In this section, for \(0<\varepsilon<1\) and \(M>1\), we set
$$b (\varepsilon )= \bigl[\exp \bigl\{ \exp \bigl\{ M\varepsilon ^{-2/d} \bigr\} \bigr\} \bigr]. $$
(7)
Without loss of generality, we assume that \(\tau=1\).
Theorem 1.2 will be proved if we show the following propositions.
Proposition 3.1
Under the conditions of Theorem
1.2, we have
$$\begin{aligned}& \lim_{\varepsilon\searrow0}\varepsilon^{2\beta/d}\sum _{n=3}^{\infty}\frac { (\log\log n )^{\beta-1}}{n\log n} \mathbb{E} \bigl\{ \vert \mathcal{N}\vert -\varepsilon (\log\log n )^{d/2} \bigr\} _{+}= \frac{d\mathbb{E}\vert \mathcal{N}\vert ^{2\beta/d+1}}{\beta (2\beta+d )}, \end{aligned}$$
(8)
$$\begin{aligned}& \lim_{M\rightarrow\infty}\varepsilon^{2\beta/d}\sum _{n>b(\varepsilon )}\frac{ (\log\log n )^{\beta-1}}{n\log n} \mathbb{E} \bigl\{ \vert \mathcal{N} \vert -\varepsilon (\log\log n )^{d/2} \bigr\} _{+}=0, \end{aligned}$$
(9)
where (9) uniformly holds true with respect to
\(0<\varepsilon<1\).
Proof
See the proof of Proposition 1 and Proposition 3 in Xiao and Yin [10]. □
Proposition 3.2
Under the conditions of Theorem
1.2, we have
$$\begin{aligned} &\lim_{\varepsilon\searrow0}\varepsilon^{2\beta/d}\sum _{n\leq b(\varepsilon)}\frac{ (\log\log n )^{\beta-1}}{n\log n} \bigl\vert \mathbb{E} \bigl\{ \vert \mathcal{N}\vert -\varepsilon (\log \log n )^{d/2} \bigr\} _{+} \\ &\quad{}- \mathbb{E} \bigl\{ \vert S_{n}\vert /\sqrt {n}-\varepsilon (\log\log n )^{d/2} \bigr\} _{+}\bigr\vert =0. \end{aligned}$$
(10)
Proof
Let \(\Delta_{n}=\sup_{x\in\mathbb {R}}\vert \mathbb{P} (\vert \mathcal{N}\vert \geq x )-\mathbb{P} (\vert S_{n}\vert /\sqrt{n}\geq x )\vert \). By Lemma 2.3, then we have \(\Delta_{n}\rightarrow0\) as \(n\rightarrow\infty\). We have
$$\begin{aligned} &\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log\log n )^{\beta-1}}{n\log n}\bigl\vert \mathbb{E} \bigl\{ \vert \mathcal {N}\vert - \varepsilon (\log\log n )^{d/2} \bigr\} _{+}-\mathbb {E} \bigl\{ \vert S_{n}\vert /\sqrt{n}-\varepsilon (\log\log n )^{d/2} \bigr\} _{+}\bigr\vert \\ &\quad=\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log \log n )^{\beta-1}}{n\log n}\biggl\vert \int_{0}^{\infty}\mathbb{P} \bigl(\vert \mathcal{N} \vert \geq x+\varepsilon (\log\log n )^{d/2} \bigr)\,\mathrm{d}x \\ &\qquad{}- \int_{0}^{\infty}\mathbb{P} \bigl(\vert S_{n}\vert /\sqrt{n}\geq x+\varepsilon (\log\log n )^{d/2} \bigr)\,\mathrm{d}x\biggr\vert \\ &\quad\leq\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log \log n )^{\beta-1+d/2}}{n\log n} \int_{0}^{\infty}\bigl\vert \mathbb{P} \bigl(| \mathcal{N}|\geq (x+\varepsilon ) (\log\log n )^{d/2} \bigr) \\ &\qquad{}- \mathbb{P} \bigl(|S_{n}|/\sqrt{n}\geq (x+\varepsilon ) (\log \log n )^{d/2} \bigr)\bigr\vert \,\mathrm{d}x \\ &\quad\leq\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log \log n )^{\beta-1+d/2}}{n\log n} \int_{0}^{l(n)}\bigl\vert \mathbb{P} \bigl(| \mathcal{N}|\geq (x+\varepsilon ) (\log\log n )^{d/2} \bigr) \\ &\qquad{}- \mathbb{P} \bigl(|S_{n}|/\sqrt{n}\geq (x+\varepsilon ) (\log \log n )^{d/2} \bigr)\bigr\vert \,\mathrm{d}x \\ &\qquad{}+\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log \log n )^{\beta-1+d/2}}{n\log n} \int_{l(n)}^{\infty}\mathbb{P} \bigl(\vert \mathcal{N} \vert \geq (x+\varepsilon ) (\log\log n )^{d/2} \bigr)\,\mathrm{d}x \\ &\qquad{}+\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log \log n )^{\beta-1+d/2}}{n\log n} \int_{l(n)}^{\infty}\mathbb{P} \bigl(\vert S_{n}\vert /\sqrt{n}\geq (x+\varepsilon ) (\log\log n )^{d/2} \bigr)\,\mathrm{d}x \\ &\quad:=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}, \end{aligned}$$
(11)
where \(l(n)= (\log\log n )^{-d/2}\Delta_{n}^{-1/2}\). Hence, for \(\mathrm{I}_{1}\), by applying Lemma 2.3, we have
$$\begin{aligned} \mathrm{I}_{1} \leq&c\varepsilon^{2\beta/d}\sum _{n\leq b(\varepsilon)}\frac { (\log\log n )^{\beta-1+d/2}}{n\log n} \int_{0}^{l(n)}\Delta_{n}\,\mathrm{d}x \leq c\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac { (\log\log n )^{\beta-1+d/2}}{n\log n}\Delta_{n}l(n) \\ =&c\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac { (\log\log n )^{\beta-1}}{n\log n}\Delta_{n}^{1/2} \leq\frac{cM^{\beta}}{ (\log\log b (\varepsilon ) )^{\beta}}\sum_{n\leq b(\varepsilon)} \frac{ (\log\log n )^{\beta-1}}{n\log n} \Delta_{n}^{1/2}. \end{aligned}$$
(12)
So, by the Toeplitz lemma, we have
$$ \lim_{\varepsilon\searrow0}\mathrm{I}_{1}=0. $$
(13)
As for \(\mathrm{I}_{2}\), by the Markov inequality, (12), and (13), we have
$$\begin{aligned} \mathrm{I}_{2} \leq&c\varepsilon^{2\beta/d}\sum _{n\leq b(\varepsilon)}\frac { (\log\log n )^{\beta-1+d/2}}{n\log n} \int_{l(n)}^{\infty}\frac{1}{ (x+\varepsilon )^{2} (\log\log n )^{d}}\,\mathrm{d}x \\ \leq&c\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log \log n )^{\beta-1-d/2}}{n\log n}l(n)^{-1} \\ \leq&c\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log \log n )^{\beta-1}}{n\log n}\Delta_{n}^{1/2} \rightarrow0, \quad\mbox{as } \varepsilon\searrow0. \end{aligned}$$
(14)
For \(\mathrm{I}_{3}\), note that \(S_{n}=\sum_{k=1}^{n}X_{k}=\sum_{i=-\infty }^{+\infty}\sum_{k=1}^{n}a_{i}\xi_{i+k}=\sum_{i=-\infty}^{+\infty}a_{i}\sum_{j=i+1}^{i+n}\xi_{j}\). By Lemma 2.1, without loss of generality, we assume \(\sum_{i=-\infty}^{+\infty} \vert a_{i}\vert \leq1\), and set \(S_{n}'=\sum_{i=-\infty}^{+\infty}a_{i}\sum_{j=i+1}^{i+n}\xi_{j}'\), where \(\xi _{j}'=\xi_{j}I (\vert \xi_{j}\vert \leq (x+\varepsilon )\sqrt{n} (\log\log n )^{d/2} )\). By \(\mathbb{E}\xi_{1}=0\), then, for all \(x>0\), we have \(\mathbb{E}\xi_{1}I (|\xi_{1}|\leq x )=-\mathbb{E}\xi_{1}I (|\xi _{1}|>x )\). So, we have
$$\begin{aligned} \bigl\vert \mathbb{E}S_{n}'\bigr\vert =&\Biggl\vert \sum_{i=-\infty}^{+\infty }a_{i} \mathbb{E}\sum_{j=i+1}^{i+n}\xi_{j}I \bigl( \vert \xi_{j}\vert \leq(x+\varepsilon)\sqrt{n} (\log\log n )^{d/2} \bigr)\Biggr\vert \\ \leq&\sum_{i=-\infty}^{+\infty} \vert a_{i}\vert \Biggl\vert \mathbb{E}\sum_{j=i+1}^{i +n} \xi_{j}I \bigl(\vert \xi_{j}\vert \leq(x+\varepsilon) \sqrt{n} (\log \log n )^{d/2} \bigr)\Biggr\vert \\ =&\sum_{i=-\infty}^{+\infty} \vert a_{i} \vert \Biggl\vert \mathbb{E}\sum_{j=i+1}^{i +n} \xi_{j}I \bigl(\vert \xi_{j}\vert >(x+\varepsilon)\sqrt{n} (\log \log n )^{d/2} \bigr)\Biggr\vert \\ \leq&n\mathbb{E}\vert \xi_{1}\vert I \bigl(\vert \xi_{1}\vert >(x+\varepsilon)\sqrt{n} (\log\log n )^{d/2} \bigr) \\ \leq&\frac{\sqrt{n}\mathbb{E}\xi_{1}^{2}}{(x+\varepsilon) (\log\log n )^{d/2}}, \end{aligned}$$
(15)
thus for \(x\geq l(n)\) and n large enough, we have
$$ \frac{\vert \mathbb{E}S_{n}'\vert }{(x+\varepsilon)\sqrt{n} (\log \log n )^{d/2}}\leq\frac{c\mathbb{E}\xi_{1}^{2}}{ (x+\varepsilon )^{2} (\log\log n )^{d}}\leq\frac{c}{l(n)^{2} (\log\log n )^{d}}\leq c\Delta_{n}< \frac{1}{4}, $$
(16)
so we obtain
$$\begin{aligned} & \int_{l(n)}^{\infty}\mathbb{P} \bigl(\vert S_{n}\vert /\sqrt{n}\geq (x+\varepsilon ) (\log\log n )^{d/2} \bigr)\,\mathrm{d}x \\ &\quad\leq c \int_{l(n)}^{\infty}\mathbb{P} \Biggl(\Biggl\vert \sum _{i=-\infty }^{+\infty}a_{i}\sum _{j=i+1}^{i+n}\xi_{j}I \bigl( \vert \xi_{j}\vert >(x+\varepsilon)\sqrt{n} (\log\log n )^{d/2} \bigr)\Biggr\vert \\ &\qquad{} >\frac{(x+\varepsilon)\sqrt{n} (\log\log n )^{d/2}}{2} \Biggr)\,\mathrm{d}x \\ &\qquad{}+c \int_{l(n)}^{\infty}\mathbb{P} \Biggl(\Biggl\vert \sum _{i=-\infty}^{+\infty }a_{i}\sum _{j=i+1}^{i+n} \bigl( \xi_{j}'- \mathbb{E}\xi_{j}' \bigr)\Biggr\vert >\frac{(x+\varepsilon)\sqrt {n} (\log\log n )^{d/2}}{4} \Biggr)\,\mathrm{d}x \\ &\quad:=\mathrm{I}_{31}+\mathrm{I}_{32}. \end{aligned}$$
(17)
By (11) and (17), we have
$$\mathrm{I}_{3}\leq c\varepsilon^{2\beta/d}\sum _{n\leq b(\varepsilon)}\frac { (\log\log n )^{\beta-1+d/2}}{n\log n} (\mathrm{I}_{31}+ \mathrm{I}_{32} ). $$
On the one hand, by (12) and (13), we obtain
$$\begin{aligned} &\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log \log n )^{\beta-1+d/2}}{n\log n}\mathrm{I}_{31} \\ &\quad\leq c\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log\log n )^{\beta-1+d/2}}{n\log n} \int_{l(n)}^{\infty}\frac {n\mathbb{E}\vert \xi_{1}\vert I (\vert \xi_{1}\vert >(x+\varepsilon )\sqrt{n} ( \log\log n )^{d/2} )}{\sqrt{n}(x+\varepsilon) ( \log\log n )^{d/2}}\,\mathrm{d}x \\ &\quad\leq c\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log\log n )^{\beta-1+d/2}}{n\log n} \int_{l(n)}^{\infty}\frac {\mathbb{E}\xi_{1}^{2}I (\vert \xi_{1}\vert >(x+\varepsilon)\sqrt{n} ( \log\log n )^{d/2} )}{ (x+\varepsilon )^{2} ( \log\log n )^{d}}\,\mathrm{d}x \\ &\quad\leq c\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log\log n )^{\beta-1-d/2}}{n\log n} \int_{l(n)}^{\infty}(x+\varepsilon )^{-2}\, \mathrm{d}x \\ &\quad\leq c\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log\log n )^{\beta-1-d/2}}{n\log n}l(n)^{-1} \\ &\quad\leq c\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log\log n )^{\beta-1}}{n\log n}\Delta_{n}^{1/2} \rightarrow0,\quad \mbox{as } \varepsilon\searrow0. \end{aligned}$$
(18)
On the other hand, by the Markov inequality, the Hölder inequality, and Lemma 2.2, and noting that \(\sum_{m=1}^{\infty }\varphi^{1/2}(2^{m})<\infty\), we have
$$\begin{aligned} & \varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log \log n )^{\beta-1+d/2}}{n\log n}\mathrm{I}_{32} \\ &\quad\leq c\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log \log n )^{\beta-1+d/2-dq/2}}{n^{1+q/2}\log n} \int_{l(n)}^{\infty}(x+\varepsilon )^{-q} \mathbb{E}\Biggl\vert \sum_{i=-\infty}^{+\infty }a_{i} \sum_{j=i+1}^{i+n} \bigl( \xi_{j}'- \mathbb{E}\xi_{j}' \bigr)\Biggr\vert ^{q}\, \mathrm{d}x \\ &\quad\leq c\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log \log n )^{\beta-1+d/2-dq/2}}{n^{1+q/2}\log n}\\ &\qquad{}\times \int_{l(n)}^{\infty}(x+\varepsilon )^{-q} \mathbb{E} \Biggl[\sum_{i=-\infty}^{+\infty } \bigl( \vert a_{i}\vert ^{1-\frac{1}{q}} \bigr) \Biggl(\vert a_{i}\vert ^{\frac{1}{q}}\Biggl\vert \sum _{j=i+1}^{i+n} \bigl(\xi_{j}'- \mathbb{E}\xi_{j}' \bigr)\Biggr\vert \Biggr) \Biggr] ^{q}\,\mathrm{d}x \\ &\quad\leq c\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log \log n )^{\beta-1+d/2-dq/2}}{n^{1+q/2}\log n}\\ &\qquad{}\times \int_{l(n)}^{\infty}(x+\varepsilon )^{-q} \mathbb{E} \Biggl[ \Biggl(\sum_{i=-\infty }^{+\infty} \vert a_{i}\vert ^{1-\frac{1}{q} \cdot\frac{1}{1-\frac{1}{q}}} \Biggr)^{1-\frac{1}{q}} \Biggl(\sum _{i=-\infty}^{+\infty} \vert a_{i}\vert ^{\frac{1}{q}\cdot q}\Biggl\vert \sum_{j=i+1}^{i+n} \bigl(\xi_{j}'-\mathbb{E}\xi_{j}' \bigr)\Biggr\vert ^{q} \Biggr)^{\frac{1}{q}} \Biggr]^{q}\, \mathrm{d}x \\ &\quad\leq c\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log \log n )^{\beta-1+d/2-dq/2}}{n^{1+q/2}\log n}\\ &\qquad{}\times \int_{l(n)}^{\infty}(x+\varepsilon )^{-q} \mathbb{E} \Biggl(\sum_{i=-\infty}^{+\infty }|a_{i}| \Biggr)^{ (1-\frac{1}{q} )q} \sum_{i=-\infty}^{+\infty} \vert a_{i}\vert \Biggl\vert \sum_{j=i+1}^{i+n} \bigl(\xi_{j}'-\mathbb{E}\xi_{j}' \bigr)\Biggr\vert ^{q}\,\mathrm{d}x \\ &\quad\leq c\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log \log n )^{\beta-1+d/2-dq/2}}{n^{1+q/2}\log n}\\ &\qquad{}\times \int_{l(n)}^{\infty}(x+\varepsilon )^{-q}\sum _{i=-\infty}^{+\infty} \vert a_{i}\vert \mathbb{E}\Biggl\vert \sum_{j=i+1}^{i+n} \bigl(\xi_{j}'-\mathbb{E}\xi_{j}' \bigr)\Biggr\vert ^{q}\,\mathrm{d}x \\ &\quad\leq c\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log \log n )^{\beta-1+d/2-dq/2}}{n^{1+q/2}\log n} \int_{l(n)}^{\infty}(x+\varepsilon )^{-q} \bigl( \bigl(n\mathbb{E} {\xi'}_{1}^{2} \bigr)^{\frac{q}{2}}+n\mathbb{E} \bigl\vert \xi_{1}' \bigr\vert ^{q} \bigr)\,\mathrm{d}x \\ &\quad:=\mathrm{I}_{321}+\mathrm{I}_{322}. \end{aligned}$$
For \(\mathrm{I}_{321}\), \(q\geq2\). By (12) and (13), we have
$$\begin{aligned} \mathrm{I}_{321} \leq&c\varepsilon^{2\beta/d}\sum _{n\leq b(\varepsilon )}\frac{ (\log\log n )^{\beta-1+d/2-dq/2}}{n\log n} \int _{l(n)}^{\infty}(x+\varepsilon )^{-q}\, \mathrm{d}x \\ \leq&c\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log\log n )^{\beta-1+d/2-dq/2}}{n\log n}l(n)^{1-q} \\ \leq&c\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log\log n )^{\beta-1}}{n\log n}\Delta_{n}^{\frac {q-1}{2}}\rightarrow0, \quad\mbox{as } \varepsilon\searrow0. \end{aligned}$$
(19)
For \(\mathrm{I}_{322}\), by (12) and (13), we have
$$\begin{aligned} \mathrm{I}_{322} \leq&c\varepsilon^{2\beta/d}\sum _{n\leq b(\varepsilon )}\frac{ (\log\log n )^{\beta-1+d/2-dq/2}}{n^{q/2}\log n} \\ &{}\times\int _{l(n)}^{\infty}(x+\varepsilon )^{-q} \mathbb{E}\vert \xi_{1}\vert ^{q}I \bigl(\vert \xi_{1}\vert \leq(x+\varepsilon) \sqrt{n} (\log\log n )^{d/2} \bigr)\,\mathrm{d}x \\ \leq&c\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log \log n )^{\beta-1-d/2}}{n\log n} \\ &{}\times \int_{l(n)}^{\infty}(x+\varepsilon )^{-2} \mathbb{E}\xi_{1}^{2}I \bigl(\vert \xi_{1} \vert \leq (x+\varepsilon) \sqrt{n} (\log\log n )^{d/2} \bigr)\, \mathrm{d}x \\ \leq&c\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log \log n )^{\beta-1-d/2}}{n\log n}l(n)^{-1} \\ =&c\varepsilon^{2\beta/d}\sum_{n\leq b(\varepsilon)} \frac{ (\log \log n )^{\beta-1}}{n\log n}\Delta_{n}^{1/2}\rightarrow0, \quad\mbox{as } \varepsilon\searrow0. \end{aligned}$$
(20)
Thus, combining with (13), (14), and (18)-(20), we complete the proof of this proposition. □
Proposition 3.3
Under the conditions of Theorem
1.2,
$$ \lim_{M\rightarrow\infty}\varepsilon^{2\beta/d}\sum _{n>b(\varepsilon )}\frac{ (\log\log n )^{\beta-1}}{n\log n} \mathbb{E} \bigl\{ \vert S_{n}\vert /\sqrt{n}-\varepsilon (\log\log n )^{d/2} \bigr\} _{+}=0 $$
(21)
uniformly holds true with respect to
\(0<\varepsilon<1\).
Proof
For \(x\geq0\), M large enough, we have
$$\frac{\vert \mathbb{E}S_{n}'\vert }{\sqrt{n}(x+\varepsilon) (\log \log n )^{d/2}}\leq\frac{c}{(x+\varepsilon)^{2} (\log\log n )^{d}}\leq\frac{c}{M^{d}}< \frac{1}{4}. $$
Then
$$\begin{aligned} &\mathbb{P} \bigl(|S_{n}|\geq\sqrt{n}(\log\log n)^{d/2}(x+\varepsilon ) \bigr) \\ &\quad\leq\mathbb{P} \Biggl(\Biggl\vert \sum _{i=-\infty}^{+\infty}a_{i}\sum _{j=i+1}^{i+n}\xi_{j}I \bigl(| \xi_{j}|> (x+\varepsilon)\sqrt{n} (\log\log n )^{d/2} \bigr) \Biggr\vert >\frac{\sqrt{n}(\log\log n)^{d/2}(x+\varepsilon)}{2} \Biggr) \\ &\qquad{}+\mathbb{P} \biggl(\bigl\vert S_{n}'\bigr\vert >\frac{\sqrt{n} (\log\log n )^{d/2}(x+\varepsilon)}{2} \biggr) \\ &\quad\leq\mathbb{P} \Biggl(\Biggl\vert \sum_{i=-\infty}^{+\infty}a_{i} \sum_{j=i+1}^{i+n}\xi_{j}I \bigl(| \xi_{j}|> (x+\varepsilon)\sqrt{n} (\log\log n )^{d/2} \bigr) \Biggr\vert >\frac{\sqrt{n}(\log\log n)^{d/2}(x+\varepsilon)}{2} \Biggr) \\ &\qquad{}+\mathbb{P} \biggl(\bigl\vert S_{n}'- \mathbb{E}S_{n}'\bigr\vert \geq\bigl\vert S_{n}'\bigr\vert -\bigl\vert \mathbb{E}S_{n}' \bigr\vert \\ &\qquad{}>\frac{\sqrt{n} (\log\log n )^{d/2}(x+\varepsilon)}{2}-\frac {\sqrt{n} (\log\log n )^{d/2}(x+\varepsilon)}{4} \biggr) \\ &\quad=\mathbb{P} \Biggl(\Biggl\vert \sum_{i=-\infty}^{+\infty}a_{i} \sum_{j=i+1}^{i+n}\xi_{j}I \bigl(| \xi_{j}|> (x+\varepsilon)\sqrt{n} (\log\log n )^{d/2} \bigr) \Biggr\vert >\frac{\sqrt{n}(\log\log n)^{d/2}(x+\varepsilon)}{2} \Biggr) \\ &\qquad{}+\mathbb{P} \Biggl(\Biggl\vert \sum_{i=-\infty}^{+\infty}a_{i} \sum_{j=i+1}^{i+n} \bigl(\xi_{j}'- \mathbb{E} \xi_{j}' \bigr)\Biggr\vert > \frac{\sqrt{n} (\log\log n )^{d/2}(x+\varepsilon)}{4} \Biggr). \end{aligned}$$
(22)
Therefore, by (22), we obtain
$$\begin{aligned} &\sum_{n>b(\varepsilon)}\frac{ (\log\log n )^{\beta -1}}{n\log n} \mathbb{E} \bigl\{ \vert S_{n}\vert /\sqrt{n}-\varepsilon (\log\log n )^{\frac{d}{2}} \bigr\} _{+} \\ &\quad=\sum_{n>b(\varepsilon)}\frac{ (\log\log n )^{\beta -1}}{n\log n} \int_{0}^{\infty}\mathbb{P} \bigl(\vert S_{n}\vert /\sqrt{n}\geq x+\varepsilon (\log\log n )^{d/2} \bigr)\,\mathrm{d}x \\ &\quad=\sum_{n>b(\varepsilon)}\frac{ (\log\log n )^{\beta-1+\frac {d}{2}}}{n\log n} \int_{0}^{\infty}\mathbb{P} \bigl( \vert S_{n}\vert \geq\sqrt{n} (\log\log n )^{\frac {d}{2}}(x+\varepsilon) \bigr)\,\mathrm{d}x \ \ \bigl(x=y (\log\log n )^{d/2} \bigr) \\ &\quad\leq\sum_{n>b(\varepsilon)}\frac{ (\log\log n )^{\beta -1+\frac{d}{2}}}{n\log n} \int_{0}^{\infty}\mathbb{P} \Biggl( \Biggl\vert \sum _{i=-\infty}^{+\infty}a_{i}\sum _{j=i+1}^{i+n}\xi_{j}I \bigl(\vert \xi_{j}\vert \\ &\qquad{} >(x+\varepsilon) \sqrt{n} (\log\log n )^{\frac{d}{2}} \bigr)\Biggr\vert >\frac {(x+\varepsilon)\sqrt{n} (\log\log n )^{\frac{d}{2}}}{2} \Biggr)\,\mathrm{d}x \\ &\qquad{}+\sum_{n>b(\varepsilon)}\frac{ (\log\log n )^{\beta-1+\frac {d}{2}}}{n\log n} \int_{0}^{\infty}\mathbb{P} \Biggl(\Biggl\vert \sum _{i=-\infty }^{+\infty}a_{i}\sum _{j=i+1}^{i+n} \bigl(\xi_{j}'- \mathbb{E} \xi_{j}' \bigr)\Biggr\vert \\ &\qquad{}> \frac{(x+\varepsilon)\sqrt{n} (\log\log n )^{\frac{d}{2}}}{4} \Biggr)\,\mathrm{d}x \\ &\quad:=J_{1}+J_{2}. \end{aligned}$$
For \(J_{1}\), when \(2\beta< d\), we have
$$\begin{aligned} \varepsilon^{2\beta/d}J_{1} \leq&c\varepsilon^{2\beta/d}\sum _{n>b(\varepsilon)}\frac{ (\log\log n )^{\beta-1+d/2}}{ n\log n} \int_{0}^{\infty}\frac{n\mathbb{E}\vert \xi_{1}\vert I (\vert \xi_{1}\vert >(x+\varepsilon)\sqrt{n} (\log\log n )^{d/2} )}{(x+\varepsilon)\sqrt{n} (\log\log n )^{d/2}}\,\mathrm{d}x \\ \leq&c\varepsilon^{2\beta/d}\sum_{n>b(\varepsilon)} \frac{ (\log \log n )^{\beta-1-d/2}}{ n\log n} \int_{0}^{\infty}\frac{\mathbb{E}\xi_{1}^{2}I (\vert \xi_{1}\vert >(x+\varepsilon)\sqrt{n} (\log\log n )^{d/2} )}{(x+\varepsilon)^{2}}\,\mathrm{d}x \\ \leq&c\varepsilon^{2\beta/d}\sum_{n>b(\varepsilon)} \frac{ (\log \log n )^{\beta-1-d/2}}{ n\log n} \int_{0}^{\infty}(x+\varepsilon)^{-2}\, \mathrm{d}x \\ \leq&c\varepsilon^{2\beta/d-1}\sum_{n>b(\varepsilon)} \frac{ (\log \log n )^{\beta-1-d/2}}{ n\log n} \\ \leq&cM^{\beta-d/2}, \end{aligned}$$
thus
$$ \lim_{M\rightarrow\infty}\varepsilon^{2\beta/d}J_{1}=0 $$
(23)
uniformly holds true with respect to \(0<\varepsilon<1\).
When \(2\beta\geq d\), we have
$$\begin{aligned} J_{1} \leq&c\sum_{n>b(\varepsilon)}\frac{ (\log\log n )^{\beta-1+d/2}}{ n\log n} \int_{0}^{\infty}\frac{n\mathbb{E}\vert \xi_{1}\vert I (\vert \xi_{1}\vert >(x+\varepsilon)\sqrt{n} (\log\log n )^{d/2} )}{(x+\varepsilon)\sqrt{n} (\log\log n )^{d/2}}\,\mathrm{d}x \\ =&c\sum_{n>b(\varepsilon)}\frac{ (\log\log n )^{\beta-1}}{ \sqrt{n}\log n} \int_{0}^{\infty}\frac{\mathbb{E}\vert \xi_{1}\vert I (\vert \xi_{1}\vert >(x+\varepsilon)\sqrt{n} (\log\log n )^{d/2} )}{x+\varepsilon}\,\mathrm{d}x \\ =&c \int_{0}^{\infty}\frac{1}{x+\varepsilon}\sum _{n>b(\varepsilon)}\frac { (\log\log n )^{\beta-1}}{ \sqrt{n}\log n}\sum_{j=n}^{\infty}\mathbb{E}|\xi_{1}|I \biggl(\sqrt{j} (\log\log j )^{d/2}< \frac{|\xi_{1}|}{x+\varepsilon}\\ &{}\leq\sqrt {j+1} \bigl(\log\log(j+1) \bigr)^{d/2} \biggr)\,\mathrm{d}x \\ =&c \int_{0}^{\infty}\frac{1}{x+\varepsilon}\sum _{j>b(\varepsilon)}\mathbb {E}|\xi_{1}|I \biggl(\sqrt{j} (\log \log j )^{\frac{d}{2}}< \frac {|\xi_{1}|}{x+\varepsilon}\leq\sqrt{j+1} \bigl(\log\log(j+1) \bigr)^{\frac {d}{2}} \biggr)\\ &{}\times\sum_{n=b(\varepsilon)+1}^{j} \frac{ (\log\log n )^{\beta-1}}{ \sqrt{n}\log n}\,\mathrm{d}x. \end{aligned}$$
Since we have \((\log\log n )^{\beta-1} (\log n )^{-1}\rightarrow0\), as \(n>b(\varepsilon)\rightarrow\infty\), then we obtain
$$\begin{aligned} J_{1} \leq&c \int_{0}^{\infty}\frac{1}{x+\varepsilon}\sum _{j>b(\varepsilon )}\sqrt{j}\mathbb{E}|\xi_{1}|I \biggl(\sqrt{j} (\log\log j )^{d/2}< \frac{|\xi_{1}|}{x+\varepsilon}\leq\sqrt{j+1} \bigl(\log\log (j+1) \bigr)^{d/2} \biggr)\,\mathrm{d}x \\ \leq&c \int_{0}^{\infty}\frac{1}{(x+\varepsilon)^{2}}\mathbb{E} \xi_{1}^{2}I \bigl(|\xi_{1}|>(x+\varepsilon) \sqrt{b (\varepsilon )} \bigl(\log\log b(\varepsilon) \bigr)^{d/2} \bigr) \,\mathrm{d}x \\ \leq&c \int_{0}^{\infty}\frac{1}{(x+\varepsilon)^{2}}\mathbb{E} \xi_{1}^{2}I \bigl(|\xi_{1}|>\varepsilon \bigl( \log\log b(\varepsilon) \bigr)^{d/2} \bigr)\,\mathrm{d}x \\ =&c\varepsilon^{-1}\mathbb{E}\xi_{1}^{2}I \bigl(|\xi_{1}|>\varepsilon \bigl(\log\log b(\varepsilon) \bigr)^{d/2}=M^{d/2} \bigr). \end{aligned}$$
Hence, when \(2\beta\geq d\),
$$ \lim_{M\rightarrow\infty}\varepsilon^{2\beta/d}J_{1} \leq\lim_{M\rightarrow\infty}\mathbb{E}\xi_{1}^{2}I \bigl( |\xi_{1}|>M^{d/2} \bigr)=0 $$
(24)
uniformly holds true with respect to \(0<\varepsilon<1\).
Next we consider \(J_{2}\), similar to the proof of \(\mathrm{I}_{32}\), we have
$$\begin{aligned}[b] \varepsilon^{2\beta/d}J_{2}&\leq c\varepsilon^{2\beta/d}\sum _{n>b(\varepsilon)}\frac{ (\log\log n )^{\beta -1+\frac{d}{2}-\frac{dq}{2}}}{n^{1+\frac{q}{2}}\log n} \int_{0}^{\infty}(x+\varepsilon )^{-q} \bigl( \bigl(n\mathbb{E} {\xi'}_{1}^{2} \bigr)^{q/2}+n\mathbb{E}\bigl\vert \xi_{1}' \bigr\vert ^{q} \bigr)\,\mathrm{d}x \\ &:=J_{21}+J_{22}. \end{aligned} $$
For \(J_{21}\), haven taken \(q>\max \{2\beta/d+1,2 \}\), then we have
$$\begin{aligned} J_{21} \leq&c\varepsilon^{2\beta/d}\sum _{n>b(\varepsilon)}\frac{ (\log\log n )^{\beta -1+\frac{d}{2}-\frac{dq}{2}}}{n\log n} \int_{0}^{\infty}(x+\varepsilon )^{-q}\, \mathrm{d}x \\ \leq&c\varepsilon^{2\beta/d-q+1}\sum_{n>b(\varepsilon)} \frac{ (\log\log n )^{\beta -1+\frac{d}{2}-\frac{dq}{2}}}{n\log n} \\ \leq&c\varepsilon^{2\beta/d-q+1} \bigl(\log\log b(\varepsilon) \bigr)^{\beta +\frac{d}{2}-\frac{dq}{2}} \\ =&M^{\beta+\frac{d}{2}-\frac{dq}{2}}. \end{aligned}$$
Thus
$$ \lim_{M\rightarrow\infty}J_{21}=0 $$
(25)
uniformly holds true with respect to \(0<\varepsilon<1\).
For \(J_{22}\), if \(2\beta< d\), we have
$$\begin{aligned} J_{22} \leq&c\varepsilon^{2\beta/d}\sum _{n>b(\varepsilon)}\frac{ (\log\log n )^{\beta -1+\frac{d}{2}-\frac{dq}{2}}}{n^{\frac{q}{2}}\log n}\\ &{}\times \int_{0}^{\infty}(x+\varepsilon )^{-q} \mathbb{E}\vert \xi_{1}\vert ^{q}I \bigl(\vert \xi_{1}\vert \leq(x+\varepsilon)\sqrt{n} (\log\log n )^{d/2} \bigr)\,\mathrm{d}x \\ \leq&c\varepsilon^{2\beta/d}\sum_{n>b(\varepsilon)} \frac{ (\log \log n )^{\beta -1-\frac{d}{2}}}{n\log n}\\ &{}\times\int_{0}^{\infty}(x+\varepsilon )^{-2} \mathbb{E} \xi_{1}^{2}I \bigl(\vert \xi_{1} \vert \leq(x+\varepsilon)\sqrt{n} (\log \log n )^{d/2} \bigr)\, \mathrm{d}x \\ \leq&c\varepsilon^{2\beta/d}\sum_{n>b(\varepsilon)} \frac{ (\log \log n )^{\beta -1-\frac{d}{2}}}{n\log n} \int_{0}^{\infty}(x+\varepsilon )^{-2}\, \mathrm{d}x \\ \leq&c\varepsilon^{2\beta/d-1}\sum_{n>b(\varepsilon)} \frac{ (\log \log n )^{\beta -1-\frac{d}{2}}}{n\log n} \\ =&cM^{\beta-d/2}. \end{aligned}$$
Thus
$$ \lim_{M\rightarrow\infty}J_{22}=0 $$
(26)
uniformly holds true with respect to \(0<\varepsilon<1\).
If \(2\beta\geq d\), we divide \(J_{22}\) into two parts:
$$\begin{aligned} J_{22} \leq&c\varepsilon^{2\beta/d}\sum _{n>b(\varepsilon)}\frac{ (\log\log n )^{\beta -1+\frac{d}{2}-\frac{dq}{2}}}{n^{\frac{q}{2}}\log n}\\ &{}\times \int_{0}^{\infty}(x+\varepsilon )^{-q} \mathbb{E}\vert \xi_{1}\vert ^{q}I \bigl(\vert \xi_{1}\vert \leq(x+\varepsilon)\sqrt{n} (\log\log n )^{d/2} \bigr)\,\mathrm{d}x \\ \leq&c\varepsilon^{2\beta/d}\sum_{n>b(\varepsilon)} \frac{ (\log \log n )^{\beta -1+\frac{d}{2}-\frac{dq}{2}}}{n^{\frac{q}{2}}\log n} \int_{0}^{\infty}(x+\varepsilon )^{-q} \mathbb{E}\vert \xi_{1}\vert ^{q} \bigl[I \bigl(\vert \xi_{1}\vert \leq\sqrt{n} (\log\log n )^{d/2}\varepsilon \bigr) \\ &{}+I \bigl(\sqrt{n} (\log\log n )^{d/2}\varepsilon< \vert \xi _{1}\vert \leq(x+\varepsilon)\sqrt{n} (\log\log n )^{d/2} \bigr) \bigr]\,\mathrm{d}x \\ :=&J_{221}+J_{222}. \end{aligned}$$
For \(J_{221}\), since \(q>2\), so we have
$$\begin{aligned} J_{221} \leq&c\varepsilon^{2\beta/d-q+1}\sum _{n>b(\varepsilon)}\frac { (\log\log n )^{\beta-1+\frac{d}{2}- \frac{dq}{2}}}{n^{\frac{q}{2}}\log n}\mathbb{E}|\xi_{1}|^{q}I \bigl(|\xi _{1}|\leq\sqrt{b(\varepsilon)} \bigl( \log\log b( \varepsilon) \bigr)^{\frac{d}{2}}\varepsilon \bigr) \\ &{}+c\varepsilon^{2\beta/d-q+1}\sum_{n>b(\varepsilon)} \frac{ (\log \log n )^{\beta-1+\frac{d}{2}- \frac{dq}{2}}}{n^{\frac{q}{2}}\log n}\\ &{}\times\sum_{j=b(\varepsilon)+1}^{n}\mathbb {E}| \xi_{1}|^{q}I \bigl(\sqrt{j-1} \bigl( \log\log (j-1 ) \bigr)^{\frac{d}{2}}\varepsilon< |\xi_{1}|\leq \sqrt{j} ( \log\log j )^{\frac{d}{2}}\varepsilon \bigr) \\ \leq&c\varepsilon^{2\beta/d-1}\frac{ (\log\log b(\varepsilon ) )^{\beta-1-\frac{d}{2}}}{\log b(\varepsilon)} \mathbb{E} \xi_{1}^{2}I \bigl(|\xi_{1}|\leq\sqrt{b( \varepsilon)} \bigl(\log\log b(\varepsilon) \bigr)^{\frac{d}{2}}\varepsilon \bigr) \\ &{}+c\varepsilon^{2\beta/d-q+1}\sum_{j=b(\varepsilon)+1}^{\infty}\mathbb {E}|\xi_{1}|^{q}I \bigl(\sqrt{j-1} \bigl( \log\log (j-1 ) \bigr)^{\frac{d}{2}}\varepsilon< |\xi_{1}|\leq \sqrt{j} ( \log \log j )^{\frac{d}{2}}\varepsilon \bigr)\\ &{}\times\sum_{n=j}^{\infty}\frac { (\log\log n )^{\beta-1+\frac{d}{2}-\frac{dq}{2}}}{n^{\frac {q}{2}}\log n} \\ \leq&c\varepsilon^{2\beta/d-1}\frac{1}{\log\log b(\varepsilon)} \\ &{}+c\varepsilon^{2\beta/d-q+1}\sum_{j=b(\varepsilon)+1}^{\infty}\mathbb {E}|\xi_{1}|^{q}I \bigl(\sqrt{j-1} \bigl( \log\log (j-1 ) \bigr)^{\frac{d}{2}}\varepsilon< |\xi_{1}|\leq \sqrt{j} ( \log \log j )^{\frac{d}{2}}\varepsilon \bigr)j^{1-\frac{q}{2}}\\ &{}\times\frac { (\log\log j )^{\beta-1+\frac{d}{2}-\frac{dq}{2}}}{\log j} \\ \leq&c\varepsilon^{2\beta/d-1+d/2}M^{-1} +c\varepsilon^{2\beta/d-1}\sum_{j=b(\varepsilon)+1}^{\infty}\frac{ (\log\log j )^{\beta-1-\frac{d}{2}}}{\log j}\\ &{}\times\mathbb{E}\xi_{1}^{2}I \bigl(\sqrt{j-1} \bigl( \log\log (j-1 ) \bigr)^{\frac{d}{2}}\varepsilon< |\xi_{1}|\leq \sqrt{j} ( \log\log j )^{\frac{d}{2}}\varepsilon \bigr) \\ \leq&cM^{-1}+c\mathbb{E}\xi_{1}^{2}I \bigl(| \xi_{1}|> \sqrt{b(\varepsilon)} \bigl( \log\log b(\varepsilon) \bigr)^{\frac{d}{2}}\varepsilon>M^{\frac {d}{2}} \bigr). \end{aligned}$$
Thus
$$ \lim_{M\rightarrow\infty}J_{221}\leq\lim _{M\rightarrow\infty }M^{-1}+\lim_{M\rightarrow\infty}\mathbb{E} \xi_{1}^{2}I \bigl(|\xi_{1}|>M^{\frac{d}{2}} \bigr)=0 $$
(27)
uniformly holds true with respect to \(0<\varepsilon<1\).
Finally we consider \(J_{222}\), since \(2\beta\geq d\) and \(q>2\), we have
$$\begin{aligned} J_{222} \leq&c\varepsilon^{2\beta/d}\sum _{n>b(\varepsilon)}\frac{ (\log\log n )^{\beta-1+\frac{d}{2}-\frac{dq}{2}}}{n^{\frac {q}{2}}\log n}\mathbb{E}|\xi_{1}|^{q}I \bigl(|\xi_{1}|>\sqrt{n} (\log\log n )^{\frac{d}{2}}\varepsilon \bigr) \\ &{}\times \int_{0}^{\infty}(x+\varepsilon )^{-q} I \bigl(|\xi_{1}|\leq(x+\varepsilon)\sqrt{n} (\log\log n )^{\frac {d}{2}} \bigr)\,\mathrm{d}x \\ \leq&c\varepsilon^{2\beta/d}\sum_{n>b(\varepsilon)} \frac{ (\log \log n )^{\beta-1+\frac{d}{2}-\frac{dq}{2}}}{n^{\frac{q}{2}}\log n}\mathbb{E}|\xi_{1}|^{q}I \bigl(| \xi_{1}|>\sqrt{n} (\log\log n )^{\frac{d}{2}}\varepsilon \bigr)\\ &{}\times \int_{\frac{|\xi_{1}|}{\sqrt{n} (\log \log n )^{\frac{d}{2}}}-\varepsilon}^{\infty}(x+\varepsilon )^{-q}\, \mathrm{d}x \\ \leq&c\varepsilon^{2\beta/d}\sum_{n>b(\varepsilon)} \frac{ (\log \log n )^{\beta-1}}{\sqrt{n}\log n} \mathbb{E}|\xi_{1}|I \bigl(|\xi_{1}|> \sqrt{n} (\log\log n )^{\frac {d}{2}}\varepsilon \bigr) \\ \leq&c\varepsilon^{2\beta/d}\sum_{n>b(\varepsilon)} \frac{ (\log \log n )^{\beta-1}}{\sqrt{n}\log n}\\ &{}\times\sum_{j=n}^{\infty}\mathbb{E}| \xi _{1}|I \bigl(\sqrt{j}(\log\log j)^{\frac{d}{2}}\varepsilon< | \xi_{1}|\leq \sqrt{j+1} \bigl(\log\log(j+1) \bigr)^{\frac{d}{2}} \varepsilon \bigr) \\ \leq&c\varepsilon^{2\beta/d}\sum_{j=b(\varepsilon)}^{\infty}\mathbb {E}|\xi_{1}|I \bigl(\sqrt{j}(\log\log j)^{\frac{d}{2}} \varepsilon< |\xi _{1}|\leq\sqrt{j+1} \bigl(\log\log(j+1) \bigr)^{\frac{d}{2}}\varepsilon \bigr) \\ &{}\times\sum_{n=b(\varepsilon)}^{j} \frac{ (\log\log n )^{\beta-1}}{\sqrt {n}\log n} \\ \leq&c\varepsilon^{2\beta/d}\sum_{j=b(\varepsilon)}^{\infty}\sqrt {j}\mathbb{E}|\xi_{1}|I \bigl(\sqrt{j}(\log\log j)^{\frac{d}{2}} \varepsilon < |\xi_{1}|\leq\sqrt{j+1} \bigl(\log\log(j+1) \bigr)^{\frac {d}{2}}\varepsilon \bigr) \\ \leq&c\varepsilon^{2\beta/d-1}\sum_{j=b(\varepsilon)}^{\infty}\mathbb {E}\xi_{1}^{2}I \bigl(\sqrt{j}(\log\log j)^{\frac{d}{2}}\varepsilon< |\xi _{1}|\leq\sqrt{j+1} \bigl(\log \log(j+1) \bigr)^{\frac{d}{2}}\varepsilon \bigr) \\ \leq&c\varepsilon^{2\beta/d-1}\mathbb{E}\xi_{1}^{2}I \bigl(|\xi_{1}|\geq \varepsilon\sqrt{b(\varepsilon)} \bigl( \log\log b( \varepsilon) \bigr)^{\frac{d}{2}}>M^{\frac{d}{2}} \bigr). \end{aligned}$$
Hence,
$$ \lim_{M\rightarrow\infty}J_{222}\leq\lim _{M\rightarrow\infty }\varepsilon^{2\beta/d-1}\mathbb{E}\xi_{1}^{2}I \bigl( |\xi_{1}|>M^{d/2} \bigr)\leq\lim_{M\rightarrow\infty} \mathbb{E}\xi _{1}^{2}I \bigl( |\xi_{1}|>M^{d/2} \bigr)=0 $$
(28)
uniformly holds true with respect to \(0<\varepsilon<1\).
Thus, combining with (23)-(28), we complete the proof of this proposition. □
Combining with Proposition 3.1, Proposition 3.2, and Proposition 3.3 we complete the proof of Theorem 1.2.