biggl\vert\sum_{i=1}^{l} v(v)_{j_{i}}\Biggr\vert\Biggr)^{2+\delta}\hsspace{-24pt}\\&\leq C\gamma^{2+/delta}_{3n},\end{aligned}$$
(4.25)
这意味着\(G{5n}\leqc\gamma{3n}^{(2+δ)/(3+δ)}).所以我们得到\(\sum_{k=2}^{6} G公司_{kn}=O(\mu_{2n}).
第2步。我们核实\(G{1n}=O(\gamma{1n{^{1/2}+\gamma_{2n}^{1/2}+\upsilon_{2n}).让\({\zeta{nm}:m=1,2,\ldots,k\})是独立的随机变量\(泽塔{nm})具有与相同的分布\(\chi_{nm}\),\(m=1,2,\ldot,k).设置\(T_{n}=\sum_{m=1}^{k}\ζ_{nm}\)和\(t_{n}^{2}=\sum_{m=1}^{k}\操作符名{Var}(\chi_{nm})\).类似于(4.17),我们很容易看到
$$开始{对齐}[b]G_{1n}&=\sup_{y}\bigl\vert P(J_{11n}\leqy)-\varPhi(y)\bigr\vert\\&\leq\sup_y}\bigl\vertP(J_{11n}\leq y)-P/T_{n})\bigr\vert+\sup_{y}\bigl\vert\varPhi。\结束{对齐}$$
(4.26)
类似于(4.13)–(4.20),我们可以获得\(|t{n}^{2}-1|\列q C(\gamma_{1n}^{1/2}+\gamma_2n}^}1/2}+u(q)),\(|G_{12n}|\leq C\gamma_{2n}^{\delta/2}\),\(|G_{13n}|\leq C(\gamma_{1n}^{1/2}+\gamma_2n}^_1/2}+u(q))、和\(|G_{11n}|\leq C\upsilon _{2n}\)。由此可见\(G{1n}=O(\gamma{1n{^{1/2}+\gamma_{2n}^{1/2}+\upsilon_{2n}).定理证明3.4已完成。 □
出租\(p=\lfloor n^{\rho}\rfloor),\(q=地板n^{2\rho-1}\rfloor),\(增量=1),何时\(1/2<ρ<θ<1),我们有\(\gamma^{1/3}_{1n}=O(n^{-(θ-\rho)/3}),\(\gamma^{1/3}_{2n}=O(n^{-(θ-\rho)/3}),\(伽玛^{3/4}_{3n}=O(n^{-3(1-\θ)/8})、和\(u^{1/3}(q)=O(n^{-(θ-\rho)/3}).因此(3.8)直接遵循定理3.4. □