A supplement to the convergence rate in a theorem of Heyde

Abstract

Let $\{X,X_n,n\ge 1\}$ be a sequence of i.i.d. random variables with zero mean, set $S_n={\sum }_{k=1}^n{X}_k$, $E{X}^2={\sigma }^2>0$, and $\lambda (ϵ)={\sum }_{n=1}^{\mathrm{\infty }}P(|S_n|\ge nϵ)$. In this paper, the authors discuss the rate of approximation of ${\sigma }^2$ by ${ϵ}^2\lambda (ϵ)$ under suitable conditions, improve the results of Klesov (Theory Probab. Math. Stat. 49:83-87, 1994), and extend the work He and Xie (Acta Math. Appl. Sin. 2012, doi:10.1007/s10255-012-0138-6).

MSC:60F15, 60G50.

1 Introduction and main results

Let $\{X,X_n,n\ge 1\}$ be a sequence of i.i.d. random variables, set $S_n={\sum }_{k=1}^n{X}_k$, and $\lambda (ϵ)={\sum }_{n=1}^{\mathrm{\infty }}P(|S_n|\ge nϵ)$. Heyde [1] proved that

$$\underset{ϵ\to 0}{lim}{ϵ}^2\lambda (ϵ)={\sigma }^2,$$

whenever $E{X}^2={\sigma }^2<\mathrm{\infty }$ and $EX=0$.

There are various extensions of this result: Chen [2], Gut and Spǎtara [3], Lanzinger and Stadtmüller [4]. Liu and Lin [5] introduced a new kind of complete moment convergence; Klesov [6] studied the rate of approximation of ${\sigma }^2$ by ${ϵ}^2\lambda (ϵ)$ and proved the following Theorem A.

Theorem A Let$\{X,X_n,n\ge 1\}$be a sequence of i.i.d. random variables with zero mean, if$E{X}^2={\sigma }^2>0$, and$E{|X|}^3<\mathrm{\infty }$, then

$${ϵ}^2\lambda (ϵ)-{\sigma }^2=o\left({ϵ}^{1/2}\right),\mathit{\text{as}}ϵ\to 0.$$

Recently, He and Xie [7] obtained Theorem B which improved Theorem A. Gut and Steinebach [8] extended the results of Klesov [6].

Theorem B Let$\{X,X_n,n\ge 1\}$be a sequence of i.i.d. random variables, and$0<\delta \le 1$, if

$$EX=0,E{X}^2={\sigma }^2>0\mathit{\text{and}}E{|X|}^{2+\delta }<\mathrm{\infty },$$

then

$${ϵ}^2\lambda (ϵ)-{\sigma }^2=\{\begin{array}{cc}O(ϵ),\hfill & \delta =1,\hfill \\ o({ϵ}^{\delta }),\hfill & 0<\delta <1.\hfill \end{array}$$

Let G be the set of functions $g(x)$ that are defined for all real x and satisfy the following conditions: (a) $g(x)$ is nonnegative, even, nondecreasing in the interval $x>0$, and $g(x)\ne 0$ for $x\ne 0$; (b) $\frac{x}{g(x)}$ is nondecreasing in the interval $x>0$.

Let $G_0$ be the set of functions $g(x)\in G$ satisfying the supplementary condition (c) ${lim}_{x\to \mathrm{\infty }}\frac{g(x^2)}{xg(x)}=0$. Obviously, the function $g(x)={|x|}^{\delta }$ with $0<\delta <1$ belongs to $G_0$ and does not belong to $G_0$ if $\delta =1$. The purpose of this paper is to generalize Theorem B to the case where the condition $E{|X|}^{2+\delta }<\mathrm{\infty }$ is replaced by a more general condition $E{|X|}^{2}g(X)<\mathrm{\infty }$ in which the function g belongs to some subset of G. Denote $T_g(v)=E{X}^{2}g(X)I(|X|>v)$, $T_g(v)$ is a nonnegative nonincreasing function in the interval $v>0$, and ${lim}_{v\to \mathrm{\infty }}{T}_g(v)=0$ with $E{X}^{2}g(X)<\mathrm{\infty }$. Now we state our results as follows.

Theorem 1.1 Let$\{X,X_n;n\ge 1\}$be a sequence of i.i.d. random variables with zero mean and$E{X}^2={\sigma }^2>0$, if$E{X}^{2}g(X)<\mathrm{\infty }$for some function$g(x)\in G$, and

$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{ng(\sqrt{n})}<\mathrm{\infty },$$

(1.1)

then

$${ϵ}^2\lambda (ϵ)-{\sigma }^2=O\left({ϵ}^{1/2}\right)+o(1)(h_1(ϵ)+f_1(ϵ)),\mathit{\text{as}}ϵ\to 0,$$

(1.2)

where$f_1(ϵ)={\sum }_{n=[\frac{1}{{ϵ}^2}]+1}^{\mathrm{\infty }}\frac{1}{ng(\sqrt{n})}$, $h_1(ϵ)={ϵ}^2{\sum }_{n=1}^{[\frac{1}{{ϵ}^2}]}\frac{1}{g(\sqrt{n})}$.

Theorem 1.2 Under the conditions of Theorem 1.1, and$g(x)\in G_0$, then

$${ϵ}^2\lambda (ϵ)-{\sigma }^2=o(1)(h_1(ϵ)+f_1(ϵ)),\mathit{\text{as}}ϵ\to 0.$$

(1.3)

Throughout this paper, we suppose that C denotes a constant which only depends on some given numbers and may be different at each appearance, and that $[x]$ denotes the integer part of x.

2 Proofs of the main results

Before we prove the main results we state some lemmas. Lemma 2.1 is from [7]. $\mathrm{\Phi }(x)$ is the standard normal distribution function, $\mathrm{\Phi }(x)=\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^x{e}^{-t^2/2}dt$.

Lemma 2.1 Let$\{X,X_n,n\ge 1\}$be a sequence of i.i.d. standard normal distribution random variables. Then

$${ϵ}^2\lambda (ϵ)={ϵ}^2\sum _{n=1}^{\mathrm{\infty }}\frac{2}{\sqrt{2\pi }}{\int }_{ϵ\sqrt{n}}^{\mathrm{\infty }}{e}^{-t^2/2}dt=1-\frac{{ϵ}^2}{2}+O\left({ϵ}^3\right),\mathit{\text{as}}ϵ\to 0.$$

(2.1)

If$\{X_n,n\ge 1\}$is a sequence of independent random variables with zero mean and finite variance, and put$E{X}_j^2={\sigma }_j^2$, $B_n={\sum }_{j=1}^n{\sigma }_j^2$, Bikelis[9]obtained the following inequality:

for every x, where$V_j(x)=P(X_j<x)$is the distribution function of the random variable$X_j$. By applying the above inequality to the sequence of i.i.d. random variables with zero mean and variance 1, and letting$|x|=ϵ\sqrt{n}$, we have the following lemma.

Lemma 2.2 Let$\{X,X_n,n\ge 1\}$be a sequence of i.i.d. random variables with zero mean and$E{X}^2=1$. Then for any given$ϵ>0$, we have

where$V(x)=P(X<x)$is the distribution function of a random variable X.

Proof of Theorem 1.1 Without loss of generality, we suppose that ${\sigma }^2=1$, $0<ϵ<1$, and write

$${ϵ}^2\lambda (ϵ)=I+{ϵ}^2\sum _{n=1}^{\mathrm{\infty }}\frac{2}{\sqrt{2\pi }}{\int }_{ϵ\sqrt{n}}^{\mathrm{\infty }}{e}^{-t^2/2}dt,$$

where

$$I={ϵ}^2\sum _{n=1}^{\mathrm{\infty }}(P(|S_n|>nϵ)-\frac{2}{\sqrt{2\pi }}{\int }_{ϵ\sqrt{n}}^{\mathrm{\infty }}{e}^{-t^2/2}dt).$$

Applying Lemma 2.1, we obtain

$${ϵ}^2\lambda (ϵ)=I+1-\frac{{ϵ}^2}{2}+O\left({ϵ}^3\right),$$

then

$${ϵ}^2\lambda (ϵ)-1=-\frac{{ϵ}^2}{2}+{ϵ}^2\sum _{n=1}^{\mathrm{\infty }}{R}_n+O\left({ϵ}^3\right),$$

here $R_n=P(|S_n|>nϵ)-\frac{2}{\sqrt{2\pi }}{\int }_{ϵ\sqrt{n}}^{\mathrm{\infty }}{e}^{-t^2/2}dt$. By Lemma 2.2,

$$|R_n|\le R_{1n}+R_{2n},$$

where

We obtain

$${ϵ}^2\lambda (ϵ)-1={ϵ}^2\sum _{n=1}^{\mathrm{\infty }}{R}_{1n}+{ϵ}^2\sum _{n=1}^{\mathrm{\infty }}{R}_{2n}+O\left({ϵ}^2\right).$$

(2.2)

Firstly, we estimate ${ϵ}^2{\sum }_{n=1}^{\mathrm{\infty }}{R}_{1n}$. Note that

$${ϵ}^2\sum _{n=1}^{\mathrm{\infty }}{R}_{1n}={ϵ}^2\sum _{n=1}^{[\frac{1}{{ϵ}^2}]}{R}_{1n}+{ϵ}^2\sum _{n=[\frac{1}{{ϵ}^2}]+1}^{\mathrm{\infty }}{R}_{1n}=:T_1+T_2.$$

Applying the condition $E{X}^{2}g(X)<\mathrm{\infty }$, we have

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_{|u|>\sqrt[4]{n}}{u}^{2}g(u)dV(u)=0.$$

Therefore, for any $\eta >0$, there is an integer $N_0$ such that ${\int }_{|u|>\sqrt[4]{n}}{u}^{2}g(u)dV(u)\le \eta$, whenever $n>N_0$. Hence

$$\begin{array}{rcl}{T}_1& \le & C{ϵ}^2\sum _{n=1}^{N_0}{\int }_{|u|>\sqrt{n}}{u}^{2}dV(u)+C{ϵ}^2\sum _{n=N_0+1}^{[\frac{1}{{ϵ}^2}]}{(1+ϵ\sqrt{n})}^{-2}{\int }_{|u|>(1+ϵ\sqrt{n})\sqrt{n}}{u}^{2}dV(u)\\ \le & C{ϵ}^2{N}_0+C{ϵ}^2\eta \sum _{n=N_0+1}^{[\frac{1}{{ϵ}^2}]}\frac{1}{{(1+ϵ\sqrt{n})}^{2}g(\sqrt{n}(1+ϵ\sqrt{n}))}\\ \le & C{ϵ}^2(N_0+\eta \sum _{n=1}^{[\frac{1}{{ϵ}^2}]}\frac{1}{g(\sqrt{n})})\\ =& C{h}_1(ϵ)(\frac{N_0}{{\sum }_{n=1}^{[\frac{1}{{ϵ}^2}]}\frac{1}{g(\sqrt{n})}}+\eta )\\ \le & C{h}_1(ϵ)(N_0ϵ+\eta )\\ =& o(h_1(ϵ)),\end{array}$$

(2.3)

where $h_1(ϵ)={ϵ}^2{\sum }_{n=1}^{[\frac{1}{{ϵ}^2}]}\frac{1}{g(\sqrt{n})}$. For $T_2$, noting that $g(x)\in G$, we have the following inequality:

$$\begin{array}{rcl}{T}_2& \le & C{ϵ}^2\sum _{n=[\frac{1}{{ϵ}^2}]+1}^{\mathrm{\infty }}\frac{1}{n{ϵ}^2}{\int }_{|u|>\sqrt{n}(1+ϵ\sqrt{n})}{u}^{2}dV(u)\\ \le & C\sum _{n=[\frac{1}{{ϵ}^2}]+1}^{\mathrm{\infty }}\frac{1}{ng(\sqrt{n}(1+ϵ\sqrt{n}))}{\int }_{|u|>\sqrt{n}(1+ϵ\sqrt{n})}{u}^{2}g(u)dV(u)\\ \le & C\sum _{n=[\frac{1}{{ϵ}^2}]+1}^{\mathrm{\infty }}\frac{1}{ng(\sqrt{n})}{\int }_{|u|>\frac{1}{ϵ}}{u}^{2}g(u)dV(u)\\ \le & C{T}_g\left(\frac{1}{ϵ}\right)f_1(ϵ).\end{array}$$

(2.4)

Next, we estimate the second term of (2.2). Note that

$$\begin{array}{rcl}{ϵ}^2\sum _{n=1}^{\mathrm{\infty }}{R}_{2n}& =& C{ϵ}^2\sum _{n=1}^{\mathrm{\infty }}{n}^{-1/2}{(1+ϵ\sqrt{n})}^{-3}{\int }_{|u|\le {(\sqrt{n}(1+ϵ\sqrt{n}))}^{1/2}}{|u|}^{3}dV(u)\\ +C{ϵ}^2\sum _{n=1}^{\mathrm{\infty }}{n}^{-1/2}{(1+ϵ\sqrt{n})}^{-3}{\int }_{{(\sqrt{n}(1+ϵ\sqrt{n}))}^{1/2}<|u|<\sqrt{n}(1+ϵ\sqrt{n})}{|u|}^{3}dV(u)\\ =:& J_1+J_2.\end{array}$$

For $J_1$, we can write

$$\begin{array}{rcl}{J}_1& =& C{ϵ}^2(\sum _{n=1}^{[\frac{1}{{ϵ}^2}]}+\sum _{n=[\frac{1}{{ϵ}^2}]+1}^{\mathrm{\infty }})n^{-1/2}{(1+ϵ\sqrt{n})}^{-3}{\int }_{|u|\le {(\sqrt{n}(1+ϵ\sqrt{n}))}^{1/2}}{|u|}^{3}dV(u)\\ =:& J_{11}+J_{12}.\end{array}$$

Noting that $\frac{x}{g(x)}$ is nondecreasing in the interval $x>0$, we have

$$\begin{array}{rcl}{J}_{11}& =& C{ϵ}^2\sum _{n=1}^{[\frac{1}{{ϵ}^2}]}\frac{1}{\sqrt{n}{(1+ϵ\sqrt{n})}^3}{\int }_{|u|\le {(\sqrt{n}(1+ϵ\sqrt{n}))}^{1/2}}{|u|}^{3}dV(u)\\ \le & C{ϵ}^2\sum _{n=1}^{[\frac{1}{{ϵ}^2}]}\frac{1}{n^{1/4}{(1+ϵ\sqrt{n})}^{5/2}g({(\sqrt{n}(1+ϵ\sqrt{n}))}^{1/2})}{\int }_{|u|\le {(\sqrt{n}(1+ϵ\sqrt{n}))}^{1/2}}{u}^{2}g(u)dV(u)\\ \le & C{ϵ}^2\sum _{n=1}^{[\frac{1}{{ϵ}^2}]}\frac{1}{n^{1/4}g(n^{1/4})}\\ =& C{h}_2(ϵ),\end{array}$$

(2.5)

where $h_2(ϵ)={ϵ}^2{\sum }_{n=1}^{[\frac{1}{{ϵ}^2}]}\frac{1}{n^{1/4}g(n^{1/4})}$.

Similarly, we can obtain

$$\begin{array}{rcl}{J}_{12}& =& C{ϵ}^2\sum _{n=[\frac{1}{{ϵ}^2}]+1}^{\mathrm{\infty }}\frac{1}{\sqrt{n}{(1+ϵ\sqrt{n})}^3}{\int }_{|u|\le {(\sqrt{n}(1+ϵ\sqrt{n}))}^{1/2}}{|u|}^{3}dV(u)\\ \le & C{ϵ}^2\sum _{n=[\frac{1}{{ϵ}^2}]+1}^{\mathrm{\infty }}\frac{1}{n^{1/4}{(1+ϵ\sqrt{n})}^{5/2}g({(\sqrt{n}(1+ϵ\sqrt{n}))}^{1/2})}{\int }_{|u|\le {(\sqrt{n}(1+ϵ\sqrt{n}))}^{1/2}}{u}^{2}g(u)dV(u)\\ \le & C{ϵ}^2\sum _{n=[\frac{1}{{ϵ}^2}]+1}^{\mathrm{\infty }}\frac{1}{{ϵ}^{5/2}{n}^{3/2}g(n^{1/4})}\\ =& C\frac{1}{\sqrt{ϵ}}{f}_2(ϵ),\end{array}$$

(2.6)

where $f_2(ϵ)={\sum }_{n=[\frac{1}{{ϵ}^2}]+1}^{\mathrm{\infty }}\frac{1}{n^{3/2}g(n^{1/4})}$.

For $J_2$, we write

$$\begin{array}{rcl}{J}_2& =& C{ϵ}^2(\sum _{n=1}^{[\frac{1}{{ϵ}^2}]}+\sum _{n=[\frac{1}{{ϵ}^2}]+1}^{\mathrm{\infty }})n^{-1/2}{(1+ϵ\sqrt{n})}^{-3}{\int }_{{(\sqrt{n}(1+ϵ\sqrt{n}))}^{1/2}<|u|<\sqrt{n}(1+ϵ\sqrt{n})}{|u|}^{3}dV(u)\\ =:& J_{21}+J_{22}.\end{array}$$

Using the properties of $g(x)$ by simple calculation, it follows that

$$\begin{array}{rcl}{J}_{21}& =& C{ϵ}^2\sum _{n=1}^{[\frac{1}{{ϵ}^2}]}{n}^{-1/2}{(1+ϵ\sqrt{n})}^{-3}{\int }_{{(\sqrt{n}(1+ϵ\sqrt{n}))}^{1/2}<|u|<\sqrt{n}(1+ϵ\sqrt{n})}{|u|}^{3}dV(u)\\ \le & C{ϵ}^2(\sum _{n=1}^{N_0}+\sum _{n=N_0+1}^{[\frac{1}{{ϵ}^2}]})\frac{1}{{(1+ϵ\sqrt{n})}^{2}g(\sqrt{n}(1+ϵ\sqrt{n}))}\\ \times {\int }_{{(\sqrt{n}(1+ϵ\sqrt{n}))}^{1/2}<|u|<\sqrt{n}(1+ϵ\sqrt{n})}{u}^{2}g(u)dV(u)\\ \le & C{ϵ}^2(\sum _{n=1}^{N_0}+\sum _{n=N_0+1}^{[\frac{1}{{ϵ}^2}]})\frac{1}{g(\sqrt{n})}{\int }_{|u|>n^{1/4}}{u}^{2}g(u)dV(u)\\ \le & C{ϵ}^2(N_0+\eta \sum _{n=1}^{[\frac{1}{{ϵ}^2}]}\frac{1}{g(\sqrt{n})})\\ =& o(h_1(ϵ)),\end{array}$$

(2.7)

and

$$\begin{array}{rcl}{J}_{22}& \le & C{ϵ}^2\sum _{n=[\frac{1}{{ϵ}^2}]+1}^{\mathrm{\infty }}{n}^{-\frac{1}{2}}{(1+ϵ\sqrt{n})}^{-3}{\int }_{{(\sqrt{n}(1+ϵ\sqrt{n}))}^{1/2}<|u|<\sqrt{n}(1+ϵ\sqrt{n})}{|u|}^{3}dV(u)\\ \le & C\sum _{n=[\frac{1}{{ϵ}^2}]+1}^{\mathrm{\infty }}\frac{1}{ng(\sqrt{n})}{\int }_{{(\sqrt{n}(1+ϵ\sqrt{n}))}^{1/2}<|u|<\sqrt{n}(1+ϵ\sqrt{n})}{u}^{2}g(u)dV(u)\\ \le & C{T}_g\left(\frac{1}{\sqrt{ϵ}}\right)\sum _{n=[\frac{1}{{ϵ}^2}]+1}^{\mathrm{\infty }}\frac{1}{ng(\sqrt{n})}\\ \le & C{T}_g\left(\frac{1}{\sqrt{ϵ}}\right)f_1(ϵ).\end{array}$$

(2.8)

From (2.2) to (2.8), we conclude that

$${ϵ}^2\lambda (ϵ)-1\le C\frac{1}{\sqrt{ϵ}}{f}_2(ϵ)+C{T}_g\left(\frac{1}{\sqrt{ϵ}}\right)f_1(ϵ)+o(1)h_1(ϵ)+C{h}_2(ϵ).$$

(2.9)

Since

$$\frac{1}{\sqrt{ϵ}}{f}_2(ϵ)\le \frac{C}{\sqrt{ϵ}}\sum _{n=[\frac{1}{{ϵ}^2}]+1}^{\mathrm{\infty }}\frac{1}{n^{3/2}}\le C\sqrt{ϵ},$$

and

$$h_2(ϵ)={ϵ}^2\sum _{n=1}^{[\frac{1}{{ϵ}^2}]}\frac{1}{\sqrt[4]{n}g(\sqrt[4]{n})}\le C{ϵ}^2\sum _{n=1}^{[\frac{1}{{ϵ}^2}]}\frac{1}{\sqrt[4]{n}}\le C\sqrt{ϵ},$$

by (2.9), we have

$${ϵ}^2\lambda (ϵ)-1=O\left({ϵ}^{1/2}\right)+o(1)(f_1(ϵ)+h_1(ϵ)).$$

This completes the proof of Theorem 1.1. □

Proof of Theorem 1.2 By the conditions $g(x)\in G_0$, and ${lim}_{x\to \mathrm{\infty }}\frac{g(x^2)}{xg(x)}=0$, for any $\eta >0$, there is an integer $N_1$ such that $\frac{g(\sqrt{n})}{\sqrt[4]{n}g(\sqrt[4]{n})}\le \eta$, whenever $n>N_1$. We have

$$\begin{array}{rcl}{h}_2(ϵ)& \le & {ϵ}^2\sum _{n=1}^{N_1}\frac{1}{\sqrt[4]{n}g(\sqrt[4]{n})}+{ϵ}^2\sum _{n=N_1}^{[\frac{1}{{ϵ}^2}]}\frac{\eta }{g(\sqrt{n})}\\ \le & C{ϵ}^2{N}_1+{ϵ}^2\sum _{n=N_1+1}^{[\frac{1}{{ϵ}^2}]}\frac{\eta }{g(\sqrt{n})}\\ \le & C{ϵ}^2{N}_1+{ϵ}^2\sum _{n=1}^{[\frac{1}{{ϵ}^2}]}\frac{\eta }{g(\sqrt{n})}\\ =& o(1)h_1(ϵ),\end{array}$$

(2.10)

and

$$\begin{array}{rcl}\frac{1}{\sqrt{ϵ}}{f}_2(ϵ)& \le & \frac{1}{\sqrt{ϵ}}\sum _{n=[\frac{1}{{ϵ}^2}]+1}^{\mathrm{\infty }}\frac{\eta }{n^{5/4}g(\sqrt{n})}\\ \le & \sum _{n=[\frac{1}{{ϵ}^2}]+1}^{\mathrm{\infty }}\frac{\eta }{ng(\sqrt{n})}=o(1)f_1(ϵ).\end{array}$$

(2.11)

By (2.9)-(2.11), note that $T_g(\frac{1}{\sqrt{ϵ}})=o(1)$, as $ϵ\to 0$, we have

$${ϵ}^2\lambda (ϵ)-{\sigma }^2=o(1)(h_1(ϵ)+f_1(ϵ)),\text{as }ϵ\to 0.$$

This completes the proof of Theorem 1.2. □

Remark 2.1 If $g(x)={|x|}^{\delta }$, $0<\delta <1$, then $f_1(ϵ)=O({ϵ}^{\delta })$, $h_1(ϵ)=O({ϵ}^{\delta })$. By Theorem 1.2, we get

$${ϵ}^2\lambda (ϵ)-{\sigma }^2=o\left({ϵ}^{\delta }\right),\text{as }ϵ\to 0.$$

Remark 2.2 If $g(x)=|x|$, $\delta =1$, then $\frac{1}{\sqrt{ϵ}}{f}_2(ϵ)=O(ϵ)$, $f_1(ϵ)=O(ϵ)$, $h_1(ϵ)=O(ϵ)$, $h_2(ϵ)=O(ϵ)$. By (2.9), we get

$${ϵ}^2\lambda (ϵ)-{\sigma }^2=O(ϵ),\text{as }ϵ\to 0.$$