Lemma 4.1
(Hall and Heyde, [31], Corollary A.2, i.e. Davydov’s lemma)
Suppose that
X
and
Y
are random variables which are
\(\mathscr{G}\)-measurable and
\(\mathscr{H}\)-measurable, respectively, and
\(E|X|^{p}<\infty\), \(E|Y|^{q}<\infty\), where
\(p,q>1\), \(p^{-1}+q^{-1}<1\). Then
$$\bigl|E(XY)-EX EY\bigr|\leq8\bigl(E|X|^{p}\bigr)^{1/p} \bigl(E|Y|^{q}\bigr)^{1/q}\bigl[\alpha(\mathscr {G}, \mathscr{H})\bigr]^{1-p^{-1}-q^{-1}}. $$
Lemma 4.2
(Liebscher [32], Proposition 5.1)
Let
\(\{X_{n},n\geq1\}\)
be a stationary
α-mixing sequence with mixing coefficient
\(\alpha(k)\). Assume that
\(EX_{i}=0\)
and
\(|X_{i}|\leq S<\infty\), a.s., \(i=1,2,\ldots,n\). Then, for
\(n,m\in N\), \(0< m\leq n/2\), and all
\(\varepsilon>0\),
$$P \Biggl( \Biggl|\sum_{i=1}^{n} X_{i} \Biggr|>\varepsilon \Biggr)\leq4\exp \biggl\{ -\frac{\varepsilon^{2}}{16(\frac{n}{m}D_{m}+\frac{1}{3}\varepsilon Sm)} \biggr\} +32 \frac{S}{\varepsilon}n\alpha(m), $$
where
\(D_{m}=\max_{1\leq j\leq2m}\operatorname{Var}(\sum_{i=1}^{j} X_{i})\).
Lemma 4.3
(Shen and Xie [1], Lemma 3.2)
Under Assumption
2.3, for
\(x\in S_{f}^{0}\), one has
$$\bigl|E\widehat{m}_{n}(x)-m(x)\bigr|=O\bigl(h^{2}\bigr). $$
Proof of Theorem 2.1
For \(x\in S_{f}\), let \(Z_{i}:=\frac{Y_{i}K_{h}(x-X_{i})}{f(X_{i})}\), \(1\leq i\leq n\). Consider now
$$ \widehat{m}_{n}(x)=\frac{1}{n}\sum_{i=1}^{n} \frac{Y_{i}K_{h}(x-X_{i})}{f(X_{i})}=\frac{1}{n}\sum_{i=1}^{n} Z_{i}, \quad n\geq 1. $$
For any \(1\leq r\leq s\) and \(s>2\), it follows from (2.4) and (2.6) that
$$\begin{aligned} h^{d(r-1)}E|Z_{1}|^{r} =&h^{d(r-1)}E \biggl| \frac{K_{h}(x-X_{1})Y_{1}}{f(X_{1})} \biggr|^{r} \\ =&h^{d(r-1)}E \biggl(\frac{|K_{h}(x-X_{1})|^{r}}{f^{r}(X_{1})}E\bigl(|Y_{1}|^{r}| X_{1} \bigr) \biggr) \\ =& \int _{S_{f}}\biggl|K\biggl(\frac{x-u}{h}\biggr)\biggr|^{r}E \bigl(|Y_{1}|^{r}|X_{1}=u\bigr)\frac {1}{h^{d}} \frac{f(u)}{f^{r}(u)}\,du \\ \leq& \int _{S_{f}}\biggl|K\biggl(\frac {x-u}{h}\biggr)\biggr|^{r} \bigl(E\bigl(|Y_{1}|^{s}|X_{1}=u\bigr)f(u) \bigr)^{r/s}\frac{1}{h^{d}}\frac {1}{f^{r-1+r/s}(u)}\,du \\ \leq& \frac{(B_{0})^{r/s}\bar{K}^{r-1}\mu}{(\inf_{x\in S_{f}} f)^{r-1+r/s}}:=\bar{\mu}(r,s)< \infty. \end{aligned}$$
(4.1)
For \(j\geq j^{*}\), by (2.5), one has
$$\begin{aligned} E|Z_{1}Z_{j+1}| =&E \biggl|\frac {K_{h}(x-X_{1})K_{h}(x-X_{j+1})Y_{1}Y_{j+1}}{f(X_{1})f(X_{j+1})} \biggr| \\ =&E \biggl(\frac {|K_{h}(x-X_{1})K_{h}(x-X_{j+1})|}{f(X_{1})f(X_{j+1})}E\bigl(|Y_{1}Y_{j+1}||X_{1},X_{j+1}\bigr) \biggr) \\ =& \int _{S_{f}} \int _{S_{f}}\biggl|K\biggl(\frac{x-u_{1}}{h}\biggr)K\biggl( \frac {x-u_{j+1}}{h}\biggr)\biggr|E\bigl(|Y_{1}Y_{j+1}||X_{1}=u_{1},X_{j}=u_{j+1}\bigr) \\ &{}\times\frac{1}{h^{2d}}\frac {1}{f(u_{1})f(u_{j+1})}f_{j}(u_{1},u_{j+1})\,du_{1}\,du_{j+1} \\ \leq&\frac{B_{1}}{(\inf_{x\in S_{f}} f)^{2}} \int _{R^{d}} \int _{R^{d}}\bigl|K(u_{1})K(u_{j+1})\bigr|\,du_{1}\,du_{j+1} \leq \frac{B_{1}\mu^{2}}{(\inf_{x\in S_{f}} f)^{2}}< \infty. \end{aligned}$$
(4.2)
Define the covariances \(\gamma_{j}=\operatorname{Cov}(Z_{1},Z_{j+1})\), \(j>0\). Assume that n is sufficiently large so that \(h^{-d}\geq j^{*}\). We now bound the \(\gamma_{j}\) separately for \(j\leq j^{*}\), \(j^{*}< j\leq h^{-d}\), and \(h^{-d}< j<\infty\). First, for \(1\leq j\leq j^{*}\), by the Cauchy-Schwarz inequality and (4.1) with \(r=2\),
$$ |\gamma_{j}|\leq\sqrt{\operatorname{Var}(Z_{1})\cdot \operatorname{Var}(Z_{j+1})}= \operatorname{Var}(Z_{1}) \leq EZ_{1}^{2} \leq\bar{\mu}(2,s) h^{-d}. $$
(4.3)
Second, for \(j^{*}< j\leq h^{-d}\), in view of (4.1) (\(r=1\)) and (4.2), we establish that
$$ |\gamma_{j}|\leq E|Z_{1}Z_{j+1}|+\bigl(E|Z_{1}|\bigr)^{2} \leq \frac{B_{1}\mu^{2}}{(\inf_{x\in S_{f}} f)^{2}}+\bar{\mu}^{2}(1,s). $$
(4.4)
Third, for \(j>h^{-d}\), we apply Lemma 4.1, (2.1) and (4.1) with \(r=s\) (\(s>2\)) and we thus obtain
$$\begin{aligned} |\gamma_{j}| \leq& 8\bigl(\alpha({j})\bigr)^{1-2/s} \bigl(E|Z_{1}|^{s}\bigr)^{2/s}\leq 8A^{-1-2/s}j^{-\beta(1-2/s)}\bigl(\bar{\mu}(s,s) h^{-d(s-1)} \bigr)^{2/s} \\ \leq& 8A^{-1-2/s}\bar{\mu}^{\frac{2}{s}}(s,s) j^{-(2-2/s)}h^{-2d(s-1)/s}. \end{aligned}$$
(4.5)
Consequently, in view of the property of second-order stationarity and (4.3)-(4.5), for n sufficiently large, we establish
$$\begin{aligned} \operatorname{Var}\bigl(\widehat{m}_{n}(x)\bigr) =&\frac{1}{n^{2}}\operatorname{Var} \Biggl(\sum _{i=1}^{n} Z_{i} \Biggr)= \frac{1}{n^{2}} \Biggl(n\gamma_{0}+2n\sum _{j=1}^{n-1}\biggl(1-\frac {j}{n}\biggr) \gamma_{j} \Biggr) \\ \leq&\frac{1}{n^{2}} \biggl(nh^{-d}\bar{\mu}(2,s)+2n\sum _{1\leq j\leq j^{*}} |\gamma_{j}|+2n\sum _{j^{*}< j\leq h^{-d}} |\gamma_{j}|+2n\sum _{h^{-d}< j} |\gamma_{j}| \biggr) \\ \leq&\frac{1}{n}\bar{\mu}(2,s)h^{-d}+\frac{2}{n}j^{*} \bar{\mu }(2,s)h^{-d}+\frac{2}{n}\bigl(h^{-d}-j^{*} \bigr) \biggl(\bar{\mu}^{2}(1,s)+\frac{B_{1}\mu ^{2}}{(\inf_{x\in S_{f}}f)^{2}} \biggr) \\ &{}+\frac{2}{n}\sum_{h^{-d}< j< \infty}8A^{1-2/s} \bar{\mu}^{\frac{2}{s}}(s,s) j^{-(2-2/s)}h^{-2d(s-1)/s} \\ \leq& \biggl(\bar{\mu}(2,s)+2j^{*}\bar{\mu}(2,s)+2 \biggl(\bar{\mu}^{2}(1,s)+\frac{B_{1}\mu^{2}}{(\inf_{x\in S_{f}}f)^{2}} \biggr) \\ &{}+\frac{16A^{1-2/s}\bar{\mu}^{\frac {2}{s}}(s,s)}{(s-2)/s} \biggr) \frac{1}{nh^{d}} \\ :=&\frac{\Theta}{nh^{d}}, \end{aligned}$$
(4.6)
where the final inequality uses the fact that \(\sum_{j=k+1}^{\infty}j^{-\delta}\leq\int _{k}^{\infty}x^{-\delta}\,dx=\frac{k^{1-\delta}}{\delta-1}\) for \(\delta>1\) and \(k\geq1\).
Thus, (2.7) is completely proved. □
Proof of Corollary 2.1
It is easy to see that
$$\bigl|\widehat{m}_{n}(x)-m(x)\bigr|\leq \bigl|\widehat{m}_{n}(x)-E \widehat{m}(x)\bigr|+\bigl|E\widehat{m}_{n}(x)-m(x)\bigr|, $$
which can be treated as ‘variance’ part and ‘bias’ part, respectively.
On the one hand, by the proof of Theorem 3.1 of Shen and Xie [1], one has \(|E\widehat{m}_{n}(x)-m(x)|\rightarrow0\). On the other hand, we apply Theorem 2.1 and obtain that \(|\widehat{m}_{n}(x)-E\widehat{m}_{n}(x)|\xrightarrow{ P } 0\). So, (2.8) is proved finally. □
Proof of Theorem 2.2
Let \(\tau_{n}=a_{n}^{-1/(s-1)}\) and define
$$ R_{n}=\widehat{m}_{n}(x)-\frac{1}{n}\sum _{i=1}^{n} \frac{Y_{i}K_{h}(x-X_{i})}{f(X_{i})}I\bigl(|Y_{i}|\leq \tau_{n}\bigr) =\frac{1}{n}\sum_{i=1}^{n} \frac{Y_{i}K_{h}(x-X_{i})}{f(X_{i})}I\bigl(|Y_{i}|>\tau_{n}\bigr). $$
Obviously, we have
$$\begin{aligned} E|R_{n}| \leq& E \biggl(\frac{|K_{h}(x-X_{1})|}{f(X_{1})}E\bigl(|Y_{1}|I\bigl(|Y_{1}|> \tau _{n}\bigr)|X_{1}\bigr) \biggr) \\ \leq&\frac{1}{(\inf_{S_{f}}f)} \int _{S_{f}}\biggl|K\biggl(\frac {x-u}{h}\biggr)\biggr|E \bigl(|Y_{1}|I\bigl(|Y_{1}|>\tau_{n}\bigr)|X_{1}=u \bigr)\frac{f(u)}{h^{d}}\,du \\ = &\frac{1}{(\inf_{S_{f}}f)} \int _{S_{f}}\bigl|K(u)\bigr|E\bigl(|Y_{1}|I\bigl(|Y_{1}|> \tau_{n}\bigr)|X_{1}=x-hu\bigr)f(x-hu)\,du \\ \leq&\frac{1}{(\inf_{S_{f}}f)}\frac{1}{\tau_{n}^{s-1}} \int _{S_{f}}\bigl|K(u)\bigr|E\bigl(|Y_{1}|^{s}|X_{1}=x-hu \bigr)f(x-hu)\,du \\ \leq&\frac{1}{(\inf_{S_{f}}f)}\frac{\mu B_{0}}{\tau_{n}^{s-1}}. \end{aligned}$$
(4.7)
Combining Markov’s inequality with (4.7), one has
$$ |R_{n}-ER_{n}|=O_{P}\bigl(\tau_{n}^{-(s-1)} \bigr)=O_{P}(a_{n}). $$
(4.8)
Denote
$$ \widetilde{m}_{n}(x)=\frac{1}{n}\sum _{i=1}^{n} \frac{Y_{i}K_{h}(x-X_{i})}{f(X_{i})}I\bigl(|Y_{i}|\leq \tau_{n}\bigr):=\frac{1}{n}\sum_{i=1}^{n} \tilde{Z}_{n},\quad n\geq1. $$
(4.9)
It can be seen that
$$\begin{aligned} \bigl|\widehat{m}_{n}(x)-m(x)\bigr| \leq& \bigl|\widehat{m}_{n}(x)-E \widehat{m}_{n}(x)\bigr|+\bigl|E\widehat{m}_{n}(x)-m(x)\bigr| \\ \leq&\bigl|\widetilde{m}_{n}(x)-E\widetilde{m}_{n}(x)\bigr|+|R_{n}-ER_{n}|+\bigl|E \widehat {m}_{n}(x)-m(x)\bigr|. \end{aligned}$$
(4.10)
Similar to the proof of (4.6), it can be argued that
$$\operatorname{Var} \Biggl(\sum_{i=1}^{j} \tilde{Z}_{i} \Biggr)\leq C_{2}jh^{-d}, $$
which implies
$$D_{m}=\max_{1\leq j\leq2m}\operatorname{Var} \Biggl(\sum _{i=1}^{j} \tilde {Z}_{i} \Biggr)\leq C_{3}mh^{-d}. $$
Meanwhile, one has \(|\tilde{Z}_{i}-E\tilde{Z}_{i}|\leq \frac{C_{1}\tau_{n}}{h^{d}}\), \(1\leq i\leq n\). Setting \(m=a_{n}^{-1}\tau_{n}^{-1}\) and using (2.9), \(h=n^{-\theta/d}\), and Lemma 4.2 with \(\varepsilon=a_{n}n\), we obtain for n sufficiently large
$$\begin{aligned} &P\bigl(\bigl|\widetilde{m}_{n}(x)-E\widetilde{m}_{n}(x)\bigr|>a_{n} \bigr) \\ &\quad=P \Biggl( \Biggl|\sum_{i=1}^{n} ( \tilde{Z}_{i}-E\tilde{Z}_{i}) \Biggr|>na_{n} \Biggr) \\ &\quad\leq4\exp \biggl\{ -\frac{na_{n}^{2}}{16(C_{3}h^{-d} +\frac{1}{3}C_{1}h^{-d})} \biggr\} +32\frac{C_{1}\tau_{n}}{a_{n}nh^{d}}nA(a_{n} \tau_{n})^{\beta} \\ &\quad\leq4\exp\biggl\{ -\frac{\ln n}{16(C_{3}+\frac{1}{3}C_{1})}\biggr\} +C_{4}h^{-d}a_{n}^{\frac{\beta(s-2)-s}{s-1}} \\ &\quad\leq o(1)+C_{5}n^{\theta}n^{(\theta-1)\frac{\beta(s-2)-s}{2(s-1)}}(\ln n)^{\frac{\beta(s-2)-s}{2(s-1)}} \\ &\quad=o(1)+C_{5}n^{\frac{\beta(\theta-1)(s-2)+\theta s+s-2\theta}{2(s-1)}}(\ln n)^{\frac{\beta(s-2)-s}{2(s-1)}} =o(1), \end{aligned}$$
(4.11)
in view of \(s>2\), \(0<\theta<1\), \(\beta>\max\{\frac{\theta s+s-2\theta}{(1-\theta)(s-2)},\frac{2s-2}{s-2}\}\), and \(\frac{\beta(\theta-1)(s-2)+\theta s+s-2\theta}{2(s-1)}<0\).
Consequently, by (4.8), (4.10), (4.11), and Lemma 4.3, we establish the result of (2.10). □
Proof of Theorem 2.3
We use some similar notation in the proof of Theorem 2.2. Obviously, one has
$$ \sup_{x\in S_{f}^{\prime}}\bigl|\widehat{m}_{n}(x)-m(x)\bigr|\leq \sup _{x\in S_{f}^{\prime}}\bigl|\widehat{m}_{n}(x)-E\widehat{m}_{n}(x)\bigr|+ \sup_{x\in S_{f}^{\prime}}\bigl|E\widehat{m}_{n}(x)-m(x)\bigr|. $$
(4.12)
By the proof of (3.21) of Shen and Xie [1], we establish that
$$ \bigl|E\widehat{m}_{n}(x)-m(x)\bigr|\leq h^{2}\frac{b}{2}\sum _{1\leq i,j\leq d} \int _{R^{d}}K(v)|v_{i}v_{j}|\,dv\leq C_{0}h^{2}, $$
which implies
$$ \sup_{x\in S_{f}^{\prime}}\bigl|E\widehat{m}_{n}(x)-m(x)\bigr|=O \bigl(h^{2}\bigr). $$
(4.13)
Since \(\hat{m}_{n}(x)=R_{n}(x)+\tilde{m}_{n}(x)\),
$$ \sup_{x\in S_{f}^{\prime}}\bigl|\widehat{m}_{n}(x)-E\widehat{m}_{n}(x)\bigr| \leq \sup_{x\in S_{f}^{\prime}}\bigl|\widetilde{m}_{n}(x)-E \widetilde{m}_{n}(x)\bigr|+\sup_{x\in S_{f}^{\prime}}|R_{n}-ER_{n}|. $$
(4.14)
It follows from the proof of (4.8) that
$$ \sup_{x\in S_{f}^{\prime}}|R_{n}-ER_{n}|=O_{p}(a_{n}). $$
(4.15)
Since \(S_{f}^{\prime}\) is a compact set, there exists a \(\xi>0\) such that \(S_{f}^{\prime}\subset B:=\{x:\|x\|\leq\xi\}\). Let \(v_{n}\) be a positive integer. Take an open covering \(\bigcup_{j=1}^{v_{n}^{d}}B_{jn}\) of B, where \(B_{jn}\subset \{x:\|x-x_{jn}\|\leq\frac{\xi}{v_{n}}\}\), \(j=1,2,\ldots,v_{n}^{d}\), and their interiors are disjoint. So it follows that
$$\begin{aligned} &\sup_{x\in S_{f}^{\prime}}\bigl|\widetilde{m}_{n}(x)-E \widetilde{m}_{n}(x)\bigr| \\ &\quad\leq \max_{1\leq j\leq v_{n}^{d}}\sup _{x\in B_{jn}\cap S_{f}^{\prime}}\bigl|\widetilde{m}_{n}(x)-E\widetilde{m}_{n}(x)\bigr| \\ &\quad\leq\max_{1\leq j\leq v_{n}^{d}}\sup_{x\in B_{jn}\cap S_{f}^{\prime}}\bigl| \widetilde{m}_{n}(x)-\widetilde{m}_{n}(x_{jn})\bigr|+ \max_{1\leq j\leq v_{n}^{d}}\bigl|\widetilde{m}_{n}(x_{jn})-E \widetilde{m}_{n}(x_{jn})\bigr| \\ &\qquad{}+\max_{1\leq j\leq v_{n}^{d}}\sup_{x\in B_{jn}\cap S_{f}^{\prime}}\bigl|E \widetilde{m}_{n}(x_{jn})-E\widetilde{m}_{n}(x)\bigr| \\ &\quad:=I_{1}+I_{2}+I_{3}. \end{aligned}$$
(4.16)
By the definition of \(\widetilde{m}_{n}(x)\) in (4.9) and the Lipschitz condition of K,
$$\begin{aligned} \bigl|\widetilde{m}_{n}(x)-\widetilde{m}_{n}(x_{jn})\bigr| \leq& \Bigl(\inf_{S_{f}^{\prime}}f \Bigr)^{-1}\frac{\tau_{n}}{nh^{d}} \sum_{i=1}^{n} \biggl|K\biggl(\frac{x-X_{i}}{h} \biggr)-K\biggl(\frac{x_{jn}-X_{i}}{h}\biggr) \biggr| \\ \leq&\frac{L\tau_{n}}{h^{d+1}\inf_{S_{f}^{\prime}} f}\|x-x_{jn}\|,\quad x\in S_{f}^{\prime}. \end{aligned}$$
Taking \(v_{n}=\lfloor\frac{\tau_{n}}{h^{d+1}a_{n}}\rfloor+1\), we obtain
$$ \sup_{x\in B_{jn}\cap S_{f}^{\prime}}\bigl|\widetilde{m}_{n}(x)- \widetilde{m}_{n}(x_{jn})\bigr|\leq \frac{L\xi}{\inf_{S_{f}^{\prime}}f}a_{n},\quad 1\leq j\leq v_{n}^{d} , $$
(4.17)
and
$$ I_{1}=\max_{1\leq j\leq v_{n}^{d}}\sup_{x\in B_{jn}\cap S_{f}^{\prime}}\bigl| \widetilde{m}_{n}(x)-\widetilde{m}_{n}(x_{jn})\bigr|=O(a_{n}). $$
(4.18)
In view of \(|E\widetilde{m}_{n}(x_{jn})-E\widetilde{m}_{n}(x)|\leq E|\widetilde{m}_{n}(x_{jn})-\widetilde{m}_{n}(x)|\), we have by (4.17)
$$ I_{3}=\max_{1\leq j\leq v_{n}^{d}}\sup_{x\in B_{jn}\cap S_{f}^{\prime}}\bigl|E \widetilde{m}_{n}(x_{jn})-E\widetilde{m}_{n}(x)\bigr| \leq \frac{L\xi}{\inf_{S_{f}^{\prime}}f}a_{n}=O(a_{n}). $$
(4.19)
For \(1\leq i\leq n\) and \(1\leq j\leq v_{n}^{d}\), denote \(\tilde{Z}_{i}(j)=\frac{Y_{i}K_{h}(x_{jn}-X_{i})}{f(X_{i})}I(|Y_{i}|\leq \tau_{n})\). Then similar to the proof of (4.11), we obtain by Lemma 4.2 with \(m=a_{n}^{-1}\tau_{n}^{-1}\) and \(\varepsilon=Mna_{n}\) for n sufficiently large
$$\begin{aligned} P\bigl(|I_{2}|>Ma_{n}\bigr) =&P\Bigl(\max_{1\leq j\leq v_{n}^{d}}\bigl| \widetilde{m}_{n}(x_{jn})-E\widetilde{m}_{n}(x_{jn})\bigr|>Ma_{n} \Bigr) \\ \leq&\sum_{j=1}^{v_{n}^{d}}P \Biggl( \Biggl|\sum _{i=1}^{n} \bigl(\tilde{Z}_{i}(j)-E \tilde{Z}_{i}(j)\bigr) \Biggr|>Mna_{n} \Biggr) \\ \leq&4v_{n}^{d}\exp \biggl\{ -\frac{M^{2}na_{n}^{2}}{16(C_{3}h^{-d}+\frac {1}{3}C_{1}Mh^{-d})} \biggr\} +32v_{n}^{d}\frac{C_{1}\tau_{n}}{Ma_{n}h^{d}}A(a_{n}\tau _{n})^{\beta} \\ =&I_{21}+I_{22}, \end{aligned}$$
(4.20)
where the value of M will be given in (4.22).
In view of \(0<\theta<1\), \(s>2\), \(h=n^{-\theta/d}\), and \(a_{n}=(\frac{\ln n}{nh^{d}})^{1/2}\), one has \(h^{-d(d+1)}=n^{\theta(d+1)}\) and \(a_{n}^{-\frac{sd}{s-1}}=(\ln n)^{-\frac{sd}{2(s-1)}}n^{\frac{sd(1-\theta)}{2(s-1)}}\). Therefore, by \(v_{n}=\lfloor\frac{\tau_{n}}{h^{d+1}a_{n}}\rfloor+1\) and \(\tau_{n}=a_{n}^{-\frac{1}{s-1}}\), we obtain for n sufficiently large
$$\begin{aligned} I_{21} =&4v_{n}^{d}\exp \biggl\{ - \frac{M^{2}na_{n}^{2}}{16(C_{3}h^{-d}+\frac {1}{3}MC_{1}h^{-d})} \biggr\} \\ \leq& C_{4}h^{-d(d+1)}a_{n}^{-\frac{sd}{s-1}}\exp \biggl\{ -\frac{M^{2}\ln n}{16(C_{3}+\frac{1}{3}MC_{1})} \biggr\} \\ \leq&C_{5}(\ln n)^{-\frac{sd}{2(s-1)}}n^{\theta(d+1) +\frac{sd(1-\theta)}{2(s-1)}-\frac {M^{2}}{16(C_{3}+\frac{1}{3}MC_{1})}} =o(1), \end{aligned}$$
(4.21)
where M is sufficiently large such that
$$ \frac{M^{2}}{16(C_{3}+\frac{1}{3}MC_{1})}\geq \theta(d+1) +\frac{sd(1-\theta)}{2(s-1)}. $$
(4.22)
Meanwhile, by (2.11) and \(h=n^{-\theta/d}\), one has for n sufficiently large
$$\begin{aligned} I_{22} =&32v_{n}^{d}\frac{C_{1}\tau_{n}}{Ma_{n}h^{d}}A(a_{n} \tau_{n})^{\beta} \leq\frac{C_{6}}{M} \biggl( \frac{\tau_{n}}{h^{d+1}a_{n}} \biggr)^{d}a_{n}^{\frac{\beta (s-2)-s}{s-1}}h^{-d} =\frac{C_{6}}{M}a_{n}^{\frac{\beta(s-2)-s(d+1)}{s-1}}h^{-d(d+2)} \\ =&\frac{C_{6}}{M}(\ln n)^{\frac{\beta(s-2)-s(d+1)}{2(s-1)}}n^{\frac{\beta(\theta -1)(s-2)+s\theta d+3\theta s+sd+s-2\theta d-4\theta}{2(s-1)}} =o(1), \end{aligned}$$
(4.23)
in which is used the fact that \(s>2\), \(0<\theta<1\),
$$\beta>\max\biggl\{ \frac{s\theta d+3\theta s+sd+s-2\theta d-4\theta}{(1-\theta)(s-2)},\frac{2s-2}{s-2}\biggr\} , $$
and
$$\frac{\beta(\theta-1)(s-2)+s\theta d+3\theta s+sd+s-2\theta d-4\theta}{2(s-1)}< 0. $$
Thus, by (4.20)-(4.23), we establish that
$$ |I_{2}|=O_{p}(a_{n}). $$
(4.24)
Finally, the result of (2.12) follows from (4.12)-(4.16), (4.18), (4.19), and (4.24) immediately. □