By a Beveridge and Nelson decomposition for a linear process, for
\(m,n,t\in\mathbb{N}\), let
$$\begin{gathered} Y_{m,n}=(2n\log\log{n})^{-\frac{1}{2}}\sum _{t=1}^{n}\sum _{j=-m}^{m} a_{j} \xi_{t-j}, \\ \widetilde{a}_{m}=0, \qquad\widetilde{a}_{j}=\sum _{i=j+1}^{m}a_{i},\quad j=0,1,\ldots,m-1, \\ \widetilde{\widetilde{a}}_{-m}=0,\qquad \widetilde{ \widetilde{a}}_{j} =\sum_{i=-m}^{j-1}a_{i}, \quad j=-m+1,-m+2,\ldots,0, \\ \widetilde{\xi}_{t}=\sum_{j=0}^{m} \widetilde{a}_{j}\xi_{t-j}, \quad\quad \widetilde{\widetilde{ \xi}}_{t}=\sum_{j=-m}^{0} \widetilde{\widetilde{a}}_{j} \xi_{t-j}. \end{gathered} $$
Obviously
$$\begin{aligned}& Y_{m,n}=\Biggl(\sum_{j=-m}^{m} a_{j}\Biggr) (2n\log\log{n})^{-\frac{1}{2}}\Biggl(\sum _{t=1}^{n}\xi_{t}\Biggr)+ (2n\log \log{n})^{-\frac{1}{2}}(\widetilde{\xi}_{0} -\widetilde{ \xi}_{n}+\widetilde{\widetilde{\xi}}_{n+1}- \widetilde{ \widetilde{\xi}}_{1}), \end{aligned}$$
(3.12)
$$\begin{aligned}& (2n\log\log{n})^{-\frac{1}{2}}\sum _{t=1}^{n}X_{t}=Y_{m,n}+(2n\log \log {n})^{-\frac{1}{2}} \Biggl(\sum_{t=1}^{n} \sum_{|j|>m}a_{j}\xi_{t-j} \Biggr). \end{aligned}$$
(3.13)
By the strictly stationarity, for every
\(\varepsilon>0\), we have
$$ \sum_{n=1}^{\infty}P\bigl\{ | \xi_{n-j}|/(2n\log\log{n})^{\frac{1}{2}}> \varepsilon\bigr\} \leq\sum _{n=1}^{\infty}P\bigl\{ |\xi_{0}|^{2}> 2{\varepsilon}^{2} n\log\log{n}\bigr\} \leq C E|\xi_{0}|^{2}< \infty. $$
(3.14)
Then by the Borel-Cantelli lemma, for any
\(j\geq0\),
$$ (2n\log\log{n})^{-\frac{1}{2}}\xi_{n-j}\rightarrow0 \quad\text{a.s. } n \rightarrow\infty. $$
Therefore
$$ (2n\log\log{n})^{-\frac{1}{2}} \cdot\widetilde{\xi}_{n}=(2n\log \log{n})^{-\frac{1}{2}}\sum_{j=0}^{m} \widetilde{{a}}_{j} \xi_{n-j} \rightarrow0 \quad\text{a.s. } n \rightarrow\infty. $$
Similarly, we obtain
$$\begin{gathered} (2n\log\log{n})^{-\frac{1}{2}} \cdot\widetilde{\xi}_{0}\rightarrow0 \quad\text{a.s. } n\rightarrow\infty, \\(2n\log\log{n})^{-\frac{1}{2}} \cdot\widetilde{\widetilde{\xi}}_{1} \rightarrow0 \quad\text{a.s. } n\rightarrow\infty, \\(2n\log\log{n})^{-\frac{1}{2}} \cdot\widetilde{\widetilde{\xi}}_{n+1} \rightarrow0 \quad\text{a.s. } n\rightarrow\infty.\end{gathered} $$
By the above statement, we have
$$ (2n\log\log{n})^{-\frac{1}{2}} (\widetilde{\xi}_{0} - \widetilde{\xi}_{n}+\widetilde{\widetilde{\xi}}_{n+1}- \widetilde{\widetilde{\xi}}_{1})\rightarrow0 \quad\text{a.s. } n \rightarrow\infty. $$
(3.15)
By Theorem
2.1
$$ \limsup_{n\rightarrow\infty}(2n\log\log{n})^{-\frac{1}{2}} \sum _{t=1}^{n}{\xi}_{t}=\sigma \quad \text{a.s.} $$
From the definition of LNQD and Lemma
2.3, it is easy to check that
\(\{-\xi_{i};i\in\mathbb{Z}\}\) is an LNQD sequence of random variables. Then, by Theorem
2.1,
$$ \limsup_{n\rightarrow\infty}(2n\log\log{n})^{-\frac{1}{2}} \sum _{t=1}^{n}{(-\xi}_{t})=\sigma \quad \text{a.s.} $$
Thus
$$ \limsup_{n\rightarrow\infty}(2n\log\log{n})^{-\frac{1}{2}} \Bigg| \sum_{t=1}^{n}{\xi}_{t}\Bigg|=\sigma \quad\text{a.s.} $$
(3.16)
Let
\(S_{n}=\sum_{t=1}^{n}X_{t}\), combining (
3.12)-(
3.16), then
$$ \begin{aligned}[b] &\limsup_{n\rightarrow \infty}(2n\log \log{n})^{-\frac{1}{2}} |S_{n}| \\ &\quad=\limsup_{n\rightarrow \infty}\Bigg|Y_{m,n}+\sum _{|j|>m}a_{j} (2n\log\log{n})^{-\frac{1}{2}}\sum _{t=1}^{n}{\xi}_{t-j}\Bigg| \\ &\quad\leq\limsup_{n\rightarrow \infty}\Bigg|\sum _{j=-m}^{m}a_{j}\Bigg| (2n\log \log{n})^{-\frac{1}{2}}\Bigg|\sum_{t=1}^{n}{ \xi}_{t}\Bigg| \\ &\qquad+\limsup_{n\rightarrow\infty} \sum_{|j|>m}|a_{j}|(2n \log\log{n})^{-\frac{1}{2}}\Bigg|\sum_{t=1}^{n}{ \xi}_{t-j}\Bigg| \\ &\quad\leq\Bigg|\sum_{j=-m}^{m}a_{j}\Bigg| \sigma+ \sum_{|j|>m}|a_{j}|\sup _{n}(2n\log\log{n})^{-\frac{1}{2}}\Bigg|\sum _{t=1}^{n}{\xi}_{t-j}\Bigg| \quad\text{a.s.}\end{aligned} $$
(3.17)
Then by the strictly stationarity, Lemma
3.2 and Lemma
3.3, we know
$$ E\sup_{n}(2n\log\log{n})^{-\frac{1}{2}}\Bigg| \sum _{t=1}^{n}{\xi}_{t-j}\Bigg| =E\sup _{n}(2n\log\log{n})^{-\frac{1}{2}}\Bigg| \sum _{t=1}^{n}{\xi}_{t}\Bigg|< \infty. $$
(3.18)
Then, by (
3.18),
$$ \sup_{n}(2n\log\log{n})^{-\frac{1}{2}}\Bigg| \sum _{t=1}^{n}{\xi}_{t-j}\Bigg|< \infty \quad\text{a.s.} $$
(3.19)
By (
3.19), letting
\(m\to\infty\) in (
3.17), we have
$$ \limsup_{n\rightarrow\infty}(2n\log\log{n})^{-\frac{1}{2}} |S_{n}|\leq4\Bigg|\sum_{j=-\infty}^{\infty}a_{j}\Bigg| \sigma \quad\text{a.s.} $$
(3.20)
On the other hand, by (
3.13), (
3.15) and (
3.16), we obtain
$$ \begin{aligned}[b] &\limsup_{n\rightarrow \infty}(2n\log \log{n})^{-\frac{1}{2}} |S_{n}| \\ &\quad\geq\limsup_{n\rightarrow\infty}\Bigg|\sum _{j=-m}^{m}a_{j}\Bigg|(2n\log \log{n})^{-\frac{1}{2}} \Bigg|\sum_{t=1}^{n}{ \xi}_{t}\Bigg| \\ &\qquad-\lim_{n\rightarrow \infty} (2n\log\log{n})^{-\frac{1}{2}}| \widetilde{\xi}_{0} -\widetilde{\xi}_{n}+\widetilde{ \widetilde{\xi}}_{n+1}- \widetilde{\widetilde{\xi}}_{1}| \\ &\qquad-\sum_{|j|>m}|a_{j}|\sup _{n} (2n\log\log{n})^{-\frac{1}{2}}\Bigg| \sum _{t=1}^{n}{\xi}_{t-j}\Bigg| \\ &\quad=\Bigg|\sum_{j=-m}^{m}a_{j}\Bigg|\sigma -\sum_{|j|>m}|a_{j}|\sup _{n} (2n\log\log{n})^{-\frac{1}{2}}\Bigg|\sum _{t=1}^{n}{\xi}_{t-j}\Bigg| \quad\text{a.s.}\end{aligned} $$
(3.21)
Then, letting
\(m\to\infty\),
$$ \limsup_{n\rightarrow\infty}(2n\log\log{n})^{-\frac{1}{2}} |S_{n}|\geq\Bigg|\sum_{j=-\infty}^{\infty}a_{j}\Bigg| \sigma \quad\text{a.s.} $$
(3.22)
Hence from (
3.20) and (
3.22) the desired conclusion (
3.1) follows. □