1 Introduction
Applications of the central limit theorem and other weak limit theorems are closely connected to the rate of convergence to the limit distribution. The rate of convergence was studied by many authors; see [6] and the references therein. The simplest result in this direction is the Berry–Esseen inequality. Let $\{\xi _{n},n\ge 1\}$ be a sequence of independent identically distributed random variables (iidrvs) with distribution function $F(x)$, $E\xi _{i}=0$, and $D\xi _{i}={\sigma }^{2}<\infty $. Let $\beta _{3}=\int _{\mathbb{R}}|x{|}^{3}dF(x)$ be the absolute 3rd moment, $\varPhi _{n}(x)=P\{\frac{\xi _{1}+\xi _{2}+\dots +\xi _{n}}{\sigma \sqrt{n}}<x\}$, and let $\varPhi (x)$, $x\in \mathbb{R},$ be the standard normal distribution function. Then the Berry–Esseen inequality states that
\[\underset{x\in \mathbb{R}}{\sup }\big|\varPhi _{n}(x)-\varPhi (x)\big|\le \frac{C\beta _{3}}{{\sigma }^{3}\sqrt{n}}.\]
This estimate gives the rate of convergence $O({n}^{-1/2})$ and the asymptotic expansions of the distribution function of the sum of iidrvs in terms of semiinvariants, presented in the book [6]. The same rate of convergence was obtained by Paulauskas [5] in terms of pseudomoments. Let $\sigma =1$. Then the “pseudomoment” function is defined as $H(x)=F(x)-\varPhi (x)$, the (absolute) third pseudomoment is defined as $\nu =\int _{\mathbb{R}}|x{|}^{3}|dH(x)|$, and we have
\[\underset{x\in \mathbb{R}}{\sup }\big|\varPhi _{n}(x)-\varPhi (x)\big|\le C\max \big(\nu ,{\nu }^{\frac{1}{4}}\big){n}^{-\frac{1}{2}}.\]
However, this rate of convergence is slow, for instance, from the point of view of financial applications. The conditions that allow one to improve the rate of convergence were formulated by several authors. After the introduction of pseudomoments in [10], they are widely used in limit theorems. Zolotarev [11] obtained very general estimates in the central limit theorem using a different type of pseudomoments. Studnyev [9] obtained the following estimate of the rate of convergence in terms of pseudomoments. Let $\{\xi _{n},n\ge 1\}$ be centered iidrvs with unit variance and characteristic function $f(t)$, $\mu _{k}=\int _{R}{x}^{k}dH(x)$ be the kth-order pseudomoment, and $V(x)={V_{-\infty }^{x}}H(z)$ be the variation of the function H.Proposition 1 ([9]).
Let $F(x)$ have finite moments up to the qth order for some $q\ge 3$ and satisfy the Cramer condition $\overline{\lim }_{|t|\to \infty }|f(t)|<1$. Then
We can see that the condition $\mu _{k}=0$, $3\le k\le r$ supplies the rate of convergence $O({n}^{\frac{-r+1}{2}})$. The rate of convergence was also studied in [1, 3, 7]. In our work, we use a different type of pseudomoments and get the same rate of convergence avoiding the Cramer condition. Instead, we impose the boundedness of the truncated pseudomoments and integrability of the characteristic function.
2 Generalization of Studnyev’s estimate. Main results
Let, as before, $\{\xi _{n},n\ge 1\}$ be a sequence of iidrvs with $E\xi _{i}=0$, $D\xi _{i}={\sigma }^{2}\in (0,\infty )$, distribution function $F(x)$, and characteristic function $f(t)$, and let $\varPhi _{n}(x)$, $x\in \mathbb{R}$, be the distribution function of the random variable
We assume that, for some $m\ge 3$, there exist the pseudomoments
where $H(x)=F(x\sigma )-\varPhi (x)$. The truncated pseudomoments are defined as
(“truncation from above”) and
(“truncation from below”).
Theorem 1.
Let the following conditions hold:
Then, for all $n\ge 2$,
\[\underset{x\in \mathbb{R}}{\sup }\big|\varPhi _{n}(x)\hspace{0.1667em}-\hspace{0.1667em}\varPhi (x)\big|\le 2{C_{m}^{(1)}}\frac{{\nu _{n}^{(1)}}(m)}{{n}^{\frac{m-1}{2}}}\hspace{0.1667em}+\hspace{0.1667em}2{C_{m}^{(2)}}\frac{{\nu _{n}^{(2)}}(m)}{{n}^{\frac{m-2}{2}}}\hspace{0.1667em}+\hspace{0.1667em}\frac{\sigma A}{\pi }{b}^{n-1}\hspace{0.1667em}+\hspace{0.1667em}\nu _{n}(m)\frac{4{e}^{\frac{3}{2}}}{\pi }\frac{{e}^{-\frac{n}{2}}}{n},\]
where
Corollary 1.
Let $\xi _{i}$ be a random variable with bounded density $p(x)\le A_{1}$. Suppose that condition (ii) of Theorem 1 holds. Then, for all $n\ge 3$,
\[\underset{x\in \mathbb{R}}{\sup }\big|\varPhi _{n}(x)\hspace{0.1667em}-\hspace{0.1667em}\varPhi (x)\big|\hspace{0.1667em}\le \hspace{0.1667em}2{C_{m}^{(1)}}\frac{{\nu _{n}^{(1)}}(m)}{{n}^{\frac{m-1}{2}}}\hspace{0.1667em}+\hspace{0.1667em}2{C_{m}^{(2)}}\frac{{\nu _{n}^{(2)}}(m)}{{n}^{\frac{m-2}{2}}}\hspace{0.1667em}+\hspace{0.1667em}2\sigma A_{1}{b_{1}^{n-2}}\hspace{0.1667em}+\hspace{0.1667em}\nu _{n}(m)\frac{4{e}^{\frac{3}{2}}}{\pi }\frac{{e}^{-\frac{n}{2}}}{n},\]
where $b_{1}=\exp \{-\frac{1}{96{A_{1}^{2}}{\sigma }^{2}{(2+\pi )}^{2}}\}<1$.
Note assumption (i) implies the existence of the density $p_{n}(x)$ of the random variable $S_{n}$. Also, let $\phi (x)$ be the density of the standard normal law.
Theorem 2.
Let the conditions of Theorem 1 hold. Then, for all $n\ge 2$,
\[\underset{x\in \mathbb{R}}{\sup }\big|p_{n}(x)-\phi (x)\big|\le {C_{m}^{(3)}}\frac{{\nu _{n}^{(1)}}(m)}{{n}^{\frac{m-1}{2}}}+{C_{m}^{(4)}}\frac{{\nu _{n}^{(2)}}(m)}{{n}^{\frac{m-2}{2}}}+{b}^{n-1}\frac{\sigma \sqrt{n}}{2\pi }A+\nu _{n}(m)\frac{{e}^{\frac{3}{2}}}{\pi }\frac{{e}^{-\frac{n}{2}}}{n},\]
where
3 Auxiliary results. Proofs of the main results
First we prove two auxiliary results. Denote $\omega (t)=|f(\frac{t}{\sigma })-{e}^{-\frac{{t}^{2}}{2}}|$.
Proof.
Recall that $f(t)={\int _{-\infty }^{\infty }}{e}^{itx}dF(x)$. Therefore, $f(\frac{t}{\sigma })={\int _{-\infty }^{\infty }}{e}^{\frac{itx}{\sigma }}dF(x)={\int _{-\infty }^{\infty }}{e}^{itx}dF(x\sigma )$. By the condition of the lemma, the pseudomoments up to order m equal zero. Hence, it is easy to deduce that
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \omega (t)& \displaystyle =\bigg|\int _{\mathbb{R}}{e}^{itx}dF(x\sigma )-\int _{\mathbb{R}}{e}^{itx}d\varPhi (x)\bigg|=\bigg|\int _{\mathbb{R}}{e}^{itx}d\big(F(x\sigma )-\varPhi (x)\big)\bigg|\\{} & \displaystyle =\Bigg|\int _{\mathbb{R}}\Bigg({e}^{itx}-\sum \limits_{k=0}^{m}\frac{{(itx)}^{k}}{k!}\Bigg)dH(x)\Bigg|\le \int _{\mathbb{R}}\Bigg|{e}^{itx}-\sum \limits_{k=0}^{m}\frac{{(itx)}^{k}}{k!}\Bigg|\big|dH(x)\big|.\end{array}\]
Using the inequality ([12], p. 372)
\[\bigg|{e}^{i\alpha }-1-\cdots -\frac{{(i\alpha )}^{m}}{m!}\bigg|\le \frac{{2}^{1-\delta }|\alpha {|}^{m+\delta }}{m!{(m+1)}^{\delta }},\hspace{1em}m=0,1,\dots ,\hspace{2.5pt}\delta \in [0,1],\]
with $\delta =1$ and $\delta =0$ we obtain
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \omega (t)& \displaystyle \le \int _{|x|\le \sigma \sqrt{n}}\Bigg|{e}^{itx}-\sum \limits_{k=0}^{m}\frac{{(itx)}^{k}}{k!}\Bigg|\big|dH(x)\big|+\int _{|x|>\sigma \sqrt{n}}\Bigg|{e}^{itx}-\sum \limits_{k=0}^{m}\frac{{(itx)}^{k}}{k!}\Bigg|\big|dH(x)\big|\\{} & \displaystyle \le \int _{|x|\le \sigma \sqrt{n}}\frac{|tx{|}^{m+1}}{(m+1)!}\big|dH(x)\big|\hspace{1em}+\int _{|x|>\sigma \sqrt{n}}\frac{2|tx{|}^{m}}{m!}\big|dH(x)\big|\\{} & \displaystyle =\frac{|t{|}^{m+1}}{(m+1)!}{\nu _{n}^{(1)}}(m)+\frac{2|t{|}^{m}}{m!}{\nu _{n}^{(2)}}(m).\end{array}\]
The lemma is proved. □Now, denote $T_{1}(n,m)=\sqrt{-2\ln (2e\nu _{n}(m))}$. Then, in turn, we have that $\nu _{n}(m)=\frac{1}{2e}\exp \{-\frac{1}{2}{T_{1}^{2}}(n,m)\}$. Note also that condition (ii) implies $T_{1}(n,m)\ge 1$.
Proof.
Evidently, $|f(\frac{t}{\sigma })|=|f(\frac{t}{\sigma })-{e}^{-\frac{{t}^{2}}{2}}+{e}^{-\frac{{t}^{2}}{2}}|\le {e}^{-\frac{{t}^{2}}{2}}+\omega (t)$. Now consider two cases.
1) Let $|t|\le T_{1}(n,m)$. Then we can deduce from Lemma 1 that
Consider the function $f_{1}(x)=\exp \{-\frac{{x}^{2}}{4}\}{x}^{m-1}$. It attains its maximal value at the point $x=\sqrt{2(m-1)}$, and this value equals
Furthermore,
(1)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \bigg|f\bigg(\frac{t}{\sigma }\bigg)\bigg|& \displaystyle \le {e}^{-\frac{{t}^{2}}{4}}\big({e}^{-\frac{{t}^{2}}{4}}+{e}^{\frac{{t}^{2}}{4}}\omega (t)\big)\\{} & \displaystyle \le {e}^{-\frac{{t}^{2}}{4}}\bigg(1+{e}^{\frac{{t}^{2}}{4}}\bigg(\frac{|t{|}^{m+1}}{(m+1)!}{\nu _{n}^{(1)}}(m)+\frac{2|t{|}^{m}}{m!}{\nu _{n}^{(2)}}(m)\bigg)\bigg)\\{} & \displaystyle \le {e}^{-\frac{{t}^{2}}{4}}\bigg(1+{e}^{\frac{{T_{1}^{2}}(n,m)}{4}}{t}^{2}\bigg(\frac{{T_{1}^{m-1}}(n,m)}{(m+1)!}{\nu _{n}^{(1)}}(m)+\frac{2{T_{1}^{m-2}}(n,m)}{m!}{\nu _{n}^{(2)}}(m)\bigg)\bigg)\\{} & \displaystyle \le {e}^{-\frac{{t}^{2}}{4}}\bigg(1+{t}^{2}{e}^{\frac{{T_{1}^{2}}(n,m)}{4}}\nu _{n}(m)\bigg(\frac{{T_{1}^{m-1}}(n,m)}{(m+1)!}+\frac{2{T_{1}^{m-2}}(n,m)}{m!}\bigg)\bigg)\\{} & \displaystyle ={e}^{-\frac{{t}^{2}}{4}}\bigg(1+\frac{1}{2e}{t}^{2}{e}^{-\frac{{T_{1}^{2}}(n,m)}{4}}\bigg(\frac{{T_{1}^{m-1}}(n,m)}{(m+1)!}+\frac{2{T_{1}^{m-2}}(n,m)}{m!}\bigg)\bigg).\end{array}\]
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \exp \bigg\{-\frac{m-1}{2}\bigg\}\frac{{(2(m-1))}^{\frac{m-1}{2}}}{(m+1)!}& \displaystyle \le \frac{\exp \{-\frac{m-1}{2}\}{(2(m-1))}^{\frac{m-1}{2}}}{m(m+1)\sqrt{2\pi (m-1)}{(m-1)}^{m-1}{e}^{-(m-1)}}\\{} & \displaystyle ={\bigg(\frac{2e}{m-1}\bigg)}^{\frac{m-1}{2}}\frac{1}{\sqrt{2\pi (m-1)}}\frac{1}{m(m+1)}\\{} & \displaystyle \le \frac{1}{m(m+1)}.\end{array}\]
The last fraction attains its maximal value at the point $m=3$. Therefore,
\[\exp \bigg\{-\frac{{T_{1}^{2}}(n,m)}{4}\bigg\}\frac{{T_{1}^{m-1}}(n,m)}{(m+1)!}\le \frac{1}{12}.\]
Similarly,
From (1) together with two last bounds it follows that
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \bigg|f\bigg(\frac{t}{\sigma }\bigg)\bigg|& \displaystyle \le {e}^{-\frac{{t}^{2}}{4}}\bigg(1+\frac{1}{2e}{t}^{2}\bigg(\frac{1}{12}+\frac{1}{3}\bigg)\bigg)\le {e}^{-\frac{{t}^{2}}{4}}\bigg(1+\frac{1}{12}{t}^{2}\bigg)\\{} & \displaystyle \le {e}^{-\frac{{t}^{2}}{4}}{e}^{\frac{{t}^{2}}{12}}\le {e}^{-\frac{{t}^{2}}{6}},\end{array}\]
and the proof of the first statement follows.2) Now, let $|t|>T_{1}(n,m)$. Then we get from Lemma 1 that
Recall that $T_{1}(n,m)>1$. Then $|t|>T_{1}(n,m)>1$, and from (2) we get that
whence the proof follows. □
(2)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \bigg|f\bigg(\frac{t}{\sigma }\bigg)\bigg|& \displaystyle \le {e}^{-\frac{{t}^{2}}{2}}+\omega (t)\\{} & \displaystyle \le {e}^{-\frac{{T_{1}^{2}}(n,m)}{2}}+\frac{|t{|}^{m+1}}{(m+1)!}{\nu _{n}^{(1)}}(m)+\frac{2|t{|}^{m}}{m!}{\nu _{n}^{(2)}}(m)\\{} & \displaystyle \le \nu _{n}(m)\bigg(2e+\frac{|t{|}^{m+1}}{(m+1)!}+\frac{2|t{|}^{m}}{m!}\bigg).\end{array}\]Now we are in position to prove the main results.
Proof of Theorem 1.
Let F and G be two distribution functions with characteristic functions f and g, respectively, and suppose that G has a density function, which we denote ${G^{\prime }}$. We shall use the following inequality from [4], p. 297:
Let $n\ge 2$. First, from the elementary inequality
and from Lemma 1 it follows that, for $t\le T_{1}(n,m)\sqrt{n}$,
Second, introduce the following notation:
\[\underset{x\in \mathbb{R}}{\sup }\big|F(x)-G(x)\big|\le \frac{2}{\pi }{\int _{0}^{T}}\big|f(t)-g(t)\big|\frac{dt}{t}+\frac{24\sup |{G}^{{^{\prime }}}|}{\pi T}.\]
Taking $F(x)=\varPhi _{n}(x)$ and $G(x)=\varPhi (x)$, we have
(3)
\[\rho _{n}:=\underset{x\in \mathbb{R}}{\sup }\big|\varPhi _{n}(x)-\varPhi (x)\big|\le \frac{2}{\pi }{\int _{0}^{T}}\bigg|{f}^{n}\bigg(\frac{t}{\sigma \sqrt{n}}\bigg)-{e}^{-\frac{{t}^{2}}{2}}\bigg|\frac{dt}{t}+\frac{24}{\pi \sqrt{2\pi }T}.\](4)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \bigg|{f}^{n}\bigg(\frac{t}{\sigma \sqrt{n}}\bigg)-{e}^{-\frac{{t}^{2}}{2}}\bigg|& \displaystyle \le \omega \bigg(\frac{t}{\sqrt{n}}\bigg)\sum \limits_{k=1}^{n}{\bigg|f\bigg(\frac{t}{\sigma \sqrt{n}}\bigg)\bigg|}^{k-1}{e}^{-\frac{{t}^{2}}{2n}(n-k)}\\{} & \displaystyle \le \omega \bigg(\frac{t}{\sqrt{n}}\bigg)\sum \limits_{k=1}^{n}{e}^{-\frac{{t}^{2}}{6}\frac{n-1}{n}}\le \omega \bigg(\frac{t}{\sqrt{n}}\bigg)n{e}^{-\frac{{t}^{2}}{12}}\\{} & \displaystyle \le n\bigg(\frac{|t{|}^{m+1}}{(m+1)!{n}^{\frac{m+1}{2}}}{\nu _{n}^{(1)}}(m)+\frac{2|t{|}^{m}}{m!{n}^{\frac{m}{2}}}{\nu _{n}^{(2)}}(m)\bigg)\exp \bigg\{-\frac{{t}^{2}}{12}\bigg\}\\{} & \displaystyle =\exp \bigg\{-\frac{{t}^{2}}{12}\bigg\}\bigg(\frac{|t{|}^{m+1}}{(m+1)!{n}^{\frac{m-1}{2}}}{\nu _{n}^{(1)}}(m)+\frac{2|t{|}^{m}}{m!{n}^{\frac{m-2}{2}}}{\nu _{n}^{(2)}}(m)\bigg).\end{array}\]
\[\begin{array}{r}\displaystyle {C_{m,n}^{(1)}}=\frac{{12}^{\frac{m-1}{2}}\varGamma (\frac{m+1}{2})}{2{n}^{\frac{m-1}{2}}(m+1)!},\hspace{2em}{C_{m,n}^{(2)}}=2{C_{m-1,n}^{(1)}},\\{} \displaystyle T_{2}(n,m)=\frac{1}{\sqrt{2\pi }({C_{m,n}^{(1)}}{\nu _{n}^{(1)}}(m)+{C_{m,n}^{(2)}}{\nu _{n}^{(2)}}(m))}.\end{array}\]
Then
Let $T_{3}(n,m)=(T_{1}(n,m)\sqrt{n})\wedge T_{2}(n,m)$. Then it follows from (3) and (5) that
Since $T_{3}(n,m)\le T_{1}(n,m)\sqrt{n}$, from (4) we get that
If $T_{3}(n,m)=T_{2}(n,m)$, then $I_{2}(n,m)=0$ and $I_{3}(n,m)=0$. Therefore, we consider the case $T_{3}(n,m)=T_{1}(n,m)\sqrt{n}$. Then
Finally, we bound $I_{3}(n,m)$. Note that $I_{3}(n,m)$ is nonzero only if $T_{1}(n,m)\sqrt{n}<T_{2}(n,m)$. Therefore,
Relations (6)–(10) supply the proof of Theorem 1. □
(6)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \rho _{n}& \displaystyle \le \frac{2}{\pi }{\int _{0}^{T_{3}(n,m)}}\bigg|{f}^{n}\bigg(\frac{t}{\sigma \sqrt{n}}\bigg)-{e}^{-\frac{{t}^{2}}{2}}\bigg|\frac{dt}{t}+\frac{2}{\pi }{\int _{T_{3}(n,m)}^{T_{2}(n,m)}}{\bigg|f\bigg(\frac{t}{\sigma \sqrt{n}}\bigg)\bigg|}^{n}\frac{dt}{t}\\{} & \displaystyle \hspace{1em}+\frac{2}{\pi }{\int _{T_{3}(n,m)}^{T_{2}(n,m)}}{e}^{-\frac{{t}^{2}}{2}}\frac{dt}{t}+\frac{24}{\pi \sqrt{2\pi }T_{2}(n,m)}\\{} & \displaystyle =I_{1}(n,m)+I_{2}(n,m)+I_{3}(n,m)+{C_{m}^{(1)}}\frac{{\nu _{n}^{(1)}}(m)}{{n}^{\frac{m-1}{2}}}+{C_{m}^{(2)}}\frac{{\nu _{n}^{(2)}}(m)}{{n}^{\frac{m-2}{2}}}.\end{array}\](7)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle I_{1}(n,m)& \displaystyle =\frac{2}{\pi }{\int _{0}^{T_{3}(n,m)}}\bigg|{f}^{n}\bigg(\frac{t}{\sigma \sqrt{n}}\bigg)-{e}^{-\frac{{t}^{2}}{2}}\bigg|\frac{dt}{t}\\{} & \displaystyle \le \frac{2}{\pi }{\int _{0}^{T_{3}(n,m)}}\bigg(\frac{{t}^{m}}{(m+1)!{n}^{\frac{m-1}{2}}}{\nu _{n}^{(1)}}(m)+\frac{2{t}^{m-1}}{m!{n}^{\frac{m-2}{2}}}{\nu _{n}^{(2)}}(m)\bigg){e}^{-\frac{{t}^{2}}{12}}dt\\{} & \displaystyle \le \frac{{12}^{\frac{m+1}{2}}\varGamma (\frac{m+1}{2})}{\pi (m+1)!{n}^{\frac{m-1}{2}}}{\nu _{n}^{(1)}}(m)+\frac{2\cdot {12}^{\frac{m}{2}}\varGamma (\frac{m}{2})}{\pi m!{n}^{\frac{m-2}{2}}}{\nu _{n}^{(2)}}(m)\\{} & \displaystyle ={C_{m}^{(1)}}\frac{{\nu _{n}^{(1)}}(m)}{{n}^{\frac{m-1}{2}}}+{C_{m}^{(2)}}\frac{{\nu _{n}^{(2)}}(m)}{{n}^{\frac{m-2}{2}}}.\end{array}\]
\[I_{2}(n,m)=\frac{2}{\pi }{\int _{T_{3}(n,m)}^{T_{2}(n,m)}}{\bigg|f\bigg(\frac{t}{\sigma \sqrt{n}}\bigg)\bigg|}^{n}\frac{dt}{t}=\frac{2}{\pi }{\int _{T_{1}(n,m)/\sigma }^{T_{2}(n,m)/\sigma \sqrt{n}}}{\big|f(t)\big|}^{n}\frac{dt}{t}.\]
Now we apply the following result of Statulevičius [8]: if a random variable with characteristic function $f(t)$ has a density $p(x)\le d<\infty $ and variance ${\sigma }^{2}$, then, for any $t\in \mathbb{R}$,
It follows from condition (i) that the density $p(x)$ of any $\xi _{n}$ can be obtained as the inverse Fourier transform $p(x)=\frac{1}{2\pi }\int _{\mathbb{R}}{e}^{-itx}f(t)dt$ and $p(x)\le \frac{1}{2\pi }{\int _{-\infty }^{\infty }}|f(t)|dt=\frac{A}{2\pi }$. Besides, the function $\frac{{t}^{2}}{{(2\sigma t+\pi )}^{2}}$ is increasing for $t>0$. Therefore, for $|t|\ge T_{1}(n,m)/\sigma $ (recall that $T_{1}(n,m)>1$),
\[\big|f(t)\big|\le \exp \bigg\{-\frac{{\pi }^{2}}{24{A}^{2}{\sigma }^{2}{(2+\pi )}^{2}}\bigg\}=:b,\]
and $0<b<1$. Then
(9)
\[I_{2}(n,m)=\frac{2}{\pi }{\int _{T_{1}(n,m)/\sigma }^{T_{2}(n,m)/\sigma \sqrt{n}}}{\big|f(t)\big|}^{n}\frac{dt}{t}\le \frac{2\sigma }{\pi }{b}^{n-1}{\int _{0}^{\infty }}\big|f(t)\big|dt=\frac{\sigma A}{\pi }{b}^{n-1}.\](10)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle I_{3}(n,m)& \displaystyle \le \frac{2}{\pi }{\int _{T_{1}(n,m)\sqrt{n}}^{\infty }}{e}^{-\frac{{t}^{2}}{2}}\frac{dt}{t}\le \frac{2}{\pi }\frac{{e}^{-\frac{n{T_{1}^{2}}(n,m)}{2}}}{n{T_{1}^{2}}(n,m)}\\{} & \displaystyle \le \frac{2{(2e\nu _{n}(m))}^{n}}{\pi n}\le \frac{4e\nu _{n}(m)}{\pi n}{\big(2e\nu _{n}(m)\big)}^{n-1}\le \nu _{n}(m)\frac{4e\cdot {e}^{-\frac{n-1}{2}}}{\pi n}\\{} & \displaystyle =\nu _{n}(m)\frac{4{e}^{\frac{3}{2}}}{\pi }\frac{{e}^{-\frac{n}{2}}}{n}.\end{array}\]Remark 1.
Let the following conditions hold: $\mu _{k}=0$, $k=3,\dots ,m$, $m\ge 3$. Then
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \underset{x\in \mathbb{R}}{\sup }\big|\varPhi _{1}(x)-\varPhi (x)\big|& \displaystyle =\underset{x\in \mathbb{R}}{\sup }\big|F(x\sigma )-\varPhi (x)\big|\\{} & \displaystyle \le \bigg(\frac{6}{\pi (m+1)!}+\frac{2}{\pi \sqrt{2\pi }}\bigg)\max \big(\nu _{1}(m),{\big(\nu _{1}(m)\big)}^{\frac{1}{m+2}}\big).\end{array}\]
Indeed, let $n=1$. Theorem is obvious when $\nu _{1}(m)>1$. Let $\nu _{1}(m)\le 1$. Put $T={(\nu _{1}(m))}^{-\frac{1}{m+2}}$ into (3). Then from Lemma 1 it follows that
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \rho _{1}& \displaystyle =\underset{x\in \mathbb{R}}{\sup }\big|\varPhi _{1}(x)-\varPhi (x)\big|=\underset{x\in \mathbb{R}}{\sup }\big|F(x\sigma )-\varPhi (x)\big|\\{} & \displaystyle \le \frac{2}{\pi }{\int _{0}^{T}}\bigg(\frac{|t{|}^{m+1}}{(m+1)!}{\nu _{1}^{(1)}}(m)+\frac{2|t{|}^{m}}{m!}{\nu _{1}^{(2)}}(m)\bigg)\frac{dt}{t}+\frac{24}{\pi \sqrt{2\pi }T}\\{} & \displaystyle \le \frac{2}{\pi }\bigg(\frac{{T}^{m+1}}{(m+1)\cdot (m+1)!}{\nu _{1}^{(1)}}(m)+\frac{2{T}^{m}}{m\cdot m!}{\nu _{1}^{(2)}}(m)\bigg)+\frac{24}{\pi \sqrt{2\pi }}{\big(\nu _{1}(m)\big)}^{\frac{1}{m+2}}\\{} & \displaystyle \le {\big(\nu _{1}(m)\big)}^{\frac{1}{m+2}}\bigg(\frac{2}{\pi }\frac{3}{(m+1)!}+\frac{24}{\pi \sqrt{2\pi }}\bigg).\end{array}\]
Proof of Corollary 1.
Proof is similar to that of Theorem 1. We apply inequality (8) and recall again that the function $\frac{{t}^{2}}{{(2\sigma t+\pi )}^{2}}$ is increasing for $t>0$. Therefore, for $|t|\ge T_{1}(n,m)/\sigma $ (recall that $T_{1}(n,m)>1$),
\[\big|f(t)\big|\le \exp \bigg\{-\frac{1}{96{A_{1}^{2}}{\sigma }^{2}{(2+\pi )}^{2}}\bigg\}=:b_{1},\]
and $0<b_{1}<1$. It follows from [2], p. 510, that ${\int _{-\infty }^{\infty }}|f(t){|}^{2}dt\le 2\pi A_{1}$. Therefore,
\[I_{2}(n,m)=\frac{2}{\pi }{\int _{T_{1}(n,m)/\sigma }^{T_{2}(n,m)/\sigma \sqrt{n}}}{\big|f(t)\big|}^{n}\frac{dt}{t}\le \frac{2\sigma }{\pi }{b_{1}^{n-2}}{\int _{0}^{\infty }}{\big|f(t)\big|}^{2}dt=2\sigma A_{1}{b_{1}^{n-2}}.\]
Corollary 1 is proved. □Proof of Theorem 2.
As it was mentioned before, condition (i) implies the existence of a density for the random variable $\xi _{k}$, so the random variable $S_{n}$ has the density
Since $\phi (x)=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{{x}^{2}}{2}}$ is the density of the standard normal law, we have $\phi (x)=\frac{1}{2\pi }{\int _{-\infty }^{\infty }}{e}^{-itx}{e}^{-\frac{{t}^{2}}{2}}dt$ and
From the conditions of the theorem, Lemmas 1 and 2, and from (4) $(n\ge 2)$ we obtain the following: for $|t|\le T_{1}(n,m)\sqrt{n}$ and $\nu _{n}(m)<\frac{1}{2}{e}^{-\frac{3}{2}}$,
From the conditions of the theorem, similarly to (9), we get
Relations (11)–(14) supply the proof of Theorem 2. □
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \big|p_{n}(x)-\phi (x)\big|& \displaystyle =\frac{1}{2\pi }\bigg|{\int _{-\infty }^{\infty }}{e}^{-itx}{f}^{n}\bigg(\frac{t}{\sigma \sqrt{n}}\bigg)dt-{\int _{-\infty }^{\infty }}{e}^{-itx}{e}^{-\frac{{t}^{2}}{2}}dt\bigg|\\{} & \displaystyle \le \frac{1}{2\pi }{\int _{-\infty }^{\infty }}\bigg|{f}^{n}\bigg(\frac{t}{\sigma \sqrt{n}}\bigg)-{e}^{-\frac{{t}^{2}}{2}}\bigg|dt.\end{array}\]
Therefore,
(11)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \big|p_{n}(x)-\phi (x)\big|& \displaystyle \le \frac{1}{2\pi }\int _{|t|\le T_{1}(n,m)\sqrt{n}}\bigg|{f}^{n}\bigg(\frac{t}{\sigma \sqrt{n}}\bigg)-{e}^{-\frac{{t}^{2}}{2}}\bigg|dt\\{} & \displaystyle \hspace{1em}+\frac{1}{2\pi }\int _{|t|>T_{1}(n,m)\sqrt{n}}{\bigg|f\bigg(\frac{t}{\sigma \sqrt{n}}\bigg)\bigg|}^{n}dt\\{} & \displaystyle \hspace{1em}+\frac{1}{2\pi }\int _{|t|>T_{1}(n,m)\sqrt{n}}{e}^{-\frac{{t}^{2}}{2}}dt\\{} & \displaystyle =I_{1}+I_{2}+I_{3}.\end{array}\](12)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle I_{1}& \displaystyle =\frac{1}{2\pi }\int _{|t|\le T_{1}(n,m)\sqrt{n}}\bigg|{f}^{n}\bigg(\frac{t}{\sigma \sqrt{n}}\bigg)-{e}^{-\frac{{t}^{2}}{2}}\bigg|dt\\{} & \displaystyle \le \frac{1}{2\pi }\int _{|t|\le T_{1}(n,m)\sqrt{n}}\bigg(\frac{|t{|}^{m+1}{\nu _{n}^{(1)}}(m)}{(m+1)!{n}^{\frac{m-1}{2}}}+\frac{2|t{|}^{m}{\nu _{n}^{(2)}}(m)}{m!{n}^{\frac{m}{2}-1}}\bigg){e}^{-\frac{{t}^{2}}{12}}dt\\{} & \displaystyle \le \frac{{12}^{\frac{m+2}{2}}\varGamma (\frac{m}{2}+1)}{4\pi (m+1)!}\frac{{\nu _{n}^{(1)}}(m)}{{n}^{\frac{m-1}{2}}}+\frac{2\cdot {12}^{\frac{m+1}{2}}\varGamma (\frac{m-1}{2})}{4\pi m!}\frac{{\nu _{n}^{(2)}}(m)}{{n}^{\frac{m-2}{2}}}\\{} & \displaystyle ={C_{m}^{(3)}}\frac{{\nu _{n}^{(1)}}(m)}{{n}^{\frac{m-1}{2}}}+{C_{m}^{(4)}}\frac{{\nu _{n}^{(2)}}(m)}{{n}^{\frac{m-2}{2}}}.\end{array}\](13)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle I_{2}& \displaystyle =\frac{1}{2\pi }\int _{|t|>T_{1}(n,m)\sqrt{n}}{\bigg|f\bigg(\frac{t}{\sigma \sqrt{n}}\bigg)\bigg|}^{n}dt\\{} & \displaystyle =\frac{\sigma \sqrt{n}}{2\pi }\int _{|z|>T_{1}(n,m)/\sigma }{\big|f(z)\big|}^{n}dz\le {b}^{n-1}\frac{\sigma \sqrt{n}}{2\pi }A.\end{array}\](14)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle I_{3}& \displaystyle =\frac{1}{2\pi }\int _{|t|>T_{1}(n,m)\sqrt{n}}{e}^{-\frac{{t}^{2}}{2}}dt\le \frac{1}{2\pi }{\int _{T_{1}(n,m)\sqrt{n}}^{\infty }}{e}^{-\frac{{t}^{2}}{2}}dt\\{} & \displaystyle \le \frac{{e}^{-\frac{n{T_{1}^{2}}(n,m)}{2}}}{2\pi \sqrt{n}T_{1}(n,m)}\le \frac{{(2e\nu _{n}(m))}^{n}}{2\pi \sqrt{n}}\\{} & \displaystyle \le \frac{e\nu _{n}(m)}{\pi \sqrt{n}}{e}^{-\frac{n-1}{2}}=\nu _{n}(m)\frac{{e}^{\frac{3}{2}}}{\pi }\frac{{e}^{-\frac{n}{2}}}{\sqrt{n}}.\end{array}\]4 Example
We give an example of application of Theorem 1. It is similar to the example of [12], p. 375, where the discrete distribution was considered. Define the distribution function $F(x)$ as
This equation has a unique solution because ${\int _{0}^{\epsilon }}{x}^{2}d\varPhi (x)\le \frac{{\epsilon }^{2}}{3}(\varPhi (\epsilon )-\frac{1}{2})$. Indeed, on one hand,
\[F(x)=\left\{\begin{array}{l@{\hskip10.0pt}l}\varPhi (x)\hspace{1em}& \text{if}\hspace{2.5pt}|x|\ge \epsilon ,\\{} \varPhi (-\epsilon )\hspace{1em}& \text{if}\hspace{2.5pt}-\epsilon <x<-\theta \epsilon ,\\{} \varPhi (\epsilon )\hspace{1em}& \text{if}\hspace{2.5pt}\theta \epsilon <x<\epsilon ,\\{} \frac{1}{2}+\frac{\varPhi (\epsilon )-\frac{1}{2}}{\theta \epsilon }x\hspace{1em}& \text{if}\hspace{2.5pt}|x|\le \theta \epsilon ,\end{array}\right.\]
where $0<\epsilon <1$, and $0<\theta <1$ is the root of the equation
(15)
\[{\int _{0}^{\epsilon }}{x}^{2}d\varPhi (x)=\frac{{(\theta \epsilon )}^{2}}{3}\bigg(\varPhi (\epsilon )-\frac{1}{2}\bigg).\]
\[\varphi (x)=\frac{1}{\sqrt{2\pi }}\bigg(1-\frac{{x}^{2}}{2}+\frac{{x}^{4}}{{2}^{2}2!}-\frac{{x}^{6}}{{2}^{3}3!}+\frac{{x}^{8}}{{2}^{4}4!}-\cdots \hspace{0.1667em}\bigg),\]
and therefore,
\[\varPhi (\epsilon )-\frac{1}{2}={\int _{0}^{\epsilon }}\varphi (x)dx=\frac{1}{\sqrt{2\pi }}\bigg(\epsilon -\frac{{\epsilon }^{3}}{6}+\frac{{\epsilon }^{5}}{40}-\frac{{\epsilon }^{7}}{7\cdot {2}^{3}3!}+\cdots \hspace{0.1667em}\bigg).\]
So
\[\varPhi (\epsilon )-\frac{1}{2}\ge \frac{1}{\sqrt{2\pi }}\bigg(\epsilon -\frac{{\epsilon }^{3}}{6}\bigg).\]
On the other hand,
\[\begin{array}{r@{\hskip0pt}l}\displaystyle {\int _{0}^{\epsilon }}{x}^{2}d\varPhi (x)& \displaystyle =\frac{1}{\sqrt{2\pi }}{\int _{0}^{\epsilon }}\bigg({x}^{2}-\frac{{x}^{4}}{2}+\frac{{x}^{6}}{{2}^{2}2!}-\frac{{x}^{8}}{{2}^{3}3!}+\cdots \hspace{0.1667em}\bigg)dx\\{} & \displaystyle =\frac{1}{\sqrt{2\pi }}\bigg(\frac{{\epsilon }^{3}}{3}-\frac{{\epsilon }^{5}}{10}+\frac{{\epsilon }^{7}}{7\cdot {2}^{2}2!}-\frac{{\epsilon }^{9}}{9\cdot {2}^{3}3!}+\cdots \hspace{0.1667em}\bigg)\\{} & \displaystyle \le \frac{1}{\sqrt{2\pi }}\bigg(\frac{{\epsilon }^{3}}{3}-\frac{{\epsilon }^{5}}{10}+\frac{{\epsilon }^{7}}{56}\bigg)\\{} & \displaystyle \le \frac{1}{\sqrt{2\pi }}\bigg(\frac{{\epsilon }^{3}}{3}-\frac{{\epsilon }^{5}}{10}+\frac{{\epsilon }^{5}}{56}\bigg)=\frac{1}{\sqrt{2\pi }}{\epsilon }^{3}\bigg(\frac{1}{3}-\frac{23}{280}{\epsilon }^{2}\bigg)\\{} & \displaystyle =\frac{1}{\sqrt{2\pi }}\frac{{\epsilon }^{3}}{3}\bigg(1-\frac{69}{280}{\epsilon }^{2}\bigg)\le \frac{1}{\sqrt{2\pi }}\frac{{\epsilon }^{3}}{3}\bigg(1-\frac{{\epsilon }^{2}}{6}\bigg)\le \frac{{\epsilon }^{2}}{3}\bigg(\varPhi (\epsilon )-\frac{1}{2}\bigg),\end{array}\]
and we immediately get that
It is obvious that the density function is bounded. Moreover, $F(x)$ is symmetric. Therefore, $\mu _{1}=0$ and $\mu _{3}=0$. Furthermore, taking into account (15), consider
\[\begin{array}{r@{\hskip0pt}l}\displaystyle {\sigma }^{2}& \displaystyle ={\int _{-\infty }^{\infty }}{x}^{2}dF(x)=\int _{|x|\ge \epsilon }{x}^{2}d\varPhi (x)+{\int _{-\theta \epsilon }^{\theta \epsilon }}{x}^{2}\frac{\varPhi (\epsilon )-\frac{1}{2}}{\theta \epsilon }dx\\{} & \displaystyle =\int _{|x|\ge \epsilon }{x}^{2}d\varPhi (x)+\frac{\varPhi (\epsilon )-\frac{1}{2}}{\theta \epsilon }\frac{2}{3}{(\theta \epsilon )}^{3}={\int _{-\infty }^{\infty }}{x}^{2}d\varPhi (x)=1.\end{array}\]
This means that $\mu _{2}=0$. Consider further the pseudomoments
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \nu _{4}& \displaystyle ={\int _{-\infty }^{\infty }}{x}^{4}\big|d\big(F(x)-\varPhi (x)\big)\big|={\int _{-\epsilon }^{\epsilon }}{x}^{4}\big|d\big(F(x)-\varPhi (x)\big)\big|\\{} & \displaystyle \le {\epsilon }^{4}{\int _{-\epsilon }^{\epsilon }}\big|d\big(F(x)-\varPhi (x)\big)\big|\le {\epsilon }^{4}4\bigg(\varPhi (\epsilon )-\frac{1}{2}\bigg),\\{} \displaystyle {\nu _{n}^{(1)}}(3)& \displaystyle =\int _{|x|\le \sigma \sqrt{n}}{x}^{4}\big|d\big(F(x)-\varPhi (x)\big)\big|={\int _{-\epsilon }^{\epsilon }}{x}^{4}\big|d\big(F(x)-\varPhi (x)\big)\big|,\end{array}\]
where ϵ can be chosen so that $\nu _{4}\le \frac{1}{2}{e}^{-\frac{3}{2}}$ and ${\nu _{n}^{(1)}}(3)\le \frac{1}{2}{e}^{-\frac{3}{2}}$. Then
\[{\nu _{n}^{(2)}}(3)=\int _{|x|>\sigma \sqrt{n}}|x{|}^{3}\big|d\big(F(x)-\varPhi (x)\big)\big|=0.\]
Hence, condition (ii) of Theorem 1 holds. Therefore, the function $F(x)$ satisfies the conditions of Theorem 1 with $m=3$.
