For fixed
\(n\ge1\), denote
$$Y_{ni}=A_{i}X_{i}I\bigl( \vert X_{i} \vert \le c_{n}\bigr)-E\bigl[A_{i}X_{i}I \bigl( \vert X_{i} \vert \le c_{n}\bigr)|\mathcal {G}_{i-1}\bigr],\quad i=1, 2, \ldots. $$
Since
$$A_{i}X_{i}=A_{i}X_{i}I\bigl( \vert X_{i} \vert >c_{n}\bigr)+Y_{ni}+E \bigl[A_{i}X_{i}I\bigl( \vert X_{i} \vert \le c_{n}\bigr)|\mathcal{G}_{i-1}\bigr], $$
we have
$$\begin{aligned}& \sum_{n=1}^{\infty}b_{n}l(n)P \Biggl(\max_{1\le j\le n} \Biggl\vert \sum _{i=1}^{j}A_{i}X_{i} \Biggr\vert \ge\epsilon c_{n} \Biggr) \\& \quad \le \sum_{n=1}^{\infty}b_{n}l(n)P \Biggl(\max_{1\le j\le n} \Biggl\vert \sum _{i=1}^{j}A_{i}X_{i}I \bigl( \vert X_{i} \vert >c_{n}\bigr) \Biggr\vert \ge \epsilon c_{n}/3 \Biggr) \\& \qquad {} +\sum_{n=1}^{\infty}b_{n}l(n)P \Biggl(\max_{1\le j\le n} \Biggl\vert \sum _{i=1}^{j}E\bigl[A_{i}X_{i}I \bigl( \vert X_{i} \vert \le c_{n}\bigr)| \mathcal{G}_{i-1}\bigr] \Biggr\vert \ge\epsilon c_{n}/3 \Biggr) \\& \qquad {} +\sum_{n=1}^{\infty}b_{n}l(n)P \Biggl(\max_{1\le j\le n} \Biggl\vert \sum _{i=1}^{j}Y_{ni} \Biggr\vert \ge \epsilon c_{n}/3 \Biggr) \\& \quad = :H+I+J. \end{aligned}$$
(3.1)
To prove (
2.5), it is enough to show
\(H<\infty, I<\infty\), and
\(J<\infty\). Obviously, it follows from Hölder’s inequality, Lyapunov’s inequality, and (
2.3) that
$$ \sum_{i=1}^{n}E|A_{i}| \le \Biggl(\sum_{i=1}^{n}E|A_{i}|^{q} \Biggr)^{\frac {1}{q}} \Biggl(\sum_{i=1}^{n}1 \Biggr)^{1-\frac{1}{q}}=O(n). $$
(3.2)
By the fact that
\(\{A_{n}, n\ge1\}\) is independent of
\(\{X_{n}, n\ge1\}\), it is easy to check by Markov’s inequality, Lemma
3.2, (
3.2), (
2.1), and (
2.4) that
$$\begin{aligned} H \le& C\sum_{n=1}^{\infty}\frac{b_{n}l(n)}{c_{n}}E \Biggl(\max_{1\le j\le n} \Biggl\vert \sum _{i=1}^{j}A_{i}X_{i}I\bigl( \vert X_{i} \vert >c_{n}\bigr) \Biggr\vert \Biggr) \\ \le& C\sum_{n=1}^{\infty}\frac{b_{n}l(n)}{c_{n}} \sum_{i=1}^{n}E \vert A_{i} \vert E\bigl[ \vert X_{i} \vert I\bigl( \vert X_{i} \vert >c_{n}\bigr)\bigr] \\ \le& C\sum_{n=1}^{\infty}\frac{nb_{n}l(n)}{c_{n}}E \bigl[ \vert X \vert I\bigl( \vert X \vert >c_{n}\bigr)\bigr] \\ =& C\sum_{n=1}^{\infty}\frac{nb_{n}l(n)}{c_{n}}\sum _{m=n}^{\infty}E\bigl[ \vert X \vert I \bigl(c_{m}< \vert X \vert \le c_{m+1}\bigr)\bigr] \\ =& C\sum_{m=1}^{\infty}E\bigl[ \vert X \vert I\bigl(c_{m}< \vert X \vert \le c_{m+1}\bigr)\bigr] \sum_{n=1}^{m}\frac {nb_{n}l(n)}{c_{n}} \\ \le& C\sum_{m=1}^{\infty}E\bigl[ \vert X \vert I\bigl(c_{m}< \vert X \vert \le c_{m+1}\bigr) \bigr]c_{m}^{p-1}l\bigl(c_{m}^{1/\alpha}\bigr) \\ \le& CE \vert X \vert ^{p}l\bigl( \vert X \vert ^{1/\alpha}\bigr)< \infty. \end{aligned}$$
(3.3)
For
I, since
\(\{X_{n}, \mathcal{F}_{n}, n\ge1\}\) is a sequence of martingale difference, we can see that
\(\{X_{n}, \mathcal{G}_{n}, n\ge1\} \) is also a sequence of martingale difference. Combining with the fact that
\(\{A_{n}, n\ge1\}\) is independent of
\(\{X_{n}, n\ge1\}\), we have
$$\begin{aligned} E(A_{n}X_{n}|\mathcal{G}_{n-1})&=E \bigl[E(A_{n}X_{n}|\mathcal{G}_{n})|\mathcal {G}_{n-1}\bigr] \\ &=E\bigl[X_{n}E(A_{n}|\mathcal{G}_{n})| \mathcal{G}_{n-1}\bigr] \\ &=EA_{n}E(X_{n}|\mathcal{G}_{n-1}) \\ &=0\quad \mbox{a.s.}, n\ge1. \end{aligned}$$
Consequently, by Markov’s inequality and the proof of (
3.3), we have
$$\begin{aligned} I \le&C\sum_{n=1}^{\infty}\frac{b_{n}l(n)}{c_{n}}E \Biggl(\max_{1\le j\le n} \Biggl\vert \sum _{i=1}^{j}E\bigl[A_{i}X_{i}I \bigl( \vert X_{i} \vert \le c_{n}\bigr)| \mathcal{G}_{i-1}\bigr] \Biggr\vert \Biggr) \\ \le&C\sum_{n=1}^{\infty}\frac{b_{n}l(n)}{c_{n}}E \Biggl(\max_{1\le j\le n} \Biggl\vert \sum _{i=1}^{j}E\bigl[A_{i}X_{i}I \bigl( \vert X_{i} \vert >c_{n}\bigr)| \mathcal{G}_{i-1}\bigr] \Biggr\vert \Biggr) \\ \le&C\sum_{n=1}^{\infty}\frac{b_{n}l(n)}{c_{n}}\sum _{i=1}^{n}E \vert A_{i} \vert E\bigl[ \vert X_{i} \vert I\bigl( \vert X_{i} \vert >c_{n}\bigr)\bigr] \\ \le&CE \vert X \vert ^{p}l\bigl( \vert X \vert ^{1/\alpha}\bigr)< \infty. \end{aligned}$$
(3.4)
Next, we shall show that
\(J<\infty\). Let
\(X_{ni}=X_{i}I(|X_{i}|\le c_{n})\) and
\(\hat{Y}_{ni}=a_{i}X_{ni}-E(a_{i}X_{ni}|\mathcal{G}_{i-1})\). It can be found that for fixed real numbers
\(a_{1}, \ldots, a_{n}\),
\(\{\hat {Y}_{ni}, \mathcal{G}_{i}, 1\le i\le n\}\) is also a sequence of martingale difference. Note that
\(\{A_{1}, \ldots, A_{n}\}\) is independent of
\(\{X_{n1}, \ldots, X_{nn}\}\). So, by Markov’s inequality and Lemma
3.1, we have
$$\begin{aligned} J \le&C\sum_{n=1}^{\infty}\frac{b_{n}l(n)}{c_{n}^{q}}E \Biggl(\max_{1\le j\le n} \Biggl\vert \sum _{i=1}^{j}Y_{ni} \Biggr\vert \Biggr)^{q} \\ =&C\sum_{n=1}^{\infty}\frac{b_{n}l(n)}{c_{n}^{q}}E \Biggl\{ E \Biggl(\max_{1\le j\le n} \Biggl\vert \sum _{i=1}^{j}\bigl[a_{i}X_{ni}-E(a_{i}X_{ni}| \mathcal {G}_{i-1})\bigr] \Biggr\vert ^{q} \Biggr) \Big|A_{1}=a_{1}, \ldots, A_{n}=a_{n} \Biggr\} \\ \le&C\sum_{n=1}^{\infty}\frac{b_{n}l(n)}{c_{n}^{q}}E \Biggl\{ E \Biggl(\sum_{i=1}^{n}E \bigl( \hat{Y}_{ni}^{2}|\mathcal{G}_{i-1} \bigr) \Biggr)^{q/2}+\sum_{i=1}^{n}E| \hat{Y}_{ni}|^{q} \Big|A_{1}=a_{1}, \ldots, A_{n}=a_{n} \Biggr\} \\ =&C\sum_{n=1}^{\infty}\frac{b_{n}l(n)}{c_{n}^{q}}\sum _{i=1}^{n}E|Y_{ni}|^{q}+C \sum_{n=1}^{\infty}\frac{b_{n}l(n)}{c_{n}^{q}}E \Biggl( \sum_{i=1}^{n}E \bigl(Y_{ni}^{2}| \mathcal{G}_{i-1} \bigr) \Biggr)^{q/2} \\ =:&J_{1}+J_{2}. \end{aligned}$$
(3.5)
For
\(J_{1}\), we have by
\(C_{r}\)-inequality, Lemma
3.2 with
\(b=c_{n}\), and (
2.3) that
$$\begin{aligned} J_{1} \le&C\sum_{n=1}^{\infty}\frac{b_{n}l(n)}{c_{n}^{q}}\sum_{i=1}^{n}E \vert A_{i} \vert ^{q}E \bigl[ \vert X_{i} \vert ^{q}I\bigl( \vert X_{i} \vert \le c_{n}\bigr) \bigr] \\ \le&C\sum_{n=1}^{\infty}\frac{b_{n}l(n)}{c_{n}^{q}}\sum _{i=1}^{n}E \vert A_{i} \vert ^{q}E \bigl[ \vert X \vert ^{q}I\bigl( \vert X \vert \le c_{n}\bigr) \bigr]+C\sum_{n=1}^{\infty}\frac{b_{n}l(n)}{c_{n}^{q}}\sum_{i=1}^{n}c_{n}^{q}P \bigl( \vert X \vert >c_{n}\bigr) \\ =&C\sum_{n=1}^{\infty}\frac{nb_{n}l(n)}{c_{n}^{q}}E \bigl[ \vert X \vert ^{q}I\bigl( \vert X \vert \le c_{n}\bigr) \bigr]+C\sum_{n=1}^{\infty}nb_{n}l(n)P\bigl( \vert X \vert >c_{n}\bigr) \\ \le&C\sum_{n=1}^{\infty}\frac{nb_{n}l(n)}{c_{n}^{q}}E \bigl[ \vert X \vert ^{q}I\bigl( \vert X \vert \le c_{n}\bigr) \bigr]+C\sum_{n=1}^{\infty}\frac{nb_{n}l(n)}{c_{n}}E\bigl[ \vert X \vert I\bigl( \vert X \vert >c_{n}\bigr)\bigr] \\ =:&CJ_{11}+CJ_{12}. \end{aligned}$$
(3.6)
For
\(J_{11}\), we have by (
2.1) and (
2.4) that
$$\begin{aligned} J_{11} =&\sum_{n=1}^{\infty}\frac{nb_{n}l(n)}{c_{n}^{q}}\sum_{m=1}^{n}E \bigl[ \vert X \vert ^{q}I\bigl(c_{m-1}< \vert X \vert \le c_{m}\bigr) \bigr] \\ =&\sum_{m=1}^{\infty}E \bigl[ \vert X \vert ^{q}I\bigl(c_{m-1}< \vert X \vert \le c_{m}\bigr) \bigr]\sum_{n=m}^{\infty}\frac{nb_{n}l(n)}{c_{n}^{q}} \\ \le&C\sum_{m=1}^{\infty}E \bigl[ \vert X \vert ^{q}I\bigl(c_{m-1}< \vert X \vert \le c_{m}\bigr) \bigr]c_{m}^{p-q}l\bigl(c_{m}^{1/\alpha} \bigr) \\ \le&CE \vert X \vert ^{p}l\bigl( \vert X \vert ^{1/\alpha}\bigr)< \infty. \end{aligned}$$
(3.7)
By the proof of (
3.3), it follows
$$\begin{aligned} J_{12}&=\sum_{n=1}^{\infty}\frac{nb_{n}l(n)}{c_{n}}E\bigl[ \vert X \vert I\bigl( \vert X \vert >c_{n}\bigr)\bigr] \\ &\le CE \vert X \vert ^{p}l\bigl( \vert X \vert ^{1/\alpha}\bigr)< \infty. \end{aligned}$$
(3.8)
Furthermore, by Hölder’s inequality and (
2.3), for any
\(1< p\le2\), we have
$$ \sum_{i=1}^{n}E|A_{i}|^{p} \le \Biggl(\sum_{i=1}^{n}E|A_{i}|^{q} \Biggr)^{\frac {p}{q}} \Biggl(\sum_{i=1}^{n}1 \Biggr)^{1-\frac{p}{q}}=O(n). $$
(3.9)
Obviously, for
\(1\le i\le n\), it has
$$\begin{aligned} E \bigl(Y_{ni}^{2}|\mathcal{G}_{i-1} \bigr) =&E \bigl[A_{i}^{2}X_{i}^{2}I\bigl( \vert X_{i} \vert \le c_{n}\bigr)|\mathcal{G}_{i-1} \bigr] \\ &{}- \bigl[E\bigl(A_{i}X_{i}I\bigl( \vert X_{i} \vert \le c_{n}\bigr)|\mathcal{G}_{i-1} \bigr) \bigr]^{2} \\ \le&E \bigl[A_{i}^{2}X_{i}^{2}I\bigl( \vert X_{i} \vert \le c_{n}\bigr)|\mathcal{G}_{i-1} \bigr] \\ \le&EA_{i}^{2}E \bigl(X_{i}^{2}| \mathcal{G}_{i-1} \bigr),\quad \mbox{a.s.} \end{aligned}$$
(3.10)
Combining (
3.9) and (
2.2), we obtain that
$$\begin{aligned} J_{2} \le&\sum_{n=1}^{\infty}\frac{b_{n}l(n)}{c_{n}^{q}}E \Biggl(\sum_{i=1}^{n}EA_{i}^{2}E \bigl(X_{i}^{2}|\mathcal{G}_{i-1} \bigr) \Biggr)^{q/2} \\ \le&\sum_{n=1}^{\infty}\frac{b_{n}l(n)}{c_{n}^{q}} \Biggl(\sum_{i=1}^{n}EA_{i}^{2} \Biggr)^{q/2}E \Bigl(\sup_{i\ge1}E \bigl(X_{i}^{2}| \mathcal{G}_{i-1} \bigr) \Bigr)^{q/2} \\ \le&C\sum_{n=1}^{\infty}\frac{n^{q/2}b_{n}l(n)}{c_{n}^{q}}< \infty. \end{aligned}$$
(3.11)
By (
3.1) and (
3.3)–(
3.11), we can get (
2.5). This completes the proof of Theorem
2.1.