在这一节中,我们给出了一些引理,它们对于证明定理非常有用1.
引理2.1
对于由定义的模型(1),我们有
$$\mathbb{E}\biggl[\frac{1}{\omega(Y_{i})}\bigr]=\frac}1}{\ mu},\qquad\mathbb}E}\biggl[\frac{\mu\varphi_{j,k}(Y_}i}[\frac{\mu\psi{j,k}(Y{i})}{\omega(Y{i}){\biggr]=\beta{j,k}$$
证明
这个引理可以由寇和郭的相同论点证明[10]. □
引理2.2
让
\(f\在B_{r,q}^{s}\(1\leqr,q<+infty,s>1/r)中,然后让
\(ω(y))
是一个非递增函数,以便
\(ω(y)\sim 1).如果
\(2^{j}\leqn)
和
\(1\leq p<+\infty),然后
$$\mathbb{E}\bigl[\vert\widehat{\alpha}_{j{0},k}-\alpha_{j}0}_{j,k}-\β{j,k}\vert^{p}\bigr]\lesssim n^{-\frac{p}{2}}$$
证明
因为这两个不等式的证明是相似的,所以我们只证明了第二个不等式。根据的定义\(\widehat{\beta}{j,k}\)我们有
$$\beart{aligned}\vert\widehat{\beta}_{j,k}-\beta_{j,k}\vert\leq\Biggl\vert\frac{\widehat{\mu}_{n}}{\muneneneep \Biggl \widehat{\mu}_{n}\Biggl(\frac{1}{\mu{-\frac}1}{\ widehat{\muneneneep{n}}\bigr)\Biggr\vert。\结束{对齐}$$
请注意\(\widehat{\mu}{n}\)和\(ω(y)\sim 1)意味着\(|\widehat{\mu}_{n}|\lesssim 1\)。我们有\(B_{r,q}^{s}(\mathbb{r})在以下情况下\(s>\frac{1}{r}\); 然后\(在B_{\infty,\infty}^{s-1/r}(\mathbb{r})中)和\(\|f\|_{\infty}\lesssim 1\)此外,\(|\beta_{j,k}|=|\langle f,\psi_{j,k}\rangle|\lesssim 1)通过Cauchy–Schwarz不等式和小波函数的正交性。因此,我们得出以下结论:
$$\开始{aligned}\mathbb{E}\bigl[\vert\widehat{\beta}_{j,k}-\beta_{j,k}\vert^{p}\bigr]\lesssim\mathbb{E}\Biggl[\Biggl \vert\frac{1}{n}\sum_{i=1}^{n}\frac{\mu\psi{j,k}(Y{i})}{\omega gl\vert\frac{1}{\mu}-\frac{1}{\widehat{\mu{n}}\Biggr\vert^{p}\bigr]。\结束{对齐}$$
(10)
然后我们需要估计\(T_{1}:=\mathbb{E}[\vert\frac{1}{n}\sum{i=1}^{n}\frac{\mu\psi{j,k}(Y{i})}{\omega(Y{i}){-\beta_{j,k}\vert^{p}]\)和\(T_{2}:=\mathbb{E}[\vert\frac{1}{\mu}-\frac{1'{\widehat{\mu{n}}\vert^{p}]\).
•的上限\(T_{1}\).服用\(eta{i}:=frac{mu\psi{j,k}(Y{i})}{omega(Y{i}){-\beta{j,k}),我们得到
$$T_{1}=\mathbb{E}\Biggl[\Biggl\vert\frac{1}{n}\sum_{i=1}^{n}\eta_{i}\Bigr\vert^{p}\Bigcr]=\Biggl(\frac{1\n}\Biggr \vert^{p}\Biggr]$$
请注意ψ是有界变化的函数(参见刘和徐[12]). 我们可以得到\(\psi:=\widetilde{\psi}-\overline{\ps2}\),其中ψ̃和ψ有界非负非递减函数。定义
$$\widetilde{\eta}_{i}:=\frac{\mu\widetelde{\psi}_{j,k}上划线{\beta}{j,k}$$
具有\(\widetilde{\beta}{j,k}:=\langlef,\widetelde{\psi}{j、k}\rangle\)和\(\overline{\beta}_{j,k}:=\langle f,\overline{\psi}_{j,k}\langle).然后\(\eta_{i}=\widetilde{\eta}_{我}-\上划线{\eta}{i}\),\(β{j,k}=widetilde{β}_{j,k}-\上划线{\beta}{j,k}\)、和
$$\begin{aligned}T_{1}=\biggl(\frac{1}{n}\biggr)^{p}\mathbb{E}\biggl[\biggl\vert\sum _{i=1}^{n}(\widetilde{\eta}_{我}-\overline{\eta}_{i})\Biggr\vert^{p}\Biggr]\lesssim\biggl(\frac{1}{n}\Biggr)^{p{\biggl\{\mathbb{E}\Bigl[\biggl \vert\sum_{i=1}^{n}\widetilde{\eta}_{i}\Biggr\vert^}\Bigcr]+\mathbb{E}\ biggl[\biggl \vert\sum_{i=1}^{n}\上划线{\eta}_{i}\Biggr\vert^{p}\Bighr]\Biggr\}。\结束{对齐}$$
(11)
与引理中的类似参数2.1向大家展示\(\mathbb{E}[\widetilde{eta}_{i}]=0\).功能\(压裂{\widetilde{\psi}_{j,k}(y)}{\omega(y){)不会因\(\widetilde{\psi}(y)\)和\(ω(y))。此外,我们也知道\({\widetilde{\eta}_{i},i=1,2,\ldots,n\}\)由引理负相关1.1另一方面,它是从(1)和\(ω(y)\sim 1)那个
$$\begin{aligned}\mathbb{E}\bigl[\vert\widetilde{\eta}_{i}\vert^{p}\bigr]\lesssim\mathbb{E}\ biggl[\biggl\vert\frac{\mu\widetilde{\psi}_{j,k}(Y{i})}{omega \bigl\vert\widetilde{\psi}_{j,k}(Y)\bigr\vert^{p} (f)(y) \,dy\lesssim 2^{j(p/2-1)}。\结束{对齐}$$
(12)
特别地,\(\mathbb{E}[|\widetilde{\eta}_{i}|^{2}]\lesssim 1\)回顾罗森塔尔不等式[12]:如果\(Y_{1},Y_{2},\ldots,Y__{n})是负相关的随机变量\(\mathbb{E}[Y_{i}]=0\)和\(\mathbb{E}[|Y_{i}|^{p}]<\infty\),然后
$$\mathbb{E}\Biggl[\Biggl \vert\sum_{i=1}^{n} Y(Y)_{i} \Biggr\vert^{p}\Biggr]\lesssim\textstyle\begin{cases}\sum_{i=1}^{n}\mathbb{E}[\vert Y_{i}\vert^}]+(\sum_{i=1}^{n}\mathbb{E{[\vertY_{i}\vert ^{2}])^{p{/{2}},&\text{$p>2$;}\\(\sum_{i=1{{n}\mathbb{E}[\vert Y_{i}\vert^{2}])^{p}/{2}},&\text{$1\leqp\leq2$.}\end{cases}$$
从这里我们清楚地看到
$$\mathbb{E}\Biggl[\Biggl\vert\sum_{i=1}^{n}\widetilde{\eta}_{i}\Biggr\vert^{p}\Biggr]\lesssim\textstyle\bbegin{cases}n2 ^{j(p/2-1)}+n ^{p/2},&&text{$p>2$;}\\n^{p/2},&&text{$1\leq p\leq 2$.}\end{cases}$$
这个,连同\(2^{j}\leqn),表明\(\mathbb{E}[\vert\sum_{i=1}^{n}\widetilde{\eta}_{i}\vert^{p}]\lesssim n^{p/2}\)同样,\(\mathbb{E}[\vert\sum_{i=1}^{n}\overline{eta}_{i}\vert^{p}]\lesssim n^{p/2}).将这些与(11),我们明白了
$$\开始{对齐}T_{1}\lesssim\biggl(\frac{1}{n}\biggr)^{p}\biggl\{\mathbb{E}\Bigl[\biggl \vert\sum_{i=1}^{n}\widetilde{\eta}_{i}\Bigr\vert^{p{\biggr]+\mathbb{E}\biggl/[\Bighl \vert\sum_{i=1}^{n}上划线{\eta}_{i}\biggr\vert^{p}\Bigr]\biggr}\lesssim n^{-\frac{p}{2}}。\结束{对齐}$$
(13)
•的上限\(T_{2}\)从定义中很容易看出\(\widehat{\mu}{n}\)那个
$$开始{对齐}T_{2}=\mathbb{E}\biggl[\biggl\vert\frac{1}{\mu}-\frac{1'{\widehat{\muneneneep _{n}}\bigr\vert^{p}\bigbr]=\biggl(\frac}{n}\bigcr)^{p{\mathbb{E}\biggl[\biggl \vert\sum_{i=1}^{n}gl(\frac{1}{\omega(Y_{i})}-\frac{1'{\mu}\biggr)\biggr\vert^{p}\biggr]。\结束{对齐}$$
(14)
定义\(\xi{i}:=\frac{1}{\omega(Y{i})}-\frac{1}}{\mu}\),我们得到\(\mathbb{E}[\xi_{i}]=0\)和\(\mathbb{E}[|\xi_{i}|^{p}]\lesssim 1\)通过引理2.1和\(ω(y)\sim 1)此外,由于\(ω(y))和引理1.1我们知道这一点\(xi{1},xi{2},ldots,xi{n})也呈负相关。然后利用罗森塔尔不等式,我们得到
$$\mathbb{E}\Biggl[\Biggl\vert\sum_{i=1}^{n}\xi_{i}\Bigr\vert^{p}\Bighr]\lesssim\textstyle\begin{cases}n+n^{p/2},&\text{$p>2$;}\\n^{p/2},&\text{$1\leqp\leq2$.}\end{cases{$$
因此
$$\begin{aligned}T_{2}=\biggl(\frac{1}{n}\biggr)^{p}\mathbb{E}\biggl[\biggl \vert\sum_{i=1}^{n}\xi_{i}\biggr \vert^{p{\biggr]\lesssim n^{-\frac}{p}{2}。\结束{对齐}$$
(15)
最后,由(10), (13)、和(15)我们有
$$\mathbb{E}\bigl[\vert\widehat{\beta}_{j,k}-\β{j,k}\vert^{p}\bigr]\lesssim n^{-\frac{p}{2}}$$
这就结束了证明。 □
引理2.3
让
\(f\在B_{r,q}^{s}\中)
\((1\leqr,q<+\infty,s>1/r)\)
和
\(\widehat{\beta}{j,k}\)
由定义(7).如果
\(ω(y))
是非递增函数,\(ω(y)\sim 1),和
\(2^{j}\leq\frac{n}{ln}\),那么对于每个
\(\lambda>0),存在一个常数
\(\kappa>1\)
这样的话
$$\mathbb{P}\bigl\{vert\widehat{beta}_{j,k}-\β{j,k}\vert\geq\kappat{n}\bigr\}\lesssim2^{-\lambdaj}$$
证明
通过相同的参数(10)我们可以得到
$$\开始{aligned}\mathbb{P}\bigl\{\vert\widehat{\beta}_{j,k}-\beta_{j,k}\vert\geq\kappa t_{n}\bigr\}\leq{}&\mathbb{P}\Biggl \{\Biggl\vert\frac{1}{n}\ sum_{i=1}^{n}\Biggl(\frac{1'{\omega(Y_{i})}-\frac}{1}{\mu}\biggr)\biggr\vert\gerc{\kappaT_{n}}{2}\biggr\}\\&{}+\mathbb{P}\Bigl\{\Biggl\vert\frac{1}{n}\sum_{i=1}^{n}\Biggl(\frac{\mu\psi_{j,k}(Y{i})}{\omega(Y{1})}-\β_{j,k}\biggr)\biggr\vert\geq\frac{\kappa t_{n}}{2}\biggr。\结束{对齐}$$
(16)
估计\(\mathbb{P}\{\vert\frac{1}{n}\sum_{i=1}^{n}(\frac{1}{\omega(Y{i})}-\frac}1}{\mu})\vert\geq\frac{\kappat_{n}}{2}\}),我们还定义\(\xi{i}:=\frac{1}{\omega(Y{i})}-\frac{1}}{\mu}\).然后引理2.1意味着\(\mathbb{E}[\xi _{i}]=0\)此外,\(|\xi_{i}|\lesssim 1)和\(\mathbb{E}[|\xi_{i}|^{2}]\lesssim 1\)多亏了\(ω(y)\sim 1)另一方面,由于\(ω(y))和引理1.1,\(xi{1},xi{2},ldots,xi{n})也有负相关。
回忆伯恩斯坦不等式[12]:如果\(Y_{1},Y_{2},\ldots,Y__{n})是负相关的随机变量\(\mathbb{E}[Y_{i}]=0\),\(|Y_{i}|\leq M<\infty\)、和\(\mathbb{E}[|Y_{i}|^{2}]=\sigma^{2{),然后针对每个\(\varepsilon>0\),
$$\mathbb{P}\Biggl\{\Biggl \vert\frac{1}{n}\sum_{i=1}^{n} 年_{i} \Biggr\vert\geq\varepsilon\Biggr \}\lesssim\exp\biggl(-\frac{n\varepsi隆^{2}}{2(\sigma^{2{+varepsilen M/3)}\Biggr)$$
因此,根据之前的论点\(xi{i})和\(t{n}=\sqrt{frac{lnn}{n}}),我们推导
$$\mathbb{P}\Biggl\{\Biggl \vert\frac{1}{n}\sum_{i=1}^{n}\Biggl kappa^{2}/4}{2(\sigma^{2]+\kappa/6)}\biggr)$$
然后就有了\(\kappa>1\)这样的话\(\exp(-\frac{(lnn)\kappa^{2}/4}{2(\sigma^{2]+\kappa/6)})\lesssim2^{-\lambdaj}\)带有固定的\(\lambda>0).因此
$$\开始{aligned}\mathbb{P}\Biggl\{\Biggl \vert\frac{1}{n}\sum_{i=1}^{n}\Biggl(\frac{1'{\omega(Y{i})}-\frac}{\mu}\bigr)\biggr\vert\geq\frac{\kappa t_{n}{2}\biggr\}\lesssim2^{-\lambda j}。\结束{对齐}$$
(17)
接下来,我们估计\(\mathbb{P}\{\vert\frac{1}{n}\sum{i=1}^{n}(\frac{\mu\psi{j,k}(Y{i})}{omega(Y{i}){-\beta{j,k})\vert\geq\frac}\kappat_{n}}{2}).By对的相同参数(11)我们得到
$$\开始{aligned}\mathbb{P}\Biggl\{\Biggl \vert\frac{1}{n}\sum_{i=1}^{n}\eta_{i}\Bigr\vert\geq\frac{\kappat_{n}}{2}\Biggr\}\leq\mathbb{P}\Biggl/vert\frac{1}{n}\sum_{i=1}^{n}\widetilde{\eta}_{i}\Biggr\vert\geq\frac{\kappat_{n}}{4}\Biggr\}+\mathbb{P}\Bigl\{\Biggl\vert\frac{1}{n}\sum_{i=1}^{n}\上划线{\eta}_{i} \Biggr\vert\geq\frac{\kappa t_{n}}{4}\Biggr\}。\结束{对齐}$$
(18)
从\(\widetilde{\eta}_{i}\)和引理2.1那个\(\mathbb{E}[\widetilde{eta}_{i}]=0\)此外,\(\mathbb{E}[|\widetilde{\eta}_{i}|^{2}]\lesssim 1\)由(12)带有\(p=2\).使用\(ω(y)\sim 1),我们得到\(vert\frac{\mu\widetilde{\psi}_{j,k}(Y{i})}{\omega(Y{i}){\vert\lesssim2^{j/2})和\(|\widetilde{\eta}{i}|\leq\vert\frac{\mu\widetelde{\psi}{j,k}(Y{i})}{\omega(Y{i}){\vert+\mathbb{E}[\vert\frac{\mu\ widetilde{\psi},k}(Y{i}然后根据伯恩斯坦不等式,\(2^{j}\leq\frac{n}{ln}\)、和\(t{n}=\sqrt{frac{lnn}{n}})那个
$$\mathbb{P}\Biggl\{\Biggl \vert\frac{1}{n}\sum_{i=1}^{n}\widetilde{\eta}_{i}\Bigr\vert\geq\frac{\kappat_{n}}{4}\Biggr\}\lesssim\exp\Biggl(-\frac}n(\kappa t_{n}/4)^{2}{2(\sigma^{2{+\卡帕特_{n} 2个^{j/2}/12)}\biggr)\lesssim\exp\biggl(-\frac{(ln n)\kappa^{2}/16}{2(\sigma^{2}+\kappa/12)}\biggr)$$
很明显,我们可以\(\kappa>1\)这样的话\(\mathbb{P}\{vert\frac{1}{n}\sum{i=1}^{n}\widetilde{\ta}{i}\vert\geq\frac{\kappat{n}}{4}\}\lesssim2^{-\lambdaj}\)。然后类似的论据表明\(\mathbb{P}\{\vert\frac{1}{n}\sum_{i=1}^{n}\上横线{\eta}_{i}\vert\geq\frac{\kappat_{n}}{4}\}\lesssim2^{-\lambdaj}\).将这些与(18),我们获得
$$\begin{aligned}\mathbb{P}\Biggl\{\Biggl\vert\frac{1}{n}\sum _{i=1}^{n}\Biggl(\frac{\mu\psi_{j,k}(Y_{i})}{\omega(Y_{i})}-\beta\ j,k}\biggr)\biggr\vert\geq\frac{\kappa t_{n}}}{2}\biggr\}\lesssim 2^{-\lambda j}。\结束{对齐}$$
(19)
由(16), (17)、和(19)我们得到
$$\mathbb{P}\bigl\{vert\widehat{beta}_{j,k}-\β{j,k}\vert\geq\kappat{n}\bigr\}\lesssim2^{-\lambdaj}$$
这就结束了证明。 □