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The moment of maximum normed randomly weighted sums of martingale differences

Abstract

By using some inequalities and properties of martingale differences, we investigate the moment of maximum normed randomly weighted sums of martingale differences under some weakly conditions. A sufficient condition to the moment of this stochastic sequence with maximum norm is presented in this paper.

Introduction

Let \(\{X_{n},n\geq1\}\) be a sequence, independent and identically distributed with \(EX_{1}=0\). Denote \(S_{n}=\sum_{i=1}^{n} X_{i}\), \(n\geq1\). For \(0< r<2\) and \(p>0\), it is well known that

$$\begin{aligned}& \left \{ \textstyle\begin{array}{l} E|X_{1}|^{r}< \infty,\quad \text{if } p< r, \\ E[|X_{1}|^{r}\log(1+|X_{1}|)]< \infty,\quad \text{if } p=r, \\ E|X_{1}|^{p}< \infty,\quad \text{if } p>r, \end{array}\displaystyle \right . \end{aligned}$$
(1.1)
$$\begin{aligned}& E \biggl(\sup_{n\geq1}\biggl\vert \frac{X_{n}}{n^{1/r}}\biggr\vert ^{p} \biggr)< \infty, \end{aligned}$$
(1.2)
$$\begin{aligned}& E \biggl(\sup_{n\geq1}\biggl\vert \frac{S_{n}}{n^{1/r}}\biggr\vert ^{p} \biggr)< \infty \end{aligned}$$
(1.3)

are all equivalent. Marcinkiewicz and Zygmund [1] obtain (1.1) (1.3) for the case \(p\geq r=1\), Burkerholder [2] gets (1.3) (1.1) for the case \(p=r=1\), Gut [3] proves that (1.1)-(1.3) are equivalent in the case \(p\geq r\), and Choi and Sung [4] show that (1.1)-(1.3) are equivalent in the case \(p< r\). For \(0< r<2\) and \(p>0\), Chen and Gan [5] prove that (1.1)-(1.3) are equivalent under the dependent case such as ρ-mixing random variables.

By using the method of dominated by a nonnegative random variable, we investigate the randomly weighted sums of martingale differences under some weakly conditions. A sufficient condition to (1.2) and (1.3) is presented. To a certain extent, we generalize the result of Chen and Gan [5] for ρ-mixing random variables to the case of randomly weighted sums of martingale differences. For the details, please see our main result of Theorem 2.1 in Section 2.

Recall that the sequence \(\{X_{n},n\geq1\}\) is stochastically dominated by a nonnegative random variable X if

$$\sup_{n\geq1}P\bigl(\vert X_{n}\vert >t\bigr)\leq CP(X>t)\quad \text{for some positive constant } C \text{ and all } t\geq0 $$

(see Adler and Rosalsky [6] and Adler et al. [7]). A bound on tail probabilities for quadratic forms in independent random variables is seen by using the following condition. There exist \(C>0\) and \(\gamma>0\) such that for all \(n\geq1\) and all \(x\geq0\), we have \(P(|X_{n}|\geq x)\leq C\int_{x}^{\infty}e^{-\gamma t^{2}}\, dt\). For details, see Hanson and Wright [8] and Wright [9].

Meanwhile, the definition of martingale differences can be found in many books such as Stout [10], Hall and Heyde [11], Shiryaev [12], Gut [13], and so on. There are many results of martingale differences. For example, Ghosal and Chandra [14] gave the complete convergence of martingale arrays; Stoica [15, 16] investigated the Baum-Katz-Nagaev-type results for martingale differences; Wang et al. [17] also studied the complete moment convergence for martingale differences; Yang et al. [18] obtained the complete convergence for the moving average process of martingale differences; Yang et al. [19] investigated the complete moment convergence for randomly weighted sums of martingale differences, etc.

On the other hand, randomly weighted sums have been an attractive research topic in the literature of applied probability. For example, Thanh and Yin [20] studied the almost sure and complete convergence of randomly weighted sums of independent random elements in Banach spaces; Thanh et al. [21] investigated the convergence analysis of double-indexed and randomly weighted sums of mixing processes and gave its application to state observers of linear-time-invariant systems; Kevei and Mason [22] and Hormann and Swan [23] studied the asymptotic properties of randomly weighted sums and self-normalized sums; Cabrera et al. [24] and Shen et al. [25] investigated the conditional convergence for randomly weighted sums; Gao and Wang [26] and Tang and Yuan [27] investigated the randomly weighted sums of random variables and have given an application to ruin theory and capital allocation; Chen [28] obtained some asymptotically results of randomly weighted sums of dependent random variables with dominated variation, and so on.

Throughout the paper, \(I(A)\) is the indicator function of set A and \(C, C_{1}, C_{2},\ldots\) denote some positive constants not depending on n. The following lemmas are our basic techniques to prove our results.

Lemma 1.1

(cf. Hall and Heyde (Theorem 2.11 in [11]))

If \(\{X_{i},\mathscr{F}_{i},1\leq i\leq n\}\) is a martingale difference and \(p>0\), then there exists a constant C depending only on p such that

$$E \Biggl(\max_{1\leq k\leq n}\Biggl\vert \sum _{i=1}^{k}X_{i}\Biggr\vert ^{p} \Biggr)\leq C \Biggl\{ E \Biggl(\sum_{i=1}^{n}E \bigl(X_{i}^{2}|\mathscr{F}_{i-1}\bigr) \Biggr)^{p/2} +E \Bigl(\max_{1\leq i\leq n}|X_{i}|^{p} \Bigr) \Biggr\} , \quad n\geq1. $$

Lemma 1.2

(cf. Adler and Rosalsky (Lemma 1 in [6]) and Adler et al. (Lemma 3 in [7]))

Let \(\{ X_{n},n\geq 1\}\) be a sequence of random variables, which is stochastically dominated by a nonnegative random variable X. Then, for any \(\alpha>0\) and \(b>0\), the following two statements hold:

  • \(E[|X_{n}|^{\alpha}I(|X_{n}|\leq b)]\leq C_{1}\{E[X^{\alpha}I(X\leq b)]+b^{\alpha}P(X>b)\}\),

  • \(E[|X_{n}|^{\alpha}I(|X_{n}|>b)]\leq C_{2}E[X^{\alpha}I(X>b)]\),

where \(C_{1}\) and \(C_{2}\) are positive constants not depending on n. Consequently, for all \(n\geq1\), one has \(E|X_{n}|^{\alpha}\leq C_{3}EX^{\alpha}\), where \(C_{3}\) is a positive constant not depending on n.

The main result and its proof

Theorem 2.1

Let \(0< r<2\), \(0< p<2\), and \(\{X_{n},\mathscr{F}_{n},n\geq1\}\) be a martingale difference sequence, which is stochastically dominated by a nonnegative random variable X such that

$$ \left \{ \textstyle\begin{array}{l} \textit{for } p< r, \quad \left \{ \textstyle\begin{array}{l} C_{1}EX< \infty,\quad \textit{if } 0< r< 1, \\ C_{2}E[X\log(1+X)]< \infty, \quad \textit{if } r=1, \\ C_{3}EX^{r}< \infty, \quad \textit{if } r>1, \end{array}\displaystyle \right . \\ \textit{for } p=r,\quad \left \{ \textstyle\begin{array}{l} C_{4}EX< \infty, \quad \textit{if } 0< r< 1, \\ C_{5}E[X\log^{2}(1+X)]< \infty,\quad \textit{if } r=1, \\ C_{6}E[X^{r}\log(1+X)]< \infty,\quad \textit{if } r>1, \end{array}\displaystyle \right . \\ \textit{for } p>r,\quad \left \{ \textstyle\begin{array}{l} C_{7}EX< \infty, \quad \textit{if } 0< p< 1, \\ C_{8}E[X\log(1+X)]< \infty,\quad \textit{if } p=1, \\ C_{9}EX^{p}< \infty, \quad \textit{if } p>1. \end{array}\displaystyle \right . \end{array}\displaystyle \right . $$
(2.1)

Assume that \(\{A_{n},n\geq1\}\) is an independent sequence of random variables, which is also independent of the sequence \(\{X_{n},n\geq1\}\). In addition, it is assumed that

$$ \sum_{i=1}^{n}EA_{i}^{2}= O(n). $$
(2.2)

Let \(S_{n}=\sum_{i=1}^{n} A_{i}X_{i}\), \(n\geq1\). Then one has the result (1.3), which implies the result (1.2).

Remark 2.1

In Theorem 2.1, \(\{A_{n}X_{n},\mathscr{F}_{n},n\geq 1\}\) may be not a martingale difference, since \(A_{n}\) is not required to be measurable with respect to \(\mathscr{F}_{n-1}\). We use the property of independence and the method of martingales to study the moments of maximum normed (1.2) and (1.3) and give a sufficient condition (2.1) for them.

Proof

It can be argued that

$$\begin{aligned} E\biggl(\sup_{n\geq1}\biggl\vert \frac{S_{n}}{n^{1/r}}\biggr\vert ^{p}\biggr) =&\int_{0}^{\infty}P\biggl(\sup _{n\geq1}\biggl\vert \frac{S_{n}}{n^{1/r}}\biggr\vert >t^{1/p} \biggr)\,dt \\ \leq& 2^{p/r}+\int_{2^{p/r}}^{\infty}\sum _{k=1}^{\infty}P\biggl(\max_{2^{k-1}\leq n< 2^{k}} \biggl\vert \frac{S_{n}}{n^{1/r}}\biggr\vert >t^{1/p}\biggr)\,dt \\ \leq& 2^{p/r}+\int_{2^{p/r}}^{\infty}\sum _{k=1}^{\infty}P\Bigl(\max_{1\leq n\leq2^{k}}| S_{n}|>2^{(k-1)/r}t^{1/p}\Bigr)\,dt \quad \bigl(\text{let } s=2^{(k-1)p/r}t\bigr) \\ =& 2^{p/r}+2^{p/r}\sum_{k=1}^{\infty}2^{-kp/r} \int_{2^{kp/r}}^{\infty}P\Bigl(\max_{1\leq n\leq2^{k}}|S_{n}|>s^{1/p} \Bigr)\,ds. \end{aligned}$$
(2.3)

Let \(\mathscr{G}_{0}=\{\emptyset,\Omega\}\), \(\mathscr{G}_{n}=\sigma(X_{1},\ldots,X_{n})\), \(n\geq1\), and \(X_{si}=X_{i}I(|X_{i}|\leq s^{1/p})\), where \(s^{1/p}>0\). It can be argued that

$$A_{i}X_{i}=A_{i}X_{i}I \bigl(|X_{i}|>s^{1/p}\bigr)+\bigl[A_{i}X_{si}-E(A_{i}X_{si}| \mathscr {G}_{i-1})\bigr]+E(A_{i}X_{si}| \mathscr{G}_{i-1}), \quad 1\leq i\leq n. $$

Then

$$\begin{aligned}& \sum_{k=1}^{\infty}2^{-kp/r}\int _{2^{kp/r}}^{\infty}P\Bigl(\max_{1\leq n\leq 2^{k}}|S_{n}|>s^{1/p} \Bigr)\,ds \\& \quad \leq \sum_{k=1}^{\infty}2^{-kp/r} \int_{2^{kp/r}}^{\infty}P\Biggl(\max_{1\leq n\leq2^{k}} \Biggl\vert \sum_{i=1}^{n} A_{i}X_{i}I\bigl(|X_{i}|>s^{1/p}\bigr) \Biggr\vert >s^{1/p}/2\Biggr)\,ds \\& \qquad {} +\sum_{k=1}^{\infty}2^{-kp/r} \int_{2^{kp/r}}^{\infty}P\Biggl(\max_{1\leq n\leq2^{k}} \Biggl\vert \sum_{i=1}^{n} \bigl[A_{i}X_{si}-E(A_{i}X_{si}| \mathscr{G}_{i-1})\bigr]\Biggr\vert >s^{1/p}/4 \Biggr)\,ds \\& \qquad {} +\sum_{k=1}^{\infty}2^{-kp/r} \int_{2^{kp/r}}^{\infty}P\Biggl(\max_{1\leq n\leq2^{k}} \Biggl\vert \sum_{i=1}^{n} E(A_{i}X_{si}|\mathscr{G}_{i-1})\Biggr\vert >s^{1/p}/4\Biggr)\,ds \\& \quad =: I_{1}+I_{2}+I_{3}. \end{aligned}$$
(2.4)

Combining Hölder’s inequality with (2.2), we have

$$ \sum_{i=1}^{n} E|A_{i}|\leq \Biggl(\sum_{i=1}^{n} EA_{i}^{2} \Biggr)^{1/2}\Biggl(\sum_{i=1}^{n} 1 \Biggr)^{1-1/2}=O(n). $$
(2.5)

In view of \(\{A_{n},n\geq1\}\) is independent of the sequence \(\{X_{n},n\geq1\}\); by Markov’s inequality, (2.5), and Lemma 1.2, we get

$$\begin{aligned} I_{1} \leq&2\sum_{k=1}^{\infty}2^{-kp/r} \int_{2^{kp/r}}^{\infty}s^{-1/p} E\Biggl(\max _{1\leq n\leq2^{k}}\Biggl\vert \sum_{i=1}^{n} A_{i}X_{i}I\bigl(|X_{i}|>s^{1/p}\bigr) \Biggr\vert \Biggr)\,ds \\ \leq&2\sum_{k=1}^{\infty}2^{-kp/r}\int _{2^{kp/r}}^{\infty}s^{-1/p}\Biggl(\sum _{i=1}^{2^{k}}E|A_{i}|E|X_{i}|I \bigl(|X_{i}|>s^{1/p}\bigr) \Biggr)\,ds \\ \leq&C_{1}\sum_{k=1}^{\infty}2^{k-kp/r}\sum_{m=k}^{\infty}\int _{2^{mp/r}}^{2^{(m+1)p/r}} s^{-1/p}E\bigl[XI \bigl(X>s^{1/p}\bigr)\bigr]\,ds \\ \leq& C_{2}\sum_{k=1}^{\infty}2^{k-kp/r}\sum_{m=k}^{\infty}2^{mp/r-m/r}E \bigl[XI\bigl(X>2^{m/r}\bigr)\bigr] \\ =&C_{2}\sum_{m=1}^{\infty}2^{mp/r-m/r}E \bigl[XI\bigl(X>2^{m/r}\bigr)\bigr]\sum_{k=1}^{m} 2^{k-kp/r} \\ \leq&\left \{ \textstyle\begin{array}{l} C_{3}\sum_{m=1}^{\infty}2^{m-m/r}E[XI(X>2^{m/r})], \quad \text{if } p< r, \\ C_{4}\sum_{m=1}^{\infty}m2^{m-m/r}E[XI(X>2^{m/r})],\quad \text{if } p=r, \\ C_{5}\sum_{m=1}^{\infty}2^{mp/r-m/r}E[XI(X>2^{m/r})],\quad \text{if } p>r. \end{array}\displaystyle \right . \end{aligned}$$
(2.6)

For the case \(p< r\), if \(0< r<1\), then

$$\begin{aligned}& \sum_{m=1}^{\infty}2^{m-m/r}E\bigl[XI \bigl(X>2^{m/r}\bigr)\bigr] \\& \quad =\sum_{m=1}^{\infty}2^{m-m/r}\sum _{k=m}^{\infty}E\bigl[XI\bigl(2^{k/r}< Y \leq2^{(k+1)/r}\bigr)\bigr] \\& \quad = \sum_{k=1}^{\infty}E\bigl[XI \bigl(2^{k/r}< X\leq2^{(k+1)/r}\bigr)\bigr]\sum _{m=1}^{k} 2^{m(1-1/r)} \\& \quad \leq C_{1}\sum _{k=1}^{\infty}E\bigl[XI\bigl(2^{k/r}< X \leq2^{(k+1)/r}\bigr)\bigr]\leq C_{1}EX. \end{aligned}$$

If \(r=1\), then

$$\begin{aligned}& \sum_{m=1}^{\infty}2^{m-m/r}E\bigl[XI \bigl(X>2^{m/r}\bigr)\bigr] \\& \quad =\sum_{m=1}^{\infty}E\bigl[XI \bigl(X>2^{m}\bigr)\bigr] = \sum_{m=1}^{\infty}\sum _{k=m}^{\infty}E\bigl[XI\bigl(2^{k}< X\leq 2^{k+1}\bigr)\bigr] \\& \quad =\sum_{k=1}^{\infty}E\bigl[XI\bigl(2^{k}< X\leq2^{k+1}\bigr)\bigr]\sum _{m=1}^{k}1 = \sum_{k=1}^{\infty}k E\bigl[XI \bigl(2^{k}< X\leq2^{k+1}\bigr)\bigr] \\& \quad \leq C_{2}\sum _{k=1}^{\infty}E\bigl[X\log(1+X)I \bigl(2^{k}< X\leq2^{k+1}\bigr)\bigr] \\& \quad \leq C_{2}E\bigl[X\log(1+X)\bigr]. \end{aligned}$$

Otherwise, for \(r>1\), one has

$$\begin{aligned}& \sum_{m=1}^{\infty}2^{m-m/r}E\bigl[XI \bigl(X>2^{m/r}\bigr)\bigr] \\& \quad =\sum_{m=1}^{\infty}2^{m-m/r}\sum _{k=m}^{\infty}E\bigl[XI\bigl(2^{k/r}< X \leq2^{(k+1)/r}\bigr)\bigr] \\& \quad = \sum_{k=1}^{\infty}E\bigl[XI \bigl(2^{k/r}< X\leq2^{(k+1)/r}\bigr)\bigr]\sum _{m=1}^{k} 2^{m-m/r} \\& \quad \leq C_{1}\sum _{k=1}^{\infty}2^{k-k/r} E\bigl[XI \bigl(2^{k/r}< X\leq 2^{(k+1)/r}\bigr)\bigr] \\& \quad \leq C_{3}\sum_{k=1}^{\infty}E\bigl[X^{r}I\bigl(2^{k/r}< X\leq2^{(k+1)/r}\bigr)\bigr] \leq C_{3}EX^{r}. \end{aligned}$$

Similarly, for the case \(p=r\), if \(0< r<1\), then

$$\begin{aligned}& \sum_{m=1}^{\infty}m2^{m-m/r}E\bigl[XI \bigl(X>2^{m/r}\bigr)\bigr] \\& \quad =\sum_{m=1}^{\infty}m2^{m-m/r} \sum_{k=m}^{\infty}E\bigl[XI \bigl(2^{k/r}< X\leq2^{(k+1)/r}\bigr)\bigr] \\& \quad = \sum_{k=1}^{\infty}E\bigl[XI \bigl(2^{k/r}< X\leq2^{(k+1)/r}\bigr)\bigr]\sum _{m=1}^{k} m2^{m(1-1/r)} \\& \quad \leq C_{4}\sum_{k=1}^{\infty}E\bigl[XI\bigl(2^{k/r}< X\leq2^{(k+1)/r}\bigr)\bigr] \leq C_{4}EX. \end{aligned}$$

If \(r=1\), then

$$\begin{aligned}& \sum_{m=1}^{\infty}m2^{m-m/r}E\bigl[XI \bigl(X>2^{m/r}\bigr)\bigr] \\& \quad =\sum_{m=1}^{\infty}m\sum _{k=m}^{\infty}E\bigl[XI\bigl(2^{k}< X \leq2^{k+1}\bigr)\bigr] \\& \quad = \sum_{k=1}^{\infty}E\bigl[XI \bigl(2^{k}< X\leq2^{k+1}\bigr)\bigr]\sum _{m=1}^{k} m \leq C_{5}\sum _{k=1}^{\infty}k^{2}E\bigl[XI \bigl(2^{k}< X\leq2^{k+1}\bigr)\bigr] \\& \quad \leq C_{5}\sum_{k=1}^{\infty}E\bigl[X\log^{2} (1+X)I\bigl(2^{k}< X\leq2^{k+1} \bigr)\bigr] \leq C_{5}E\bigl[X\log^{2} (1+X)\bigr]. \end{aligned}$$

Otherwise, for \(r>1\), it follows that

$$\begin{aligned}& \sum_{m=1}^{\infty}m2^{m-m/r}E\bigl[XI \bigl(X>2^{m/r}\bigr)\bigr] \\& \quad =\sum_{m=1}^{\infty}m2^{m-m/r} \sum_{k=m}^{\infty}E\bigl[XI \bigl(2^{k/r}< X\leq2^{(k+1)/r}\bigr)\bigr] \\& \quad \leq \sum_{k=1}^{\infty}E\bigl[XI \bigl(2^{k/r}< X\leq 2^{(k+1)/r}\bigr)\bigr]k\sum _{m=1}^{k} 2^{m-m/r} \\& \quad \leq C_{6}\sum_{k=1}^{\infty}k2^{k-k/r} E\bigl[XI\bigl(2^{k/r}< X\leq 2^{(k+1)/r}\bigr) \bigr] \leq C_{7}E\bigl[X^{r}\log(1+X)\bigr]. \end{aligned}$$

On the other hand, for the case \(p>r\), if \(0< p<1\), then

$$\begin{aligned}& \sum_{m=1}^{\infty}2^{mp/r-m/r}E\bigl[XI \bigl(X>2^{m/r}\bigr)\bigr] \\& \quad =\sum_{m=1}^{\infty}2^{m(p-1)/r}\sum _{k=m}^{\infty}E\bigl[XI\bigl(2^{k/r}< X \leq2^{(k+1)/r}\bigr)\bigr] \\& \quad = \sum_{k=1}^{\infty}E\bigl[XI \bigl(2^{k/r}< X\leq2^{(k+1)/r}\bigr)\bigr]\sum _{m=1}^{k} 2^{m(p-1)/r} \\& \quad \leq C_{8}\sum _{k=1}^{\infty}E\bigl[XI\bigl(2^{k/r}< X \leq 2^{(k+1)/r}\bigr)\bigr] \leq C_{8}EX. \end{aligned}$$

If \(p=1\), then

$$\begin{aligned}& \sum_{m=1}^{\infty}2^{mp/r-m/r}E\bigl[XI \bigl(X>2^{m/r}\bigr)\bigr] \\& \quad =\sum_{m=1}^{\infty}\sum _{k=m}^{\infty}E\bigl[XI\bigl(2^{k/r}< X \leq2^{(k+1)/r}\bigr)\bigr] \\& \quad = \sum_{k=1}^{\infty}E\bigl[XI \bigl(2^{k/r}< X\leq2^{(k+1)/r}\bigr)\bigr]\sum _{m=1}^{k} 1 \\& \quad =\sum_{k=1}^{\infty}kE\bigl[XI\bigl(2^{k/r}< X\leq 2^{(k+1)/r}\bigr)\bigr] \\& \quad \leq C_{9}EX\log(1+X). \end{aligned}$$

For \(p>1\), one has

$$\begin{aligned}& \sum_{m=1}^{\infty}2^{mp/r-m/r}E\bigl[XI \bigl(X>2^{m/r}\bigr)\bigr] \\& \quad =\sum_{m=1}^{\infty}2^{m(p-1)/r}\sum _{k=m}^{\infty}E\bigl[XI\bigl(2^{k/r}< X \leq2^{(k+1)/r}\bigr)\bigr] \\& \quad = \sum_{k=1}^{\infty}E\bigl[XI \bigl(2^{k/r}< X\leq2^{(k+1)/r}\bigr)\bigr]\sum _{m=1}^{k} 2^{m(p-1)/r} \\& \quad \leq C_{10}\sum_{k=1}^{\infty}2^{k(p-1)/r}E \bigl[XI\bigl(2^{k/r}< X\leq 2^{(k+1)/r}\bigr)\bigr] \\& \quad \leq C_{10}\sum_{k=1}^{\infty}E\bigl[X^{p}I\bigl(2^{k/r}< X\leq2^{(k+1)/r}\bigr)\bigr] \leq C_{10}EX^{p}. \end{aligned}$$

Consequently, in view of (2.6), the conditions of Theorem 2.1, and the inequalities above, we have

$$\begin{aligned} I_{1} \leq&\left \{ \textstyle\begin{array}{l} C_{1}\sum_{m=1}^{\infty}2^{m-m/r}E[XI(X>2^{m/r})], \quad \text{if } p< r, \\ C_{2}\sum_{m=1}^{\infty}m2^{m-m/r}E[XI(X>2^{m/r})],\quad \text{if } p=r, \\ C_{3}\sum_{m=1}^{\infty}2^{mp/r-m/r}E[XI(X>2^{m/r})], \quad \text{if } p>r \end{array}\displaystyle \right . \\ \leq&\left \{ \textstyle\begin{array}{l} \text{for } p< r,\quad \left \{ \textstyle\begin{array}{l} C_{4}EX< \infty,\quad \text{if } 0< r< 1, \\ C_{5}E[X\log(1+X)]< \infty, \quad \text{if } r=1, \\ C_{6}EX^{r}< \infty, \quad \text{if } r>1, \end{array}\displaystyle \right . \\ \text{for } p=r,\quad \left \{ \textstyle\begin{array}{l} C_{7}EX< \infty, \quad \text{if } 0< r< 1, \\ C_{8}E[X\log^{2}(1+X)]< \infty,\quad \text{if } r=1, \\ C_{9}E[X^{r}\log(1+X)]< \infty, \quad \text{if } r>1, \end{array}\displaystyle \right . \\ \text{for } p>r,\quad \left \{ \textstyle\begin{array}{l} C_{10}EX< \infty,\quad \text{if } 0< p< 1, \\ C_{11}E[X\log(1+X)]< \infty, \quad \text{if } p=1, \\ C_{12}EX^{p}< \infty, \quad \text{if } p>1. \end{array}\displaystyle \right . \end{array}\displaystyle \right . \end{aligned}$$
(2.7)

It can be checked that for the fixed real numbers \(a_{1}, \ldots, a_{n}\),

$$\bigl\{ a_{i}X_{si}-E(a_{i}X_{si}| \mathscr{G}_{i-1}),\mathscr{G}_{i},1\leq i\leq n\bigr\} $$

is also a martingale difference. So one has by Lemma 1.1

$$\begin{aligned}& E \Biggl\{ \max_{1\leq n\leq2^{k}}\sum_{i=1}^{n} \bigl[A_{i} X_{si}-E(A_{i}X_{si}| \mathscr{G}_{i-1})\bigr] \Biggr\} ^{2} \\& \quad = E \Biggl\{ E \Biggl(\max_{1\leq n\leq2^{k}}\sum _{i=1}^{n}\bigl[a_{i} X_{si}-E(a_{i}X_{si}| \mathscr{G}_{i-1})\bigr] \Biggr)^{2} \Big|A_{1}=a_{1}, \ldots ,A_{n}=a_{n} \Biggr\} \\& \quad \leq C_{1}E \Biggl\{ \sum_{i=1}^{2^{k}}E \bigl(a_{i}^{2}X_{si}^{2}\bigr) \Big|A_{1}=a_{1},\ldots,A_{n}=a_{n} \Biggr\} \\& \quad = C_{1}\sum_{i=1}^{2^{k}}EA_{i}^{2}EX_{si}^{2}, \end{aligned}$$
(2.8)

by using the fact that \(\{A_{1},\ldots, A_{n}\}\) is independent of \(\{X_{s1}, \ldots, X_{sn}\}\). Consequently, by Markov’s inequality, (2.2), (2.8), and Lemma 1.2, one can check that

$$\begin{aligned} I_{2} \leq&C_{1}\sum_{k=1}^{\infty}2^{-kp/r} \int_{2^{kp/r}}^{\infty}s^{-2/p}E \Biggl\{ \max _{1\leq n\leq2^{k}}\sum_{i=1}^{n} \bigl[A_{i} X_{si}-E(A_{i}X_{si}| \mathscr{G}_{i-1})\bigr] \Biggr\} ^{2}\,ds \\ \leq&C_{2}\sum_{k=1}^{\infty}2^{-kp/r} \int_{2^{kp/r}}^{\infty}s^{-2/p} \Biggl(\sum _{i=1}^{2^{k}}EA_{i}^{2}EX_{si}^{2} \Biggr)\,ds \\ \leq&C_{3}\sum_{k=1}^{\infty}2^{-kp/r+k}\int_{2^{kp/r}}^{\infty}s^{-2/p}E \bigl[X^{2}I\bigl(X\leq s^{1/p}\bigr)\bigr]\,ds \\ &{}+C_{4}\sum_{k=1}^{\infty}2^{-kp/r+k} \int_{2^{kp/r}}^{\infty}P\bigl(X>s^{1/p}\bigr)\,ds \\ =:&C_{3}I_{21}+C_{4}I_{22}. \end{aligned}$$
(2.9)

For \(I_{21}\), it follows that

$$\begin{aligned} I_{21} =&\sum_{k=1}^{\infty}2^{-kp/r+k}\sum_{m=k}^{\infty}\int _{2^{mp/r}}^{2^{(m+1)p/r}} s^{-2/p}E\bigl[X^{2}I \bigl(X\leq s^{1/p}\bigr)\bigr]\,ds \\ \leq&\sum_{k=1}^{\infty}2^{-kp/r+k}\sum _{m=k}^{\infty}2^{mp/r-2m/r}E \bigl[X^{2}I\bigl(X\leq 2^{(m+1)/r}\bigr)\bigr] \\ =&\sum_{m=1}^{\infty}2^{m(p-2)/r}E \bigl[X^{2}I\bigl(X\leq 2^{(m+1)/r}\bigr)\bigr]\sum _{k=1}^{m} 2^{k(1-p/r)}. \end{aligned}$$

If \(p< r\), one has by \(r<2\)

$$\begin{aligned}& \sum_{m=1}^{\infty}2^{m(p-2)/r}E \bigl[X^{2}I\bigl(X\leq 2^{(m+1)/r}\bigr)\bigr]\sum _{k=1}^{m} 2^{k(1-p/r)} \\& \quad \leq C_{1}\sum_{m=1}^{\infty}2^{m(r-2)/r}E \bigl[X^{2}I\bigl(X\leq2^{(m+1)/r}\bigr)\bigr] \\& \quad = C_{1}\sum_{m=1}^{\infty}2^{m(r-2)/r}E \bigl[X^{2}I\bigl(X\leq2^{1/r}\bigr)\bigr] \\& \qquad {} +C_{1}\sum_{m=1}^{\infty}2^{m(r-2)/r}\sum_{i=1}^{m}E \bigl[X^{2}I\bigl(2^{i/r}< X\leq2^{(i+1)/r}\bigr)\bigr] \\& \quad \leq C_{2}+C_{1}\sum_{i=1}^{\infty}E \bigl[X^{2}I\bigl(2^{i/r}< X\leq2^{(i+1)/r}\bigr)\bigr]\sum _{m=i}^{\infty}2^{m(r-2)/r} \\& \quad \leq C_{2}+C_{3}2^{(2-r)/r}\sum _{i=1}^{\infty }2^{(i+1)(r-2)/r}E\bigl[X^{2}I \bigl(2^{i/r}< X\leq2^{(i+1)/r}\bigr)\bigr] \\& \quad \leq C_{4}+C_{5}EX^{r}. \end{aligned}$$

For the case \(p=r\), by \(r<2\), one has

$$\begin{aligned}& \sum_{m=1}^{\infty}2^{m(p-2)/r}E \bigl[X^{2}I\bigl(X\leq 2^{(m+1)/r}\bigr)\bigr]\sum _{k=1}^{m} 2^{k(1-p/r)} \\& \quad \leq C_{1}\sum_{m=1}^{\infty}m 2^{m(r-2)/r}E\bigl[X^{2}I\bigl(X\leq2^{(m+1)/r}\bigr)\bigr] \\& \quad = C_{1}\sum_{m=1}^{\infty}m2^{m(r-2)/r}E\bigl[X^{2}I\bigl(X\leq2^{1/r}\bigr)\bigr] \\& \qquad {} +C_{1}\sum_{m=1}^{\infty}m 2^{m(r-2)/r}\sum_{i=1}^{m}E \bigl[X^{2}I\bigl(2^{i/r}< X\leq2^{(i+1)/r}\bigr)\bigr] \\& \quad \leq C_{2}+C_{1}\sum_{i=1}^{\infty}E \bigl[X^{2}I\bigl(2^{i/r}< X\leq2^{(i+1)/r}\bigr)\bigr]\sum _{m=i}^{\infty}m 2^{m(r-2)/r} \\& \quad \leq C_{2}+C_{3}2^{(2-r)/r}\sum _{i=1}^{\infty }i2^{(i+1)(r-2)/r}E\bigl[X^{2}I \bigl(2^{i/r}< X\leq2^{(i+1)/r}\bigr)\bigr] \\& \quad \leq C_{4}+C_{5}E\bigl[X^{r}\log(1+X) \bigr]. \end{aligned}$$

For the case \(p>r\), it can be checked by \(p<2\) that

$$\sum_{m=1}^{\infty}2^{m(p-2)/r}E \bigl[X^{2}I\bigl(Y\leq 2^{(m+1)/r}\bigr)\bigr]\leq CEX^{p}. $$

Consequently, it follows that

$$\begin{aligned} I_{21} \leq&\left \{ \textstyle\begin{array}{l} C_{1}\sum_{m=1}^{\infty}2^{m(r-2)/r}E[X^{2}I(X\leq2^{(m+1)/r})], \quad \text{if } p< r, \\ C_{2}\sum_{m=1}^{\infty}m2^{m(r-2)/r}E[X^{2}I(X\leq2^{(m+1)/r})],\quad \text{if } p=r, \\ C_{3}\sum_{m=1}^{\infty}2^{m(p-2)/r}E[X^{2}I(Y\leq 2^{(m+1)/r})],\quad \text{if } p>r \end{array}\displaystyle \right . \\ \leq&\left \{ \textstyle\begin{array}{l} C_{4}+C_{5}EX^{r}< \infty,\quad \text{if } p< r, \\ C_{6}+C_{7}E[X^{r}\log(1+X)]< \infty, \quad \text{if } p=r, \\ C_{8}EX^{p}< \infty,\quad \text{if } p>r. \end{array}\displaystyle \right . \end{aligned}$$
(2.10)

On the other hand, similar to the proofs of (2.6) and (2.7), we obtain

$$\begin{aligned} I_{22} \leq& \sum_{k=1}^{\infty}2^{k-kp/r}\int_{2^{kp/r}}^{\infty}s^{-1/p}E \bigl[XI\bigl(X>s^{1/p}\bigr)\bigr]\,ds \\ \leq&C_{2}\sum_{m=1}^{\infty}2^{mp/r-m/r}E \bigl[XI\bigl(X>2^{m/r}\bigr)\bigr]\sum_{k=1}^{m} 2^{k-kp/r} \\ \leq&\left \{ \textstyle\begin{array}{l} C_{3}\sum_{m=1}^{\infty}2^{m-m/r}E[XI(X>2^{m/r})], \quad \text{if } p< r, \\ C_{4}\sum_{m=1}^{\infty}m2^{m-m/r}E[XI(X>2^{m/r})], \quad \text{if } p=r, \\ C_{5}\sum_{m=1}^{\infty}2^{mp/r-m/r}E[XI(X>2^{m/r})], \quad \text{if } p>r. \end{array}\displaystyle \right . \end{aligned}$$
(2.11)

For the case \(p< r\), if \(0< r<1\), then

$$\begin{aligned}& \sum_{m=1}^{\infty}2^{m-m/r}E\bigl[XI \bigl(X>2^{m/r}\bigr)\bigr] \\& \quad = \sum_{m=1}^{\infty}2^{m-m/r}\sum _{k=m}^{\infty}E\bigl[XI\bigl(2^{k/r}< X \leq 2^{(k+1)/r}\bigr)\bigr] \\& \quad = \sum_{k=1}^{\infty}E\bigl[XI \bigl(2^{k/r}< X\leq2^{(k+1)/r}\bigr)\bigr]\sum _{m=1}^{k} 2^{m(1-1/r)} \\& \quad \leq C_{1}\sum_{k=1}^{\infty}E\bigl[XI\bigl(2^{k/r}< X\leq2^{(k+1)/r}\bigr)\bigr] \leq C_{1}EX. \end{aligned}$$

If \(r=1\), then

$$\begin{aligned}& \sum_{m=1}^{\infty}2^{m-m/r}E\bigl[XI \bigl(X>2^{m/r}\bigr)\bigr] \\& \quad =\sum_{m=1}^{\infty}E\bigl[XI \bigl(X>2^{m}\bigr)\bigr] = \sum_{m=1}^{\infty}\sum _{k=m}^{\infty}E\bigl[XI\bigl(2^{k}< X\leq 2^{k+1}\bigr)\bigr] \\& \quad =\sum_{k=1}^{\infty}E\bigl[XI\bigl(2^{k}< X\leq 2^{k+1}\bigr)\bigr]\sum _{m=1}^{k}1 \leq \sum_{k=1}^{\infty}k E\bigl[XI \bigl(2^{k}< X\leq2^{k+1}\bigr)\bigr] \\& \quad \leq C_{2}\sum _{k=1}^{\infty}E\bigl[X\log(1+X)I \bigl(2^{k}< X\leq2^{k+1}\bigr)\bigr] \leq C_{2}E\bigl[X\log(1+X)\bigr]. \end{aligned}$$

Otherwise, for \(r>1\), we have

$$\begin{aligned}& \sum_{m=1}^{\infty}2^{m-m/r}E\bigl[XI \bigl(X>2^{m/r}\bigr)\bigr] \\& \quad = \sum_{m=1}^{\infty}2^{m-m/r}\sum _{k=m}^{\infty}E\bigl[XI\bigl(2^{k/r}< X \leq 2^{(k+1)/r}\bigr)\bigr] \\& \quad = \sum_{k=1}^{\infty}E\bigl[XI \bigl(2^{k/r}< X\leq 2^{(k+1)/r}\bigr)\bigr]\sum _{m=1}^{k} 2^{m-m/r} \\& \quad \leq C_{3}\sum_{k=1}^{\infty}2^{k-k/r} E\bigl[XI\bigl(2^{k/r}< X\leq 2^{(k+1)/r}\bigr)\bigr] \\& \quad \leq C_{3}\sum_{k=1}^{\infty}E\bigl[X^{r}I\bigl(2^{k/r}< X\leq2^{(k+1)/r}\bigr)\bigr] \leq C_{3}EX^{r}. \end{aligned}$$

Similarly, for the case \(p=r\), if \(0< r<1\), then

$$\begin{aligned}& \sum_{m=1}^{\infty}m2^{m-m/r}E\bigl[XI \bigl(X>2^{m/r}\bigr)\bigr] \\& \quad =\sum_{m=1}^{\infty}m2^{m-m/r} \sum_{k=m}^{\infty}E\bigl[XI \bigl(2^{k/r}< X\leq2^{(k+1)/r}\bigr)\bigr] \\& \quad = \sum_{k=1}^{\infty}E\bigl[XI \bigl(2^{k/r}< X\leq2^{(k+1)/r}\bigr)\bigr]\sum _{m=1}^{k} m2^{m(1-1/r)} \\& \quad \leq C_{4}\sum_{k=1}^{\infty}E\bigl[XI\bigl(2^{k/r}< X\leq2^{(k+1)/r}\bigr)\bigr] \leq C_{4}EX. \end{aligned}$$

For \(r=1\), it follows that

$$\begin{aligned}& \sum_{m=1}^{\infty}m2^{m-m/r}E\bigl[XI \bigl(X>2^{m/r}\bigr)\bigr] \\& \quad =\sum_{m=1}^{\infty}m\sum _{k=m}^{\infty}E\bigl[XI\bigl(2^{k}< X \leq2^{k+1}\bigr)\bigr] \\& \quad = \sum_{k=1}^{\infty}E\bigl[XI \bigl(2^{k}< X\leq2^{k+1}\bigr)\bigr]\sum _{m=1}^{k} m \leq C_{5}\sum _{k=1}^{\infty}k^{2}E\bigl[XI \bigl(2^{k}< X\leq 2^{k+1}\bigr)\bigr] \\& \quad \leq C_{5}\sum_{k=1}^{\infty}E\bigl[X\log^{2} (1+X)I\bigl(2^{k}< X\leq2^{k+1} \bigr)\bigr] \leq C_{5}E\bigl[X\log^{2} (1+X)\bigr]. \end{aligned}$$

Otherwise, for \(r>1\), it follows that

$$\begin{aligned}& \sum_{m=1}^{\infty}m2^{m-m/r}E\bigl[XI \bigl(X>2^{m/r}\bigr)\bigr] \\& \quad =\sum_{m=1}^{\infty}m2^{m-m/r} \sum_{k=m}^{\infty}E\bigl[XI \bigl(2^{k/r}< X\leq2^{(k+1)/r}\bigr)\bigr] \\& \quad \leq \sum_{k=1}^{\infty}E\bigl[XI \bigl(2^{k/r}< X\leq 2^{(k+1)/r}\bigr)\bigr]\sum _{m=1}^{k} m2^{m-m/r} \\& \quad \leq C_{6}\sum_{k=1}^{\infty}k 2^{k-k/r} E\bigl[XI\bigl(2^{k/r}< X\leq 2^{(k+1)/r}\bigr) \bigr] \leq C_{7}E\bigl[X^{r}\log(1+X)\bigr]. \end{aligned}$$

Therefore, by (2.7) for case \(p>r\), (2.11), and the inequalities above, we obtain

$$\begin{aligned} I_{22} \leq&\left \{ \textstyle\begin{array}{l} C_{1}\sum_{m=1}^{\infty}2^{m-m/r}E[XI(X>2^{m/r})],\quad \text{if } p< r, \\ C_{2}\sum_{m=1}^{\infty}m2^{m-m/r}E[XI(X>2^{m/r})],\quad \text{if } p=r, \\ C_{3}\sum_{m=1}^{\infty}2^{mp/r-m/r}E[XI(X>2^{m/r})], \quad \text{if } p>r \end{array}\displaystyle \right . \\ \leq&\left \{ \textstyle\begin{array}{l} \text{for } p< r, \quad \left \{ \textstyle\begin{array}{l} C_{4}EX< \infty, \quad \text{if } 0< r< 1, \\ C_{5}E[X\log(1+X)]< \infty, \quad \text{if } r=1, \\ C_{6}EX^{r}< \infty, \quad \text{if } r>1, \end{array}\displaystyle \right . \\ \text{for } p=r,\quad \left \{ \textstyle\begin{array}{l} C_{7}EX< \infty,\quad \text{if } 0< r< 1, \\ C_{8}E[X\log^{2}(1+X)]< \infty,\quad \text{if } r=1, \\ C_{9}E[X^{r}\log(1+X)]< \infty, \quad \text{if } r>1, \end{array}\displaystyle \right . \\ \text{for } p>r,\quad \left \{ \textstyle\begin{array}{l} C_{10}EX< \infty,\quad \text{if } 0< p< 1, \\ C_{11}E[X\log(1+X)]< \infty,\quad \text{if } p=1, \\ C_{12}EX^{p}< \infty, \quad \text{if } p>1. \end{array}\displaystyle \right . \end{array}\displaystyle \right . \end{aligned}$$
(2.12)

Obviously, it can be seen that \(\{X_{n},\mathscr{G}_{n},n\geq1\}\) is also a martingale difference, since \(\{X_{n},\mathscr{F}_{n},n\geq 1\}\) is a martingale difference. Combining with the fact that \(\{A_{n},n\geq1\}\) is independent of \(\{X_{n},n\geq1\}\), we have

$$\begin{aligned} E(A_{i}X_{i}|\mathscr{G}_{i-1}) =&E \bigl[E(A_{i}X_{i}|\mathscr{G}_{i})|\mathscr {G}_{i-1}\bigr]=E\bigl[X_{i}E(A_{i}| \mathscr{G}_{i})|\mathscr{G}_{i-1}\bigr] \\ =&EA_{i}E[X_{i}|\mathscr{G}_{i-1}]=0,\quad \mbox{a.s.}, 1\leq i\leq n. \end{aligned}$$

In view of the proofs of (2.6), (2.7), and the inequality above, we obtain by Markov’s inequality and (2.5)

$$\begin{aligned} I_{3} \leq&4\sum_{k=1}^{\infty}2^{-kp/r} \int_{2^{kp/r}}^{\infty}s^{-1/p} E\Biggl(\max _{1\leq n\leq2^{k}}\Biggl\vert \sum_{i=1}^{n} E(A_{i}X_{si}|\mathscr{G}_{i-1})\Biggr\vert \Biggr)\,ds \\ =&4\sum_{k=1}^{\infty}2^{-kp/r}\int _{2^{kp/r}}^{\infty}s^{-1/p} E\Biggl(\max _{1\leq n\leq2^{k}}\Biggl\vert \sum_{i=1}^{n} E\bigl(A_{i}X_{i}I\bigl(|X_{i}|\leq s^{1/p}\bigr)|\mathscr{G}_{i-1}\bigr)\Biggr\vert \Biggr)\,ds \\ =&4\sum_{k=1}^{\infty}2^{-kp/r}\int _{2^{kp/r}}^{\infty}s^{-1/p} E\Biggl(\max _{1\leq n\leq2^{k}}\Biggl\vert \sum_{i=1}^{n} E\bigl(A_{i}X_{i}I\bigl(|X_{i}|> s^{1/p}\bigr)|\mathscr{G}_{i-1}\bigr)\Biggr\vert \Biggr)\,ds \\ \leq&4\sum_{k=1}^{\infty}2^{-kp/r}\int _{2^{kp/r}}^{\infty}s^{-1/p} \sum _{i=1}^{2^{k}} E|A_{i}|E|X_{i}|I \bigl(|X_{i}|> s^{1/p}\bigr)\,ds \\ \leq&C_{1}\sum_{k=1}^{\infty}2^{k-kp/r} \int_{2^{kp/r}}^{\infty}s^{-1/p}E\bigl[XI \bigl(X>s^{1/p}\bigr)\bigr]\,ds \\ \leq&\left \{ \textstyle\begin{array}{l} \text{for } p< r, \quad \left \{ \textstyle\begin{array}{l} C_{1}EX< \infty,\quad \text{if } 0< r< 1, \\ C_{2}E[X\log(1+X)]< \infty,\quad \text{if } r=1, \\ C_{3}EX^{r}< \infty, \quad \text{if } r>1, \end{array}\displaystyle \right . \\ \text{for } p=r,\quad \left \{ \textstyle\begin{array}{l} C_{4}EX< \infty, \quad \text{if } 0< r< 1, \\ C_{5}E[X\log^{2}(1+X)]< \infty,\quad \text{if } r=1, \\ C_{6}E[X^{r}\log(1+X)]< \infty, \quad \text{if } r>1, \end{array}\displaystyle \right . \\ \text{for } p>r, \quad \left \{ \textstyle\begin{array}{l} C_{7}EX< \infty, \quad \text{if } 0< p< 1, \\ C_{8}E[X\log(1+X)]< \infty,\quad \text{if } p=1, \\ C_{9}EX^{p}< \infty, \quad \text{if } p>1. \end{array}\displaystyle \right . \end{array}\displaystyle \right . \end{aligned}$$
(2.13)

Consequently, in view of (2.1), (2.3), (2.4), (2.7)-(2.10), (2.12), and (2.13), one has (1.3). By (1.3), it is easy to get (1.2). □

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Acknowledgements

The authors are deeply grateful to the editor and two anonymous referees, whose insightful comments and suggestions have contributed substantially to the improvement of this paper. This work is supported by the National Natural Science Foundation of China (11171188), National Social Science Fund of China (14ATJ005), the Humanities and Social Science Planning Foundation of Ministry of Education of China (14YJCZH155) and the Fundamental Scientific Research Funds for the Central Universities (2015HGZX0018).

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Yao, M., Lin, L. The moment of maximum normed randomly weighted sums of martingale differences. J Inequal Appl 2015, 264 (2015). https://doi.org/10.1186/s13660-015-0786-1

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MSC

  • 60F15
  • 60F25

Keywords

  • random weighted
  • maximum normed
  • martingale differences