1 Introduction and main result
The concept of complete convergence was first introduced by Hsu and Robbins [1] and has played a very important role in probability theory. A sequence of random variables \(\{U_{n},n\geq1\}\) is said to converge completely to a constant C if \(\sum^{\infty}_{n=1}P\{|U_{n}-C|>\varepsilon\}<\infty\) for any \(\varepsilon>0\). Hsu and Robbins [1] proved that the sequence of arithmetic means of independent and identically distributed (i.i.d.) random variables converges completely to the expected value if the variance of the summands is finite. Their result has been generalized and extended by many authors.
The following result is well known.
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Theorem A
Let
\(r\geq1\)
and
\(0< p<2\). Let
\(\{X, X_{n},n\geq1\}\)
be a sequence of i.i.d. random variables with partial sums
\(S_{n}=\sum_{k=1}^{n} X_{k}\), \(n\geq1\). Then the following statements are equivalent:
where
\(b=0\)
if
\(0< rp<1\)
and
\(b=EX\)
if
\(rp\geq1\).
$$\begin{aligned} &E|X|^{rp}< \infty, \end{aligned}$$
(1.1)
$$\begin{aligned} &\sum^{\infty}_{n=1}n^{r-2}P \bigl\{ \vert S_{n}-nb\vert >\varepsilon n^{1/p} \bigr\} < \infty \quad\forall \varepsilon>0, \end{aligned}$$
(1.2)
$$\begin{aligned} &\sum^{\infty}_{n=1}n^{r-2}P \Bigl\{ \max_{1\leq m\leq n}\vert S_{m}-mb\vert >\varepsilon n^{1/p} \Bigr\} < \infty\quad \forall \varepsilon>0, \end{aligned}$$
(1.3)
When \(r=1\), each of (1.1)∼(1.3) is equivalent to
$$ \frac{S_{n} -nb}{n^{1/p}}\to0 \quad \mbox{a.s.} $$
(1.4)
When \(r=1\), the equivalence of (1.1) and (1.4) is known as the Marcinkiewicz and Zygmund strong law of large numbers. Katz [2] proved the equivalence of (1.1) and (1.2) for the case of \(p=1\). Baum and Katz [3] proved the equivalence of (1.1) and (1.2) for the case of \(0< p<2\). The result of Baum and Katz was generalized and extended in several directions. Some versions of the Baum and Katz theorem under higher-order moment conditions were established by Lanzinger [4], Gut and Stadtmüller [5], and Chen and Sung [6]. When \(p=1\), \(1\leq r<3\), and \(\{X_{n}, n\geq1\}\) is a sequence of pairwise independent, but not necessarily identically distributed, random variables, Spătaru [7] gave sufficient conditions for (1.2).
It is interesting to find more general moment conditions such that the complete convergence holds. In fact, Li et al. [8] and Sung [9] have done something. In particular, it is worth pointing out that Sung [9] obtained the following complete convergence for pairwise i.i.d. random variables \(\{X,X_{n},n\geq1\}\): provided that \(\sum^{\infty}_{n=1}P\{|X|>a_{n}\}<\infty\), where \(0< a_{n}/n\uparrow\).
$$\sum^{\infty}_{n=1}n^{-1}P \Biggl\{ \Biggl\vert \sum^{n}_{k=1}X_{k}-nEXI\bigl(|X| \leq a_{n}\bigr)\Biggr\vert >\varepsilon a_{n} \Biggr\} < \infty \quad \forall\varepsilon>0, $$
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Motivated by the work of Sung [9], the aim of this paper is to obtain the complete convergence under more general moment conditions. Our main result includes the Baum and Katz [3] complete convergence and the Marcinkiewicz and Zygmund strong law of large numbers.
Now we state the main result. Some lemmas and the proof of the main result will be detailed in next section.
Theorem 1.1
Let
\(r\geq1\)
and
\(0< p<2\). Let
\(\{X, X_{n},n\geq1\}\)
be a sequence of i.i.d. random variables with partial sums
\(S_{n}=\sum_{k=1}^{n} X_{k}\), \(n\geq1\), and
\(\{a_{n},n\geq1\}\)
a sequence of positive real numbers with
\(0< a_{n}/n^{1/p}\uparrow\). Then the following statements are equivalent:
where
\(b_{n}=0\)
if
\(0< p<1\)
and
\(b_{n}=EXI(|X|\leq a_{n})\)
if
\(1\leq p<2\).
$$\begin{aligned} &\sum^{\infty}_{n=1}n^{r-1}P \bigl\{ |X|>a_{n}\bigr\} < \infty, \end{aligned}$$
(1.5)
$$\begin{aligned} &\sum^{\infty}_{n=1}n^{r-2}P \bigl\{ \vert S_{n} -nb_{n}\vert >\varepsilon a_{n} \bigr\} < \infty \quad\forall \varepsilon>0, \end{aligned}$$
(1.6)
$$\begin{aligned} &\sum^{\infty}_{n=1}n^{r-2}P \Bigl\{ \max_{1\leq m\leq n}\vert S_{m} -mb_{n} \vert >\varepsilon a_{n} \Bigr\} < \infty\quad \forall \varepsilon>0, \end{aligned}$$
(1.7)
When \(r=1\), each of (1.5)-(1.7) is equivalent to
$$ a_{n}^{-1} (S_{n} -nb_{n} )\rightarrow0 \quad\mbox{a.s. } $$
(1.8)
Remark 1.1
When \(a_{n}=n^{1/p}\) for \(n\geq1\), (1.5) is equivalent to (1.1). In this case, \(a_{n}^{-1}\cdot n EXI(|X|>a_{n})\to0\) if \(1\leq p<2\) and (1.1) holds. Hence, (1.1) ⇒ (1.2), (1.1) ⇒ (1.3), and (1.1) ⇒ (1.4) (in this case, \(r=1\)) follow from Theorem 1.1. Although the converses do not follow directly from Theorem 1.1, the proofs can be done easily. When \(a_{n}=n^{1/p}(\ln n)^{\alpha}\) for \(n\geq1\), where \(\alpha>0\), (1.5) is equivalent to \(E|X|^{rp}/(\ln(|X|+2))^{\alpha rp}<\infty\).
Throughout this paper, the symbol C denotes a positive constant that is not necessarily the same one in each appearance, and \(I(A)\) denotes the indicator function of an event A.
2 Lemmas and proofs
To prove the main result, the following lemmas are needed. Lemma 2.1 is the Rosenthal inequality for the sum of independent random variables; see, for example, Petrov [10].
Lemma 2.1
Let
\(\{Y_{n}, n\ge1 \}\)
be a sequence of independent random variables with
\(EY_{n}=0\)
and
\(E|Y_{n}|^{s}<\infty\)
for some
\(s\ge2\)
and all
\(n\ge1\). Then there exists a positive constant
C
depending only on
s
such that for all
\(n\ge1\),
$$E\max_{1\leq m\leq n}\Biggl\vert \sum_{k=1}^{m} Y_{k}\Biggr\vert ^{s}\le C \Biggl\{ \sum _{k=1}^{n} E|Y_{k}|^{s} + \Biggl( \sum_{k=1}^{n} EY_{k}^{2} \Biggr)^{s/2} \Biggr\} . $$
Lemma 2.2
Proof
Lemma 2.3
Proof
By \(0< a_{n}/n^{1/p}\uparrow\) we have \(a_{k}/a_{n}\leq (k/n)^{1/p}\) for any \(1\leq k\leq n\). Hence, Set \(C=1+ r2^{r-1}\sum^{\infty}_{k=1}k^{r-1}P\{|X|>a_{k}\}\). By (1.5), \(C<\infty\). So we complete the proof. □
$$\begin{aligned} a_{n}^{-2}\cdot nE|X|^{2}I\bigl(|X|\leq a_{n}\bigr) &=a_{n}^{-2}\cdot n\sum ^{n}_{k=1}E|X|^{2}I\bigl(a_{k-1}< |X|\leq a_{k}\bigr) \quad (\mbox{set } a_{0}=0) \\ &\leq a_{n}^{-2}\cdot n\sum^{n}_{k=1}a_{k}^{2}P \bigl\{ a_{k-1}< |X|\leq a_{k}\bigr\} \\ &\leq n^{1-2/p}\sum^{n}_{k=1}k^{2/p}P \bigl\{ a_{k-1}< |X|\leq a_{k}\bigr\} \\ &\leq n^{1-2/p}\sum^{n}_{k=1}k^{r}P \bigl\{ a_{k-1}< |X|\leq a_{k}\bigr\} \quad (\mbox{by }rp\geq2) \\ &\leq n^{1-2/p}\sum^{\infty}_{k=0} \bigl[(k+1)^{r}-k^{r}\bigr]P\bigl\{ |X|>a_{k}\bigr\} \\ &\leq n^{1-2/p}\cdot \Biggl(1+ r2^{r-1}\sum ^{\infty}_{k=1}k^{r-1}P\bigl\{ |X|>a_{k}\bigr\} \Biggr). \end{aligned}$$
Lemma 2.4
Proof
By \(0< a_{n}/n^{1/p}\uparrow\) we have \(a_{k}/a_{n}\leq (k/n)^{1/p}\) for any \(n\geq k\). Hence, Therefore, the proof is completed. □
$$\begin{aligned} \sum^{\infty}_{n=1}n^{r-2} \cdot a_{n}^{-s}nE|X|^{s}I\bigl(|X|\leq a_{n}\bigr) &= \sum^{\infty}_{n=1}n^{r-1}a_{n}^{-s} \sum^{n}_{k=1}E|X|^{s}I\bigl(a_{k-1}< |X| \leq a_{k}\bigr) \\ &\leq\sum^{\infty}_{n=1}n^{r-1}a_{n}^{-s} \sum^{n}_{k=1}a_{k}^{s}P \bigl\{ a_{k-1}< |X|\leq a_{k}\bigr\} \\ &=\sum^{\infty}_{k=1}a_{k}^{s}P \bigl\{ a_{k-1}< |X|\leq a_{k}\bigr\} \sum^{\infty}_{n=k}n^{r-1}a_{n}^{-s} \\ &\leq\sum^{\infty}_{k=1}k^{s/p}P \bigl\{ a_{k-1}< |X|\leq a_{k}\bigr\} \sum^{\infty}_{n=k}n^{r-1-s/p} \\ &\leq C\sum^{\infty}_{k=1}k^{r}P \bigl\{ a_{k-1}< |X|\leq a_{k}\bigr\} < \infty. \end{aligned}$$
Lemma 2.5
Let
\(\{X,X_{n}\geq1\}\)
be a sequence of i.i.d. symmetric random variables, and
\(\{a_{n},n\geq1\}\)
a sequence of real numbers with
\(0< a_{n}\uparrow \infty\). Suppose that
Then
$$ \sum^{\infty}_{n=1}n^{-1}P \Biggl\{ \Biggl\vert \sum^{n}_{k=1}X_{k} \Biggr\vert >\varepsilon a_{n} \Biggr\} < \infty \quad\forall \varepsilon>0. $$
(2.1)
$$ a_{n}^{-1}\sum^{n}_{k=1}X_{k} \rightarrow0 \textit{ in probability}. $$
(2.2)
Proof
Set \(S_{n}=\sum^{n}_{k=1}X_{k}\), \(n\geq1\). Note that for all \(\varepsilon>0\), and \(P\{|X|>\varepsilon a_{2n+1}/2 \}\rightarrow0\) as \(n\rightarrow \infty\). Hence, to prove (2.2), it suffices to prove that We will prove (2.3) by contradiction. Suppose that there exist a constant \(\varepsilon>0\) and a sequence of integers \(\{n_{i},i\geq1\}\) with \(n_{i}\uparrow\infty\) such that Without loss of generality, we can assume that \(2n_{i}< n_{i+1}\). By the Lévy inequality (see, for example, formula (2.6) in Ledoux and Talagrand [11]) we have which leads a contradiction to (2.1). Hence, (2.3) holds, and so the proof is completed. □
$$\begin{aligned}[b] P\bigl\{ |S_{2n+1}|>\varepsilon a_{2n+1}\bigr\} &\leq P \bigl\{ |S_{2n}|>\varepsilon a_{2n+1}/2\bigr\} +P\bigl\{ |X_{2n+1}|> \varepsilon a_{2n+1}/2 \bigr\} \\ &\leq P\bigl\{ |S_{2n}|>\varepsilon a_{2n}/2\bigr\} +P\bigl\{ |X|>\varepsilon a_{2n+1}/2 \bigr\} \end{aligned} $$
$$ a_{2n}^{-1}S_{2n}\rightarrow0 \mbox{ in probability}. $$
(2.3)
$$P\bigl\{ |S_{2n_{i}}|>\varepsilon a_{2n_{i}}\bigr\} \geq\varepsilon \quad \mbox{for all } i\geq1. $$
$$\begin{aligned} \sum^{\infty}_{n=1}n^{-1}P \bigl\{ |S_{n}|>\varepsilon a_{n}/2\bigr\} &\geq\frac{1}{2}\sum ^{\infty}_{n=1}n^{-1}P\Bigl\{ \max _{1\leq k\leq n}|S_{k}|>\varepsilon a_{n}/2\Bigr\} \\ &\geq\frac{1}{2}\sum^{\infty}_{i=2}\sum ^{2n_{i}}_{n=n_{i}+1}n^{-1}P\Bigl\{ \max _{1\leq k\leq n}|S_{k}|>\varepsilon a_{n}/2\Bigr\} \\ &\geq\frac{1}{2}\sum^{\infty}_{i=2}\sum ^{2n_{i}}_{n=n_{i}+1}n^{-1}P\Bigl\{ \max _{1\leq k\leq n_{i}}|S_{k}|>\varepsilon a_{2n_{i}}/2\Bigr\} \\ &\geq\frac{1}{2}\sum^{\infty}_{i=2}\sum ^{2n_{i}}_{n=n_{i}+1}n^{-1}P\bigl\{ |S_{n_{i}}|>\varepsilon a_{2n_{i}}/2\bigr\} \\ &=\frac{1}{4}\sum^{\infty}_{i=2}\sum ^{2n_{i}}_{n=n_{i}+1}n^{-1} \bigl(P\bigl\{ |S_{n_{i}}|>\varepsilon a_{2n_{i}}/2\bigr\} +P\bigl\{ |S_{2n_{i}}-S_{n_{i}}|> \varepsilon a_{2n_{i}}/2\bigr\} \bigr) \\ &\geq\frac{1}{4}\sum^{\infty}_{i=2}\sum ^{2n_{i}}_{n=n_{i}+1}n^{-1}P\bigl\{ |S_{2n_{i}}|>\varepsilon a_{2n_{i}}\bigr\} \\ &\geq\frac{\varepsilon}{4}\sum^{\infty}_{i=2}\sum ^{2n_{i}}_{n=n_{i}+1}n^{-1}=\infty, \end{aligned}$$
Proof of Theorem 1.1
We first prove that (1.5) implies (1.7). By Lemma 2.2, to prove (1.7), it suffices to prove that Note that Hence, by (1.5), to prove (2.4) it suffices to prove that for all \(\varepsilon>0\), For any \(s\geq2\), by the Markov inequality and Lemma 2.1, If \(rp\geq2\), taking s large enough such that \(r-2-s/p+s/2<-1\), by Lemma 2.3 we have Since \(s>rp\), \(I_{2}<\infty\) by Lemma 2.4. If \(0< rp<2\), taking \(s=2\) (in this case \(I_{1}=I_{2}\)), we have \(I_{1}=I_{2}<\infty\) by Lemma 2.4 again. Hence, (2.5) holds for all \(\varepsilon>0\).
$$ \sum^{\infty}_{n=1}n^{r-2}P \Biggl\{ \max_{1\leq m\leq n}\Biggl\vert \sum^{m}_{k=1} \bigl(X_{k}-EX_{k}I\bigl(|X_{k}|\leq a_{n}\bigr) \bigr)\Biggr\vert >\varepsilon a_{n} \Biggr\} < \infty\quad \forall \varepsilon>0. $$
(2.4)
$$\begin{aligned} & \Biggl\{ \max_{1\leq m\leq n}\Biggl\vert \sum ^{m}_{k=1}\bigl(X_{k}-EX_{k}I\bigl(|X_{k}| \leq a_{n}\bigr)\bigr)\Biggr\vert >\varepsilon a_{n} \Biggr\} \\ &\quad\subset\bigcup^{n}_{k=1} \bigl\{ |X_{k}|>a_{n}\bigr\} \cup \Biggl\{ \max_{1\leq m\leq n} \Biggl\vert \sum^{m}_{k=1} \bigl(X_{k}I\bigl(|X_{k}|\leq a_{n}\bigr)-EX_{k}I\bigl(|X_{k}| \leq a_{n}\bigr)\bigr)\Biggr\vert >\varepsilon a_{n} \Biggr\} . \end{aligned}$$
$$ \sum^{\infty}_{n=1}n^{r-2}P \Biggl\{ \max_{1\leq m\leq n}\Biggl\vert \sum^{m}_{k=1} \bigl(X_{k}I\bigl(|X_{k}|\leq a_{n}\bigr)-EX_{k}I\bigl(|X_{k}| \leq a_{n}\bigr)\bigr)\Biggr\vert >\varepsilon a_{n} \Biggr\} < \infty. $$
(2.5)
$$\begin{aligned} &\sum^{\infty}_{n=1}n^{r-2}P \Biggl\{ \max_{1\leq m\leq n}\Biggl\vert \sum^{m}_{k=1} \bigl(X_{k}I\bigl(|X_{k}|\leq a_{n}\bigr)-EX_{k}I\bigl(|X_{k}| \leq a_{n}\bigr)\bigr)\Biggr\vert >\varepsilon a_{n} \Biggr\} \\ &\quad\leq C\sum^{\infty}_{n=1}n^{r-2} a_{n}^{-s}E\max_{1\leq m\leq n}\Biggl\vert \sum ^{m}_{k=1}\bigl(X_{k}I\bigl(|X_{k}| \leq a_{n}\bigr)-EX_{k}I\bigl(|X_{k}|\leq a_{n}\bigr) \bigr)\Biggr\vert ^{s} \\ &\quad\leq C \Biggl(\sum^{\infty}_{n=1}n^{r-2} \bigl\{ a_{n}^{-2}nEX^{2}I\bigl(|X|\leq a_{n}\bigr) \bigr\} ^{s/2}+\sum^{\infty}_{n=1}n^{r-1}a_{n}^{-s}E|X|^{s}I\bigl(|X| \leq a_{n}\bigr) \Biggr) \\ &\quad=C(I_{1}+I_{2}). \end{aligned}$$
$$I_{1}\leq C\sum^{\infty}_{n=1}n^{r-2-s/p+s/2}< \infty. $$
It is trivial that (1.7) implies (1.6). Now we prove that (1.6) implies (1.5). Let \(\{X', X_{n}', n\geq1\}\) be an independent copy of \(\{X,X_{n},n\geq1\}\). Then we also have Hence, from which it follows that Then, by Lemma 2.5, By the Lévy inequality (see, for example, formula (2.7) in Ledoux and Talagrand [11]), for any fixed \(\varepsilon>0\), as \(n\rightarrow\infty\). Then, for all n large enough, Therefore, by Lemma 2.6 in Ledoux and Talagrand [11] and the Lévy inequality (see formula (2.7) in Ledoux and Talagrand [11]) we have that for all n large enough, Therefore, Since \(P\{|X|>a_{n}/2\}\to0\) as \(n\to\infty\), \(|\operatorname{med}(X)/(a_{n}/2)|\leq1\) for all n large enough. By the weak symmetrization inequality we have that for all n large enough, which, together with (2.6), implies that (1.5) holds.
$$\sum^{\infty}_{n=1}n^{r-2}P \Biggl\{ \Biggl\vert \sum^{n}_{k=1}X_{k}'-nb_{n} \Biggr\vert >\varepsilon a_{n} \Biggr\} < \infty \quad \forall \varepsilon>0. $$
$$\sum^{\infty}_{n=1}n^{r-2}P \Biggl\{ \Biggl\vert \sum^{n}_{k=1} \bigl(X_{k}-X_{k}'\bigr)\Biggr\vert > \varepsilon a_{n} \Biggr\} < \infty \quad \forall \varepsilon>0, $$
$$\sum^{\infty}_{n=1}n^{-1}P \Biggl\{ \Biggl\vert \sum^{n}_{k=1} \bigl(X_{k}-X_{k}'\bigr)\Biggr\vert > \varepsilon a_{n} \Biggr\} < \infty \quad \forall \varepsilon>0. $$
$$a_{n}^{-1}\sum^{n}_{k=1} \bigl(X_{k}-X_{k}'\bigr)\rightarrow0 \mbox{ in probability}. $$
$$P \Bigl\{ \max_{1\leq k\leq n}\bigl|X_{k}-X_{k}'\bigr|> \varepsilon a_{n} \Bigr\} \leq 2P \Biggl\{ \Biggl\vert \sum ^{n}_{k=1}\bigl(X_{k}-X_{k}' \bigr)\Biggr\vert >\varepsilon a_{n} \Biggr\} \rightarrow0 $$
$$P\Bigl\{ \max_{1\leq k\leq n}\bigl|X_{k}-X_{k}'\bigr|> \varepsilon a_{n}\Bigr\} \leq1/2. $$
$$\begin{aligned} nP\bigl\{ \bigl|X-X'\bigr|>\varepsilon a_{n}\bigr\} &=\sum ^{n}_{k=1}P\bigl\{ \bigl|X_{k}-X_{k}'\bigr|> \varepsilon a_{n}\bigr\} \\ &\leq2P\Bigl\{ \max_{1\leq k\leq n}\bigl|X_{k}-X_{k}'\bigr|> \varepsilon a_{n}\Bigr\} \leq4P \Biggl\{ \Biggl\vert \sum ^{n}_{k=1}\bigl(X_{k}-X_{k}' \bigr)\Biggr\vert >\varepsilon a_{n} \Biggr\} . \end{aligned}$$
$$ \sum^{\infty}_{n=1}n^{r-1}P\bigl\{ \bigl|X-X'\bigr|>\varepsilon a_{n}\bigr\} < \infty \quad\forall \varepsilon>0. $$
(2.6)
$$ P\bigl\{ |X|>a_{n}\bigr\} \leq P\bigl\{ \bigl|X-\operatorname{med}(X)\bigr|>a_{n}/2 \bigr\} \leq2P\bigl\{ \bigl|X-X'\bigr|> a_{n}/2\bigr\} , $$
(2.7)
Finally, we prove that (1.5) and (1.8) are equivalent when \(r=1\). Assume that (1.5) holds for \(r=1\). Since \(\sum^{\infty}_{i=1}iP\{a_{i}<|X|\leq a_{i+1}\}=\sum^{\infty}_{n=1} P\{ |X|>a_{n}\}<\infty\), for any fixed \(\varepsilon>0\), there exists a positive integer N such that \(\sum^{\infty}_{i=N+1}iP\{a_{i}<|X|\leq a_{i+1}\}<\varepsilon\). Then, for \(n>N+1\), It follows that as \(n\to\infty\). Hence, to prove (1.8), by Lemma 2.2 it suffices to prove that Since \(0< a_{n}/n^{1/p}\uparrow\) and \(0< p<2\), Then by the Kolmogorov convergence criterion and the Kronecker lemma, (2.8) holds, and so (1.8) also holds.
$$\begin{aligned} &\Biggl\vert a_{n}^{-1}\cdot nEXI\bigl(|X|\leq a_{n}\bigr)-a_{n}^{-1}\sum^{n}_{k=1}EXI\bigl(|X| \leq a_{k}\bigr)\Biggr\vert \\ &\quad\leq a_{n}^{-1}\sum^{n-1}_{k=1}E|X|I\bigl(a_{k}< |X| \leq a_{n}\bigr) \\ &\quad= a_{n}^{-1}\sum^{n-1}_{k=1} \sum^{n-1}_{i=k}E|X|I\bigl(a_{i}< |X|\leq a_{i+1}\bigr) \\ &\quad= a_{n}^{-1}\sum^{n-1}_{i=1}iE|X|I\bigl(a_{i}< |X| \leq a_{i+1}\bigr) \\ &\quad\leq a_{n}^{-1}\sum^{N}_{i=1}iE|X|I\bigl(a_{i}< |X|< a_{i+1}\bigr)+ \sum^{n-1}_{i=N+1}iP\bigl\{ a_{i}< |X|\leq a_{i+1}\bigr\} \\ &\quad\leq a_{n}^{-1}\sum^{N}_{i=1}iE|X|I\bigl(a_{i}< |X|< a_{i+1}\bigr)+ \varepsilon \\ &\quad\rightarrow\varepsilon \quad\mbox{as } n\to\infty. \end{aligned}$$
$$\Biggl\vert a_{n}^{-1}\cdot nEXI\bigl(|X|\leq a_{n}\bigr)-a_{n}^{-1}\sum^{n}_{k=1}EXI\bigl(|X| \leq a_{k}\bigr)\Biggr\vert \to0 $$
$$ a^{-1}_{n}\sum^{n}_{k=1} \bigl(X_{k}I\bigl(|X|\leq a_{k}\bigr)-EX_{k}I\bigl(|X|\leq a_{k}\bigr)\bigr)\rightarrow0\quad \mbox{a.s.} $$
(2.8)
$$\begin{aligned} &\sum^{\infty}_{n=1}a_{n}^{-2}Var \bigl(X_{n}I\bigl(|X|\leq a_{n}\bigr)\bigr) \\ &\quad\leq\sum^{\infty}_{n=1}a_{n}^{-2}EX^{2}I\bigl(|X| \leq a_{n}\bigr) \\ &\quad=\sum^{\infty}_{n=1}a_{n}^{-2} \sum^{n}_{k=1}EX^{2}I\bigl(a_{k-1}< |X| \leq a_{k}\bigr) \quad (\mbox{set } a_{0}=0) \\ &\quad\leq\sum^{\infty}_{k=1}a_{k}^{2}P \bigl\{ a_{k-1}< |X|\leq a_{k}\bigr\} \sum^{\infty}_{n=k}a_{n}^{-2} \\ &\quad\leq C\sum^{\infty}_{k=1}kP \bigl\{ a_{k-1}< |X|\leq a_{k}\bigr\} < \infty. \end{aligned}$$
Conversely, assume that (1.8) holds. Let \(\{X', X_{n}', n\geq1\}\) be an independent copy of \(\{X,X_{n},n\geq1\}\). Then we also have Hence, we have So, we have by \(0< a_{n}\uparrow\) that By the Borel-Cantelli lemma, which, together with (2.7), implies that (1.5) holds for \(r=1\). So we complete the proof. □
$$a_{n}^{-1} \Biggl(\sum^{n}_{k=1}X_{k}'-nb_{n} \Biggr)\rightarrow0 \quad\mbox{a.s.} $$
$$a_{n}^{-1}\sum^{n}_{k=1} \bigl(X_{k}-X_{k}'\bigr)\rightarrow0 \quad \mbox{a.s.} $$
$$a^{-1}_{n}\bigl(X_{n}-X_{n}' \bigr)=a_{n}^{-1}\sum^{n}_{k=1} \bigl(X_{k}-X_{k}'\bigr)-\bigl(a_{n-1}a_{n}^{-1} \bigr)a_{n-1}^{-1}\sum^{n-1}_{k=1} \bigl(X_{k}-X_{k}'\bigr) \rightarrow0 \quad \mbox{a.s.} $$
$$\sum^{\infty}_{n=1}P\bigl\{ \bigl|X_{n}-X_{n}'\bigr|> \varepsilon a_{n}\bigr\} < \infty \quad \forall \varepsilon>0, $$
Acknowledgements
The authors would like to thank the referees for the helpful comments. The research of Pingyan Chen and Jiaming Yi is supported by the National Natural Science Foundation of China (No. 11271161). The research of Soo Hak Sung is supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2014R1A1A2058041).
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors read and approved the manuscript.