Self-Normalized Moderate Deviations for Degenerate U-Statistics

Abstract

In this paper, we study self-normalized moderate deviations for degenerate U-statistics of order 2. Let $\{X_i,i\ge 1\}$ be i.i.d. random variables and consider symmetric and degenerate kernel functions in the form $h(x,y)={\sum }_{l=1}^{\infty }{\lambda }_l{g}_l\left(x\right)g_l\left(y\right)$, where ${\lambda }_l>0$, $E{g}_l\left(X_1\right)=0$, and $g_l\left(X_1\right)$ is in the domain of attraction of a normal law for all $l\ge 1$. Under the condition ${\sum }_{l=1}^{\infty }{\lambda }_l<\infty$ and some truncated conditions for $\{g_l\left(X_1\right):l\ge 1\}$, we show that $\log P({\displaystyle \frac{{\sum }_{1\le i\ne j\le n}h(X_i,X_j)}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2}}\ge x_n^2)\sim -{\displaystyle \frac{x_n^2}{2}}$ for $x_n\to \infty$ and $x_n=o\left(\sqrt{n}\right)$, where $V_{n,l}^2={\sum }_{i=1}^n{g}_l^2\left(X_i\right)$. As application, a law of the iterated logarithm is also obtained.

1. Introduction and Main Results

The recent three decades have witnessed significant developments on self-normalized limit theory, especially on large deviations, Cramér-type moderate deviations, and the law of the iterated logarithm. Compared with the classical limit theorems, these self-normalized limit theorems usually require much less moment assumptions.

Let $X,X_1,X_2,\cdots$ be independent identically distributed (i.i.d.) random variables. Set

$$\begin{array}{c}\hfill S_n=\sum _{i=1}^n{X}_i\mathrm{and}{V}_n^2=\sum _{i=1}^n{X}_i^2.\end{array}$$

Griffin and Kuelbs [1] obtained a law of the iterated logarithm (LIL) for the self-normalized sum of i.i.d. random variables with distributions in the domain of attraction of a normal or stable law. They proved that

• If $EX=0$ and X is in the domain of attraction of a normal law, then

  $$\underset{n\to \infty }{lim\; sup}{\displaystyle \frac{S_n}{V_n{\left(2\log \log n\right)}^{1/2}}}=1a.s.$$

  (1)
• If X is symmetric and is in the domain of attraction of a stable law, then there exists a positive constant C such that

  $$\underset{n\to \infty }{lim\; sup}{\displaystyle \frac{S_n}{V_n{\left(2\log \log n\right)}^{1/2}}}=Ca.s.$$

  (2)

Shao [2] obtained the following self-normalized moderate deviations and specified the constant C in (2). Let $\{x_n,n\ge 1\}$ be a sequence of positive numbers such that $x_n\to \infty$ and $x_n=o\left(\sqrt{n}\right)$ as $n\to \infty$.

• If $EX=0$ and X is in the domain of attraction of a normal law, then

  $$\underset{n\to \infty }{lim}{x}_n^{-2}\log P\left({\displaystyle \frac{S_n}{V_n}}\ge x_n\right)=-{\displaystyle \frac{1}{2}}.$$

  (3)
• If X is in the domain of attraction of a stable law such that $EX=0$ with index $1<\alpha <2$, or $X_1$ is symmetric with index $\alpha =1$, then

  $$\underset{n\to \infty }{lim}{x}_n^{-2}\log P\left({\displaystyle \frac{S_n}{V_n}}\ge x_n\right)=-\beta (\alpha ,c_1,c_2),$$

where $\beta (\alpha ,c_1,c_2)$ is a constant depending on the tail distribution; see [2] for an explicit definition.

Shao [3] refined (3) and obtained the following Cramér-type moderate deviation theorem under a finite third moment: if $EX=0$ and ${E\left|X\right|}^3<\infty$, then

$${\displaystyle \frac{P(S_n/V_n\ge x_n)}{P(Z\ge x_n)}}\to 1$$

(4)

for any $x_n\in [0,o\left(n^{1/6}\right))$, where Z is the standard normal random variable.

Jing, Shao and Wang [4] further extended (4) to general independent random variables under a Lindeberg-type condition, while Shao and Zhou [5] established the result for self-normalized non-linear statistics, which include U-statistics as a special case.

The U-statistics were introduced by Halmos [6] and Hoeffding [7]. The LIL for nondegenerate U-statistics was obtained by Serfling [8]. The LIL for degenerate U-statistics was studied by Dehling, Denker and Philipp ([9,10]), Dehling [11], Arcones and Giné [12], Teicher [13], Giné and Zhang [14], and others. Giné, Kwapień, Latała and Zinn [15] provided necessary and sufficient conditions for the LIL of degenerate U-statistics of order 2, which was extended to any order by Adamczak and Latała [16].

The main purpose of this paper is to study the self-normalized moderate deviations and the LIL for degenerate U-statistics of order 2. Let

$$\begin{array}{c}\hfill U_n={\displaystyle \frac{1}{n(n-1)}}\sum _{1\le i\ne j\le n}h(X_i,X_j),\end{array}$$

where

$$\begin{array}{c}\hfill h(x,y)=\sum _{l=1}^{\infty }{\lambda }_l{g}_l\left(x\right)g_l\left(y\right).\end{array}$$

(5)

A motivation example for the LIL is the one with the kernel $h(x,y)=xy$. Obviously, $V_n^2/\left(2V_n^2\log \log n\right)\to 0$. Then, via (1) and (2), we have

• If $EX=0$ and X is in the domain of attraction of a normal law, then

  $$\begin{array}{cc}\hfill & \underset{n\to \infty }{lim\; sup}{\displaystyle \frac{1}{2V_n^2\log \log n}}\left|\sum _{1\le i\ne j\le n}{X}_i{X}_j\right|\hfill \\ \hfill & =\underset{n\to \infty }{lim\; sup}{\left[{\displaystyle \frac{S_n}{V_n{\left(2\log \log n\right)}^{1/2}}}\right]}^2=1a.s.\hfill \end{array}$$

  (6)
• If X is symmetric and is in the domain of attraction of a stable law, then there exists a positive constant C such that

  $$\begin{array}{cc}\hfill & \underset{n\to \infty }{lim\; sup}{\displaystyle \frac{1}{2V_n^2\log \log n}}\left|\sum _{1\le i\ne j\le n}{X}_i{X}_j\right|\hfill \\ \hfill & =\underset{n\to \infty }{lim\; sup}{\left[{\displaystyle \frac{S_n}{V_n{\left(2\log \log n\right)}^{1/2}}}\right]}^2=C^{2}a.s.\hfill \end{array}$$

For the general degenerate kernel h defined in (5), we have

$$\begin{array}{ccc}\hfill n(n-1)U_n& =& \sum _{l=1}^{\infty }{\lambda }_l\sum _{1\le i\ne j\le n}{g}_l\left(X_i\right)g_l\left(X_j\right)\hfill \\ & =& \sum _{l=1}^{\infty }{\lambda }_l\left({\left(\sum _{i=1}^n{g}_l\left(X_i\right)\right)}^2-\sum _{i=1}^n{g}_l^2\left(X_i\right)\right).\hfill \end{array}$$

Suppose that $g_l\left(X\right)$ is in the domain of attraction of a normal law for every $l\ge 1$. Then, $L_l\left(x\right):=E{g}_l^2\left(X_1\right)I\left(\right|g_l\left(X_1\right)|\le x)$ is a slowly varying function for all $l\ge 1$ as $x\to \infty$. Let $\{x_n,n\ge 1\}$ be a sequence of positive numbers such that $x_n\to \infty$ and $x_n=o\left(\sqrt{n}\right)$ as $n\to \infty$. For each $l\ge 1$, set

$$\begin{array}{ccc}\hfill b_l& =& inf\left\{x\ge 1:L_l\left(x\right)>0\right\},\hfill \\ \hfill z_{n,l}& =& inf\left\{s:s\ge b_l+1,{\displaystyle \frac{L_l\left(s\right)}{s^2}}\le {\displaystyle \frac{x_n^2}{n}}\right\}.\hfill \end{array}$$

(7)

Write

$$\begin{array}{c}\hfill W_n={\displaystyle \frac{n(n-1)U_n}{{\max }_{1\le l<\infty }{\lambda }_l{\sum }_{i=1}^n{g}_l^2\left(X_i\right)}}.\end{array}$$

We have the following self-normalized moderate deviation:

Theorem 1.

Let $E{g}_l\left(X\right)=0$ and ${\lambda }_l\ge 0$ for every $l\ge 1$.

$$\begin{array}{c}\hfill \underset{x\in R}{sup}{\displaystyle \frac{{\sum }_{l=m+1}^{\infty }{\lambda }_l{g}_l^2\left(x\right)}{{\sum }_{1\le l<\infty }{\lambda }_l{g}_l^2\left(x\right)}}\to 0asm\to \infty \end{array}$$

(8)

and

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}{\displaystyle \frac{E{g}_l\left(X\right)I\left(\right|g_l\left(X\right)|\le z_{n,l})g_k\left(X\right)I\left(\right|g_k\left(X\right)|\le z_{n,k})}{\sqrt{L_l\left(z_{n,l}\right)L_k\left(z_{n,k}\right)}}}\to 0\end{array}$$

(9)

for any $l\ne k$. Then, for $x_n\to \infty$ and $x_n=o\left(\sqrt{n}\right)$,

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}{x}_n^{-2}\log P\left(W_n\ge x_n^2\right)=-{\displaystyle \frac{1}{2}}.\end{array}$$

As an application, we have the following self-normalized LIL:

Theorem 2.

Under the assumptions in Theorem 1, and instead of (8), we assume that for each $l\in [1,\infty )$, there is a constant $c_l>0$ such that

$$\begin{array}{c}\hfill \underset{x\in R}{sup}{\displaystyle \frac{{\lambda }_l{g}_l^2\left(x\right)}{{\sum }_{l=1}^{\infty }{\lambda }_l{g}_l^2\left(x\right)}}\le c_{l}and\sum _{l=1}^{\infty }{c}_l<\infty .\end{array}$$

(10)

Then,

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim\; sup}{\displaystyle \frac{W_n}{\log \log n}}=2a.s.\end{array}$$

(11)

Remark 1.

We use an example to show that (9) cannot be removed. Let $g_1\left(x\right)=x$, $g_2\left(x\right)=x^3$ and ${\lambda }_1={\lambda }_2=1$, ${\lambda }_l=0$ for $l\ge 3$. Let X be a Rademacher random variable. Then, $n(n-1)U_n={\sum }_{1\le i\ne j\le n}(X_i{X}_j+X_i^3{X}_j^3)=2{\sum }_{1\le i\ne j\le n}{X}_i{X}_j$ and $W_n=2{\sum }_{1\le i\ne j\le n}{X}_i{X}_j/{\sum }_{i=1}^n{X}_i^2$. By (6), ${lim\; sup}_{n\to \infty }{W}_n/\log \log n=4$ a.s. which contradicts (11).

2. Proofs

In the proofs of theorems, we will use the following properties for the slowly varying functions $g_l$ (e.g., Bingham et al. [17]). As $x\to \infty$,

$$\begin{array}{c}\hfill P\left(\right|g_l\left(X\right)|\ge x)=o(L_l\left(x\right)/x^2),\end{array}$$

(12)

$$\begin{array}{c}\hfill E|g_l\left(X\right)\left|I\right(|g_l\left(X\right)|\ge x)=o(L_l\left(x\right)/x),\end{array}$$

(13)

$$\begin{array}{c}\hfill E|g_l{\left(X\right)|}^{p}I\left(\right|g_l\left(X\right)|\le x)=o\left(x^{p-2}{L}_l\left(x\right)\right),p>2.\end{array}$$

(14)

Since $L_l\left(s\right)/s^2\to 0$ as $s\to \infty$ and $L_l\left(x\right)$ is right continuous, in (7), $z_{n,l}\to \infty$ and for all sufficiently large n values, we have

$$\begin{array}{c}\hfill n{L}_l\left(z_{n,l}\right)=x_n^2{z}_{n,l}^2.\end{array}$$

(15)

2.1. The Upper Bound of Theorem 1

For each $l\ge 1$ and $i\ge 1$, denote the truncated function

$$\begin{array}{c}\hfill {\overline{g}}_l\left(X_i\right)=g_l\left(X_i\right)I\left(|g_l\left(X_i\right)|\le z_{n,l}\right).\end{array}$$

Since $E{g}_l\left(X_i\right)=0$, we have

$$\begin{array}{c}\hfill E{\overline{g}}_l\left(X_i\right)=o(L_l\left(z_{n,l}\right)/z_{n,l})=o\left(x_n\sqrt{L_l(z_{n,l}}\right)/\sqrt{n})\end{array}$$

(16)

by (13) and (15). For each $l\ge 1$, write

$$\begin{array}{c}\hfill Y_{n,l}=\sum _{i=1}^n{g}_l\left(X_i\right),{\overline{Y}}_{n,l}=\sum _{i=1}^n{\overline{g}}_l\left(X_i\right)\mathrm{and}{V}_{n,l}^2=\sum _{i=1}^n{g}_l^2\left(X_i\right).\end{array}$$

By Condition (8), for each $0<\epsilon <1$, there exists $1\le m<\infty$ such that

$$\begin{array}{c}\hfill m\underset{1\le l<\infty }{\max }{\lambda }_l{V}_{n,l}^2\ge \sum _{l=1}^m{\lambda }_l{V}_{n,l}^2\ge (1-\epsilon )\sum _{l=1}^{\infty }{\lambda }_l{V}_{n,l}^2.\end{array}$$

(17)

Hence, for $x_n\to \infty$,

$$\begin{array}{c}\hfill P\left(W_n\ge (1+\epsilon )x_n^2\right)\le P\left({\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{Y}_{n,l}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2}}\ge x_n^2\right).\end{array}$$

(18)

Observe that for any random variables ${\left\{U_l\right\}}_{l=1}^{\infty }$ and ${\left\{Z_l\right\}}_{l=1}^{\infty }$ and any constants $x>0$ and $0<a<1/3$, we have the following via the Cauchy inequality:

$$\begin{array}{ccc}\hfill & & P\left(\sum _l{(U_l+Z_l)}^2\ge x\right)\hfill \\ & \le & P\left(\sum _l{U}_l^2\ge {(1-a)}^{2}x\right)+P\left(\sum _l{Z}_l^2\ge a^{2}x\right)\hfill \\ & \le & P\left(\sum _l{U}_l^2\ge (1-2a)x\right)+P\left(\sum _l{Z}_l^2\ge a^{2}x\right).\hfill \end{array}$$

(19)

For any $0<\epsilon <1/3$, by (18) and (19), we have

$$\begin{array}{ccc}\hfill & & P\left(W_n\ge (1+\epsilon )x_n^2\right)\hfill \\ & \le & P\left({\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\left({\sum }_{i=1}^n{g}_l\left(X_i\right)I\left(|g_l\left(X_i\right)|>z_{n,l}\right)\right)}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2}}\ge {\epsilon }^2{x}_n^2\right)\hfill \\ & & +P\left({\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\overline{Y}}_{n,l}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2}}\ge (1-2\epsilon )x_n^2\right).\hfill \end{array}$$

(20)

For any integer $m\ge 1$ and any constant $C_1>0$ with $C_1\epsilon <1$, we have

$$\begin{array}{ccc}\hfill & & P\left({\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\overline{Y}}_{n,l}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2}}\ge (1-2\epsilon )x_n^2\right)\hfill \\ & \le & P\left(\underset{1\le l<\infty }{\max }{\lambda }_l{V}_{n,l}^2\le {\displaystyle \frac{n}{\epsilon }}\sum _{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)\right)\hfill \\ & & +P\left(\underset{1\le l<\infty }{\max }{\lambda }_l{V}_{n,l}^2\le (1-\epsilon )n\underset{1\le l\le m}{\max }{\lambda }_l{L}_l\left(z_{n,l}\right)\right)\hfill \\ & & +P\left({\displaystyle \frac{{\sum }_{l=m+1}^{\infty }{\lambda }_l{\overline{Y}}_{n,l}^2}{(n/\epsilon ){\sum }_{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)}}\ge C_1\epsilon (1-2\epsilon )x_n^2\right)\hfill \\ & & +P\left({\displaystyle \frac{{\sum }_{l=1}^m{\lambda }_l{\overline{Y}}_{n,l}^2}{(1-\epsilon )n{\max }_{1\le l\le m}{\lambda }_l{L}_l\left(z_{n,l}\right)}}\ge (1-C_1\epsilon )(1-2\epsilon )x_n^2\right).\hfill \end{array}$$

(21)

Applying (21) to (20), we have

$$\begin{array}{ccc}\hfill & & P\left(W_n\ge (1+\epsilon )x_n^2\right)\hfill \\ & \le & P\left(\underset{1\le l<\infty }{\max }{\lambda }_l{V}_{n,l}^2\le {\displaystyle \frac{n}{\epsilon }}\sum _{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)\right)\hfill \\ & & +P\left(\underset{1\le l<\infty }{\max }{\lambda }_l{V}_{n,l}^2\le (1-\epsilon )n\underset{1\le l\le m}{\max }{\lambda }_l{L}_l\left(z_{n,l}\right)\right)\hfill \\ & & +P\left({\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\left({\sum }_{i=1}^n{g}_l\left(X_i\right)I\left(|g_l\left(X_i\right)|>z_{n,l}\right)\right)}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2}}\ge {\epsilon }^2{x}_n^2\right)\hfill \\ & & +P\left({\displaystyle \frac{{\sum }_{l=m+1}^{\infty }{\lambda }_l{\overline{Y}}_{n,l}^2}{n{\sum }_{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)}}\ge C_1(1-2\epsilon )x_n^2\right)\hfill \\ & & +P\left({\displaystyle \frac{{\sum }_{l=1}^m{\lambda }_l{\overline{Y}}_{n,l}^2}{n{\max }_{1\le l\le m}{\lambda }_l{L}_l\left(z_{n,l}\right)}}\ge (1-C_1\epsilon ){(1-2\epsilon )}^2{x}_n^2\right)\hfill \\ & :=& I_{1,1}+I_{1,2}+I_2+I_3+I_4.\hfill \end{array}$$

(22)

2.2. Estimation of $I_{1,1}$ and $I_{1,2}$

Proposition 1.

For $m\ge 1$ that is sufficiently large,

$$\begin{array}{c}\hfill I_{1,1}=P\left(\underset{1\le l<\infty }{\max }{\lambda }_l{V}_{n,l}^2\le {\displaystyle \frac{n}{\epsilon }}\sum _{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)\right)\le exp(-2x_n^2)\end{array}$$

(23)

and for any constants $\delta >0$ and $0<\eta <1$,

$$\begin{array}{c}\hfill P\left(2m\underset{1\le l<\infty }{\max }{\lambda }_l{V}_{n,l}^2\le (1-\eta )n\sum _{1\le l<\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)\right)\le exp(-2x_n^2),\end{array}$$

(24)

$$\begin{array}{ccc}\hfill & & P\left(\underset{1\le l<\infty }{\max }{\lambda }_l\sum _{i=1}^n{g}_l^2\left(X_i\right)I\left(\right|g_l\left(X_i\right)|\le \delta z_{n,l})\le (1-\eta )n\underset{1\le l<\infty }{\max }{\lambda }_l{L}_l\left(z_{n,l}\right)\right)\hfill \\ & \le & exp(-2x_n^2).\hfill \end{array}$$

(25)

In particular,

$$\begin{array}{c}\hfill I_{1,2}\le P\left(\underset{1\le l<\infty }{\max }{\lambda }_l{V}_{n,l}^2\le (1-\epsilon )n\underset{1\le l<\infty }{\max }{\lambda }_l{L}_l\left(z_{n,l}\right)\right)\le exp(-2x_n^2).\end{array}$$

Proof. 

We shall apply the following exponential inequality (see, e.g., Theorem 2.19 of de la Peña, Lai and Shao [18]). If $Y_1,\dots ,Y_n$ are independent random variables with $Y_i\ge 0$, ${\mu }_n={\sum }_{i=1}^{n}E{Y}_i$ and $B_n^2={\sum }_{i=1}^{n}E{Y}_i^2<\infty$, then for $0<x<{\mu }_n$,

$$\begin{array}{c}\hfill P\left(\sum _{i=1}^n{Y}_i\le x\right)\le exp\left(-{\displaystyle \frac{{({\mu }_n-x)}^2}{2B_n^2}}\right).\end{array}$$

By (8), ${\sum }_{l=m+1}^{\infty }{\lambda }_l{V}_{n,l}^2/{\sum }_{l=1}^{\infty }{\lambda }_l{V}_{n,l}^2\to 0$ as $m\to \infty$. Then, by (17),

$$\begin{array}{c}\hfill {\displaystyle \frac{{\sum }_{l=m+1}^{\infty }{\lambda }_l{V}_{n,l}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2}}\to 0asm\to \infty .\end{array}$$

(26)

Hence,

$$\begin{array}{c}\hfill \epsilon \underset{1\le l<\infty }{\max }{\lambda }_l{V}_{n,l}^2\ge 2\sum _{l=m+1}^{\infty }{\lambda }_l{V}_{n,l}^2\ge 2\sum _{l=m+1}^{\infty }{\lambda }_l\sum _{i=1}^n{\overline{g}}_l^2\left(X_i\right).\end{array}$$

Then,

$$\begin{array}{ccc}\hfill I_{1,1}& \le & P\left(\sum _{l=m+1}^{\infty }{\lambda }_l\sum _{i=1}^n{\overline{g}}_l^2\left(X_i\right)\le {\displaystyle \frac{n}{2}}\sum _{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)\right)\hfill \\ & \le & exp\left(-{\displaystyle \frac{{(n{\sum }_{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)/2)}^2}{2nE{\left({\sum }_{l=m+1}^{\infty }{\lambda }_l{\overline{g}}_l^2\left(X_1\right)\right)}^2}}\right).\hfill \end{array}$$

(27)

By Minkowski’s integral inequality, (14) and (15),

$$\begin{array}{ccc}\hfill E{\left(\sum _{l=m+1}^{\infty }{\lambda }_l{\overline{g}}_l^2\left(X_1\right)\right)}^2& \le & {\left\{\sum _{l=m+1}^{\infty }{\lambda }_l{\left(E{\overline{g}}_l^4\left(X_1\right)\right)}^{1/2}\right\}}^2\hfill \\ & =& o{\left({\displaystyle \frac{\sqrt{n}}{x_n}}\sum _{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)\right)}^2.\hfill \end{array}$$

(28)

Therefore, (23) follows from (27) and (28). To show (24), notice that by (17),

$$\begin{array}{c}\hfill m\underset{1\le l<\infty }{\max }{\lambda }_l{V}_{n,l}^2\ge (1-\eta )\sum _{l=1}^{\infty }{\lambda }_l{V}_{n,l}^2\ge (1-\eta )\sum _{l=1}^{\infty }{\lambda }_l\sum _{i=1}^n{\overline{g}}_l^2\left(X_i\right).\end{array}$$

Then,

$$\begin{array}{ccc}& & P\left(2m\underset{1\le l<\infty }{\max }{\lambda }_l{V}_{n,l}^2\le (1-\eta )n\sum _{1\le l<\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)\right)\hfill \\ & \le & P\left(\sum _{l=1}^{\infty }{\lambda }_l\sum _{i=1}^n{\overline{g}}_l^2\left(X_i\right)\le {\displaystyle \frac{n}{2}}\sum _{l=1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)\right)\hfill \\ & \le & exp\left(-{\displaystyle \frac{({\left(\frac{n}{2}{\sum }_{l=1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)\right)}^2}{2nE{\left({\sum }_{l=1}^{\infty }{\lambda }_l{\overline{g}}_l^2\left(X_1\right)\right)}^2}}\right).\hfill \end{array}$$

Similar to the proof of $I_{1,1}$ as in (27) and (28), we have (24).

To show (25), let

$$\begin{array}{c}\hfill l_n=min\left\{l:{\lambda }_l{L}_l\left(z_{n,l}\right)=\underset{1\le l<\infty }{\max }{\lambda }_l{L}_l\left(z_{n,l}\right)\right\}.\end{array}$$

Then,

$$\begin{array}{ccc}\hfill & & P\left(\underset{1\le l<\infty }{\max }{\lambda }_l\sum _{i=1}^n{g}_l^2\left(X_i\right)I\left(\right|g_l\left(X_i\right)|\le \delta z_{n,l})\le (1-\eta )n\underset{1\le l<\infty }{\max }{\lambda }_l{L}_l\left(z_{n,l}\right)\right)\hfill \\ & \le & P\left(\sum _{i=1}^n{\lambda }_{l_n}{g}_{l_n}^2\left(X_i\right)I\left(\right|g_{l_n}\left(X_i\right)|\le \delta z_{n,l_n})\le (1-\eta )n{\lambda }_{l_n}{L}_{l_n}\left(z_{n,l_n}\right)\right)\hfill \\ & \le & exp\left(-{\displaystyle \frac{{(1-(1-\eta ))}^2{\left(n{\lambda }_{l_n}{L}_{l_n}\left(z_{n,l_n}\right)\right)}^2}{2n{\lambda }_{l_n}^{2}E{g}_{l_n}^4\left(X_1\right)I\left(\right|g_{l_n}\left(X_1\right)|\le \delta z_{n,l_n})}}\right)\hfill \\ & =& exp\left(-{\displaystyle \frac{{\eta }^2{\left(n{\lambda }_{l_n}{L}_{l_n}\left(z_{n,l_n}\right)\right)}^2}{2n{\lambda }_{l_n}^{2}o(n{L}_{l_n}^2\left(\delta z_{n,l_n}\right)/x_n^2)}}\right).\hfill \end{array}$$

Since $L_l\left(\delta z_{n,l}\right)/L_l\left(z_{n,l}\right)\to 1$, (25) follows. □

2.3. Estimation of $I_2$

Proposition 2.

$$\begin{array}{c}\hfill I_2=P\left({\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\left\{{\sum }_{i=1}^n{g}_l\left(X_i\right)I\left(|g_l\left(X_i\right)|>z_{n,l}\right)\right\}}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2}}\ge {\epsilon }^2{x}_n^2\right)\le exp(-2x_n^2).\end{array}$$

Proof. 

Via the Cauchy–Schwarz inequality,

$$\begin{array}{ccc}& & {\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\left\{{\sum }_{i=1}^n\left|g_l\left(X_i\right)\right|I\left(|g_l\left(X_i\right)|>z_{n,l}\right)\right\}}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2}}\hfill \\ & \le & {\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\sum }_{i=1}^n{g}_l^2\left(X_i\right){\sum }_{i=1}^{n}I\left(|g_l\left(X_i\right)|>z_{n,l}\right)}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2}}.\hfill \end{array}$$

By (17), the sum of the diagonal terms is as follows:

$$\begin{array}{c}\hfill {\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\sum }_{i=1}^n{g}_l^2\left(X_i\right)I\left(|g_l\left(X_i\right)|>z_{n,l}\right)}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2}}\le {\displaystyle \frac{{\epsilon }^2{x}_n^2}{2}}.\end{array}$$

$$\begin{array}{ccc}\hfill I_2& \le & P\left({\displaystyle \frac{{\sum }_{1\le i\ne j\le n}{\sum }_{l=1}^{\infty }{\lambda }_l{g}_l^2\left(X_i\right)I\left(\right|g_l\left(X_j\right)|>z_{n,l})}{{\max }_{1\le l<\infty }{\lambda }_l{\sum }_{i=1}^n{g}_l^2\left(X_i\right)}}\ge {\displaystyle \frac{{\epsilon }^2{x}_n^2}{2}}\right)\hfill \\ & \le & P\left({\displaystyle \frac{{\sum }_{1\le i<j\le n}{\sum }_{l=1}^{\infty }{\lambda }_l{g}_l^2\left(X_i\right)I\left(\right|g_l\left(X_j\right)|>z_{n,l})}{{\max }_{1\le l<\infty }{\lambda }_l{\sum }_{i=1}^n{g}_l^2\left(X_i\right)}}\ge {\displaystyle \frac{{\epsilon }^2{x}_n^2}{4}}\right)\hfill \\ & +& P\left({\displaystyle \frac{{\sum }_{1\le j<i\le n}{\sum }_{l=1}^{\infty }{\lambda }_l{g}_l^2\left(X_i\right)I\left(\right|g_l\left(X_j\right)|>z_{n,l})}{{\max }_{1\le l<\infty }{\lambda }_l{\sum }_{i=1}^n{g}_l^2\left(X_i\right)}}\ge {\displaystyle \frac{{\epsilon }^2{x}_n^2}{4}}\right)\hfill \\ & \le & P\left(\sum _{2\le j\le n}{\displaystyle \frac{{\sum }_{1\le i<j}{\sum }_{l=1}^{\infty }{\lambda }_l{g}_l^2\left(X_i\right)I\left(\right|g_l\left(X_j\right)|>z_{n,l})}{{\max }_{1\le l<\infty }{\lambda }_l{\sum }_{1\le i<j}{g}_l^2\left(X_i\right)}}\ge {\displaystyle \frac{{\epsilon }^2{x}_n^2}{4}}\right)\hfill \\ & +& P\left(\sum _{1\le j<n}{\displaystyle \frac{{\sum }_{j<i\le n}{\sum }_{l=1}^{\infty }{\lambda }_l{g}_l^2\left(X_i\right)I\left(\right|g_l\left(X_j\right)|>z_{n,l})}{{\max }_{1\le l<\infty }{\lambda }_l{\sum }_{j<i\le n}{g}_l^2\left(X_i\right)}}\ge {\displaystyle \frac{{\epsilon }^2{x}_n^2}{4}}\right)\hfill \\ & =& I_{2,1}+I_{2,2}.\hfill \end{array}$$

(29)

Let

$$\begin{array}{c}\hfill {\varphi }_j={\displaystyle \frac{{\sum }_{1\le i<j}{\sum }_{l=1}^{\infty }{\lambda }_l{g}_l^2\left(X_i\right)I\left(\right|g_l\left(X_j\right)|>z_{n,l})}{{\max }_{1\le l<\infty }{\lambda }_l{\sum }_{1\le i<j}{g}_l^2\left(X_i\right)}}.\end{array}$$

Then, for any constant $t>0$,

$$\begin{array}{c}\hfill I_{2,1}\le E{e}^{t{\sum }_{j=2}^n{\varphi }_j}{e}^{-t{\epsilon }^2{x}_n^2/4}.\end{array}$$

(30)

Let $E_j$ be the expectation of $X_j$ for $2\le j\le n$. Then,

$$\begin{array}{c}\hfill E{e}^{t{\sum }_{j=2}^n{\varphi }_j}=E\left(e^{t{\sum }_{j=2}^{n-1}{\varphi }_j}{E}_n{e}^{t{\varphi }_n}\right).\end{array}$$

(31)

Since $|e^s-1|\le e^{0\vee s}\left|s\right|$ for any $s\in R$ and $0\le {\varphi }_n\le m$ for some sufficiently large m value, then

$$\begin{array}{ccc}\hfill \left|E_n{e}^{t{\varphi }_n}-1\right|& \le & e^{mt}t{E}_n{\varphi }_n\hfill \\ & =& {\displaystyle \frac{e^{mt}t{\sum }_{1\le i<n}{\sum }_{l=1}^{\infty }{\lambda }_l{g}_l^2\left(X_i\right)P\left(\right|g_l\left(X_n\right)|>z_{n,l})}{{\max }_{1\le l<\infty }{\lambda }_l{\sum }_{1\le i<n}{g}_l^2\left(X_i\right)}}.\hfill \end{array}$$

By (12) and (15), we have $P\left(\right|g_l\left(X_n\right)|>z_{n,l})=o(x_n^2/n)$. Then, together with (17),

$$\begin{array}{c}\hfill E_n{e}^{t{\varphi }_n}=1+o(x_n^2/n)=e^{o(x_n^2/n)}.\end{array}$$

(32)

Applying (32) to (31), we have

$$\begin{array}{c}\hfill E{e}^{t{\sum }_{j=2}^n{\varphi }_j}=e^{o(x_n^2/n)}E{e}^{t{\sum }_{j=2}^{n-1}{\varphi }_j}.\end{array}$$

Similarly,

$$\begin{array}{ccc}\hfill E{e}^{t{\sum }_{j=2}^{n-1}{\varphi }_j}& =& E\left(e^{t{\sum }_{j=2}^{n-2}{\varphi }_j}{E}_{n-1}{e}^{t{\varphi }_{n-1}}\right)\hfill \\ & =& e^{o(x_n^2/n)}E{e}^{t{\sum }_{j=2}^{n-2}{\varphi }_j}.\hfill \end{array}$$

Continue this process from $X_n$ to $X_1$. We conclude that

$$\begin{array}{c}\hfill E{e}^{t{\sum }_{j=2}^n{\varphi }_j}=e^{n\times o(x_n^2/n)}=e^{o\left(x_n^2\right)}.\end{array}$$

(33)

Applying (33) to (30) and letting $t=16/{\epsilon }^2$, we have

$$\begin{array}{c}\hfill I_{2,1}\le exp(-3x_n^2).\end{array}$$

(34)

By the same argument,

$$\begin{array}{c}\hfill I_{2,2}\le exp(-3x_n^2).\end{array}$$

(35)

Combining (29), (34) and (35), we obtain the proposition. □

2.4. Estimation of $I_3$

Let $Y_1,\dots ,Y_n$ be an independent copy of $X_1,\dots ,X_n$. We will use the following lemma which is a Bernstein-type exponential inequality for degenerate U-statistics.

Lemma 1

((3.5) of Giné, Latała and Zinn [19]). For bounded degenerate kernel $h_{i,j}(X_i,Y_j)$, let

$$\begin{array}{c}\hfill A=\underset{i,j}{\max }{∥h_{i,j}(X_i,Y_j)∥}_{\infty },C^2=\sum _{i,j}E{h}_{i,j}^2(X_i,Y_j),\\ \hfill B^2=\underset{i,j}{\max }\left\{\parallel \sum _{i}E{h}_{i,j}^2(X_i,y){\parallel }_{\infty },\parallel \sum _{j}E{h}_{i,j}^2(x,Y_j){\parallel }_{\infty }\right\}.\end{array}$$

Then, there is a universal constant K such that

$$\begin{array}{c}\hfill Pr\left\{\left|\sum _{i,j}{h}_{i,j}(X_i,Y_j)\right|>x\right\}\le Kexp\left\{-{\displaystyle \frac{1}{K}}min\left[{\displaystyle \frac{x}{C}},{\left({\displaystyle \frac{x}{B}}\right)}^{2/3},{\left({\displaystyle \frac{x}{A}}\right)}^{1/2}\right]\right\}.\end{array}$$

Recall (16). Hence, by (19) and the definition of $I_3$ in (22),

$$\begin{array}{c}\hfill I_3\le P\left({\displaystyle \frac{{\sum }_{l=m+1}^{\infty }{\lambda }_l{({\overline{Y}}_{n,l}-E{\overline{Y}}_{n,l})}^2}{{\sum }_{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)}}\ge C_{1}n{(1-2\epsilon )}^2{x}_n^2\right).\end{array}$$

(36)

Let

$$\begin{array}{c}\hfill h_m(X_i,Y_j)=\sum _{l=m+1}^{\infty }{\lambda }_l({\overline{g}}_l\left(X_i\right)-E{\overline{g}}_l\left(X_i\right))({\overline{g}}_l\left(Y_j\right)-E{\overline{g}}_l\left(Y_j\right)).\end{array}$$

(37)

In addition to the estimate of $I_3$, we include (40) in the following proposition which will be used in the proof of Theorem 2, where

$$\begin{array}{ccc}\hfill & & h_{\left(\beta \right)}(X_i,Y_j)\hfill \\ & =& \sum _{l=1}^{\infty }{\lambda }_l\left\{(g_l\left(X_i\right)I\left(\right|g_l\left(X_i\right)|\le \beta z_{n,l})-E{g}_l\left(X_i\right)I\left(\right|g_l\left(X_i\right)|\le \beta z_{n,l})\right\}\hfill \\ & & \times \left\{(g_l\left(Y_j\right)I\left(\right|g_l\left(Y_j\right)|\le \beta z_{n,l})-E{g}_l\left(Y_j\right)\left)I\right(|g_l\left(Y_j\right)|\le \beta z_{n,l})\right\}.\hfill \end{array}$$

(38)

Proposition 3.

For a sufficiently large constant $C_2>0$,

$$\begin{array}{c}\hfill P\left({\displaystyle \frac{{\sum }_{1\le i,j\le n}{h}_m(X_i,Y_j)}{n{\sum }_{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)}}\ge C_2{x}_n^2\right)\le exp(-3x_n^2),\end{array}$$

(39)

Then, by (36) and the decoupling inequalities of de la Peña and Montgomery-Smith [20], for a sufficiently large $C_1>0$,

$$\begin{array}{c}\hfill I_3\le P\left({\displaystyle \frac{{\sum }_{1\le i,j\le n}{h}_m(X_i,X_j)}{n{\sum }_{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)}}\ge C_1{(1-2\epsilon )}^2{x}_n^2\right)\le exp(-2x_n^2).\end{array}$$

Suppose that ${\lambda }_l>0$ for all $1\le l<\infty$ and $d>0$ is a constant. For constants $0<\alpha ,\beta \le 1$ that are sufficiently small,

$$\begin{array}{c}\hfill P\left({\displaystyle \frac{{\sum }_{1\le i,j\le \left[\alpha n\right]}{h}_{\left(\beta \right)}(X_i,Y_j)}{n{\sum }_{l=1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)}}\ge d{x}_n^2\right)\le exp(-2x_n^2).\end{array}$$

(40)

Proof. 

We will prove (39) and (40) simultaneously. By (15) and (37),

$$\begin{array}{ccc}\hfill A_n& :=& {∥h_m(X_i,Y_j)∥}_{\infty }\le 4\sum _{l=m+1}^{\infty }{\lambda }_l{z}_{n,l}^2=4\sum _{l=m+1}^{\infty }{\lambda }_l{\displaystyle \frac{n{L}_l\left(z_{n,l}\right)}{x_n^2}}.\hfill \end{array}$$

(41)

By (15) and (38),

$$\begin{array}{c}\hfill A_{n,\left(\beta \right)}:={∥h_{\left(\beta \right)}(X_i,Y_j)∥}_{\infty }\le 4\sum _{l=1}^{\infty }{\lambda }_l{\beta }^2{z}_{n,l}^2={\displaystyle \frac{4{\beta }^{2}n{\sum }_{l=1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)}{x_n^2}}.\end{array}$$

(42)

Let

$$B_n^2:=\max \left\{{\parallel \sum _{1\le i\le n}E{h}_m^2(X_i,y)\parallel }_{\infty },{\parallel \sum _{1\le j\le n}E{h}_m^2(x,Y_j)\parallel }_{\infty }\right\}.$$

Since $|{\overline{g}}_l\left(Y_j\right)|\le z_{n,l}$, then by Cauchy–Schwarz inequality, (15) and (37),

$$\begin{array}{ccc}\hfill & & {\parallel \sum _{1\le i\le n}E{h}_m^2(X_i,y)\parallel }_{\infty }\le nE{\left(2\sum _{l=m+1}^{\infty }{\lambda }_l|{\overline{g}}_l\left(X_1\right)-E{\overline{g}}_l\left(X_1\right)|z_{n,l}\right)}^2\hfill \\ & \le & 4nE\sum _{l=m+1}^{\infty }{\lambda }_l{\left({\overline{g}}_l\left(X_1\right)-E{\overline{g}}_l\left(X_1\right)\right)}^2\sum _{l=m+1}^{\infty }{\lambda }_l{z}_{n,l}^2\hfill \\ & \le & 4n\sum _{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)\sum _{l=m+1}^{\infty }{\lambda }_l{\displaystyle \frac{n{L}_l\left(z_{n,l}\right)}{x_n^2}}.\hfill \end{array}$$

The same result can be obtained for $\parallel {\sum }_{1\le j\le n}E{h}_m^2(x,Y_j){\parallel }_{\infty }$. Therefore,

$$\begin{array}{c}\hfill B_n^2\le {\displaystyle \frac{4n^2{\left({\sum }_{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)\right)}^2}{x_n^2}}.\end{array}$$

(43)

Similarly,

$$\begin{array}{ccc}\hfill B_{n,\alpha ,\left(\beta \right)}^2& :=& \max \left\{{\parallel \sum _{1\le i\le \left[\alpha n\right]}E{h}_{\left(\beta \right)}^2(X_i,y)\parallel }_{\infty },{\parallel \sum _{1\le j\le \left[\alpha n\right]}E{h}_{\left(\beta \right)}^2(x,Y_j)\parallel }_{\infty }\right\}\hfill \\ & \le & 4\alpha n\sum _{l=1}^{\infty }{\lambda }_l{L}_l\left(\beta z_{n,l}\right)\sum _{l=1}^{\infty }{\lambda }_l{\beta }^2{\displaystyle \frac{n{L}_l\left(z_{n,l}\right)}{x_n^2}}.\hfill \end{array}$$

Since $0<\beta \le 1$, then $L_l\left(\beta z_{n,l}\right)/L_l\left(z_{n,l}\right)\le 1$. Hence,

$$\begin{array}{c}\hfill B_{n,\alpha ,\left(\beta \right)}^2\le {\displaystyle \frac{4\alpha {\beta }^2{n}^2{\left({\sum }_{l=1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)\right)}^2}{x_n^2}}.\end{array}$$

By (37) and the Cauchy–Schwarz inequality,

$$\begin{array}{ccc}\hfill C_n^2& :=& \sum _{1\le i,j\le n}E{h}_m^2(X_i,Y_j)\hfill \\ & \le & \sum _{1\le i,j\le n}\sum _{l=m+1}^{\infty }{\lambda }_{l}E{\left({\overline{g}}_l\left(X_i\right)-E{\overline{g}}_l\left(X_i\right)\right)}^2\sum _{l=m+1}^{\infty }{\lambda }_{l}E{\left({\overline{g}}_l\left(Y_j\right)-E{\overline{g}}_l\left(Y_j\right)\right)}^2\hfill \\ & \le & n^2{\left(\sum _{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)\right)}^2.\hfill \end{array}$$

(44)

Similarly,

$$\begin{array}{ccc}\hfill C_{n,\alpha ,\left(\beta \right)}^2& :=& \sum _{1\le i\ne j\le \left[\alpha n\right]}E{h}_{\left(\beta \right)}^2(X_i,Y_j)\hfill \\ & \le & {\alpha }^2{n}^2{\left(\sum _{l=1}^{\infty }{\lambda }_l{L}_l\left(\beta z_{n,l}\right)\right)}^2\hfill \\ & \le & {\alpha }^2{n}^2{\left(\sum _{l=1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)\right)}^2.\hfill \end{array}$$

(45)

Now let

$$\begin{array}{c}\hfill x=C_{2}n{x}_n^2\sum _{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right).\end{array}$$

(46)

By (41) and (46),

$$\begin{array}{c}\hfill {\left({\displaystyle \frac{x}{A_n}}\right)}^{1/2}\ge {\left({\displaystyle \frac{C_{2}n{x}_n^2{\sum }_{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)}{4n{\sum }_{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)/x_n^2}}\right)}^{1/2}={\left(C_2/4\right)}^{1/2}{x}_n^2.\end{array}$$

By (43) and (46),

$$\begin{array}{c}\hfill {\left({\displaystyle \frac{x}{B_n}}\right)}^{2/3}\ge {\left({\displaystyle \frac{C_{2}n{x}_n^2{\sum }_{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)}{2n{\sum }_{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)/x_n}}\right)}^{2/3}={(C_2/2)}^{2/3}{x}_n^2.\end{array}$$

By (44) and (46),

$$\begin{array}{c}\hfill {\displaystyle \frac{x}{C_n}}\ge {\displaystyle \frac{C_{2}n{x}_n^2{\sum }_{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)}{n{\sum }_{l=m+1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)}}=C_2{x}_n^2.\end{array}$$

Then, (39) follows from Lemma 1 for a sufficiently large $C_2$ value. Similarly, let

$$\begin{array}{c}\hfill x_d=dn{x}_n^2\sum _{l=1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right).\end{array}$$

(47)

By (42) and (47),

$$\begin{array}{c}\hfill {\left({\displaystyle \frac{x_d}{A_{n,\left(\beta \right)}}}\right)}^{1/2}\ge {\left({\displaystyle \frac{dn{x}_n^2{\sum }_{l=1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)}{4{\beta }^{2}n{\sum }_{l=1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)/x_n^2}}\right)}^{1/2}={\displaystyle \frac{\sqrt{d}}{2\beta }}{x}_n^2.\end{array}$$

By (44) and (47),

$$\begin{array}{c}\hfill {\left({\displaystyle \frac{x_d}{B_{n,\alpha ,\left(\beta \right)}}}\right)}^{2/3}\ge {\left({\displaystyle \frac{dn{x}_n^2{\sum }_{l=1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)}{2\sqrt{\alpha }\beta n{\sum }_{l=1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)/x_n}}\right)}^{2/3}={\left({\displaystyle \frac{d}{2\sqrt{\alpha }\beta }}\right)}^{2/3}{x}_n^2.\end{array}$$

By (45) and (47),

$$\begin{array}{c}\hfill {\displaystyle \frac{x_d}{C_{n,\alpha ,\left(\beta \right)}}}\ge {\displaystyle \frac{dn{x}_n^2{\sum }_{l=1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)}{\alpha n{\sum }_{l=1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)}}={\displaystyle \frac{d}{\alpha }}{x}_n^2.\end{array}$$

Therefore, (40) follows from Lemma 1 for $\alpha$ and $\beta$ values that are sufficiently small. □

2.5. Estimation of $I_4$

Lemma 2 below follows Corollary 1(b) of Einmahl [21] and Lemma 4.2 of Lin and Liu [22]. However, we add the condition ${\sum }_{i=1}^{k_n}E{\parallel {\xi }_{n,i}\parallel }^2\le b_n^2$, and our result is in a form of an exponential inequality for independent random vectors. We use the same positive constants $c_{17},c_{20}$ and $c_{22}$ (depending only on the vector dimension d) in Einmahl [21].

Lemma 2.

Let ${\xi }_{n,1},\cdots ,{\xi }_{n,k_n}$ be independent random vectors with a mean of zero and values in ${\mathbb{R}}^d$ such that $\parallel {\xi }_{n,i}\parallel \le A_n$ and ${\sum }_{i=1}^{k_n}E{\parallel {\xi }_{n,i}\parallel }^2\le b_n^2$, where $\parallel ·\parallel$ denotes the Euclidean norm. Let $S_n={\sum }_{i=1}^{k_n}{\xi }_{n,i}$. Suppose that

$$\begin{array}{c}\hfill Cov\left(S_n\right)=B_n{I}_d\end{array}$$

(48)

where $B_n>0$, $I_d$ is a $d\times d$ identity matrix, and ${\alpha }_n$ is a positive sequence such that ${\alpha }_n{B}_n^{1/2}\to \infty$ and

$$\begin{array}{c}\hfill {\alpha }_n\sum _{i=1}^{k_n}E\left\{\parallel {\xi }_{n,i}{\parallel }^{3}exp\left({\alpha }_n\parallel {\xi }_{n,i}\parallel \right)\right\}\le B_n.\end{array}$$

(49)

Let

$$\begin{array}{c}\hfill {\beta }_n=B_n^{-3/2}\sum _{i=1}^{k_n}E\left\{\parallel {\xi }_{n,i}{\parallel }^{3}exp\left({\alpha }_n\parallel {\xi }_{n,i}\parallel \right)\right\}.\end{array}$$

(50)

Then, for any $0<\gamma <1$, there exists $n_{\gamma }$ such that for all $n\ge n_{\gamma }$,

$$\begin{array}{ccc}\hfill P\left(\parallel S_n\parallel \ge x\right)& \le & exp\left\{c_{20}{\beta }_n\left(c_{17}^3{\alpha }_n^3{B}_n^{3/2}+1\right)\right\}\hfill \\ & & \times \left\{exp\left(-{\displaystyle \frac{{(1-\gamma )}^6{x}^2}{2B_n}}\right)+exp\left(-{\displaystyle \frac{{\gamma }^3{(1-\gamma )}^3{x}^2}{2c_{22}{B}_n{\beta }_n^2\log (1/{\beta }_n)}}\right)\right\}\hfill \\ & & +2dexp\left(-{\displaystyle \frac{{(1-\gamma )}^2{c}_{17}^2{\alpha }_n^2{B}_n^2}{2(d^2{b}_n^2+d{c}_{17}{A}_n{\alpha }_n{B}_n)}}\right)\hfill \end{array}$$

uniformly for $x\in [e_n{B}_n^{1/2},c_{17}{\alpha }_n{B}_n]$, where ${\left\{e_n\right\}}_{n\ge 1}$ can be any sequence with $e_n\to \infty$ and $e_n\le c_{17}{\alpha }_n{B}_n^{1/2}$.

Proof. 

Let ${\eta }_{n,i}$, $1\le i\le k_n$, be independent $N(0,{\sigma }^{2}Cov\left({\xi }_{n,i}\right))$ random vectors, which are independent of the ${\xi }_{n,i}$s, where

$$\begin{array}{c}\hfill {\sigma }^2=c_{22}{\beta }_n^2\log (1/{\beta }_n).\end{array}$$

(51)

By (49) and (50), we have ${\beta }_n\le {\alpha }_n^{-1}{B}_n^{-1/2}\to 0$ as $n\to \infty$. Hence, $\sigma \to 0$ as $n\to \infty$. Let $p_n\left(y\right)$ be the probability density of $B_n^{-1/2}{\sum }_{i=1}^{k_n}({\xi }_{n,i}+{\eta }_{n,i})$ and ${\varphi }_{(1+{\sigma }^2)I_d}$ be the density of $N(0,(1+{\sigma }^2)I_d)$. By Corollary 1(b) in Einmahl [21] (together with the Remark on page 32), for $\parallel y\parallel \le c_{17}{\alpha }_n{B}_n^{1/2}$,

$$\begin{array}{c}\hfill p_n\left(y\right)={\varphi }_{(1+{\sigma }^2)I_d}\left(y\right)exp\left(T_n\left(y\right)\right)\mathrm{with}|T_n\left(y\right)|\le c_{20}{\beta }_n{(\parallel y\parallel }^3+1).\end{array}$$

(52)

For any $0<\gamma <1$ and $x\in [e_n{B}_n^{1/2},c_{17}{\alpha }_n{B}_n]$,

$$\begin{array}{ccc}\hfill & & P\left(\parallel S_n\parallel \ge x\right)\hfill \\ & \le & P\left(\parallel S_n+\sum _{i=1}^{k_n}{\eta }_{n,i}\parallel \ge (1-\gamma )x\right)+P\left(\parallel \sum _{i=1}^{k_n}{\eta }_{n,i}\parallel \ge \gamma x\right)\hfill \\ & =& P\left((1-\gamma )x\le \parallel S_n+\sum _{i=1}^{k_n}{\eta }_{n,i}\parallel <c_{17}{\alpha }_n{B}_n\right)\hfill \\ & & +P\left(\parallel S_n+\sum _{i=1}^{k_n}{\eta }_{n,i}\parallel \ge c_{17}{\alpha }_n{B}_n\right)+P\left(\parallel \sum _{i=1}^{k_n}{\eta }_{n,i}\parallel \ge \gamma x\right)\hfill \\ & \le & P\left((1-\gamma )x\le \parallel S_n+\sum _{i=1}^{k_n}{\eta }_{n,i}\parallel <c_{17}{\alpha }_n{B}_n\right)\hfill \\ & & +P\left(\parallel S_n\parallel \ge (1-\gamma )c_{17}{\alpha }_n{B}_n\right)\hfill \\ & & +P\left(\parallel \sum _{i=1}^{k_n}{\eta }_{n,i}\parallel \ge \gamma c_{17}{\alpha }_n{B}_n\right)+P\left(\parallel \sum _{i=1}^{k_n}{\eta }_{n,i}\parallel \ge \gamma x\right)\hfill \\ & & \hfill \\ & \le & P\left((1-\gamma )x\le \parallel S_n+\sum _{i=1}^{k_n}{\eta }_{n,i}\parallel <c_{17}{\alpha }_n{B}_n\right)+2P\left(\parallel \sum _{i=1}^{k_n}{\eta }_{n,i}\parallel \ge \gamma x\right)\hfill \\ & & +P\left(\parallel S_n\parallel \ge (1-\gamma )c_{17}{\alpha }_n{B}_n\right)\hfill \\ & :=& J_1+J_2+J_3.\hfill \end{array}$$

(53)

Let N denote a centered normal random vector with covariance matrix $I_d$. Then, by (52),

$$\begin{array}{ccc}\hfill & & J_1={\int }_{(1-\gamma )x/B_n^{1/2}<\parallel y\parallel \le c_{17}{\alpha }_n{B}_n^{1/2}}{\varphi }_{(1+{\sigma }^2)I_d}\left(y\right)exp\left(T_n\left(y\right)\right)dy\hfill \\ & & \le exp\left\{c_{20}{\beta }_n\left(c_{17}^3{\alpha }_n^3{B}_n^{3/2}+1\right)\right\}{\int }_{\parallel y\parallel \ge (1-\gamma )x/B_n^{1/2}}{\varphi }_{(1+{\sigma }^2)I_d}\left(y\right)dy\hfill \\ & & \le exp\left\{c_{20}{\beta }_n\left(c_{17}^3{\alpha }_n^3{B}_n^{3/2}+1\right)\right\}\times \hfill \\ & & \left\{P\left(\parallel N\parallel \ge {(1-\gamma )}^{2}x/B_n^{1/2}\right)+P\left(\sigma \parallel N\parallel \ge \gamma (1-\gamma )x/B_n^{1/2}\right)\right\}.\hfill \end{array}$$

(54)

Observe that ${\parallel N\parallel }^2$ has a ${\chi }_d^2$ distribution. It is well known that for a ${\chi }_d^2$ random variable Y, $P(Y>y)\le {(y{e}^{1-y/d}/d)}^{d/2}$ for $y>d$. Hence,

$$\begin{array}{cc}\hfill & P\left(\parallel N\parallel \ge {(1-\gamma )}^{2}x/B_n^{1/2}\right)\hfill \\ \hfill & \le {\left({\displaystyle \frac{{(1-\gamma )}^4{x}^2}{d{B}_n}}exp\left(1-{\displaystyle \frac{{(1-\gamma )}^4{x}^2}{d{B}_n}}\right)\right)}^{d/2}\hfill \\ \hfill & ={\displaystyle \frac{{(1-\gamma )}^{2d}{x}^d}{d^{d/2}{B}_n^{d/2}}}exp\left(d/2-{\displaystyle \frac{{(1-\gamma )}^4{x}^2}{2B_n}}\right)\hfill \\ \hfill & \le exp\left(-{\displaystyle \frac{{(1-\gamma )}^6{x}^2}{2B_n}}\right)\hfill \end{array}$$

(55)

by $x^2/B_n\to \infty$. Similarly,

$$\begin{array}{ccc}\hfill & & P\left(\sigma \parallel N\parallel \ge \gamma (1-\gamma )x/B_n^{1/2}\right)\le exp\left(-{\displaystyle \frac{{\gamma }^3{(1-\gamma )}^3{x}^2}{2B_n{\sigma }^2}}\right)\hfill \\ & =& exp\left(-{\displaystyle \frac{{\gamma }^3{(1-\gamma )}^3{x}^2}{2c_{22}{B}_n{\beta }_n^2\log (1/{\beta }_n)}}\right)\hfill \end{array}$$

(56)

by (51). Then, by (54)–(56), we have

$$\begin{array}{ccc}\hfill J_1& \le & exp\left\{c_{20}{\beta }_n\left(c_{17}^3{\alpha }_n^3{B}_n^{3/2}+1\right)\right\}\hfill \\ & & \times \left\{exp\left(-{\displaystyle \frac{{(1-\gamma )}^6{x}^2}{2B_n}}\right)+exp\left(-{\displaystyle \frac{{\gamma }^3{(1-\gamma )}^3{x}^2}{2c_{22}{B}_n{\beta }_n^2\log (1/{\beta }_n)}}\right)\right\}.\hfill \end{array}$$

(57)

Since the distribution of ${\eta }_{n,i}$ is $N(0,{\sigma }^{2}Cov\left({\xi }_{n,i}\right))$, the distribution of ${\sum }_{i=1}^{k_n}{\eta }_{n,i}$ is $N(0,{\sigma }^2{\sum }_{i=1}^{k_n}Cov\left({\xi }_{n,i}\right))$. Since the ${\xi }_{n,i}$s are independent, ${\sum }_{i=1}^{k_n}Cov\left({\xi }_{n,i}\right)=Cov\left({\sum }_{i=1}^{k_n}{\xi }_{n,i}\right)=B_n{I}_d$ by (48). Hence, the distribution of ${\sum }_{i=1}^{k_n}{\eta }_{n,i}$ is $N(0,{\sigma }^2{B}_n{I}_d)$. Then, similar to (56),

$$\begin{array}{ccc}\hfill J_2& =& 2P\left(\parallel \sum _{i=1}^{k_n}{\eta }_{n,i}\parallel \ge \gamma x\right)=2P\left(\sigma \parallel N\parallel \ge \gamma x/B_n^{1/2}\right)\hfill \\ & \le & 2exp\left(-{\displaystyle \frac{{\gamma }^3{x}^2}{2B_n{c}_{22}{\beta }_n^2\log (1/{\beta }_n)}}\right).\hfill \end{array}$$

(58)

By (57) and (58), we have

$$\begin{array}{ccc}\hfill & & J_1+J_2\le exp\left\{c_{20}{\beta }_n\left(c_{17}^3{\alpha }_n^3{B}_n^{3/2}+1\right)\right\}\hfill \\ & & \times \left\{exp\left(-{\displaystyle \frac{{(1-\gamma )}^6{x}^2}{2B_n}}\right)+3exp\left(-{\displaystyle \frac{{\gamma }^3{(1-\gamma )}^3{x}^2}{2c_{22}{B}_n{\beta }_n^2\log (1/{\beta }_n)}}\right)\right\}.\hfill \end{array}$$

(59)

Next, we estimate $J_3$. For each $1\le i\le n$, let ${\xi }_{n,i}={({\xi }_{n,i}^{\left(1\right)},\cdots ,{\xi }_{n,i}^{\left(d\right)})}^T$, where ${\mathit{a}}^T$ denote the transpose of a vector $\mathit{a}$. Then

$$\begin{array}{c}\hfill ∥S_n∥=\parallel \sum _{i=1}^{k_n}{\xi }_{n,i}\parallel ={\left(\sum _{l=1}^d{\left(\sum _{i=1}^{k_n}{\xi }_{n,i}^{\left(l\right)}\right)}^2\right)}^{1/2}\le \sum _{l=1}^d\left|\sum _{i=1}^{k_n}{\xi }_{n,i}^{\left(l\right)}\right|.\end{array}$$

Hence,

$$\begin{array}{ccc}\hfill J_3& =& P\left(\parallel S_n\parallel \ge (1-\gamma )c_{17}{\alpha }_n{B}_n\right)\hfill \\ & \le & P\left(\sum _{l=1}^d\left|\sum _{i=1}^{k_n}{\xi }_{n,i}^{\left(l\right)}\right|\ge (1-\gamma )c_{17}{\alpha }_n{B}_n\right)\hfill \\ & \le & \sum _{l=1}^{d}P\left(\left|\sum _{i=1}^{k_n}{\xi }_{n,i}^{\left(l\right)}\right|\ge {\displaystyle \frac{(1-\gamma )c_{17}{\alpha }_n{B}_n}{d}}\right).\hfill \end{array}$$

Since $\parallel {\xi }_{n,i}\parallel \le A_n$ and ${\sum }_{i=1}^{k_n}E{\parallel {\xi }_{n,i}\parallel }^2\le b_n^2$, then $|{\xi }_{n,i}^{\left(l\right)}|\le \parallel {\xi }_{n,i}\parallel \le A_n$ and ${\sum }_{i=1}^{k_n}E{\left({\xi }_{n,i}^{\left(l\right)}\right)}^2\le {\sum }_{i=1}^{k_n}E{\parallel {\xi }_{n,i}\parallel }^2\le b_n^2$ for each $1\le l\le d$. By Bernstein’s inequality (e.g., (2.17) of de la Peña, Lai and Shao [18]),

$$\begin{array}{ccc}\hfill J_3& \le & 2dexp\left(-{\displaystyle \frac{{(1-\gamma )}^2{c}_{17}^2{\alpha }_n^2{B}_n^2}{2d^2(b_n^2+c_{17}{A}_n{\alpha }_n{B}_n/d)}}\right).\hfill \end{array}$$

(60)

Then, the lemma follows by applying (59) and (60) to (53). □

Now, we estimate $I_4$ in the following proposition, which uses some ideas in Liu and Shao [23]:

Proposition 4.

$$\begin{array}{ccc}\hfill I_4& \le & P\left(\sum _{l=1}^m{\lambda }_l{\left({\overline{Y}}_{n,l}-E{\overline{Y}}_{n,l}\right)}^2\ge (1-C_1\epsilon ){(1-2\epsilon )}^{3}n{x}_n^2\underset{1\le l\le m}{\max }{\lambda }_l{L}_l\left(z_{n,l}\right)\right)\hfill \\ & \le & exp\left(-{\displaystyle \frac{(1-C_1\epsilon ){(1-2\epsilon )}^4{x}_n^2}{2(1+\epsilon )}}\right).\hfill \end{array}$$

Proof. 

For each $1\le i\le n$, let

$$\begin{array}{c}\hfill {\mathit{G}}_{n,i}={\left(\sqrt{{\lambda }_1}({\overline{g}}_1\left(X_i\right)-E{\overline{g}}_1\left(X_i\right)),\cdots ,\sqrt{{\lambda }_m}({\overline{g}}_m\left(X_i\right)-E{\overline{g}}_m\left(X_i\right))\right)}^T.\end{array}$$

Let $B_n=n$ and $\Sigma$ be the covariance matrix of ${\mathit{G}}_{n,1}$. For $1\le i\le n$, let

$$\begin{array}{c}\hfill {\xi }_{n,i}={\Sigma }^{-1/2}{\mathit{G}}_{n,i}.\end{array}$$

(61)

Then,

$$\begin{array}{ccc}& & Cov\left({\xi }_{n,1}+\cdots +{\xi }_{n,n}\right)\hfill \\ & =& E\left\{\left({\Sigma }^{-1/2}({\mathit{G}}_{n,1}+\cdots +{\mathit{G}}_{n,n})\right){\left({\Sigma }^{-1/2}({\mathit{G}}_{n,1}+\cdots +{\mathit{G}}_{n,n})\right)}^T\right\}\hfill \\ & =& {\Sigma }^{-1/2}E\left\{{\left({\mathit{G}}_{n,1}+\cdots +{\mathit{G}}_{n,n}\left)\right({\mathit{G}}_{n,1}+\cdots +{\mathit{G}}_{n,n}\right)}^T\right\}{\Sigma }^{-1/2}\hfill \\ & =& {\Sigma }^{-1/2}\sum _{1\le i,j\le n}E\left\{{\mathit{G}}_{n,i}{\mathit{G}}_{n,j}^T\right\}{\Sigma }^{-1/2}.\hfill \end{array}$$

Since the $X_i$s are independent, then

$$\begin{array}{ccc}\hfill Cov\left({\xi }_{n,1}+\cdots +{\xi }_{n,n}\right)& =& {\Sigma }^{-1/2}\sum _{i=1}^{n}E\left\{{\mathit{G}}_{n,i}{\mathit{G}}_{n,i}^T\right\}{\Sigma }^{-1/2}\hfill \\ & =& n{I}_m=B_n{I}_m.\hfill \end{array}$$

Hence, Condition (48) in Lemma 2 is satisfied. Let

$$\begin{array}{c}\hfill {\alpha }_n={\displaystyle \frac{C_m{x}_n}{n^{1/2}}}\end{array}$$

where $C_m>0$ is a finite constant depending only on m. We shall verify Condition (49). By (61),

$$\begin{array}{c}\hfill \parallel {\xi }_{n,i}{\parallel }^2={\left({\Sigma }^{-1/2}{\mathit{G}}_{n,i}\right)}^T\left({\Sigma }^{-1/2}{\mathit{G}}_{n,i}\right)={\mathit{G}}_{n,i}^T{\Sigma }^{-1}{\mathit{G}}_{n,i}.\end{array}$$

(62)

Observe that $\Sigma$ is positive definite by Assumption (9). Then, by the identity

$$\begin{array}{c}\hfill {\mathit{x}}^{\mathit{T}}{\mathit{A}}^{-\mathbf{1}}\mathit{x}=\underset{\parallel \vartheta \parallel =1}{\max }{\displaystyle \frac{{\left(x^T\vartheta \right)}^2}{{\vartheta }^T\mathit{A}\vartheta }}\end{array}$$

(63)

for any $m\times m$ positive definite matrix $\mathit{A}$, we have

$$\begin{array}{c}\hfill \parallel {\xi }_{n,i}{\parallel }^2={\mathit{G}}_{n,i}^T{\Sigma }^{-1}{\mathit{G}}_{n,i}=\underset{\parallel \vartheta \parallel =1}{\max }{\displaystyle \frac{{\left({\mathit{G}}_{n,i}^T\vartheta \right)}^2}{{\vartheta }^T\Sigma \vartheta }}.\end{array}$$

(64)

Let ${\vartheta }^{*}=({\vartheta }_1^{*},\cdots ,{\vartheta }_m^{*})$ such that $\parallel {\vartheta }^{*}\parallel =1$ and ${\left({\mathit{G}}_{n,i}^T{\vartheta }^{*}\right)}^2={\max }_{\parallel \vartheta \parallel =1}{\left({\mathit{G}}_{n,i}^T\vartheta \right)}^2$. Then, for any $\vartheta =({\vartheta }_1,\cdots ,{\vartheta }_m)\in l^2$, by the Cauchy–Schwarz inequality,

$$\begin{array}{ccc}\hfill {\left({\mathit{G}}_{n,i}^T\vartheta \right)}^2& =& {\left(\sum _{l=1}^m\sqrt{{\lambda }_l}({\overline{g}}_l\left(X_i\right)-E{\overline{g}}_l\left(X_i\right)){\vartheta }_l\right)}^2\hfill \\ & =& {\left(\sum _{l=1}^m{\displaystyle \frac{{\overline{g}}_l\left(X_i\right)-E{\overline{g}}_l\left(X_i\right)}{\sqrt{L_l\left(z_{n,l}\right)}}}{\vartheta }_l\sqrt{{\lambda }_l{L}_l\left(z_{n,l}\right)}\right)}^2\hfill \\ & \le & \sum _{l=1}^m{\displaystyle \frac{{({\overline{g}}_l\left(X_i\right)-E{\overline{g}}_l\left(X_i\right))}^2}{L_l\left(z_{n,l}\right)}}\sum _{l=1}^m{\vartheta }_l^2{\lambda }_l{L}_l\left(z_{n,l}\right).\hfill \end{array}$$

(65)

Since $E{g}_l\left(X_1\right)=0$ for all $l\ge 1$, then $E{\overline{g}}_l\left(X_1\right)=o(x_n\sqrt{L_l\left(z_{n,l}\right)}/\sqrt{n})$ by (13) and (15). By Assumption (9),

$$\begin{array}{ccc}\hfill {\vartheta }^T\Sigma \vartheta & =& \sum _{1\le l,l^{\prime }\le m}{\vartheta }_l{\vartheta }_{l^{\prime }}\sqrt{{\lambda }_l{\lambda }_{l^{\prime }}}E\left({\overline{g}}_l\left(X_1\right)-E{\overline{g}}_l\left(X_1\right)\right)\left({\overline{g}}_{l^{\prime }}\left(X_1\right)-E{\overline{g}}_{l^{\prime }}\left(X_1\right)\right)\hfill \\ & =& \sum _{l=1}^m{\vartheta }_l^2{\lambda }_{l}E{\overline{g}}_l^2\left(X_1\right)-\sum _{l=1}^m{\vartheta }_l^2{\lambda }_l{\left(E{\overline{g}}_l\left(X_1\right)\right)}^2\hfill \\ & & +\sum _{1\le l\ne l^{\prime }\le m}{\vartheta }_l{\vartheta }_{l^{\prime }}\sqrt{{\lambda }_l{\lambda }_{l^{\prime }}}E{\overline{g}}_l\left(X_1\right){\overline{g}}_{l^{\prime }}\left(X_1\right)\hfill \\ & & -\sum _{1\le l\ne l^{\prime }\le m}{\vartheta }_l{\vartheta }_{l^{\prime }}\sqrt{{\lambda }_l{\lambda }_{l^{\prime }}}E{\overline{g}}_l\left(X_1\right)E{\overline{g}}_{l^{\prime }}\left(X_1\right)\hfill \\ & =& \sum _{l=1}^m{\vartheta }_l^2{\lambda }_l{L}_l\left(z_{n,l}\right)-\sum _{l=1}^m{\vartheta }_l^2{\lambda }_l\times o\left({\displaystyle \frac{x_n^2{L}_l\left(z_{n,l}\right)}{n}}\right)\hfill \\ & & +o\left(1\right)\sum _{1\le l\ne l^{\prime }\le m}{\vartheta }_l{\vartheta }_{l^{\prime }}\sqrt{{\lambda }_l{\lambda }_{l^{\prime }}{L}_l\left(z_{n,l}\right)L_{l^{\prime }}\left(z_{n,l^{\prime }}\right)}\hfill \\ & & -\sum _{1\le l\ne l^{\prime }\le m}{\vartheta }_l{\vartheta }_{l^{\prime }}\sqrt{{\lambda }_l{\lambda }_{l^{\prime }}}\times o\left({\displaystyle \frac{x_n\sqrt{L_l\left(z_{n,l}\right)}}{\sqrt{n}}}\right)o\left({\displaystyle \frac{x_n\sqrt{L_{l^{\prime }}\left(z_{n,l^{\prime }}\right)}}{\sqrt{n}}}\right).\hfill \end{array}$$

By the Cauchy–Schwarz inequality,

$$\begin{array}{c}\hfill \sum _{1\le l\ne l^{\prime }\le m}{\vartheta }_l{\vartheta }_{l^{\prime }}\sqrt{{\lambda }_l{\lambda }_{l^{\prime }}{L}_l\left(z_{n,l}\right)L_{l^{\prime }}\left(z_{n,l^{\prime }}\right)}\le m\sum _{l=1}^m{\vartheta }_l^2{\lambda }_l{L}_l\left(z_{n,l}\right).\end{array}$$

Hence,

$$\begin{array}{c}\hfill {\vartheta }^T\Sigma \vartheta =(1+o\left(1\right))\sum _{l=1}^m{\vartheta }_l^2{\lambda }_l{L}_l\left(z_{n,l}\right).\end{array}$$

(66)

Applying (65) and (66) to (64), we have

$$\begin{array}{c}\hfill \parallel {\xi }_{n,i}{\parallel }^2\le 2\sum _{l=1}^m{\displaystyle \frac{{({\overline{g}}_l\left(X_i\right)-E{\overline{g}}_l\left(X_i\right))}^2}{L_l\left(z_{n,l}\right)}}.\end{array}$$

(67)

Since $|{\overline{g}}_l\left(X_i\right)|\le z_{n,l}=\sqrt{n{L}_l\left(z_{n,l}\right)}/x_n$ by (15), and (16), we have

$$\begin{array}{c}\hfill \parallel {\xi }_{n,i}{\parallel }^2\le {\displaystyle \frac{4mn}{x_n^2}}.\end{array}$$

(68)

By (67),

$$\begin{array}{c}\hfill E\parallel {\xi }_{n,i}{\parallel }^2\le 2\sum _{l=1}^m{\displaystyle \frac{E{({\overline{g}}_l\left(X_i\right)-E{\overline{g}}_l\left(X_i\right))}^2}{L_l\left(z_{n,l}\right)}}\le 2m\end{array}$$

(69)

and

$$\begin{array}{ccc}\hfill E\parallel {\xi }_{n,i}{\parallel }^3& \le & 2^{3/2}E{\left(\sum _{l=1}^m{\displaystyle \frac{{({\overline{g}}_l\left(X_i\right)-E{\overline{g}}_l\left(X_i\right))}^2}{L_l\left(z_{n,l}\right)}}\right)}^{3/2}.\hfill \end{array}$$

(70)

By Hölder’s inequality,

$$\begin{array}{c}\hfill {\left(\sum _{l=1}^m{\displaystyle \frac{{({\overline{g}}_l\left(X_i\right)-E{\overline{g}}_l\left(X_i\right))}^2}{L_l\left(z_{n,l}\right)}}\right)}^{3/2}\le m^{1/2}\sum _{l=1}^m{\displaystyle \frac{|{\overline{g}}_l\left(X_i\right)-E{\overline{g}}_l\left(X_i\right){|}^3}{L_l^{3/2}\left(z_{n,l}\right)}}.\end{array}$$

(71)

Combining (70) and (71), we have

$$\begin{array}{ccc}\hfill E\parallel {\xi }_{n,i}{\parallel }^3& \le & 2^{3/2}{m}^{1/2}\sum _{l=1}^m{\displaystyle \frac{8E|{\overline{g}}_l\left(X_i\right){|}^3}{L_l^{3/2}\left(z_{n,l}\right)}}\hfill \\ & =& \sum _{l=1}^m{\displaystyle \frac{o\left(z_{n,l}{L}_l\left(z_{n,l}\right)\right)}{L_l^{3/2}\left(z_{n,l}\right)}}=o\left({\displaystyle \frac{n^{1/2}}{x_n}}\right)\hfill \end{array}$$

(72)

by (14) and (15). Since $B_n=n$ and ${\alpha }_n=C_m{x}_n/n^{1/2}$, then by (68) and (72), we have

$$\begin{array}{ccc}\hfill & & {\alpha }_n\sum _{i=1}^{n}E\left\{\parallel {\xi }_{n,i}{\parallel }^{3}exp\left({\alpha }_n\parallel {\xi }_{n,i}\parallel \right)\right\}\hfill \\ & \le & {\displaystyle \frac{C_m{x}_n}{n^{1/2}}}n\times o\left({\displaystyle \frac{n^{1/2}}{x_n}}\right)exp\left({\displaystyle \frac{C_m{x}_n}{n^{1/2}}}{\left({\displaystyle \frac{4mn}{x_n^2}}\right)}^{1/2}\right)=o\left(n\right)=o\left(B_n\right).\hfill \end{array}$$

Hence, Condition (49) in Lemma 2 is satisfied. Similarly,

$$\begin{array}{ccc}\hfill {\beta }_n& :=& B_n^{-3/2}\sum _{i=1}^{n}E\left\{\parallel {\xi }_{n,i}{\parallel }^{3}exp\left({\alpha }_n\parallel {\xi }_{n,i}\parallel \right)\right\}\hfill \\ & =& n^{-3/2}n\times o\left({\displaystyle \frac{n^{1/2}}{x_n}}\right)exp{\left({\displaystyle \frac{C_m{x}_n}{n^{1/2}}}({\displaystyle \frac{4mn}{x_n^2}}\right)}^{1/2})\hfill \\ & =& o(1/x_n).\hfill \end{array}$$

(73)

Then, ${\beta }_n^2\log (1/{\beta }_n)=o(1/x_n)$. By (68), we have $\parallel {\xi }_{n,i}\parallel \le {(4mn/x_n^2)}^{1/2}:=A_n$. By (69), we have ${\sum }_{i=1}^{n}E{\parallel {\xi }_{n,i}\parallel }^2\le 2mn:=b_n^2$. Then, by Lemma 2 and (73) with $B_n=n$ and ${\alpha }_n=C_m{x}_n/n^{1/2}$ for a sufficiently large $C_m$ value, we have

$$\begin{array}{ccc}\hfill & & P\left(\parallel S_n\parallel \ge n^{1/2}{x}_n\right)\hfill \\ & \le & exp\left\{o\left(x_n^2\right)\right\}\left\{exp\left(-{\displaystyle \frac{{(1-\gamma )}^{6}n{x}_n^2}{2n}}\right)+exp\left(-{\displaystyle \frac{{\gamma }^3{(1-\gamma )}^{3}n{x}_n^2}{n\times o(1/x_n)}}\right)\right\}\hfill \\ & & +2mexp\left(-{\displaystyle \frac{{(1-\gamma )}^2{c}_{17}^2{C}_m^2{x}_n^{2}n}{2(2m^{3}n+m{(4mn/x_n^2)}^{1/2}{c}_{17}{C}_m{x}_n{n}^{1/2})}}\right)\hfill \\ & \le & exp\left\{o\left(x_n^2\right)\right\}\left\{exp\left(-{\displaystyle \frac{{(1-\gamma )}^6{x}_n^2}{2}}\right)+exp\left(-4x_n^2\right)\right\}+exp\left(-4x_n^2\right)\hfill \\ & \le & exp\left(-{\displaystyle \frac{{(1-\gamma )}^7{x}_n^2}{2}}\right).\hfill \end{array}$$

Letting $\gamma =1-{(1-\epsilon )}^{1/7}$, we have

$$\begin{array}{c}\hfill P\left(\parallel S_n\parallel \ge n^{1/2}{x}_n\right)\le exp\left(-{\displaystyle \frac{(1-\epsilon )x_n^2}{2}}\right).\end{array}$$

(74)

Similar to (62),

$$\begin{array}{ccc}\hfill \parallel S_n{\parallel }^2=\parallel \sum _{i=1}^n{\xi }_{n,i}{\parallel }^2& =& {\left({\Sigma }^{-1/2}\sum _{i=1}^n{\mathit{G}}_{n,i}\right)}^T\left({\Sigma }^{-1/2}\sum _{i=1}^n{\mathit{G}}_{n,i}\right)\hfill \\ & =& {\left(\sum _{i=1}^n{\mathit{G}}_{n,i}\right)}^T{\Sigma }^{-1}\left(\sum _{i=1}^n{\mathit{G}}_{n,i}\right).\hfill \end{array}$$

(75)

We will use Identity (63) to estimate (75). Let ${\vartheta }^{*}=({\vartheta }_1^{*},\cdots ,{\vartheta }_m^{*})$ be such that $\parallel {\vartheta }^{*}\parallel =1$ and

$$\begin{array}{c}\hfill {\left(\sum _{i=1}^n{\mathit{G}}_{n,i}\right)}^T{\vartheta }^{*}=\underset{\parallel \vartheta \parallel =1}{\max }{\left(\sum _{i=1}^n{\mathit{G}}_{n,i}\right)}^T\vartheta .\end{array}$$

(76)

Observe that

$$\begin{array}{c}\hfill \underset{\parallel \vartheta \parallel =1}{\max }{\left(\sum _{i=1}^n{\mathit{G}}_{n,i}\right)}^T\vartheta ={\left(\sum _{l=1}^m{\left(\sum _{i=1}^n\sqrt{{\lambda }_l}({\overline{g}}_l\left(X_i\right)-E{\overline{g}}_l\left(X_i\right))\right)}^2\right)}^{1/2}.\end{array}$$

(77)

By (66),

$$\begin{array}{ccc}\hfill {\left({\vartheta }^{*}\right)}^T\Sigma {\vartheta }^{*}& =& (1+o\left(1\right))\sum _{l=1}^m{\left({\vartheta }_l^{*}\right)}^2{\lambda }_l{L}_l\left(z_{n,l}\right)\hfill \\ & \le & (1+o\left(1\right))\underset{1\le l\le m}{\max }{\lambda }_l{L}_l\left(z_{n,l}\right)\hfill \end{array}$$

(78)

because $\parallel {\vartheta }^{*}\parallel =1$. By Identity (63) and by (76)–(78),

$$\begin{array}{c}\hfill {\left(\sum _{i=1}^n{\mathit{G}}_{n,i}\right)}^T{\Sigma }^{-1}\left(\sum _{i=1}^n{\mathit{G}}_{n,i}\right)\ge {\displaystyle \frac{{\sum }_{l=1}^m{\left({\sum }_{i=1}^n\sqrt{{\lambda }_l}({\overline{g}}_l\left(X_i\right)-E{\overline{g}}_l\left(X_i\right))\right)}^2}{(1+\epsilon ){\max }_{1\le l\le m}{\lambda }_l{L}_l\left(z_{n,l}\right)}}.\end{array}$$

(79)

By (74), (75) and (79), with the application of (16) and (19),

$$\begin{array}{ccc}\hfill I_4& \le & P\left({\displaystyle \frac{{\sum }_{l=1}^m{\lambda }_l{\left({\sum }_{i=1}^n({\overline{g}}_l\left(X_i\right)-E{\overline{g}}_l\left(X_i\right))\right)}^2}{{\max }_{1\le l\le m}{\lambda }_l{L}_l\left(z_{n,l}\right)}}\ge (1-C_1\epsilon ){(1-2\epsilon )}^{3}n{x}_n^2\right)\hfill \\ & \le & P\left({\left(\sum _{i=1}^n{\mathit{G}}_{n,i}\right)}^T{\Sigma }^{-1}\left(\sum _{i=1}^n{\mathit{G}}_{n,i}\right)\ge {\displaystyle \frac{(1-C_1\epsilon ){(1-2\epsilon )}^{3}n{x}_n^2}{1+\epsilon }}\right)\hfill \\ & =& P\left(\parallel S_n{\parallel }^2\ge {\displaystyle \frac{(1-C_1\epsilon ){(1-2\epsilon )}^{3}n{x}_n^2}{1+\epsilon }}\right)\hfill \\ & \le & exp\left(-{\displaystyle \frac{(1-C_1\epsilon ){(1-2\epsilon )}^4{x}_n^2}{2(1+\epsilon )}}\right).\hfill \end{array}$$

□

Since $\epsilon$ is arbitrary, then the upper bound of Theorem 1 follows from (22) and the estimates of $I_1$, $I_2$, $I_3$ and $I_4$.

3. The Lower Bound of Theorem 1

Let $0<\epsilon <1$ be sufficiently small. For $1\le m<\infty$ that is sufficiently large, by (26), ${\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2={\max }_{1\le l\le m}{\lambda }_l{V}_{n,l}^2$. Together with (17), we have

$$\begin{array}{ccc}\hfill & & P(W_n\ge (1-\epsilon )x_n^2)\hfill \\ & =& P\left({\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l\left({\left\{{\sum }_{i=1}^n{g}_l\left(X_i\right)\right\}}^2-{\sum }_{i=1}^n{g}_l^2\left(X_i\right)\right)}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2}}\ge (1-\epsilon )x_n^2\right)\hfill \\ & \ge & P\left({\displaystyle \frac{{\sum }_{l=1}^m{\lambda }_l{\left\{{\sum }_{i=1}^n{g}_l\left(X_i\right)\right\}}^2}{{\max }_{1\le l\le m}{\lambda }_l{V}_{n,l}^2}}\ge x_n^2\right).\hfill \end{array}$$

(80)

Let ${\tilde{g}}_l\left(X_i\right)$ be the random variable with distribution which is of the distribution of $g_l\left(X_i\right)$ conditioned on $|g_l\left(X_i\right)|\le z_{n,l}$. Define ${\tilde{Y}}_{n,l}={\sum }_{i=1}^n{\tilde{g}}_l\left(X_i\right)$ and ${\tilde{V}}_{n,l}^2={\sum }_{i=1}^n{\tilde{g}}_l^2\left(X_i\right)$. By the definition of $L_l\left(x\right)$ and (12),

$$\begin{array}{cc}\hfill & E{\tilde{g}}_l^2\left(X_i\right)=E{\overline{g}}_l^2\left(X_i\right)/P\left(\right|g_l\left(X_i\right)|\le z_{n,l})\hfill \\ \hfill & =L_l\left(z_{n,l}\right)/P\left(\right|g_l\left(X_i\right)|\le z_{n,l})=L_l\left(z_{n,l}\right)(1+o\left(1\right)).\hfill \end{array}$$

Notice that (13) implies $E{\tilde{g}}_l\left(X_1\right)=o(L_l\left(z_{n,l}\right)/z_{n,l})$. Then, we have

$$\begin{array}{cc}\hfill & {\sigma }_l^2:=E{({\tilde{Y}}_{n,l}-E{\tilde{Y}}_{n,l})}^2=nE{({\tilde{g}}_l\left(X_1\right)-E{\tilde{g}}_l\left(X_1\right))}^2\hfill \\ \hfill & =nE{\tilde{g}}_l^2\left(X_1\right)(1+o\left(1\right))=n{L}_l\left(z_{n,l}\right)(1+o\left(1\right))\hfill \end{array}$$

(81)

and

$$\begin{array}{c}\hfill E{\tilde{V}}_{n,l}^2=n{L}_l\left(z_{n,l}\right)(1+o\left(1\right)).\end{array}$$

Then, for $0<\delta <1$,

$$\begin{array}{ccc}& & P\left({\displaystyle \frac{{\sum }_{l=1}^m{\lambda }_l{\left\{{\sum }_{i=1}^n{g}_l\left(X_i\right)\right\}}^2}{{\max }_{1\le l\le m}{\lambda }_l{V}_{n,l}^2}}\ge x_n^2\right)\hfill \\ & \ge & P\left({\displaystyle \frac{{\sum }_{l=1}^m{\lambda }_l{\left\{{\sum }_{i=1}^n{g}_l\left(X_i\right)\right\}}^2}{{\max }_{1\le l\le m}{\lambda }_l{V}_{n,l}^2}}\ge x_n^2,\underset{1\le i\le n}{\max }\left|g_l\left(X_i\right)\right|\le z_{n,l},,1\le l\le m\right)\hfill \\ & =& P\left({\displaystyle \frac{{\sum }_{l=1}^m{\lambda }_l{\left\{{\sum }_{i=1}^n{\tilde{g}}_l\left(X_i\right)\right\}}^2}{{\max }_{1\le l\le m}{\lambda }_l{\tilde{V}}_{n,l}^2}}\ge x_n^2\right)P\left(\underset{1\le i\le n}{\max }\left|g_l\left(X_i\right)\right|\le z_{n,l},,1\le l\le m\right)\hfill \\ & \ge & P\left({\displaystyle \frac{{\sum }_{l=1}^m{\lambda }_l{\left\{{\sum }_{i=1}^n{\tilde{g}}_l\left(X_i\right)\right\}}^2}{{\max }_{1\le l\le m}{\lambda }_l{\tilde{V}}_{n,l}^2}}\ge x_n^2,{\tilde{V}}_{n,l}^2\le (1+2\delta ){\sigma }_l^2,1\le l\le m\right)\hfill \\ & & \times P\left(\underset{1\le i\le n}{\max }\left|g_l\left(X_i\right)\right|\le z_{n,l},,1\le l\le m\right)\hfill \\ & \ge & P\left({\displaystyle \frac{{\sum }_{l=1}^m{\lambda }_l{\left\{{\sum }_{i=1}^n{\tilde{g}}_l\left(X_i\right)\right\}}^2}{{\max }_{1\le l\le m}{\lambda }_l{\sigma }_l^2}}\ge (1+2\delta )x_n^2\right)\hfill \\ & & \times P\left(\underset{1\le i\le n}{\max }\left|g_l\left(X_i\right)\right|\le z_{n,l},,1\le l\le m\right)-\sum _{l=1}^{m}P({\tilde{V}}_{n,l}^2\ge (1+2\delta ){\sigma }_l^2).\hfill \end{array}$$

(82)

Without the loss of generality, assume that ${\max }_{1\le l\le m}{\lambda }_l{\sigma }_l^2={\lambda }_1{\sigma }_1^2$. Then,

$$\begin{array}{cc}\hfill & P\left({\displaystyle \frac{{\sum }_{l=1}^m{\lambda }_l{\left\{{\sum }_{i=1}^n{\tilde{g}}_l\left(X_i\right)\right\}}^2}{{\max }_{1\le l\le m}{\lambda }_l{\sigma }_l^2}}\ge (1+2\delta )x_n^2\right)\hfill \\ \hfill & \ge P\left({\displaystyle \frac{{\lambda }_1{\left\{{\sum }_{i=1}^n{\tilde{g}}_1\left(X_i\right)\right\}}^2}{{\lambda }_1{\sigma }_1^2}}\ge (1+2\delta )x_n^2\right)\hfill \\ \hfill & \ge P\left(\sum _{i=1}^n{\tilde{g}}_1\left(X_i\right)\ge {(1+2\delta )}^{1/2}{\sigma }_1{x}_n\right).\hfill \end{array}$$

Recall (15) and (81). Take $c=1/x_n$; thus, we have $|{\tilde{g}}_1\left(X_1\right)|\le z_{n,l}=c{\sigma }_l$. Therefore, by Theorem 5.2.2 in Stout [24], for any $\gamma >0$, we have

$$\begin{array}{c}\hfill P\left(\sum _{i=1}^n{\tilde{g}}_1\left(X_i\right)\ge {(1+2\delta )}^{1/2}{\sigma }_1{x}_n\right)\ge exp(-(x_n^2/2)(1+2\delta )(1+\gamma )).\end{array}$$

(83)

On the other hand,

$$\begin{array}{cc}\hfill & P\left(\underset{1\le i\le n}{\max }\left|g_l\left(X_i\right)\right|\le z_{n,l},1\le l\le m\right)={\left[P\left(|g_l\left(X_1\right)|\le z_{n,l},1\le l\le m\right)\right]}^n\hfill \\ \hfill & ={[1-P\left(|g_l\left(X_1\right)|\ge z_{n,l},\exists 1\le l\le m\right)]}^n\ge {[1-\sum _{l=1}^{m}P\left(|g_l\left(X_1\right)|\ge z_{n,l}\right)]}^n\hfill \\ \hfill & \ge exp(-2n\sum _{l=1}^{m}P\left(|g_l\left(X_1\right)|\ge z_{n,l}\right))=exp(-o\left(x_n^2\right)).\hfill \end{array}$$

(84)

We apply the following exponential inequality (see Lemma 2.1, Csörgő, Lin and Shao [25]; see also Pruitt [26] and Griffin and Kuelbs [1]) for the rest of the proof.

Lemma 3.

Let $\xi ,{\xi }_1,\cdots ,{\xi }_n$ be i.i.d. random variables. Then, for any $b,v,s>0$,

$$\begin{array}{cc}\hfill & P\left(\left|\sum _{i=1}^n({\xi }_{i}I\left(\right|{\xi }_i|\le b)-E{\xi }_{i}I\left(\right|{\xi }_i|\le b))\right|\ge {\displaystyle \frac{v{e}^{v}nE{\xi }_i^{2}I\left(\right|{\xi }_i|\le b)}{2b}}+{\displaystyle \frac{sb}{v}}\right)\le 2e^{-s}.\hfill \end{array}$$

By (14),

$$\begin{array}{c}\hfill E{\tilde{g}}_1^4\left(X_i\right)=o\left(z_{n,1}^2{L}_1\left(z_{n,1}\right)\right).\end{array}$$

(85)

In Lemma 3, we take ${\xi }_i={\tilde{g}}_1^2\left(X_i\right),s=x_n^2,b=z_{n,1}^2$ and $v=1/\delta$. Notice that $sb/v=\delta {\sigma }_1^2$ and ${\displaystyle \frac{v{e}^{v}nE{\xi }_i^{2}I\left(\right|{\xi }_i|\le b)}{2b}}=o\left({\sigma }_1^2\right)$ by (81) and (85). Then,

$$\begin{array}{cc}\hfill & P({\tilde{V}}_{n,1}^2\ge (1+2\delta ){\sigma }_1^2)=P({\tilde{V}}_{n,1}^2-E{\tilde{V}}_{n,1}^2\ge (1+2\delta ){\sigma }_1^2-E{\tilde{V}}_{n,1}^2)\hfill \\ \hfill & \le P\left(\sum _{i=1}^n({\tilde{g}}_1^2\left(X_i\right)-E{\tilde{g}}_1^2\left(X_i\right))\ge \delta (1+\delta ){\sigma }_1^2\right)\hfill \\ \hfill & \le 2exp(-x_n^2).\hfill \end{array}$$

(86)

Combining (80), (82), (83), (84) and (86) and letting $\lambda ,\delta \to 0$, we have

$$\begin{array}{c}\hfill P(W_n\ge (1-\epsilon )x_n^2)\ge exp(-x_n^2/2).\end{array}$$

4. The Upper Bound of Theorem 2

Lemma 4

(Lemma 2.3 of Giné, Kwapień, Latała, and Zinn [15]). There exists a universal constant $C_3<\infty$ such that for any kernel h and any two sequences of i.i.d. random variables, we have

$$\begin{array}{c}\hfill P\left(\underset{k\le m,l\le n}{\max }\left|\sum _{i\le k,j\le l}h(X_i,Y_j)\right|\ge t\right)\le C_{3}P\left(\left|\sum _{i\le m,j\le n}h(X_i,Y_j)\right|\ge t/C_3\right)\end{array}$$

for all $m,n\in \mathbb{N}$ and all $t>0$.

Proposition 5.

Under the assumptions of Theorem 1,

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim\; sup}{\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\left({\sum }_{i=1}^n{g}_l\left(X_i\right)\right)}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2\log \log n}}\le 2a.s.\end{array}$$

Consequently,

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim\; sup}{\displaystyle \frac{W_n}{\log \log n}}\le 2a.s.\end{array}$$

Proof. 

Let $x_n\to \infty$ as $n\to \infty$. Let $\theta >1$ with $\theta -1$ be sufficiently small. For any positive integer $k\in (n,\theta n]$, via a similar idea as in (19) with $0<\eta <1$,

$$\begin{array}{ccc}\hfill & & P\left(\underset{n<k\le \theta n}{\max }{\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\left({\sum }_{i=1}^k{g}_l\left(X_i\right)\right)}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{k,l}^2}}\ge 2{(1+\eta )}^3{x}_n^2\right)\hfill \\ & \le & P\left({\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\left({\sum }_{i=1}^n{g}_l\left(X_i\right)\right)}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2}}\ge 2(1-\eta ){(1+\eta )}^3{x}_n^2\right)\hfill \\ & & +P\left(\underset{n<k\le \theta n}{\max }{\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\left({\sum }_{i=n+1}^k{g}_l\left(X_i\right)\right)}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{k,l}^2}}\ge {\displaystyle \frac{{\eta }^2{(1+\eta )}^3{x}_n^2}{2}}\right)\hfill \\ & :=& H_1+H_2.\hfill \end{array}$$

(87)

Notice that (10) implies (8). By (17) and the upper bound of Theorem 1,

$$\begin{array}{c}\hfill H_1\le exp\left(-{(1-\eta )}^{3/2}{(1+\eta )}^3{x}_n^2\right).\end{array}$$

(88)

Let $0<\delta <1$ be a sufficiently small constant. By (19),

$$\begin{array}{ccc}\hfill & & H_2\le P\left(2m\underset{1\le l<\infty }{\max }{\lambda }_l{V}_{n,l}^2\le (1-\eta )n\sum _{1\le l<\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)\right)\hfill \\ & & +P\left(\underset{n<k\le \theta n}{\max }{\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\left({\sum }_{i=n+1}^k{g}_l\left(X_i\right)I\left(|g_l\left(X_i\right)|>\delta \eta z_{n,l}\right)\right)}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{k,l}^2}}\ge {\displaystyle \frac{{\eta }^4{(1+\eta )}^3{x}_n^2}{2}}\right)\hfill \\ & & +P({\displaystyle \frac{{\max }_{n<k\le \theta n}{\sum }_{l=1}^{\infty }{\lambda }_l{\left({\sum }_{i=n+1}^k{g}_l\left(X_i\right)I\left(\right|g_l\left(X_i\right)|\le \delta \eta z_{n,l})\right)}^2}{n{\sum }_{1\le l<\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)}}\hfill \\ & & \ge {\displaystyle \frac{(1-\eta )(1-2\eta ){\eta }^2{(1+\eta )}^3{x}_n^2}{4m}})\hfill \\ & & :=H_{2,1}+H_{2,2}+H_{2,3}.\hfill \end{array}$$

(89)

By (24) in Proposition 1,

$$\begin{array}{c}\hfill H_{2,1}\le exp\left(-2x_n^2\right).\end{array}$$

(90)

By the Cauchy–Schwarz inequality, for each k,

$$\begin{array}{ccc}\hfill & & {\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\left\{{\sum }_{i=n+1}^k\left|g_l\left(X_i\right)\right|I\left(|g_l\left(X_i\right)|>\delta \eta z_{n,l}\right)\right\}}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{k,l}^2}}\hfill \\ & \le & {\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\sum }_{i=n+1}^k{g}_l^2\left(X_i\right){\sum }_{i=n+1}^{k}I\left(|g_l\left(X_i\right)|>\delta \eta z_{n,l}\right)}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{k,l}^2}}.\hfill \end{array}$$

(91)

By (17), for some m value that is sufficiently large, the sum of the diagonal terms is as follows:

$$\begin{array}{c}\hfill {\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\sum }_{i=n+1}^k{g}_l^2\left(X_i\right)I\left(|g_l\left(X_i\right)|>\delta \eta z_{n,l}\right)}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{k,l}^2}}\le m.\end{array}$$

(92)

By (91) and (92),

$$\begin{array}{ccc}\hfill & & H_{2,2}\le P\left(\underset{n<k\le \theta n}{\max }{\displaystyle \frac{{\sum }_{n+1\le i\ne j\le k}{\sum }_{l=1}^{\infty }{\lambda }_l{g}_l^2\left(X_i\right)I\left(\right|g_l\left(X_j\right)|>\delta \eta z_{n,l})}{{\max }_{1\le l<\infty }{\lambda }_l{\sum }_{i=n+1}^k{g}_l^2\left(X_i\right)}}\ge {\displaystyle \frac{{\eta }^5{(1+\eta )}^3{x}_n^2}{2}}\right)\hfill \\ & \le & P\left(\underset{n<k\le \theta n}{\max }{\displaystyle \frac{{\sum }_{n+1\le i<j\le k}{\sum }_{l=1}^{\infty }{\lambda }_l{g}_l^2\left(X_i\right)I\left(\right|g_l\left(X_j\right)|>\delta \eta z_{n,l})}{{\max }_{1\le l<\infty }{\lambda }_l{\sum }_{i=n+1}^k{g}_l^2\left(X_i\right)}}\ge {\displaystyle \frac{{\eta }^5{(1+\eta )}^3{x}_n^2}{4}}\right)\hfill \\ & +& P\left(\underset{n<k\le \theta n}{\max }{\displaystyle \frac{{\sum }_{n+1\le j<i\le k}{\sum }_{l=1}^{\infty }{\lambda }_l{g}_l^2\left(X_i\right)I\left(\right|g_l\left(X_j\right)|>\delta \eta z_{n,l})}{{\max }_{1\le l<\infty }{\lambda }_l{\sum }_{i=n+1}^k{g}_l^2\left(X_i\right)}}\ge {\displaystyle \frac{{\eta }^5{(1+\eta )}^3{x}_n^2}{4}}\right)\hfill \\ & \le & P\left(\underset{n<k\le \theta n}{\max }\sum _{n+2\le j\le k}{\displaystyle \frac{{\sum }_{n+1\le i<j}{\sum }_{l=1}^{\infty }{\lambda }_l{g}_l^2\left(X_i\right)I\left(\right|g_l\left(X_j\right)|>\delta \eta z_{n,l})}{{\max }_{1\le l<\infty }{\lambda }_l{\sum }_{n+1\le i<j}{g}_l^2\left(X_i\right)}}\ge {\displaystyle \frac{{\eta }^5{(1+\eta )}^3{x}_n^2}{4}}\right)\hfill \\ & +& P\left(\underset{n<k\le \theta n}{\max }\sum _{n+1\le j<k}{\displaystyle \frac{{\sum }_{j<i\le k}{\sum }_{l=1}^{\infty }{\lambda }_l{g}_l^2\left(X_i\right)I\left(\right|g_l\left(X_j\right)|>\delta \eta z_{n,l})}{{\max }_{1\le l<\infty }{\lambda }_l{\sum }_{j<i\le k}{g}_l^2\left(X_i\right)}}\ge {\displaystyle \frac{{\eta }^5{(1+\eta )}^3{x}_n^2}{4}}\right)\hfill \\ & =& H_{2,2,1}+H_{2,2,2}.\hfill \end{array}$$

(93)

Let

$$\begin{array}{c}\hfill {\varphi }_j={\displaystyle \frac{{\sum }_{n+1\le i<j}{\sum }_{l=1}^{\infty }{\lambda }_l{g}_l^2\left(X_i\right)I\left(\right|g_l\left(X_j\right)|>\delta \eta z_{n,l})}{{\max }_{1\le l<\infty }{\lambda }_l{\sum }_{n+1\le i<j}{g}_l^2\left(X_i\right)}}.\end{array}$$

Then, for any constant $t>0$,

$$\begin{array}{ccc}\hfill H_{2,2,1}& \le & \left(\sum _{n+2\le j\le \left[\theta n\right]}{\displaystyle \frac{{\sum }_{n+1\le i<j}{\sum }_{l=1}^{\infty }{\lambda }_l{g}_l^2\left(X_i\right)I\left(\right|g_l\left(X_j\right)|>\delta \eta z_{n,l})}{{\max }_{1\le l<\infty }{\lambda }_l{\sum }_{n+1\le i<j}{g}_l^2\left(X_i\right)}}\ge {\displaystyle \frac{{\eta }^5{(1+\eta )}^3{x}_n^2}{4}}\right)\hfill \\ & \le & E{e}^{t{\sum }_{j=n+2}^{\left[\theta n\right]}{\varphi }_j}{e}^{-t{\eta }^5{(1+\eta )}^3{x}_n^2/4}.\hfill \end{array}$$

(94)

Let $E_j$ be the expectation of $X_j$ for $n+2\le j\le \left[\theta n\right]$. Then,

$$\begin{array}{c}\hfill E{e}^{t{\sum }_{j=n+2}^{\left[\theta n\right]}{\varphi }_j}=E\left(e^{t{\sum }_{j=n+2}^{\left[\theta n\right]-1}{\varphi }_j}{E}_{\left[\theta n\right]}{e}^{t{\varphi }_{\left[\theta n\right]}}\right).\end{array}$$

(95)

Since $|e^s-1|\le e^{0\vee s}\left|s\right|$ for any $s\in R$ and $0\le {\varphi }_{\left[\theta n\right]}\le m$ for some m value that is sufficiently large, then

$$\begin{array}{ccc}\hfill \left|E_{\left[\theta n\right]}{e}^{t{\varphi }_{\left[\theta n\right]}}-1\right|& \le & e^{mt}t{E}_{\left[\theta n\right]}{\varphi }_{\left[\theta n\right]}\hfill \\ & =& {\displaystyle \frac{e^{mt}t{\sum }_{n+1\le i<\left[\theta n\right]}{\sum }_{l=1}^{\infty }{\lambda }_l{g}_l^2\left(X_i\right)P\left(\right|g_l\left(X_{\left[\theta n\right]}\right)|>\delta \eta z_{n,l})}{{\max }_{1\le l<\infty }{\lambda }_l{\sum }_{n+1\le i<\left[\theta n\right]}{g}_l^2\left(X_i\right)}}.\hfill \end{array}$$

By (12) and (15), we have $P\left(\right|g_l\left(X_{\left[\theta n\right]}\right)|>\delta \eta z_{n,l})=o(x_n^2/n)$. Then, together with (17),

$$\begin{array}{c}\hfill E_{\left[\theta n\right]}{e}^{t{\varphi }_{\left[\theta n\right]}}=1+o(x_n^2/n)=e^{o(x_n^2/n)}.\end{array}$$

(96)

Applying (96) to (95), we have

$$\begin{array}{c}\hfill E{e}^{t{\sum }_{j=2}^{\left[\theta n\right]}{\varphi }_j}=e^{o(x_n^2/n)}E{e}^{t{\sum }_{j=n+2}^{\left[\theta n\right]-1}{\varphi }_j}.\end{array}$$

Similarly,

$$\begin{array}{ccc}\hfill E{e}^{t{\sum }_{j=n+2}^{\left[\theta n\right]-1}{\varphi }_j}& =& E\left(e^{t{\sum }_{j=n+2}^{\left[\theta n\right]-2}{\varphi }_j}{E}_{n-1}{e}^{t{\varphi }_{\left[\theta n\right]-1}}\right)\hfill \\ & =& e^{o(x_n^2/n)}E{e}^{t{\sum }_{j=n+2}^{\left[\theta n\right]-2}{\varphi }_j}.\hfill \end{array}$$

Continue this process from $X_{\left[\theta n\right]}$ to $X_{n+1}$. Thus, we conclude that

$$\begin{array}{c}\hfill E{e}^{t{\sum }_{j=n+2}^{\left[\theta n\right]}{\varphi }_j}=e^{\left[(\theta -1)n\right]\times o(x_n^2/n)}=e^{o\left(x_n^2\right)}.\end{array}$$

(97)

Applying (97) to (94) and letting $t=8/\left({\eta }^5{(1+\eta )}^3\right)$, we have

$$\begin{array}{c}\hfill H_{2,2,1}\le exp(-2x_n^2).\end{array}$$

(98)

To estimate $H_{2,2,2}$, let

$$\begin{array}{c}\hfill {\psi }_{j,k}={\displaystyle \frac{{\sum }_{j<i\le k}{\sum }_{l=1}^{\infty }{\lambda }_l{g}_l^2\left(X_i\right)I\left(\right|g_l\left(X_j\right)|>\delta \eta z_{n,l})}{{\max }_{1\le l<\infty }{\lambda }_l{\sum }_{j<i\le k}{g}_l^2\left(X_i\right)}}.\end{array}$$

Then, for any constant $t>0$,

$$\begin{array}{ccc}\hfill H_{2,2,2}& \le & P\left(\sum _{n+1\le j<\left[\theta n\right]}\underset{j<k\le \theta n}{\max }{\psi }_{j,k}\ge {\displaystyle \frac{{\eta }^5{(1+\eta )}^3{x}_n^2}{4}}\right)\hfill \\ & \le & E{e}^{t{\sum }_{j=n+1}^{\left[\theta n\right]-1}{\max }_{j<k\le \theta n}{\psi }_{j,k}}{e}^{-t{\eta }^5{(1+\eta )}^3{x}_n^2/4}.\hfill \end{array}$$

(99)

Let $E_j$ be the expectation of $X_j$ for $n+1\le j\le \left[\theta n\right]$. Note $k>j$. Then,

$$\begin{array}{ccc}\hfill & & E{e}^{t{\sum }_{j=n+1}^{\left[\theta n\right]-1}{\max }_{j<k\le \theta n}{\psi }_{j,k}}\hfill \\ & & =E\left(e^{t{\sum }_{j=n+2}^{\left[\theta n\right]-1}{\max }_{j<k\le \theta n}{\psi }_{j,k}}{E}_{n+1}{e}^{t{\max }_{n+1<k\le \theta n}{\psi }_{n+1,k}}\right).\hfill \end{array}$$

(100)

Observe that

$$\begin{array}{c}\hfill {\psi }_{j,k}={\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l\left({\sum }_{j<i\le k}{g}_l^2\left(X_i\right)\right)I\left(\right|g_l\left(X_j\right)|>\delta \eta z_{n,l})}{{\max }_{1\le l<\infty }{\lambda }_l{\sum }_{j<i\le k}{g}_l^2\left(X_i\right)}}.\end{array}$$

(101)

Then, by (17), $0\le {\psi }_{n+1,k}\le m$ for some m value that is sufficiently large. Since $|e^s-1|\le e^{0\vee s}\left|s\right|$ for any $s\in R$,

$$\begin{array}{ccc}\hfill & & \left|E_{n+1}{e}^{t{\max }_{n+1<k\le \theta n}{\psi }_{n+1,k}}-1\right|\le e^{mt}t{E}_{n+1}\underset{n+1<k\le \theta n}{\max }{\psi }_{n+1,k}.\hfill \end{array}$$

(102)

Under Assumption (10), for each $l\in [1,\infty )$,

$$\begin{array}{c}\hfill {\lambda }_l{V}_{n,l}^2=\sum _{i=1}^n{\lambda }_l{g}_l^2\left(X_i\right)\le c_l\sum _{\nu =1}^{\infty }\sum _{i=1}^n{\lambda }_{\nu }{g}_{\nu }^2\left(X_i\right)=c_l\sum _{\nu =1}^{\infty }{\lambda }_{\nu }{V}_{n,\nu }^2.\end{array}$$

Recall that (17); then, for each $l\in [1,\infty )$,

$$\begin{array}{c}\hfill {\displaystyle \frac{{\lambda }_l{V}_{n,l}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2}}\le {\displaystyle \frac{m{c}_l}{1-\epsilon }}.\end{array}$$

Hence, by (101),

$$\begin{array}{c}\hfill {\psi }_{j,k}\le {\displaystyle \frac{m}{1-\epsilon }}\sum _{l=1}^{\infty }{c}_{l}I\left(\right|g_l\left(X_j\right)|>\delta \eta z_{n,l}).\end{array}$$

Then,

$$\begin{array}{c}\hfill E_{n+1}\underset{n+1<k\le \theta n}{\max }{\psi }_{n+1,k}\le {\displaystyle \frac{m}{1-\epsilon }}\sum _{l=1}^{\infty }{c}_{l}P\left(\right|g_l\left(X_{n+1}\right)|>\delta \eta z_{n,l}).\end{array}$$

By (12) and (15), we have $P\left(\right|g_l\left(X_{n+1}\right)|>\delta \eta z_{n,l})=o(x_n^2/n)$. Then, together with (10),

$$\begin{array}{c}\hfill E_{n+1}\underset{n+1<k\le \theta n}{\max }{\psi }_{n+1,k}=o(x_n^2/n).\end{array}$$

(103)

Then, by (102) and (103),

$$\begin{array}{c}\hfill E_{n+1}{e}^{t{\max }_{n+1<k\le \theta n}{\psi }_{n+1,k}}=1+o(x_n^2/n)=e^{o(x_n^2/n)}.\end{array}$$

Continue this process from $j=n+2$ to $j=\left[\theta n\right]-1$ and by (100),

$$\begin{array}{c}\hfill E{e}^{t{\sum }_{j=n+1}^{\left[\theta n\right]-1}{\max }_{j<k\le \theta n}{\psi }_{j,k}}=e^{o\left(x_n^2\right)}.\end{array}$$

(104)

Applying (104) to (99) and letting $t=8/\left({\eta }^5{(1+\eta )}^3\right)$, we have

$$\begin{array}{c}\hfill H_{2,2,2}\le exp(-2x_n^2).\end{array}$$

(105)

By (93), (98) and (105),

$$\begin{array}{c}\hfill H_{2,2}\le 2exp(-2x_n^2).\end{array}$$

(106)

By the definition of $H_{2,3}$ in (89), and by Lemma 4, there is a constant $0<C^{\prime }<\infty$ such that

$$\begin{array}{c}\hfill H_{2,3}\le C^{\prime }P\left({\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\left({\sum }_{i=n+1}^{\left[\theta n\right]}{g}_l\left(X_i\right)I\left(\right|g_l\left(X_i\right)|\le \delta \eta z_{n,l})\right)}^2}{n{\sum }_{l=1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)}}\ge {\displaystyle \frac{(1-3\eta ){\eta }^2{x}_n^2}{4m{C}^{\prime }}}\right).\end{array}$$

Similar to (36),

$$\begin{array}{ccc}\hfill H_{2,3}& \le & C^{\prime }P({\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\left({\sum }_{i=n+1}^{\left[\theta n\right]}\left\{g_l\left(X_i\right)I\left(\right|g_l\left(X_i\right)|\le \delta \eta z_{n,l})-E{g}_l\left(X_i\right)I\left(\right|g_l\left(X_i\right)|\le \delta \eta z_{n,l})\right\}\right)}^2}{n{\sum }_{l=1}^{\infty }{\lambda }_l{L}_l\left(z_{n,l}\right)}}\hfill \\ & & \ge {\displaystyle \frac{{(1-3\eta )}^2{\eta }^2{x}_n^2}{4m{C}^{\prime }}}).\hfill \end{array}$$

By the decoupling version of (40) in Proposition 3,

$$\begin{array}{c}\hfill H_{2,3}\le C^{\prime }exp\left(-2x_n^2\right).\end{array}$$

(107)

Combining (89), (90), (106) and (107), we have

$$\begin{array}{c}\hfill H_2\le (3+C^{\prime })exp\left(-2x_n^2\right).\end{array}$$

(108)

By (87), (88) and (108),

$$\begin{array}{ccc}& & P\left(\underset{n<k\le \theta n}{\max }{\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\left({\sum }_{i=1}^k{g}_l\left(X_i\right)\right)}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{k,l}^2}}\ge 2{(1+\eta )}^3{x}_n^2\right)\hfill \\ & \le & exp\left(-{(1-\eta )}^{7/4}{(1+\eta )}^3{x}_n^2\right).\hfill \end{array}$$

Let $n=\left[{\theta }^j\right]$ for some $j\in \mathbb{N}$. We have

$$\begin{array}{ccc}& & P\left(\underset{{\theta }^j<k\le {\theta }^{j+1}}{\max }{\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\left({\sum }_{i=1}^k{g}_l\left(X_i\right)\right)}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{k,l}^2}}\ge 2{(1+\eta )}^3{x}_{\left[{\theta }^j\right]}^2\right)\hfill \\ & & \le exp\left(-{(1-\eta )}^{7/4}{(1+\eta )}^3{x}_{\left[{\theta }^j\right]}^2\right).\hfill \end{array}$$

Let $x_n^2=\log \log n$. Then,

$$\begin{array}{ccc}& & \sum _{j=1}^{\infty }P\left(\underset{{\theta }^j<k\le {\theta }^{j+1}}{\max }{\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\left({\sum }_{i=1}^k{g}_l\left(X_i\right)\right)}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{k,l}^2\log \log k}}\ge 2{(1+\eta )}^3\right)\hfill \\ & \le & \sum _{j=1}^{\infty }P\left(\underset{{\theta }^j<k\le {\theta }^{j+1}}{\max }{\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\left({\sum }_{i=1}^k{g}_l\left(X_i\right)\right)}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{k,l}^2}}\ge 2{(1+\eta )}^3{x}_{\left[{\theta }^j\right]}^2\right)\hfill \\ & \le & \sum _{j=1}^{\infty }exp\left(-{(1-\eta )}^{7/4}{(1+\eta )}^3\log \log \left[{\theta }^j\right]\right)\hfill \\ & \le & K\sum _{j=1}^{\infty }exp\left(-{(1-\eta )}^2{(1+\eta )}^3\log j\right)<\infty .\hfill \end{array}$$

By the Borel–Cantelli lemma,

$$\begin{array}{ccc}& & \underset{n\to \infty }{lim\; sup}{\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\left({\sum }_{i=1}^n{g}_l\left(X_i\right)\right)}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2\log \log n}}\le 2a.s.\hfill \end{array}$$

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5. The Lower Bound of Theorem 2

Proof. 

By the definition of $W_n$,

$$\begin{array}{ccc}\hfill {\displaystyle \frac{W_n}{\log \log n}}& =& {\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l\left({\left({\sum }_{i=1}^n{g}_l\left(X_i\right)\right)}^2-{\sum }_{i=1}^n{g}_l^2\left(X_i\right)\right)}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2\log \log n}}\hfill \\ & =& {\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\left({\sum }_{i=1}^n{g}_l\left(X_i\right)\right)}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2\log \log n}}-{\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{V}_{n,l}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2\log \log n}}.\hfill \end{array}$$

By (17), ${\sum }_{l=1}^{\infty }{\lambda }_l{V}_{n,l}^2/\left({\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2\log \log n\right)\le m/\left((1-\epsilon )\log \log n\right)\to 0$ as $n\to \infty$, we have

$$\begin{array}{c}\hfill {\displaystyle \frac{W_n}{\log \log n}}\ge {\displaystyle \frac{{\sum }_{l=1}^{\infty }{\lambda }_l{\left({\sum }_{i=1}^n{g}_l\left(X_i\right)\right)}^2}{{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2\log \log n}}.\end{array}$$

Then,

$$\begin{array}{c}\hfill {\displaystyle \frac{W_n}{\log \log n}}\ge \sum _{k=1}^{\infty }{\displaystyle \frac{{\left({\sum }_{i=1}^n{g}_k\left(X_i\right)\right)}^2}{V_{n,k}^2\log \log n}}{I}_{k=min\{j:{\max }_{1\le l<\infty }{\lambda }_l{V}_{n,l}^2={\lambda }_j{V}_{n,j}^2\}}.\end{array}$$

Hence, by (6),

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim\; sup}{\displaystyle \frac{W_n}{\log \log n}}\ge 2a.s.\end{array}$$

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