Let $\{X_n, n\ge 1\}$ be a sequence of random variables defined on a probability space $(\Omega, \mathscr F, \mathbb P)$ and let $\{a_{ni}, 1 \le i \le n, n\ge 1 \}$ be a triangular array of constants. Weighted sums of the form $\sum_{i=1}^n a_{ni}X_i$ arise in a variety of applications, such as least squares estimation, nonparametric regression function estimate and jackknife estimate. Because of the wide applications of weighted sums in statistics, many researchers have paid much attention to its properties.

When $\{X_n, n\ge 1\}$ is a sequence of independent identically distributed (i.i.d.) random variables, the almost sure convergence of weighted sums $\sum_{i=1}^n a_{ni}X_i$ has been founded in Choi and Sung [1], Chow [2], Chow and Lai [3], Li et al. [7], Stout [8], Sung [9, 10], Teicher [11], Thrum [12] and so on.

This paper focus on the above almost sure convergence of the weighted sums for the case which $\{X_n, n\ge 1\}$ is a martingale difference sequence. To describe the results of this paper, suppose that $(\Omega, \mathscr F, \mathbb P)$ is a probability space and let $\{\mathscr F_n, n\ge 0\}$ be a family of $\sigma$-algebras, such that $\mathscr F_0 \subseteq \mathscr F_1 \subseteq \cdots \subseteq \mathscr F.$

Let $\{X_n, n\ge 1\}$ be a martingale difference sequence with respect to the filtration $\{\mathscr F_n, n\ge0\}$ with $X_0=0$, i.e.,

ⅰ) $\mathbb E |X_n| < \infty$, for all $n \ge 1$,

ⅱ) $X_n$ is $\mathscr F_n$-measurable,

ⅲ) $\mathbb E (X_n|\mathscr F_{n-1})=0$ a.s. for every $n \ge 1$.

Throughout this paper, $C$ denotes a positive constant, which may take different values whenever it appears in different expressions. $1_{A}$ denotes the indicator function of the event $A$. $\log x$ denotes $\ln \max \{x, e\}, $ where ln is the natural logarithm.

Now we state our main result as follows.

Theorem 1.1  Let $\{X_n, n\ge 1\}$ be a stationary martingale difference sequence with respect to the filtration $\{\mathscr F_n, n\ge0\}$ with $X_0=0$. In addition, assume that

$ \begin{eqnarray} \label{aa1} \mathbb E (X_n^2| \mathscr F_{n-1}) \le 1 \ \ \text{a.s.}. \end{eqnarray} $

(1.1)

Let $\{a_{ni}, 1\le i \le n, n\ge 1\}$ be a triangular array of constants satisfying

$ \begin{eqnarray} \label{aa2} \max\limits_{1\le i \le n}|a_{ni}|\le \frac{1}{\sqrt{2n \log n}}. \end{eqnarray} $

(1.2)

Then we have

$ \begin{eqnarray} \label{aa3} \limsup\limits_{n\to \infty}\sum\limits_{i=1}^n a_{ni}X_i \le 1\ \ \text{a.s.}. \end{eqnarray} $

(1.3)

We will obtain the following better consequence than Theorem 1.1 as condition (1.2) is replaced by the stronger condition (1.4).

Corollary 1.1  Let $\{X_n, n\ge 1\}$ be a stationary martingale difference sequence with respect to the filtration $\{\mathscr F_n, n\ge0\}$ with $X_0=0$. In addition, assume that there exists a constant $C>0$ such that $ \mathbb E (X_n^2| \mathscr F_{n-1}) < C \ \ \text{a.s.}. $ If $\{a_{ni}, 1\le i \le n, n\ge 1\}$ is a triangular array of constants satisfying

$ \begin{eqnarray}\label{aa4} \max\limits_{1\le i \le n}|a_{ni}|= o\left(\frac{1}{\sqrt{n \log n}}\right), \end{eqnarray} $

(1.4)

then we have $ \sum\limits_{i=1}^n a_{ni}X_i \to 0\ \ \text{a.s.}. $

In order to prove our main result, we need the following lemma.

Lemma 2.1  If $\mathbb E X^2 < \infty$, then for any $\epsilon >0, $ $\sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{n \log n}} \mathbb E|X|1_{\{|X|> \epsilon \sqrt{\frac{n}{\log n}}\}} <\infty.$

Proof  Noting that $\left\{\frac{n}{\log n}\right\}$ is an increasing sequence, we have

$ \begin{eqnarray} &&\sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{n\log n}}\mathbb{E} |X| 1_{\{|X| > \epsilon \sqrt{\frac{n}{\log n}}\}} = \sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{n\log n}}\sum\limits_{i=n}^{\infty} \mathbb{E} |X| 1_{\{\epsilon \sqrt{\frac{i}{\log i}} <|X| \le \epsilon \sqrt{\frac{i+1}{\log (i+1)}}\}}\nonumber\\ &=& \sum\limits_{i=1}^{\infty} \mathbb{E} |X| 1_{\{\epsilon \sqrt{\frac{i}{\log i}} <|X| \le \epsilon \sqrt{\frac{i+1}{\log (i+1)}}\}}\sum\limits_{n=1}^{i} \frac{1}{\sqrt{n\log n}} \le C \sum\limits_{i=1}^{\infty} \mathbb{E} |X| 1_{\{\epsilon \sqrt{\frac{i}{\log i}} <|X| \le \epsilon \sqrt{\frac{i+1}{\log (i+1)}}\}} \\ &&\begin{matrix}\sqrt{\frac{i}{\log i}}\end{matrix}\nonumber\\ & \le&C \sum\limits_{i=1}^{\infty} \mathbb{P} \left(\epsilon \begin{matrix}\sqrt{\frac{i}{\log i}}\end{matrix} <|X| \le \epsilon \begin{matrix}\sqrt{\frac{i+1}{ \log (i+1)}}\end{matrix}\right) \frac{i}{\log i} \le C \mathbb{E} X^2 < \infty, \nonumber \end{eqnarray} $

since the first inequality follows from the following fact:

$ \sum\limits_{n=1}^{i}\frac{1}{\sqrt{n\log n}} \le C \begin{matrix} \int_{1}^{i} \frac{1}{\sqrt{x\log x}}\, dx \end{matrix} \le C \begin{matrix} \sqrt{\frac{i}{\log i}}\end{matrix}. $

Proof of Theorem 1.1  By Lemma 2.1 there exists a sequence of positive real numbers $\{\epsilon_n\}$ satisfying $\epsilon_n \downarrow 0$ such that

$ \begin{eqnarray}\label{a1} \sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{n \log n}} \mathbb E|X_n|1_{\{|X_n|> \epsilon_n \sqrt{\frac{n}{\log n}}\}} <\infty. \end{eqnarray} $

(2.1)

For $1\le i \le n$, define $ Y_i=X_i1_{\{|X_i|\le \epsilon_i \sqrt{\frac{i}{\log i}}\}}, \ \ \ Z_i=X_i1_{\{|X_i|> \epsilon_i \sqrt{\frac{i}{\log i}}\}}, $ and

$ X_i^{'}=Y_i- \mathbb E (Y_i| \mathscr F_{i-1}), \ \ X_i^{''}=Z_i- \mathbb E (Z_i| \mathscr F_{i-1}), $

then $X_i=X_i^{'}+X_i^{''}$ and $\{X_i^{'}, i\ge 1\}$, $\{X_i^{''}, i\ge 1\}$ are two martingale difference sequences with respect to the filtration $\{\mathscr F_i, i\ge0\}$.

For any $1 \le i \le n$, let us define $ X_{ni}= a_{ni} b_n X_i^{'}, $ where $b_n=\sqrt{2n \log n}. $ By (1.1) and (1.2), it is easy to check

$ \begin{align} \mathbb E (X_{ni}^2|\mathscr F_{i-1})\le& \mathbb E ({X_i^{'}}^2|\mathscr F_{i-1}) = \mathbb E (Y_i^2| \mathscr F_{i-1})-(\mathbb E (Y_i| \mathscr F_{i-1}))^2 \\ \le&\mathbb E (Y_i^2| \mathscr F_{i-1}) \le \mathbb E (X_i^2| \mathscr F_{i-1}) \le 1 \ \ \text{a.s.} \end{align} $

(2.2)

and

$\begin{eqnarray*} |X_{ni}|\le 2 \max\limits_{1\le i \le n}\epsilon_i \sqrt{\frac{i}{\log i}}=o\left(\sqrt{\frac{n}{\log n}}\right).\end{eqnarray*} $

So there exists a positive decreasing sequence $k_n\to 0$ such that

$\begin{eqnarray*} |X_{ni}|\le 2 \max\limits_{1\le i \le n}\epsilon_i \sqrt{\frac{i}{\log i}}= k_n \sqrt{\frac{n}{\log n}}.\end{eqnarray*} $

For any $\lambda >0, $ from the following elementary inequalities

$ e^x \le 1+x+\frac{x^2}{2}e^{|x|}, \ \ \forall x\in\mathbb{R}\ \ \text{and}\ \ e^x \ge 1+x, \ \ \forall \ x>0, $

and the definition of martingale difference, we have

$ \begin{align} &\ \mathbb{E} \left(\exp \left\{ \frac{\lambda X_{ni}}{b_n} \right\} |\mathscr F_{i-1}\right)\\ \le &\ \mathbb{E} \left(1+\frac{\lambda X_{ni}}{b_n}+\frac{\lambda^2X_{ni}^2}{2b_n^2}\exp\left\{ \frac{\lambda |X_{ni}|}{b_n}\right\}|\mathscr F_{i-1}\right)\\ =&\ 1+ \frac{\lambda^2}{2b_n^2} \mathbb{E} \left( X_{ni}^2 \exp\left\{ \frac{\lambda |X_{ni}|}{b_n}\right\}|\mathscr F_{i-1} \right)\\ \le &\ 1+ \frac{\lambda^2}{2b_n^2}\exp\left\{\frac{\lambda k_n}{b_n} \sqrt{\frac{n}{\log n}}\right\}\\ \le &\ \exp\left\{\frac{\lambda^2}{2b_n^2}\exp\left\{\frac{\lambda k_n}{b_n} \sqrt{\frac{n}{\log n}}\right\}\right\}. \end{align} $

By iterating the same procedure and using the properties of conditional expectation, we can give

$ \begin{eqnarray} \mathbb{E} \exp \left\{\frac{\lambda}{b_n} \sum\limits_{i=1}^n X_{ni}\right\} &=&\mathbb{E} \left\{\mathbb E \left(\exp \left\{\frac{\lambda}{b_n} \sum\limits_{i=1}^n X_{ni}\right\}|\mathscr F_{n-1}\right)\right\}\nonumber\\ &=&\mathbb{E} \left\{\exp \left\{\frac{\lambda}{b_n} \sum\limits_{i=1}^{n-1} X_{ni}\right\}\mathbb E \left(\exp \left\{\lambda \frac{X_{nn}}{b_n}\right\}|\mathscr F_{n-1}\right)\right\}\nonumber\\ &\le& \exp\left\{n \frac{\lambda^2}{2b_n^2}\exp\left\{\frac{\lambda k_n}{b_n} \sqrt{\frac{n}{\log n}}\right\}\right\}\nonumber. \end{eqnarray} $

Hence, by Markov's inequality, it follows that

$ \begin{align} &\ \mathbb{P} \left( \frac{1}{b_n}\sum\limits_{i=1}^n X_{ni}>1+r \right)\\ \le&\ \inf\limits_{\lambda>0}e^{-\lambda(1+r)}\mathbb{E} \exp \left\{\frac{\lambda}{b_n} \sum\limits_{i=1}^n X_{ni}\right\} \\ \le&\ \inf\limits_{\lambda>0} \exp\left\{-\lambda(1+r)+n \frac{\lambda^2}{2b_n^2}\exp\left\{\frac{\lambda k_n}{b_n} \sqrt{\frac{n}{\log n}}\right\}\right\}\\ =&\ n^{ -(1+r)^2(2-\exp\{\sqrt 2(1+r)k_n\})}, \end{align} $

where we choose $\lambda =2(1+r)\log n$. Since $k_n \to 0$ implies that

$ (1+r)^2(2-\exp\{\sqrt 2(1+r)k_n\}) > 1, $

for $n$ sufficiently large. Then we have

$\begin{eqnarray*} \sum\limits_{n=1}^{\infty} \mathbb P \left(\sum\limits_{i=1}^n a_{ni}X_i^{'} > 1+r \right)=\sum\limits_{n=1}^{\infty} \mathbb{P} \left(\frac{1}{b_n}\sum\limits_{i=1}^n X_{ni} > 1+r \right) < \infty.\end{eqnarray*} $

Therefore by the Borel-Cantelli lemma, we have

$\begin{eqnarray*} \limsup\limits_{n\to \infty}\sum\limits_{i=1}^n a_{ni}X_i^{'} \le 1\ \ \text{a.s.}.\end{eqnarray*} $

To finish the proof, it is enough to show that

$ \begin{eqnarray} \label{a2} \sum\limits_{i=1}^n a_{ni}X_i^{''} \to 0\ \ \text{a.s.}. \end{eqnarray} $

(2.3)

From Markov's inequality, for any $r >0$, we can obtain

$ \begin{eqnarray} &&\sum\limits_{k=1}^{\infty} \mathbb P \left( \frac{1}{\sqrt{2^k \log 2^k}} \sum\limits_{i=1}^{2^k}( |Z_i|+ \mathbb E (|Z_i| | \mathscr F_{i-1})) > r \right)\nonumber\\ &\le&\frac{2}{r} \sum\limits_{k=1}^{\infty} \frac{1}{\sqrt{2^k \log 2^k}} \sum\limits_{i=1}^{2^k} \mathbb E |Z_i| \nonumber\\ &=& \frac{2}{r} \sum\limits_{i=1}^{\infty} \mathbb E |Z_i| \sum\limits_{\{k:2^k \ge i\}} \frac{1}{\sqrt{2^k \log 2^k}}.\nonumber \end{eqnarray} $

Note the fact

$\begin{eqnarray*} \sum\limits_{\{k:2^k \ge i\}} \frac{1}{\sqrt{2^k \log 2^k}} \le \frac{1}{\sqrt{\log i}}\sum\limits_{\{k:2^k \ge i\}} \frac{1}{\sqrt{2^k}} \le \frac{\sqrt{2}}{\sqrt{2}-1}\frac{1}{\sqrt{i \log i}}.\end{eqnarray*} $

This implies

$\begin{eqnarray*} \sum\limits_{k=1}^{\infty} \mathbb P \left( \frac{1}{\sqrt{2^k \log 2^k}} \sum\limits_{i=1}^{2^k}( |Z_i|+ \mathbb E (|Z_i| | \mathscr F_{i-1})) > r \right) \le C \sum\limits_{i=1}^{\infty} \frac{\mathbb E |Z_i|}{\sqrt{i \log i}} < \infty, \end{eqnarray*} $

where the last inequality follows from (2.1). Thus, by the Borel-Cantelli lemma, we have

$ \begin{eqnarray}\label{a3} \frac{1}{\sqrt{2^k \log 2^k}} \sum\limits_{i=1}^{2^k}( |Z_i|+ \mathbb E (|Z_i| | \mathscr F_{i-1})) \to 0 \ \ \text{a.s.}. \end{eqnarray} $

(2.4)

From the condition (1.2), we get the following inequality

$ \begin{eqnarray*}\label{a4} &&\max\limits_{2^k \le n < 2^{k+1}} |\sum\nolimits_{\mathit{i}{\rm{ = 1}}}^\mathit{n} {} a_{ni}X_i^{''}|\nonumber\\ &\le&\frac{1}{\sqrt{2^{k+1} \log 2^k}} \sum\limits_{i=1}^{2^{k+1}-1}( |Z_i|+ \mathbb E (|Z_i| | \mathscr F_{i-1})) \nonumber\\ &\le&\frac{C}{\sqrt{2^{k+1} \log 2^{k+1}}} \sum\limits_{i=1}^{2^{k+1}}( |Z_i|+ \mathbb E (|Z_i| | \mathscr F_{i-1})) \ \ \text{a.s.}. \end{eqnarray*} $

(2.5)

From the inequalities (2.4) and (2.5), we get

$\begin{eqnarray*} \max\limits_{2^k \le n < 2^{k+1}} |\sum\nolimits_{\mathit{i}{\rm{ = 1}}}^\mathit{n} {} a_{ni}X_i^{''}| \to 0 \ \ \text{a.s.}, \end{eqnarray*} $

which implies (2.3). So the proof is completed.

Proof of Corollary 1.1  From the condition (1.4), there exists a sequence of real numbers denoted by $\{t_n\}$ such that $ t_n \downarrow 0$ and

$\begin{eqnarray*} \max\limits_{1 \le i \ n} | a_{ni} | \le \frac{t_n}{\sqrt{2n \log n}}.\end{eqnarray*} $

By Theorem 1.1, we get

$\begin{eqnarray*} \limsup\limits_{n\to \infty}\frac{\sum\nolimits_{\mathit{i}{\rm{ = 1}}}^\mathit{n} {} a_{ni}X_i}{t_n} \le 1\ \ \ \text{a.s.}, \end{eqnarray*} $

which implies

$\begin{eqnarray*} \limsup\limits_{n\to \infty}\sum\limits_{i=1}^n a_{ni}X_i \le 0\ \ \ \text{a.s.}.\end{eqnarray*} $

Replacing $X_i$ by $- X_i$, we can get

$\begin{eqnarray*} \liminf\limits_{n\to \infty}\sum\limits_{i=1}^n a_{ni}X_i \ge 0\ \ \ \text{a.s.}.\end{eqnarray*} $

Thus the desired result can be obtained.