我们需要以下引理。
引理2.1
([7])
让
\(在n}中为X,X_{n})
是一个严格稳定的序列
\(\rho^{-}\)-将随机变量与
\(\mathrm{E} X(X)_{1}=0\),\(0<\mathrm{E} X(X)_{1} ^{2}<\infty\),\(西格玛{1}^{2}=\mathrm{E} X(X)_{1} ^{2}+2\sum_{k=2}^{\infty}\operatorname{Cov}(X_{1},X_{k})>0\)
和
\(\sum_{k=2}^{\infty}|\操作员名称{Cov}(X_{1},X_{k})|<\infty),然后,对于
\(0<p<2),我们有
$$\frac{S_{n}}{n^{\frac{1}{p}}\rightarrow0,\quad\textit{a.S.},n\rightarrow\infty$$
引理2.2
([9])
让
\(在n}中为X,X_{n})
是一系列
\(\rho^{-}\)-混合随机变量,具有
$$\mathrm美元{E} X(X)_{n} =0,\qquad\mathrm{E}|X_{n}|^{q}<\infty,\quad\forall{n}\geq1,q\geq2$$
那么就有一个正常数
\(C=C(q,\rho^{-}(\cdot))
仅取决于
q个
和
\(\rho^{-}(\cdot)\)
这样的话
$$\mathrm{E}\Bigl(最大{1\leq{j}\leq{n}}|S_{j}|^{q}\Bigr){E} X(X)_{i} ^{2}\Biggr)^{\frac{q}{2}}\Bigr\}$$
引理2.3
([10])
假设
\(f_{1}(x)\)
和
\(f{2}(y)\)
都是真的,有界的,上的绝对连续函数
R(右)
具有
\(|f'{1}(x)|\leq C_{1}\)
和
\(|f'{2}(y)|\leq C_{2}\),然后,对于任何随机变量
X(X)
和
Y(Y),
$$\bigl\vert\operatorname{Cov}\bigl(f_{1}(X),f_{2}(Y)\biger)\bigr\vert\leq C_{1} C类_{2} \bigl\{-\operatorname{Cov}(X,Y)+8\rho^{-}(X,Y)\|X\|{2,1}\|Y\|{2,1}\bigr\}$$
哪里
\(\|X\|_{2,1}=\int_{0}^{\infty}(P(|X|>X))^{\frac{1}{2}}\,dx\).
引理2.4
让
\(在n}中为{xi,\xi_{n})
是一致有界随机变量序列.如果
\(存在增量>1),\(\rho^{-}(n)=O(\log^{-\delta}n)\),存在常量
\(C>0\)
和
\(\varepsilon>0\),这样的话
$$\vert\mathrm{E}\xi_{k}\xi_{l}\vert\leq C\biggl(\rho^{-}(k)+\biggl(\frac{k}{l}\ biggr)^{varepsilon}\biggr,\quad 1\leq2k<l$$
(3)
然后
$$\lim_{n\rightarrow\infty}\frac{1}{D_{n}}\sum_{k=1}^{n} d日_{k} \xi_{k}=0,\quad\textit{a.s.}$$
证明
参见中定理1的证明[7]. □
引理2.5
如果定理的假设1
持有,然后
$$开始{aligned}&\lim_{n\to\infty}\frac{1}{D_{n}}\sum_{k=1}^{n} d日_{k} \mathrm{I}\biggl[\frac{{T}(T)_{k,k}-\数学{E}({T}(T)_{k,k})}{\beta\delta_{k}\sqrt{k}}\le{x}\biggr]=\Phi(x)\quad\textit{a.s.},对于R中的所有x\,\end{aligned}$$
(4)
$$开始{aligned}&\lim_{n\to\infty}\frac{1}{D_{n}}\sum_{k=1}^{n} d日_{k} \biggl[f\biggl(\frac{\bar{V}_{k,l}^{2}}{k\delta_{k,l}^{2}}\biggr)-\mathrm{E} (f)\biggl(\frac{\bar{V}_{k,l}^{2}}{k\delta_{k,l}^{2}}\biggr)\biggr]=0\quad\textit{a.s.},l=1,2,\end{aligned}$$
(5)
哪里
\(d_{k}\)
和
\(D_{k}\)
定义为(1)和
(f)
是真实的,有界的,R上的绝对连续函数.
证明
首先,我们证明(4),根据的财产\(\rho^{-}\)-混合序列,我们知道\(\{\bar{Y}(Y)_{ni}{n\geq1,i\leqn})是一个\(\rho^{-}\)-混合序列;使用引理2.1英寸[7],条件(a2),(a)三)、和\(测试版>0),\(\delta_{k}^{2}\rightarrow\mathrm{E} Y(Y)^{2}>0\),因此
$$\压裂{{T}(T)_{k,k}-\数学{E}({T}(T)_{k,k})}{\beta\delta_{k}\sqrt{{k}}\stackrel{\mathrm{d}}{\rightarrow}\mathscr{N},\quad k\rightarrow\infty$$
因此,对于任何\(克(x)\)它是一个具有有界连续导数的有界函数,我们有
$$\mathrm美元{E} 克\bigl(\frac{{T}(T)_{k,k}-E({T}(T)_{k,k})}{\beta\delta{k}\sqrt{{k}}\biggr)\rightarrow\mathrm{E} 克(\mathscr{N}),\quad k\rightarrow\infty$$
通过Toeplitz引理,我们得到
$$\lim_{n\to\infty}\frac{1}{D_{n}}\sum_{k=1}^{n} d日_{k} \mathrm{E}\biggl[g\biggl(\frac{{T}(T)_{k,k}-\数学{E}({T}(T)_{k,k})}{\beta\delta_{k}\sqrt{k}}\biggr)\biggr]=\mathrm{E}\bigl(g(\mathscr{N})\bigr)$$
另一方面,根据[11]和Sect。第2页,共2页[12],我们知道(4)等于
$$\lim_{n\to\infty}\frac{1}{D_{n}}\sum_{k=1}^{n} d日_{k} 克\biggl(\压裂{{T}(T)_{k,k}-\数学{E}({T}(T)_{k,k})}{\beta\delta_{k}\sqrt{k}}\biggr)=\mathrm{E}\bigl(g(\mathscr{N})\bigr)\quad\mbox{a.s.}$$
因此,为了证明(4),这足以证明
$$\lim_{n\to\infty}\frac{1}{D_{n}}\sum_{k=1}^{n} d日_{k} \biggl[g\biggl(\frac{{T}(T)_{k,k}-\数学{E}({T}(T)_{k,k})}{\beta\delta_{k}\sqrt{k}}\biggr)-\mathrm{E}\bigl(g\frac{{T}(T)_{k,k}-\数学{E}({T}(T)_{k,k})}{\beta\delta{k}\sqrt{k}}\biggr)\biggr]=0\quad\mbox{a.s.}$$
(6)
注意到了
$$\xi_{k}=g\bigl(\frac{{T}(T)_{k,k}-\数学{E}({T}(T)_{k,k})}{\beta\delta_{k}\sqrt{k}}\biggr)-\mathrm{E}\bigl(g\biggl(\frac{{T}(T)_{k,k}-\数学{E}({T}(T)_{k,k})}{\beta\delta{k}\sqrt{k}}\biggr)$$
对于每个\(1\leq2k<l\),我们有
$$\begin{aligned}\vert\mathrm{E}\xi_{k}\xi_{l}\vert=&\biggl\vert\operatorname{Cov}\biggl(\frac{{T}(T)_{k,k}-\数学{E}{T}(T)_{k,k}}{\beta\delta{k}\sqrt{k}}\biggr),g\biggl(\frac{{T}(T)_{我,我}-\数学{E}{T}(T)_{l,l}}{\beta\delta_{l}\sqrt{l}}\biggr)\biggr)\biggr\vert\\leq&\biggl\vert\operatorname{Cov}\biggl(g\biggl(\frac{{T}(T)_{k,k}-\数学{E}{T}(T)_{k,k}}{\beta\delta{k}\sqrt{k}}\biggr),g\biggl(\frac{{T}(T)_{我,我}-\数学{E}{T}(T)_{l,l}}{\beta\delta{l}\sqrt{l}}\biggr)-g\biggl(\frac{{T}(T)_{我,我}-\数学{E}{T}(T)_{我,我}-({T}(T)_{2k,l}-\数学{E}{T}(T)_{2k,l})}{\beta\delta{l}\sqrt{l}}\biggr)\biggr\vert\\&{}+\biggl\vert\operatorname{Cov}\bigl(g\biggl(\frac{{T}(T)_{k,k}-\数学{E}{T}(T)_{k,k}}{\beta\delta{k}\sqrt{k}}\biggr),g\biggl(\frac{{T}(T)_{我,我}-\数学{E}{T}(T)_{我,我}-({T}(T)_{2k,l}-\数学{E}{T}(T)_{2k,l})}{β\delta_{l}\sqrt{l}}\biggr)\biggr)\biggr\vert\\=&I_{1}+I_{2}。\结束{对齐}$$
(7)
首先我们估计\(I_{1}\); 我们知道这一点克是一个有界Lipschitz函数,即存在一个常数C类这样的话
$$\bigl\vert g(x)-g(y)\bigr\vert\leq C|x-y|$$
对于任何\(R中的x,y\),自\(\{\bar{Y}(Y)_{ni}{n\geq1,i\leqn})也是一个\(\rho^{-}\)-混合序列;我们使用条件\(\delta_{l}^{2}\rightarrow\mathrm{E}(Y^{2{)<\infty\),\(l\rightarrow\infty\)、和引理2.2,以获取
$$\开始{对齐}I_{1}\leq&C\frac{\mathrm{E}|{T}(T)_{2k,l}-\数学{E}{T}(T)_{2k,l}|}{\sqrt{l}}\leqC\frac{\sqrt{\mathrm{E}({T}(T)_{2k,l}-\数学{E}{T}(T)_{2k,l})^{2}}{\sqrt{l}}\\leq&\frac{C}{\sqlt{l}{\sqrt{i=1}^{2k}\mathrm{E}\bar{Y}(Y)_{l,i}^{2}}\leq\frac{C}{\sqrt{l}}\sqrt{\sum_{i=1}^{2k}\mathrm{E} Y(Y)^{2} }\leq C\biggl(\frac{k}{l}\biggr)^{\frac{1}{2}}。\结束{对齐}$$
(8)
接下来我们估计\(I_{2}\); 通过引理2.2,我们有
$$\开始{aligned}\operatorname{Var}\biggl(\frac{{T}(T)_{k,k}-\数学{E}{T}(T)_{k,k}}{\beta\delta_{k}\sqrt{k}}\biggr)和\leq\frac{C}{k}\operatorname{Var}({T}(T)_{k,k}-\数学{E}{T}(T)_{k,k})\\&\leq\frac{C}{k}\sum_{i=1}^{k}\mathrm{E}(\bar{Y}(Y)_{ki}-\数学{E}\bar{Y}(Y)_{ki})^{2}\leq\frac{C}{k}\sum_{i=1}^{k}\mathrm{E}(\bar{Y}(Y)_{ki})^{2}\leq\frac{C}{k}\cdot k\leq C\end{aligned}$$
和
$$\开始{aligned}\begin{aligned}\operatorname{Var}\biggl(\frac{{T}(T)_{我,我}-\数学{E}{T}(T)_{我,我}-({T}(T)_{2k,l}-\数学{E}{T}(T)_{2k,l})}{\beta\delta_{l}\sqrt{l}}\biggr)&\leq\frac{C}{l}\operatorname{Var}\bigl({T}(T)_{我,我}-\数学{E}{T}(T)_{我,我}-({T}(T)_{2k,l}-\mathrm{E}{T}(T)_{2k,l})\biger)\\&\leq\frac{C}{l}\sum_{i=2k+1}^{l}\mathrm{E}(\bar{Y}(Y)_{李}-\数学{E}\bar{Y}(Y)_{li})^{2}\leq\frac{C}{l}\Biggl(\sum_{i=1}^{l}\mathrm{E}\bar{Y}(Y)_{li}^{2}\Biggr)\\&&\leq\frac{C}{l}\cdot l\leq C。\end{aligned}\end{alinged}$$
根据定义\(\rho^{-}\)-混合顺序,\(\mathrm{E} Y(Y)^{2} <\infty\)、和引理2.3,我们有
$$\开始{aligned}I_{2}\leq&\biggl(-\operatorname{Cov}\biggl(\frac{{T}(T)_{k,k}-\数学{E}{T}(T)_{k,k}}{\beta\delta{k}\sqrt{k}},\frac{{T}(T)_{我,我}-\数学{E}{T}(T)_{我,我}-({T}(T)_{2k,l}-\数学{E}{T}(T)_{2k,l})}{\beta\delta{l}\sqrt{l}}\biggr)\\&{}+8\rho^{-}\bigl(\frac{{T}(T)_{k,k}-\数学{E}{T}(T)_{k,k}}{\beta\delta{k}\sqrt{k}},\frac{{T}(T)_{我,我}-\数学{E}{T}(T)_{我,我}-({T}(T)_{2k,l}-\数学{E}{T}(T)_{2k,l})}{\beta\delta_{l}\sqrt{l}}\biggr)\\&{}\cdot\biggl\Vert\frac{{T}(T)_{k,k}-\数学{E}{T}(T)_{k,k}}{\beta\delta_{k}\sqrt{k}}\biggr\Vert_{2,1}\cdot\biggl\Vert\frac{{T}(T)_{我,我}-\数学{E}{T}(T)_{我,我}-({T}(T)_{2k,l}-\数学{E}{T}(T)_{2k,l})}{\beta\delta_{l}\sqrt{l}}\biggr\Vert_{2,1}\bigr)\\leq&C\rho^{-}(k)\biggl(\operatorname{Var}\bigl(\frac{{T}(T)_{k,k}-\数学{E}{T}(T)_{k,k}}{\beta\delta_{k}\sqrt{k}}\biggr)^{frac{1}{2}}\cdot\biggl(\operatorname{Var}\bigl(\frac{{T}(T)_{l,l}-\数学{E}{T}(T)_{我,我}-({T}(T)_{2k,l}-\数学{E}{T}(T)_{2k,l})}{\beta\delta_{l}\sqrt{l}}\biggr)\bigger)^{\frac{1}{2}}\\&{}+8\rho^{-}(k)\cdot\biggl\Vert\frac{{T}(T)_{k,k}-\数学{E}{T}(T)_{k,k}}{\beta\delta_{k}\sqrt{k}}\biggr\Vert_{2,1}\cdot\biggl\Vert\frac{{T}(T)_{我,我}-\数学{E}{T}(T)_{我,我}-({T}(T)_{2k,l}-\数学{E}{T}(T)_{2k,l})}{\beta\delta{l}\sqrt{l}}\biggr\Vert_{2,1}。\结束{对齐}$$
由\(\|X\|_{2,1}\leqr/(r-2)\|X\ |_{r}\),\(r>2\)(见第254页[10]或第251页[13]),Minkowski不等式,引理2.2,和Hölder不等式,我们得到
$$\开始{aligned}\biggl\Vert\frac{{T}(T)_{k,k}-\数学{E}{T}(T)_{k,k}}{\beta\delta_{k}\sqrt{k}}\biggr\Vert_{2,1}\leq&\frac{r}{r-2}\bigl\Vert\frac{{T}(T)_{k,k}-\数学{E}{T}(T)_{k,k}}{\beta\delta{k}\sqrt{k}}\biggr\Vert_{r}\\=&\frac{r}{r-2}\frac}{1\beta\delta{k{}\sqrt{k}{bigl(\mathrm{E}\Vert{T}(T)_{k,k}-\数学{E}{T}(T)_{k,k}\vert^{r}\biger)^{\frac{1}{r}}\\leq&\frac{C}{\sqrt{k}}\Biggl(\sum_{i=1}^{k}\mathrm{E}|\bar{Y}(Y)_{ki}|^{r}+\Biggl(\sum_{i=1}^{k}\mathrm{E}\bar{Y}(Y)_{ki}^{2}\Biggr)^{r/2}\bigr)^{1/r}\\leq&\frac{C}{\sqrt{k}}\bigl(k+k^{r/2}\biger)^{1/r}\leqC,\end{aligned}$$
同样地
$$\biggl\垂直\压裂{{T}(T)_{l,l}-\数学{E}{T}(T)_{我,我}-({T}(T)_{2k,l}-\数学{E}{T}(T)_{2k,l})}{\beta\delta_{l}\sqrt{l}}\biggr\Vert_{2,1}\leqC$$
因此
$$I_{2}\leq C\rho^{-}(k)$$
(9)
与结合(7)–(9), (三)持有,并通过(a)4),引理2.4, (6)保持,然后(4)是真的。
其次,我们证明(5); 对于\(对于所有k\geq1),\(eta_{k}=f({\bar{V}_{k,1}^{2}}/({k\delta_{k,1}^{2}}))-\mathrm{E}(f({\bar{V}_{k,1}^{2}}/({k\delta_{k,1}^{2}})),我们有
$$\开始{aligned}\vert\mathrm{E}\eta_{k}\eta_{l}\vert=&\biggl\vert\operatorname{Cov}\biggl(f\biggal(\frac{bar{V}_{k,1}^{2}}{k\delta{k,1}^{2}\biggr),f\biggl(\frac{\bar{V}_{l,1}^{2}}{l\delta_{l,1}^{2}\biggr)\biggr\vert\\\leq&\biggl\vert\operatorname{Cov}\bigl(f\biggl(\frac{\bar{V}_{k,1}^{2}}{k\delta_{k,1}^{2}\biggr),f\biggl(\frac{\bar{V}_{l,1}^{2}}{l\delta{l,1}^{2}\biggr)-f\biggl(\frac{\sum{i=2k+1}^{l}\bar{Y}(Y)_{l,i}^{2} 我({Y}(Y)_{i} \geq 0)}{l\delta_{l,1}^{2}}\biggr)\biggr\vert\\&{}+\biggl\vert\operatorname{Cov}\bigl(f\biggl(\frac{\bar{V}_{k,1}^{2}}{k\delta{k,1}^{2}\biggr),f\biggl(\frac{\sum{i=2k+1}^{l}\bar{Y}(Y)_{l,i}^{2} 我({Y}(Y)_{i} \geq0)}{l\delta_{l,1}^{2}}\biggr)\biggr\vert\\=&J_{1}+J_{2},\end{aligned}$$
(10)
根据…的财产(f),我们知道
$$J_{1}\leq C\Biggl(\mathrm{E}\Biggl(\sum_{i=1}^{2k}\bar{Y}(Y)_{ki}^{2}\mathrm{I}({Y}(Y)_{i} \geq 0)\Biggr)\Big/l\Bigger)\leq C\biggl(\frac{k}{l}\Biggr)$$
(11)
现在我们估计\(J_{2}\),
$$\开始{aligned}\operatorname{Var}\biggl(\frac{\bar{V}_{k,1}^{2}}{k\delta_{k,1}^{2}\biggr)=&\operatorname{Var}\bigl(\frac{\sum_{i=1}^{k}\bar{Y}(Y)_{ki}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)}{k\delta_{k,1}^{2}}\biggr)\\leq&\frac{C}{k^{2{}\mathrm{E}\Biggl(\sum_{i=1}^{k}\bar{Y}(Y)_{ki}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)\Biggr)^{2}\\=&&frac{C}{k^{2}}\mathrm{E}\Biggl(\sum_{i=1}^{k}\bar{Y}(Y)_{ki}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)-\mathrm{E}\Biggl(\sum_{i=1}^{k}\bar{Y}(Y)_{ki}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)\Biggr)+\mathrm{E}\Biggl(\sum_{i=1}^{k}\bar{Y}(Y)_{ki}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)\Biggr)\Biggr)^{2}\\leq&\frac{C}{k^{2{}}\mathrm{E}\Biggl(\sum_{i=1}^{k}\bigl(\bar{Y}(Y)_{ki}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)-\mathrm{E}\bigl(\bar{Y}(Y)_{ki}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)\bigr)\Biggr)\Biggr)^{2}\\&{}+\frac{C}{k^{2{}}\Biggl(\sum_{i=1}^{k}\mathrm{E}\bigl(\bar{Y}(Y)_{ki}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)\bigr)\Biggr)^{2}\\leq&\frac{C}{k^{2{}}\sum_{i=1}^{k}\mathrm{E}\bar{Y}(Y)_{ki}^{4}\mathrm{I}({Y}(Y)_{i} \geq0)+\frac{C}{k^{2}}\bigl(k\mathrm{E}\bigle(\bar{Y}(Y)_{k1}^{2}\mathrm{I}({Y}(Y)_{1} \geq0)\bigr)\biger)^{2}\\leq&\frac{C}{k^{2{}}\sum_{i=1}^{k}\mathrm{E} k个(Y_{i})^{2}\leq C,\end{aligned}$$
和类似的\(\operatorname{Var}(\sum_{i=2k+1}^{l}\bar){Y}(Y)_{li}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)/(l\delta_{l,1}^{2}))\leq C\)另一方面,我们有
$$\开始{aligned}\biggl\Vert\frac{\bar{V}_{k,1}^{2}}{k \delta _{k,1}^{2}}\biggr\Vert _{2,1}\leq&&\frac{r}{r-2}\cdot\frac{C}{k}\bigl(\mathrm{E}\bigl\Vert\bar{V}_{k,1}^{2}\bigr\vert^{r}\biger)^{1/r}\\leq&\frac{C}{k}\Biggl(\mathrm{E}\bigl\vert\sum_{i=1}^{k}\bigl(\bar{Y}(Y)_{ki}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)-\mathrm{E}\bigl(\bar{Y}(Y)_{ki}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)\bigr)\biger)\Biggr\vert^{r}+\Biggl\vert\sum_{i=1}^{k}\mathrm{E}\bigl(\bar{Y}(Y)_{ki}^{2}\mathrm{I}({Y}(Y)_{i} \geq 0)\bigr)\Biggr\vert^{r}\Biggr)^{1/r}\\leq&\frac{C}{k}\bigl(\sum_{i=1}^{k}\ mathrm{E}\bigl\vert\bigl(\bar{Y}(Y)_{ki}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)-\mathrm{E}\bigl(\bar{Y}(Y)_{ki}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)\biger)\bigr)\biger\vert^{r}\\&{}+\Biggl(\sum_{i=1}^{k}\mathrm{E}\bigl(\bar{Y}(Y)_{ki}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)-\mathrm{E}\bigl(\bar{Y}(Y)_{ki}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)\bigr)\biger)^{2}\Biggr)^{r/2}\Biggr)p{1/r}\\&{}+\frac{C}{k}\bigl\vert\sum_{i=1}^{k}\ mathrm{E}\bigl(\bar{Y}(Y)_{ki}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)\bigr)\Biggr\vert\\\leq&\frac{C}{k}\Biggl(\sum_{i=1}^{k}\mathrm{E}\bigl\vert\bar{Y}(Y)_{ki}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)\bigr\vert^{r}+\Biggl(\sum_{i=1}^{k}\mathrm{E}\bigl\vert\bar{Y}(Y)_{ki}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)\bigr\vert^{2}\Biggr)^{r/2}\Biggr)p{1/r}\\&{}+\frac{C}{k}\bigl\vert k\mathrm{E}\bigl(\bar{Y}(Y)_{k1}^{2}\mathrm{I}({Y}(Y)_{1} \geq0)\bigr)\biger\vert\\\leq&\frac{C}{k}\Biggl(\sum_{i=1}^{k}\ mathrm{E}\vert\sqrt{k}{Y}(Y)_{i} \vert^{r}+\Biggl(\sum_{i=1}^{k}\mathrm{E}\vert\sqrt{k}{Y}(Y)_{i} \vert^{2}\Biggr)^{r/2}\Biggr)(^{1/r}+C_{1}\\leq&\frac{C}{k}\bigl(k^{1+{r/2{}+k^{r}\bigr)^{1/r}+C_1}\leqC,\end{aligned}$$
同样地
$$\Biggl\Vert\sum_{i=2k+1}^{l}\bar{Y}(Y)_{li}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)/\bigl(l\delta_{l,1}^{2}\bigr)\Biggr\Vert_{2,1}\leq C$$
因此,通过引理2.3,我们有
$$\开始{aligned}J_{2}\leq&C\biggl\{-\operatorname{Cov}\biggl(\frac{\bar{V}_{k,1}^{2}}{k\delta_{k,1}^{2}},\frac{\sum_{i=2k+1}^{l}\bar{Y}(Y)_{li}^{2}\mathrm{I}({Y}(Y)_{i} \geq 0)}{l\delta_{l,1}^{2}\biggr)\\&{}+8\rho^{-}\bigl(\frac{\bar{V}_{k,1}^{2}}{k\delta{k,1}^{2}},\frac{\sum{i=2k+1}^{l}\bar{Y}(Y)_{li}^{2}\数学{I}({Y}(Y)_{i} \geq0)}{l\delta_{l,1}^{2}}\biggr)\cdot\biggl\Vert\frac{\bar{V}_{k,1}^{2}}{k\delta_{k,1}^{2}}\biggr\Vert_{2,1}\cdot\biggl\Vert\frac{\sum_{i=2k+1}^{l}\bar{Y}(Y)_{li}^{2}\数学{I}({Y}(Y)_{i} \geq0)}{l\delta_{l,1}^{2}}\biggr\Vert_{2,1}\bigbr\}\\leq&C\biggl\{\rho^{-}(k)\biggl(\operatorname{Var}\bigl(\frac{\bar{V}_{k,1}^{2}}{k\delta_{k,1}^{2})\biggr)^{1/2}\cdot\operatorname{Var}\biggl(\frac{\sum_{i=2k+1}^{l}\bar{Y}(Y)_{li}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)}{l\delta_{l,1}^{2}}\biggr)^{1/2}\\&{}+\rho^{-}(k)\cdot\biggl\Vert\frac{\bar{V}_{k,1}^{2}}{k\delta_{k,1}^{2}}\biggr\Vert_{2,1}\cdot\biggl\Vert\frac{\sum_{i=2k+1}^{l}\bar{Y}(Y)_{li}^{2}\mathrm{I}({Y}(Y)_{i} \geq0)}{l\delta_{l,1}^{2}}\biggr\Vert_{2,1}\bigr\}\\leq&C\rho^{-}(k),\end{aligned}$$
(12)
因此,结合(11)和(12), (三)保持不变,通过引理2.4, (5)持有。□