Local limit theorems without assuming finite third moment

Abstract

One of the most fundamental probabilities is the probability at a particular point. The local limit theorem is the well-known theorem that estimates this probability. In this paper, we estimate this probability by the density function of normal distribution in the case of lattice integer-valued random variables. Our technique is the characteristic function method. We complete to relax the third moment condition of Siripraparat and Neammanee (J. Inequal. Appl. 2021:57, 2021) and the references therein and also obtain explicit constants of the error bound.

1 Introduction and main results

Let X be an integer-valued random variable. One of the most fundamental probabilities is the probability at a particular point, i.e., $P(X = k)$ for some $k\in \mathbb{Z}$. The local limit theorem is one of the theorems that estimate this probability and describe how $P(X=k)$ approaches the normal density, $\frac {1}{\sigma \sqrt{2\pi}}e^{-\frac{(k-\mu )^{2}}{2\sigma ^{2}}}$ where μ and $\sigma ^{2}$ are mean and variance of X, respectively. There are two well-known techniques for deriving this theorem: the method of characteristic function and the Bernoulli part extraction method. The characteristic function method is to estimate the bound for the characteristic function of a random variable. This method has been used in a number of studies such as in the case of bounded random variables (see [3–6], and [7] for examples) and in the case of lattice random variables (see [7–9], and [10] for examples).

Let $X_{1},X_{2},\ldots , X_{n}$ be independent integer-valued random variables with mean $\mu _{j}$ and variance $\sigma _{j}^{2}$ for all $j=1,2,\ldots ,n$. Then let

$$ S_{n} = \sum_{j=1}^{n}X_{j}, \qquad \mu = \sum_{j=1}^{n} \mu _{j}, \qquad \sigma ^{2} = \sum _{j=1}^{n}\sigma _{j}^{2}. $$

If $P(X_{j}=1)=p_{j}=1-P(X_{j}=0)$, then $X_{j}$ is called a Bernoulli random variable with parameter $p_{j}$ and $S_{n}$ is said to be a Poisson binomial random variable. In addition, when we provide $p_{1} = p_{2} = \cdots = p_{n} = p$, we call $S_{n}$ a binomial random variable with parameters n and p and use the notation $S_{n}\sim B(n,p)$. The first local limit theorem was proved by De Moivre and Laplace ([11], 1754) for a binomial random variable. We call X a lattice random variable with parameter $(a,d)$ if its values belong to $\mathcal{L}(a,d) = \{ a + md : m \in \mathbb{Z} \}$, where a and $d>0$ are integers. In addition, d is said to be maximal if there are no other numbers $a'$ and $d' > d$ for which $P(X\in \mathcal{L}(a',d')) = 1$, and we call X a maximal lattice random variable with parameter $(a,d)$. Observe that the Bernoulli random variable is a maximal lattice random variable with parameter $(0,1)$. In the case that $X_{j}$s are common lattice $\mathcal{L}(a,d)$ and identically distributed, Ibragimov and Linnik [12] gave the rate of convergence $O ( \frac {1}{n^{\frac{1}{2}+\alpha}} )$, where $0 < \alpha < \frac{1}{2}$ in 1971. For further information, they showed that if d is maximal and

$$ \int _{|x|\geq u} x^{2}F(dx) = O \biggl(\frac{1}{u^{2\alpha}} \biggr)\quad \text{as } u \to \infty , $$

where F the distribution function of $X_{1}$, then

$$ \sup_{k\in \mathbb{Z}} \biggl\vert \sqrt{n}P(S_{n}=na+kd) - \frac{d}{\sqrt{2\pi}\sigma _{1}}e^{- \frac{(na+kd-n\mu _{1})^{2}}{2n\sigma _{1}^{2}}} \biggr\vert = O \biggl( \frac{1}{n^{\alpha}} \biggr), \quad 0 < \alpha < \frac{1}{2}. $$

(1.1)

A few years later, Petrov [13] proved that if $E|X_{1}|^{3} < \infty$, then (1.1) holds with $\alpha = \frac{1}{2}$. Moreover, for the case that $X_{j}$s are nonidentically distributed lattice random variables with parameter $(0,1)$ that satisfy the third moment condition and some properties, Petrov [13] gave the following result:

$$\begin{aligned} \sup_{k\in \mathbb{Z}} \biggl\vert P(S_{n}=k) - \frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{(k-\mu )^{2}}{2\sigma ^{2}}} \biggr\vert \leq \frac{C}{\sigma ^{2}}. \end{aligned}$$

The previous studies had not given explicit constants of the error bound until Korolev and Zhukov ([14], 2000). In 2017, Giuliano and Weber [15] used the Bernoulli part extraction method to give an error bound with explicit constants in the case of nonidentically distributed square integrable random variables taking values in a common lattice $\mathcal{L}(a,d)$. By assuming finite third moment, Siripraparat and Neammanee [1] used the characteristic function technique to illustrate the rate of convergence to $O ( \frac {1}{\sigma ^{2}} )$ in 2021. For a special case, one can see [2] and [16] in the case of Poisson binomial and binomial, respectively.

In this paper, we relax the third moment condition to find the local limit theorems for sums of independent lattice integer-valued random variables and also give explicit constants of the error bound. Our technique is the characteristic function method inspired by Petrov [13] and Siripraparat and Neammanee [1]. Throughout this paper, let $X_{1},X_{2},\ldots ,X_{n}$ be independent common lattice random variables with parameter $(a,d)$ such that $E|X_{j}|^{2+\alpha} < \infty$, where $0<\alpha < 1$, for $j=1,2,\ldots ,n$, and let $S_{n} = \sum_{j=1}^{n}X_{j}$ with mean μ and variance $\sigma ^{2}$. The following are our main results.

Theorem 1.1

Let $\beta = \sum_{j=1}^{n}\beta _{j}$, where $\beta _{j} = 2\sum_{m=-\infty}^{\infty} p_{jm}p_{j(m+1)}$ and $p_{jm} = P(X_{j}= a+md)$. If $\beta >0$ and $\sigma ^{2}>d^{2}$, then

$$ \begin{aligned} &\sup_{k\in \mathbb{Z}} \biggl\vert P(S_{n} = na + kd) - \frac{d}{\sigma \sqrt{2\pi}}e^{- \frac{(na +kd -\mu )^{2}}{2\sigma ^{2}}} \biggr\vert \\ &\quad \leq \frac{0.0020d^{\frac{(2+\alpha )(1+\alpha )}{2}}}{ (\sum_{j=1}^{n}E \vert X_{j}-a \vert ^{2+\alpha} )^{\frac{1+\alpha}{2}}} + \frac{4.6171\cdot 3^{\frac{1}{\alpha}}d^{\frac{\alpha ^{2}-\alpha +2}{2}} (\sum_{j=1}^{n}E \vert X_{j}-a \vert ^{2+\alpha} )^{\frac{3-\alpha}{2}}}{\sigma ^{4}} \\ &\quad \quad {}+ \frac{0.3184}{\sigma ^{2}\tau}e^{-\frac{\sigma ^{2}\tau ^{2}}{2}} + \frac{1.5708}{\tau \beta} e^{-\frac{\tau ^{2}\beta}{\pi ^{2}}}, \end{aligned} $$

where $\tau = \frac{d}{3^{\frac{1}{\alpha}} (\sum_{j=1}^{n}E|X_{j}-a|^{2+\alpha} )^{\frac{1}{2+\alpha}}}$.

Furthermore, if $X_{1},X_{2},\ldots ,X_{n}$ are identically distributed and $\beta _{1} >0$, then

$$\begin{aligned} \sup_{k\in \mathbb{Z}} \biggl\vert P(S_{n} = na + kd) - \frac{d}{\sigma _{1} \sqrt{2n\pi}}e^{- \frac{(na + kd-n\mu _{1})^{2}}{2n\sigma _{1}^{2}}} \biggr\vert \leq \frac{C_{1}}{n^{\frac{1+\alpha}{2}}}, \end{aligned}$$

where

$$\begin{aligned} C_{1} & = \frac{0.0020d^{\frac{(2+\alpha )(1+\alpha )}{2}}}{(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{1+\alpha}{2}}} + \frac{4.6171\cdot 3^{\frac{1}{\alpha}}d^{\frac{\alpha ^{2}-\alpha +2}{2}}(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{3-\alpha}{2}}}{\sigma _{1}^{4}} \\ & \quad {}+ \frac{0.6368\cdot 3^{\frac{3}{\alpha}}\,d(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{3}{2+\alpha}}}{\sigma _{1}^{4}} + \frac{15.5032\cdot 3^{\frac{3}{\alpha}}(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{3}{2+\alpha}}}{d^{3}\beta _{1}^{2} }. \end{aligned}$$

We note that $\beta _{j} > 0$ if $X_{j}$ is a maximal lattice random variable. So, we can apply this result when d is maximal.

Theorem 1.2

Let $\upsilon := \min_{1\leq j \leq n}\upsilon _{j}$, where $\upsilon _{j} = 2\sum_{m=-\infty}^{\infty}p_{jm}p_{j(m+j)}$. If $\upsilon _{j} > 0$ for all $j=1,2,\ldots ,n$ and $\sigma ^{2}>d^{2}$, then

$$ \begin{aligned} &\sup_{k\in \mathbb{Z}} \biggl\vert P(S_{n} = na + kd) - \frac{d}{\sigma \sqrt{2\pi}}e^{- \frac{(na +kd -\mu )^{2}}{2\sigma ^{2}}} \biggr\vert \\ &\quad \leq \frac{0.0020d^{\frac{(2+\alpha )(1+\alpha )}{2}}}{ (\sum_{j=1}^{n}E \vert X_{j}-a \vert ^{2+\alpha} )^{\frac{1+\alpha}{2}}} + \frac{4.6171\cdot 3^{\frac{1}{\alpha}}d^{\frac{\alpha ^{2}-\alpha +2}{2}} (\sum_{j=1}^{n}E \vert X_{j}-a \vert ^{2+\alpha} )^{\frac{3-\alpha}{2}}}{\sigma ^{4}} \\ &\quad \quad {}+ \frac{0.3184}{\sigma ^{2}\tau}e^{-\frac{\sigma ^{2}\tau ^{2}}{2}} + \exp \biggl(-\frac{n\upsilon}{4} \min \biggl( 1, \biggl( \frac{n\tau}{2\pi} \biggr)^{2} \biggr) \biggr), \end{aligned} $$

where $\tau = \frac{d}{3^{\frac{1}{\alpha}} (\sum_{j=1}^{n}E|X_{j}-a|^{2+\alpha} )^{\frac{1}{2+\alpha}}}$.

Furthermore, if $X_{1},X_{2},\ldots ,X_{n}$ are identically distributed and $\upsilon _{j} > 0$ for all $j=1,2,\ldots ,n$, then for $n\geq ( \frac{2\pi \cdot 3^{\frac{1}{\alpha}}(E|X_{1}-a|^{2+\alpha})^{\frac{1}{2+\alpha}}}{d} )^{\frac{2+\alpha}{1+\alpha}}$,

$$ \begin{aligned} \sup_{k\in \mathbb{Z}} \biggl\vert P(S_{n} = na + kd) - \frac{d}{\sigma _{1} \sqrt{2n\pi}}e^{- \frac{(na + kd-n\mu _{1})^{2}}{2n\sigma _{1}^{2}}} \biggr\vert \leq \frac{C_{2}}{n^{\frac{1+\alpha}{2}}} + e^{- \frac{n\upsilon}{4}}, \end{aligned} $$

where

$$\begin{aligned} C_{2} & = \frac{0.0020d^{\frac{(2+\alpha )(1+\alpha )}{2}}}{(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{1+\alpha}{2}}} + \frac{4.6171\cdot 3^{\frac{1}{\alpha}}d^{\frac{\alpha ^{2}-\alpha +2}{2}}(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{3-\alpha}{2}}}{\sigma _{1}^{4}} \\ & \quad {}+ \frac{0.6368\cdot 3^{\frac{3}{\alpha}}\,d(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{3}{2+\alpha}}}{\sigma _{1}^{4}}. \end{aligned}$$

We organize this paper as follows: First, we give auxiliary results in Sect. 2 that will be used to prove the main theorems in Sect. 3. Finally, we give some examples in Sect. 4.

2 Auxiliary results

In the following lemmas, we use an idea from [17] to give bounds of a characteristic function to prove Theorem 1.1.

Lemma 2.1

Let X be any integer-valued random variable with mean $\mu _{X}$, variance $\sigma _{X}^{2}$, and characteristic function $\psi _{X}$. If $E|X|^{2+\alpha} < \infty$ for some $0 < \alpha < 1$, then there exists a function $g_{X}$ such that, for all $|t|\leq (\frac{1}{3E|X|^{\alpha}} )^{\frac{1}{\alpha}}$,

$(i)$:

$|\psi _{X}(t)| \geq \frac{1}{3}$ and

$(\mathit{ii})$:

$\psi _{X}(t) = \exp \{i\mu _{X} t - \frac{1}{2}\sigma _{X}^{2}t^{2} + \int _{0}^{t}\frac{g_{X}(s)}{\psi _{X}(s)}\,\mathrm{d}s \}$ and $\int _{0}^{t} | \frac{g_{X}(s)}{\psi _{X}(s)} |\,\mathrm{d}s \leq 9E|X|^{2+\alpha}|t|^{2+\alpha}$.

Proof

$(i)$ Using the fact that for $x\in \mathbb{R}$, $e^{ix} = 1 + 2^{1-\alpha}|x|^{\alpha}\Theta$ for some complex function Θ such that $|\Theta | \leq 1$ ([17], p. 359), we get that

$$\begin{aligned} Ee^{itX} = E\bigl(1+\Theta _{1}2^{1-\alpha} \vert tX \vert ^{\alpha}\bigr) = 1 + 2^{1- \alpha}E\bigl(\Theta _{1} \vert X \vert ^{\alpha}\bigr) \vert t \vert ^{\alpha}, \end{aligned}$$

(2.1)

where $\Theta _{1}$ is a complex random variable such that $|\Theta _{1}| \leq 1$. From this fact and the inequality $|z_{1} +z_{2}| \geq |z_{1}|-|z_{2}|$ for complex numbers $z_{1}$ and $z_{2}$, we can see that

$$\begin{aligned} \bigl\vert Ee^{itX} \bigr\vert & = \bigl\vert 1 + 2^{1-\alpha}E\bigl(\Theta _{1} \vert X \vert ^{\alpha} \bigr) \vert t \vert ^{ \alpha } \bigr\vert \\ & \geq 1 - 2^{1-\alpha}E\bigl( \vert \Theta _{1} \vert \vert X \vert ^{\alpha}\bigr) \vert t \vert ^{\alpha } \\ & \geq 1 - 2^{1-\alpha}E \vert X \vert ^{\alpha} \vert t \vert ^{\alpha } \\ & \geq 1 - 2E \vert X \vert ^{\alpha} \vert t \vert ^{\alpha}. \end{aligned}$$

Then, for all $|t|\leq (\frac{1}{3E|X|^{\alpha}} )^{\frac{1}{\alpha}}$, we have $|\psi _{X}(t)| = |Ee^{itX}| \geq \frac{1}{3}$.

$(\mathit{ii})$ Let $t\in \mathbb{R}$ be such that $|t| \leq (\frac{1}{3E|X|^{\alpha}} )^{\frac{1}{\alpha}}$. Since $\psi _{X}(t) = Ee^{itX}$, we obtain $\psi _{X}'(t) = iE(Xe^{itX})$, which implies that

$$\begin{aligned} \psi _{X}'(t) & = \biggl(\bigl(i\mu _{X}- \sigma _{X}^{2}t\bigr) + \frac{g_{X}(t)}{\psi _{X}(t)}\biggr)\psi _{X}(t), \end{aligned}$$

where

$$\begin{aligned} g_{X}(t) & = - \bigl(i\mu _{X}-\sigma _{X}^{2}t\bigr)Ee^{itX} + iE\bigl(Xe^{itX} \bigr). \end{aligned}$$

Hence

$$\begin{aligned} \frac{\psi _{X}'(t)}{\psi _{X}(t)} & = i\mu _{X}-\sigma _{X}^{2}t + \frac{g_{X}(t)}{\psi _{X}(t)} \end{aligned}$$

and then

$$\begin{aligned} \ln \psi _{X}(t) = \int _{0}^{t} \frac{\psi _{X}'(s)}{\psi _{X}(s)} \,\mathrm{d}s = i \mu _{X} t - \frac{1}{2}\sigma _{X}^{2}t^{2} + \int _{0}^{t} \frac{g_{X}(s)}{\psi _{X}(s)}\,\mathrm{d}s, \end{aligned}$$

that is,

$$\begin{aligned} \psi _{X}(t) = \exp \biggl\{ i\mu _{X} t - \frac{1}{2}\sigma _{X}^{2}t^{2} + \int _{0}^{t}\frac{g_{X}(s)}{\psi _{X}(s)}\,\mathrm{d}s \biggr\} . \end{aligned}$$

From the fact that for $x\in \mathbb{R}$, $e^{ix} = 1 + ix + \frac{2^{1-\alpha}}{1+\alpha}|x|^{1+\alpha}\Theta$ for some complex function Θ such that $|\Theta | \leq 1$ ([17], p. 359), we have that

$$\begin{aligned} &Ee^{itX} = 1 + itEX + \frac{2^{1-\alpha}}{1+\alpha}E\bigl(\Theta _{2} \vert X \vert ^{1+ \alpha}\bigr) \vert t \vert ^{1+\alpha} \end{aligned}$$

(2.2)

$$\begin{aligned} &\text{and }\quad iE\bigl(Xe^{itX}\bigr) = i\mu _{X} - tEX^{2} + \frac{2^{1-\alpha}}{1+\alpha}E\bigl(i\Theta _{2} \vert X \vert ^{2+\alpha}\bigr) \vert t \vert ^{1+ \alpha}, \end{aligned}$$

(2.3)

where $\Theta _{2}$ is a complex random variable such that $|\Theta _{2}| \leq 1$. From (2.1)–(2.3), we have

$$\begin{aligned} g_{X}(t) & = -i\mu _{X}Ee^{itX} + \sigma _{X}^{2}tEe^{itX} + iE\bigl(Xe^{itX} \bigr) \\ &= \frac{2^{1-\alpha}}{1+\alpha}\mu _{X} E\bigl(i\Theta _{2} \vert X \vert ^{1+\alpha}\bigr) \vert t \vert ^{1+ \alpha} + 2^{1-\alpha} \sigma _{X}^{2}E\bigl(\Theta _{1} \vert X \vert ^{\alpha}\bigr) \vert t \vert ^{1+ \alpha} \\ & \quad {}+ \frac{2^{1-\alpha}}{1+\alpha}E\bigl(i\Theta _{2} \vert X \vert ^{2+\alpha}\bigr) \vert t \vert ^{1+ \alpha}. \end{aligned}$$

(2.4)

According to Lyapunov’s inequality: $(E|X|^{r})^{\frac{1}{r}} \leq (E|X|^{s})^{\frac{1}{s}}$, where $0 < r \leq s$, we have that $E|X| \leq (E|X|^{2+\alpha})^{\frac{1}{2+\alpha}}$ and $E|X|^{1+\alpha} \leq (E|X|^{2+\alpha})^{\frac{1+\alpha}{2+\alpha}}$, which imply that

$$ \mu _{X}E \vert X \vert \leq E \vert X \vert E \vert X \vert ^{1+\alpha} \leq E \vert X \vert ^{2+\alpha}. $$

We can use the same technique to show that

$$\begin{aligned} \sigma _{X}^{2}E \vert X \vert ^{\alpha }& \leq EX^{2}E \vert X \vert ^{\alpha }\leq E \vert X \vert ^{2+ \alpha}. \end{aligned}$$

From these facts and (2.4), we have

$$\begin{aligned} \bigl\vert g_{X}(t) \bigr\vert & \leq \frac{2^{1-\alpha}}{1+\alpha}\mu _{X} E\bigl( \vert i \Theta _{2} \vert \vert X \vert ^{1+\alpha}\bigr) \vert t \vert ^{1+\alpha} + 2^{1-\alpha} \sigma _{X}^{2}E\bigl( \vert \Theta _{1} \vert \vert X \vert ^{\alpha}\bigr) \vert t \vert ^{1+\alpha} \\ & \quad {}+ \frac{2^{1-\alpha}}{1+\alpha}E\bigl( \vert i\Theta _{2} \vert \vert X \vert ^{2+\alpha}\bigr) \vert t \vert ^{1+ \alpha} \\ & \leq \frac{2^{1-\alpha}}{1+\alpha}\mu _{X} E\bigl( \vert X \vert ^{1+\alpha}\bigr) \vert t \vert ^{1+ \alpha} + 2^{1-\alpha} \sigma _{X}^{2}E\bigl( \vert X \vert ^{\alpha}\bigr) \vert t \vert ^{1+\alpha} \\ & \quad {}+ \frac{2^{1-\alpha}}{1+\alpha}E\bigl( \vert X \vert ^{2+\alpha}\bigr) \vert t \vert ^{1+\alpha} \\ & \leq \frac{2^{1-\alpha}}{1+\alpha} E\bigl( \vert X \vert ^{2+\alpha}\bigr) \vert t \vert ^{1+\alpha} + 2^{1-\alpha} E\bigl( \vert X \vert ^{2+\alpha}\bigr) \vert t \vert ^{1+\alpha} \\ & \quad {}+ \frac{2^{1-\alpha}}{1+\alpha}E\bigl( \vert X \vert ^{2+\alpha}\bigr) \vert t \vert ^{1+\alpha} \\ & = \biggl( 2^{1-\alpha} + \frac{2^{2-\alpha}}{1+\alpha} \biggr)E \vert X \vert ^{2+ \alpha} \vert t \vert ^{1+\alpha} \\ & \leq 6E \vert X \vert ^{2+\alpha} \vert t \vert ^{1+\alpha}. \end{aligned}$$

(2.5)

Hence we can conclude from $(i)$ and (2.5) that for all $|t|\leq (\frac{1}{3E|X|^{\alpha}} )^{\frac{1}{\alpha}}$ we have

$$\begin{aligned} \biggl\vert \frac{g_{X}(t)}{\psi _{X}(t)} \biggr\vert &\leq 18E \vert X \vert ^{2+\alpha} \vert t \vert ^{1+ \alpha}, \end{aligned}$$

which implies that

$$\begin{aligned} \int _{0}^{t} \biggl\vert \frac{g_{X}(s)}{\psi _{X}(s)} \biggr\vert \,\mathrm{d}s &\leq \frac{18}{2+\alpha}E \vert X \vert ^{2+\alpha} \vert t \vert ^{2+\alpha} \leq 9E \vert X \vert ^{2+\alpha} \vert t \vert ^{2+\alpha}. \end{aligned}$$

 □

Lemma 2.2

Let $\tau = \frac{1}{3^{\frac{1}{\alpha}}} ( \frac{1}{\sum_{j=1}^{n}E|X_{j}|^{2+\alpha}} )^{ \frac{1}{2+\alpha}}$. Then

$$ \bigl\vert \psi (t) - e^{it\mu -\frac{1}{2}\sigma ^{2}t^{2}} \bigr\vert \leq 12.5606 \sum _{j=1}^{n}E \vert X_{j} \vert ^{2+\alpha} \vert t \vert ^{2+\alpha}e^{-\frac{1}{2} \sigma ^{2}t^{2}} $$

for all $|t|\leq \tau$.

Proof

From Lyapunov’s inequality, we have

$$\begin{aligned} \bigl(E \vert X_{l} \vert ^{\alpha}\bigr)^{\frac{1}{\alpha}} \leq \bigl(E \vert X_{l} \vert ^{2+\alpha} \bigr)^{ \frac{1}{2+\alpha}}, \end{aligned}$$

which implies that

$$\begin{aligned} \biggl(\frac{1}{\sum_{j=1}^{n}E \vert X_{j} \vert ^{2+\alpha}} \biggr)^{ \frac{1}{2+\alpha}} \leq \biggl(\frac{1}{E \vert X_{l} \vert ^{2+\alpha}} \biggr)^{ \frac{1}{2+\alpha}} \leq \biggl(\frac{1}{E \vert X_{l} \vert ^{\alpha}} \biggr)^{ \frac{1}{\alpha}} \end{aligned}$$

for all $l=1,2,\ldots ,n$. This provides that

$$ \biggl( \frac{1}{3^{\frac{2+\alpha}{\alpha}}\sum_{j=1}^{n}E \vert X_{j} \vert ^{2+\alpha}} \biggr)^{\frac{1}{2+\alpha}} \leq \biggl(\frac{1}{3E \vert X_{l} \vert ^{\alpha}} \biggr)^{\frac{1}{\alpha}} $$

for all $l=1,2,\ldots ,n$. From this fact and Lemma 2.1, we have for all $|t| \leq \tau$,

$$\begin{aligned} \psi (t) = \exp \Biggl\{ i\mu t - \frac{1}{2}\sigma ^{2}t^{2} + \sum_{j=1}^{n} G_{j}(t) \Biggr\} , \end{aligned}$$

(2.6)

where

$$\begin{aligned} G_{j}(t) = \int _{0}^{t}\frac{g_{X_{j}}(s)}{\psi _{X_{j}}(s)} \,\mathrm{d}s \quad \text{and}\quad \bigl\vert G_{j}(t) \bigr\vert \leq 9E \vert X_{j} \vert ^{2+\alpha} \vert t \vert ^{2+ \alpha}. \end{aligned}$$

From (2.6) and the inequality $|e^{z} - 1| \leq |z|e^{|z|}$ for a complex number z, we get that for all $|t|\leq \tau$,

$$\begin{aligned} \bigl\vert \psi (t) - e^{i\mu t - \frac{1}{2}\sigma ^{2}t^{2}} \bigr\vert & = \bigl\vert e^{ i\mu t - \frac{1}{2}\sigma ^{2}t^{2} + \sum _{j=1}^{n} G_{j}(t)} - e^{i\mu t - \frac{1}{2}\sigma ^{2}t^{2}} \bigr\vert \\ & = \bigl\vert e^{i\mu t - \frac{1}{2}\sigma ^{2}t^{2}} \bigr\vert \bigl\vert e^{ \sum _{j=1}^{n} G_{j}(t)} - 1 \bigr\vert \\ & \leq \Biggl\vert \sum_{j=1}^{n} G_{j}(t) \Biggr\vert e^{- \frac{1}{2} \sigma ^{2}t^{2} + \vert \sum _{j=1}^{n} G_{j}(t) \vert } \\ & \leq \sum_{j=1}^{n} \bigl\vert G_{j}(t) \bigr\vert e^{- \frac{1}{2} \sigma ^{2}t^{2} + \sum _{j=1}^{n} \vert G_{j}(t) \vert } \\ & \leq 9\sum_{j=1}^{n}E \vert X_{j} \vert ^{2+\alpha} \vert t \vert ^{2+\alpha}\times \exp \Biggl\{ -\frac{1}{2}\sigma ^{2}t^{2} + 9\sum _{j=1}^{n}E \vert X_{j} \vert ^{2+ \alpha} \vert t \vert ^{2+\alpha} \Biggr\} \\ & \leq 9\sum_{j=1}^{n}E \vert X_{j} \vert ^{2+\alpha} \vert t \vert ^{2+\alpha}\times \exp \biggl\{ -\frac{1}{2}\sigma ^{2}t^{2} + \frac{9}{3^{\frac{2+\alpha}{\alpha}}} \biggr\} \\ & \leq 9\sum_{j=1}^{n}E \vert X_{j} \vert ^{2+\alpha} \vert t \vert ^{2+\alpha}\times \exp \biggl\{ -\frac{1}{2}\sigma ^{2}t^{2} + \frac{9}{3^{3}} \biggr\} \\ & \leq 12.5606\sum_{j=1}^{n}E \vert X_{j} \vert ^{2+\alpha} \vert t \vert ^{2+\alpha}e^{ - \frac{1}{2}\sigma ^{2}t^{2}}. \end{aligned}$$

 □

3 Proof of the main results

3.1 Proof of Theorem 1.1

Proof

First, we will prove the theorem in the case of $a=0$ and $d=1$. Let $Y_{1},Y_{2},\ldots ,Y_{n}$ be independent common lattice random variables with parameter $(0,1)$, and let

$$ W_{n} = Y_{1} + Y_{2} + \cdots + Y_{n} $$

with $E(W_{n})=\mu _{W}$, $Var(W_{n})=\sigma _{W}^{2}$ and the characteristic function $\psi _{W}$. Suppose that $\beta _{Y_{j}} = 2\sum_{m=-\infty}^{\infty}P(Y_{j} = m)P(Y_{j} =m+1) > 0$ for all $j=1,2,\ldots ,n$, and let $\tau = \frac{1}{3^{\frac{1}{\alpha}}} ( \frac{1}{\sum_{j=1}^{n}E|Y_{j}|^{2+\alpha}} )^{ \frac{1}{2+\alpha}}$. Since $P(W_{n} = k) = \frac{1}{2\pi}\int _{-\pi}^{\pi} e^{-ikt} \psi _{W}(t) \,\mathrm{d}t$ ([18], p. 511), we have

$$\begin{aligned} & \biggl\vert P(W_{n} = k) - \frac{1}{\sigma _{W} \sqrt{2\pi}}e^{- \frac{(k-\mu _{W})^{2}}{2\sigma _{W}^{2}}} \biggr\vert \\ &\quad = \biggl\vert \frac{1}{2\pi} \int _{-\pi}^{\pi} e^{-ikt}\psi _{ W}(t) \,\mathrm{d}t - \frac{1}{\sigma _{W} \sqrt{2\pi}}e^{- \frac{(k-\mu _{W})^{2}}{2\sigma _{W}^{2}}} \biggr\vert \\ &\quad \leq \frac{1}{2\pi} \biggl\vert \int _{ \vert t \vert < \tau} e^{-ikt}\psi _{ W}(t) \,\mathrm{d}t - \int _{ \vert t \vert < \tau}e^{it(\mu _{ W} -k )-\frac{1}{2}\sigma _{ W}^{2}t^{2}} \,\mathrm{d}t \biggr\vert \\ &\quad \quad {}+ \frac{1}{2\pi} \biggl\vert \int _{ \vert t \vert < \tau}e^{it(\mu _{ W}-k)-\frac{1}{2}\sigma _{ W}^{2}t^{2}} \,\mathrm{d}t - \frac{\sqrt{2\pi}}{\sigma _{W}}e^{- \frac{(k-\mu _{W})^{2}}{2\sigma _{W}^{2}}} \biggr\vert \\ &\quad \quad {}+ \frac{1}{2\pi} \biggl\vert \int _{\tau \leq \vert t \vert \leq \pi} e^{-ikt} \psi _{W}(t) \,\mathrm{d}t \biggr\vert \\ &\quad := \vert A \vert + \vert B \vert + \vert C \vert . \end{aligned}$$

(3.1)

From Lemma 2.2, we have

$$\begin{aligned} \vert A \vert & \leq \frac{1}{2\pi} \int _{ \vert t \vert < \tau} \bigl\vert e^{-ikt} \bigr\vert \bigl\vert \psi _{ W}(t) - e^{it\mu _{W} - \frac{1}{2} \sigma _{W}^{2}t^{2}} \bigr\vert \,\mathrm{d}t \\ & = \frac{1}{2\pi} \int _{ \vert t \vert < \tau} \bigl\vert \psi _{W}(t) - e^{it\mu _{W} - \frac{1}{2}\sigma _{ W}^{2}t^{2}} \bigr\vert \,\mathrm{d}t \\ & \leq \frac{12.5606}{\pi}\sum_{j=1}^{n}E \vert Y_{j} \vert ^{2+\alpha} \int _{0}^{\tau} \vert t \vert ^{2+\alpha}e^{ - \frac{1}{2}\sigma _{ W}^{2}t^{2}} \,\mathrm{d}t. \end{aligned}$$

To bound $\int _{0}^{\tau} |t|^{2+\alpha}e^{ - \frac{1}{2}\sigma _{ W}^{2}t^{2}} \,\mathrm{d}t$, we let $\tilde{\tau} = \tau ^{\frac{2+\alpha}{2}}$ and note that

$$\begin{aligned} \int _{0}^{\tilde{\tau}} t^{2+\alpha}e^{ - \frac{1}{2}\sigma _{ W}^{2}t^{2}} \,\mathrm{d}t & \leq \int _{0}^{\tilde{\tau}} t^{2+\alpha} \,\mathrm{d}t \\ & = \frac{\tau ^{\frac{(2+\alpha )(3+\alpha )}{2}}}{3+\alpha} \\ & \leq \frac{1}{3^{\frac{\alpha ^{2} + 7\alpha +6}{2\alpha}} (\sum_{j=1}^{n}E \vert Y_{j} \vert ^{2+\alpha} )^{\frac{3+\alpha}{2}}} \\ & \leq \frac{0.0005}{ (\sum_{j=1}^{n}E \vert Y_{j} \vert ^{2+\alpha} )^{\frac{3+\alpha}{2}}} \end{aligned}$$

and

$$\begin{aligned} \int _{\tilde{\tau}}^{\tau} t^{2+\alpha}e^{ - \frac{1}{2} \sigma _{W}^{2}t^{2}} \,\mathrm{d}t & = \int _{\tilde{\tau}}^{\tau} \frac{t^{3}}{t^{1-\alpha}}e^{ - \frac{1}{2}\sigma _{W}^{2}t^{2}} \,\mathrm{d}t \\ & \leq \frac{1}{\tau ^{\frac{(2+\alpha )(1-\alpha )}{2}}} \int _{\tilde{\tau}}^{\tau} t^{3} e^{ - \frac{1}{2}\sigma _{ W}^{2}t^{2}} \,\mathrm{d}t \\ & \leq \frac{1}{\tau ^{\frac{(2+\alpha )(1-\alpha )}{2}}} \int _{0}^{\infty} t^{3} e^{ - \frac{1}{2}\sigma _{ W}^{2}t^{2}} \,\mathrm{d}t \\ & = 3^{\frac{(2+\alpha )(1-\alpha )}{2\alpha}} \Biggl(\sum_{j=1}^{n}E \vert Y_{j} \vert ^{2+ \alpha} \Biggr)^{\frac{1-\alpha}{2}} \biggl( \frac{2}{\sigma _{W}^{4}} \biggr) \\ & \leq \frac{1.1548\cdot 3^{\frac{1}{\alpha}}}{\sigma _{W}^{4}} \Biggl(\sum_{j=1}^{n}E \vert Y_{j} \vert ^{2+\alpha} \Biggr)^{ \frac{1-\alpha}{2}}. \end{aligned}$$

Hence,

$$\begin{aligned} \vert A \vert & \leq \frac{12.5606}{\pi}\sum_{j=1}^{n}E \vert Y_{j} \vert ^{2+ \alpha} \biggl( \int _{0}^{\tilde{\tau}} t^{2+\alpha}e^{ - \frac{1}{2}\sigma _{W}^{2}t^{2}} \,\mathrm{d}t + \int _{\tilde{\tau}}^{\tau} t^{2+\alpha}e^{ - \frac{1}{2}\sigma _{W}^{2}t^{2}} \,\mathrm{d}t \biggr) \\ & \leq \frac{0.0020}{ (\sum_{j=1}^{n}E \vert Y_{j} \vert ^{2+\alpha} )^{\frac{1+\alpha}{2}}} + \frac{4.6171\cdot 3^{\frac{1}{\alpha}}}{\sigma _{W}^{4}} \Biggl(\sum _{j=1}^{n}E \vert Y_{j} \vert ^{2+\alpha} \Biggr)^{ \frac{3-\alpha}{2}}. \end{aligned}$$

(3.1)

By the fact that

$$\begin{aligned} \int _{|t| < \tau}e^{it(\mu _{W} -k ) - \frac{1}{2}\sigma _{W}^{2}t^{2}} \,\mathrm{d}t & = \int _{\mathbb{R}}e^{it(\mu _{ W} -k ) - \frac{1}{2}\sigma _{ W}^{2}t^{2}} \,\mathrm{d}t - \int _{|t| \geq \tau}e^{it(\mu _{W} -k ) - \frac{1}{2}\sigma _{W}^{2}t^{2}} \,\mathrm{d}t \\ & = \frac{1}{\sigma _{W}} \int _{ \mathbb{R}}e^{ \frac{it(\mu _{W} -k )}{\sigma _{W}} - \frac{t^{2}}{2}} \,\mathrm{d}t - \frac{1}{\sigma _{W}} \int _{|t| \geq \sigma _{W}\tau}e^{ \frac{it(\mu _{W} -k )}{\sigma _{W}} - \frac{t^{2}}{2}} \,\mathrm{d}t \\ & = \frac{\sqrt{2\pi}}{\sigma _{W}}e^{- \frac{(k-\mu _{W})^{2}}{2\sigma _{W}^{2}}} - \frac{1}{\sigma _{W}} \int _{|t| \geq \sigma _{W}\tau}e^{ \frac{it(\mu _{W} -k )}{\sigma _{W}} - \frac{t^{2}}{2}} \,\mathrm{d}t, \end{aligned}$$

we have

$$\begin{aligned} B & = \frac{1}{2\pi} \int _{|t| < \tau}e^{it(\mu _{ W} -k ) - \frac{1}{2}\sigma _{ W}^{2}t^{2}} \,\mathrm{d}t - \frac{1}{\sigma _{W} \sqrt{2\pi}}e^{- \frac{(k-\mu _{W})^{2}}{2\sigma _{W}^{2}}} \\ & = -\frac{1}{2\pi \sigma _{W}} \int _{|t| \geq \sigma _{W}\tau}e^{ \frac{it(\mu _{W} -k )}{\sigma _{W}} - \frac{t^{2}}{2}} \,\mathrm{d}t, \end{aligned}$$

and hence,

$$\begin{aligned} \vert B \vert & \leq \frac{1}{2\pi \sigma _{W}} \int _{ \vert t \vert \geq \sigma _{W} \tau}e^{- \frac{t^{2}}{2}} \,\mathrm{d}t \\ & \leq \frac{1}{\pi \sigma _{W}^{2}\tau} \int _{\sigma _{W}\tau}^{\infty} te^{- \frac{t^{2}}{2}} \,\mathrm{d}t \\ & = \frac{0.3184}{\sigma _{W}^{2}\tau}e^{ \frac{-\sigma _{W}^{2}\tau ^{2}}{2}}. \end{aligned}$$

(3.2)

Using the fact that $|\psi _{W}(t)| \leq e^{-\frac{1}{\pi ^{2}}\beta _{ W}t^{2}}$, where $\beta _{W} = \sum_{j=1}^{n}\beta _{ Y_{j}}$, for $t\in [0,\pi )$ ([1], p. 5), we have

$$\begin{aligned} \vert C \vert & = \biggl\vert \frac{1}{2\pi} \int _{\tau \leq \vert t \vert \leq \pi} e^{-ikt} \psi _{W}(t) \,\mathrm{d}t \biggr\vert \\ & \leq \frac{1}{2\pi} \int _{\tau \leq \vert t \vert \leq \pi} \bigl\vert \psi _{ W}(t) \bigr\vert \,\mathrm{d}t \\ & = \frac{1}{\pi} \int _{\tau}^{\pi} \bigl\vert \psi _{W}(t) \bigr\vert \,\mathrm{d}t \\ & \leq \frac{1}{\pi} \int _{\tau}^{\pi} e^{- \frac{1}{\pi ^{2}}\beta _{W} t^{2}} \,\mathrm{d}t \\ & \leq \frac{1}{\pi \tau} \int _{\tau}^{\infty} te^{- \frac{1}{\pi ^{2}}\beta _{W} t^{2}} \,\mathrm{d}t \\ & = \frac{1}{\pi \tau} \biggl( \frac{\pi ^{2}e^{-\frac{\tau ^{2}\beta _{W}}{\pi ^{2}}}}{2\beta _{W}} \biggr) \\ & \leq \frac{1.5708}{\tau \beta _{W}} e^{- \frac{\tau ^{2}\beta _{W}}{\pi ^{2}}}. \end{aligned}$$

(3.3)

From (3.1)–(3.3), we have

$$\begin{aligned} & \biggl\vert P(W_{n} = k) - \frac{1}{\sigma _{W} \sqrt{2\pi}}e^{- \frac{(k-\mu _{W})^{2}}{2\sigma _{W}^{2}}} \biggr\vert \\ &\quad \leq \frac{0.0020}{ (\sum_{j=1}^{n}E \vert Y_{j} \vert ^{2+\alpha} )^{\frac{1+\alpha}{2}}} + \frac{4.6171\cdot 3^{\frac{1}{\alpha}}}{\sigma _{W}^{4}} \Biggl(\sum _{j=1}^{n}E \vert Y_{j} \vert ^{2+\alpha} \Biggr)^{ \frac{3-\alpha}{2}} \\ &\quad \quad {}+ \frac{0.3184}{\sigma _{W}^{2}\tau}e^{ \frac{-\sigma _{W}^{2}\tau ^{2}}{2}} + \frac{1.5708}{\tau \beta _{W}}e^{- \frac{\tau ^{2}\beta _{W}}{\pi ^{2}}}. \end{aligned}$$

(3.4)

In general, let $X_{1},X_{2},\ldots ,X_{n}$ be independent lattice random variables with parameter $(a,d)$. For $j=1,2,\ldots ,n$, let $Y_{j} = \frac{X_{j}-a}{d}$ and $W_{n} = Y_{1} + Y_{2} + \cdots + Y_{n}$. Observe that $Y_{1},Y_{2},\ldots ,Y_{n}$ are independent common lattice random variables with parameter $(0,1)$ and

$$\begin{aligned} & \mu _{W} = \frac{\mu -na}{d},\qquad \sigma _{ W}^{2}= \frac{\sigma ^{2}}{d^{2}}, \qquad P(Y_{j} = m) = P(X_{j} = a+dm), \end{aligned}$$

(3.5)

$$\begin{aligned} & E \vert Y_{j} \vert ^{2+\alpha} = \frac{E \vert X_{j}-a \vert ^{2+\alpha}}{d^{2+\alpha}}, \qquad \tau = \frac{d}{3^{\frac{1}{\alpha}} (\sum_{j=1}^{n}E \vert X_{j}-a \vert ^{2+\alpha} )^{\frac{1}{2+\alpha}}}. \end{aligned}$$

(3.6)

From (3.4)–(3.6), we have

$$ \begin{aligned} & \biggl\vert P(W_{n} = k) - \frac{1}{\sigma _{W} \sqrt{2\pi}}e^{- \frac{(k-\mu _{W})^{2}}{2\sigma _{W}^{2}}} \biggr\vert \\ &\quad \leq \frac{0.0020d^{\frac{(2+\alpha )(1+\alpha )}{2}}}{ (\sum_{j=1}^{n}E \vert X_{j}-a \vert ^{2+\alpha} )^{\frac{1+\alpha}{2}}} + \frac{4.6171\cdot 3^{\frac{1}{\alpha}}d^{\frac{\alpha ^{2}-\alpha +2}{2}} (\sum_{j=1}^{n}E \vert X_{j}-a \vert ^{2+\alpha} )^{\frac{3-\alpha}{2}}}{\sigma ^{4}} \\ &\quad \quad {}+ \frac{0.3184}{\sigma ^{2}\tau}e^{-\frac{\sigma ^{2}\tau ^{2}}{2}} + \frac{1.5708}{\tau \beta} e^{-\frac{\tau ^{2}\beta}{\pi ^{2}}}. \end{aligned} $$

From this fact and the fact that

$$\begin{aligned} & \biggl\vert P(S_{n} = na +kd) - \frac{d}{\sigma \sqrt{2\pi}}e^{- \frac{(na +kd -\mu )^{2}}{2\sigma ^{2}}} \biggr\vert = \biggl\vert P(W_{n} = k) - \frac{1}{\sigma _{W} \sqrt{2\pi}}e^{- \frac{(k -\mu _{W})^{2}}{2\sigma _{W}^{2}}} \biggr\vert , \end{aligned}$$

we have the conclusion of the theorem.

Furthermore, if $X_{1},X_{2},\ldots ,X_{n}$ are identically distributed, then

$$\begin{aligned} \mu = n\mu _{1},\qquad \sigma = \sigma _{1} \sqrt{n}, \qquad \sum_{j=1}^{n}E \vert X_{j}-a \vert ^{2+\alpha} = nE \vert X_{1}-a \vert ^{2+\alpha}, \quad \text{and}\quad \beta = n\beta _{1}, \end{aligned}$$

which imply that

$$\begin{aligned} &\sup_{k\in \mathbb{Z}} \biggl\vert P(S_{n} = na + kd) - \frac{d}{\sigma _{1} \sqrt{2n\pi}}e^{- \frac{(na + kd-n\mu _{1})^{2}}{2n\sigma _{1}^{2}}} \biggr\vert \\ &\quad \leq \frac{0.0020d^{\frac{(2+\alpha )(1+\alpha )}{2}}}{(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{1+\alpha}{2}}n^{\frac{1+\alpha}{2}}} + \frac{4.6171\cdot 3^{\frac{1}{\alpha}}d^{\frac{\alpha ^{2}-\alpha +2}{2}}(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{3-\alpha}{2}}}{\sigma _{1}^{4}n^{\frac{1+\alpha}{2}}} \\ &\quad \quad {}+ \frac{0.3184\cdot 3^{\frac{1}{\alpha}}\,d(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{1}{2+\alpha}}}{\sigma _{1}^{2}n^{\frac{1+\alpha}{2+\alpha}}} \exp \biggl( \frac{-\sigma _{1}^{2}n^{\frac{\alpha}{2+\alpha}}}{2\cdot 3^{\frac{2}{\alpha}}(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{2}{2+\alpha}}} \biggr) \\ &\quad \quad {}+ \frac{1.5708\cdot 3^{\frac{1}{\alpha}}(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{1}{2+\alpha}}}{d\beta _{1}n^{\frac{1+\alpha}{2+\alpha}}} \exp \biggl( \frac{-d^{2}\beta _{1}n^{\frac{\alpha}{2+\alpha}}}{3^{\frac{2}{\alpha}}\pi ^{2}(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{2}{2+\alpha}}} \biggr). \end{aligned}$$

(3.7)

Since $\frac{1+2\alpha}{2+\alpha} \geq \frac{1+\alpha}{2}$ and $e^{-x} \leq \frac{1}{x}$ for a real number $x>0$, we obtain that

$$\begin{aligned} & \frac{0.3184\cdot 3^{\frac{1}{\alpha}}\,d(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{1}{2+\alpha}}}{\sigma _{1}^{2}n^{\frac{1+\alpha}{2+\alpha}}} \exp \biggl( \frac{-\sigma _{1}^{2}n^{\frac{\alpha}{2+\alpha}}}{2\cdot 3^{\frac{2}{\alpha}}(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{2}{2+\alpha}}} \biggr) \\ &\quad \leq \frac{0.3184\cdot 3^{\frac{1}{\alpha}}\,d(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{1}{2+\alpha}}}{\sigma _{1}^{2}n^{\frac{1+\alpha}{2+\alpha}}} \biggl( \frac{2\cdot 3^{\frac{2}{\alpha}}(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{2}{2+\alpha}}}{\sigma _{1}^{2}n^{\frac{\alpha}{2+\alpha}}} \biggr) \\ &\quad\leq \frac{0.6368\cdot 3^{\frac{3}{\alpha}}\,d(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{3}{2+\alpha}}}{\sigma _{1}^{4}n^{\frac{1+\alpha}{2}}} \end{aligned}$$

(3.8)

and

$$\begin{aligned} & \frac{1.5708\cdot 3^{\frac{1}{\alpha}}(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{1}{2+\alpha}}}{d\beta _{1}n^{\frac{1+\alpha}{2+\alpha}}} \exp \biggl( \frac{-d^{2}\beta _{1}n^{\frac{\alpha}{2+\alpha}}}{3^{\frac{2}{\alpha}}\pi ^{2}(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{2}{2+\alpha}}} \biggr) \\ &\quad \leq \frac{1.5708\cdot 3^{\frac{1}{\alpha}}(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{1}{2+\alpha}}}{d\beta _{1}n^{\frac{1+\alpha}{2+\alpha}}} \biggl( \frac{3^{\frac{2}{\alpha}}\pi ^{2}(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{2}{2+\alpha}}}{d^{2}\beta _{1}n^{\frac{\alpha}{2+\alpha}}} \biggr) \\ &\quad \leq \frac{15.5032\cdot 3^{\frac{3}{\alpha}}(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{3}{2+\alpha}}}{d^{3}\beta _{1}^{2} n^{\frac{1+\alpha}{2}}}. \end{aligned}$$

(3.9)

From (3.7)–(3.9), we have

$$\begin{aligned} \sup_{k\in \mathbb{Z}} \biggl\vert P(S_{n} = na + kd) - \frac{d}{\sigma _{1} \sqrt{2n\pi}}e^{- \frac{(na + kd-n\mu _{1})^{2}}{2n\sigma _{1}^{2}}} \biggr\vert \leq \frac{C_{1}}{n^{\frac{1+\alpha}{2}}}, \end{aligned}$$

where

$$\begin{aligned} C_{1} & = \frac{0.0020d^{\frac{(2+\alpha )(1+\alpha )}{2}}}{(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{1+\alpha}{2}}} + \frac{4.6171\cdot 3^{\frac{1}{\alpha}}d^{\frac{\alpha ^{2}-\alpha +2}{2}}(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{3-\alpha}{2}}}{\sigma _{1}^{4}} \\ & \quad {}+ \frac{0.6368\cdot 3^{\frac{3}{\alpha}}\,d(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{3}{2+\alpha}}}{\sigma _{1}^{4}} + \frac{15.5032\cdot 3^{\frac{3}{\alpha}}(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{3}{2+\alpha}}}{d^{3}\beta _{1}^{2} }. \end{aligned}$$

 □

3.2 Proof of Theorem 1.2

Proof

By the same reason of Theorem 1.1, it suffices to prove the theorem in case $a=0$ and $d=1$. Let $Y_{1},Y_{2},\ldots ,Y_{n}$ be independent common lattice random variables with parameter $(0,1)$ with the characteristic functions $\psi _{Y_{j}}$, and let

$$ W_{n} = Y_{1} + Y_{2} + \cdots + Y_{n} $$

with $E(W_{n})=\mu _{W}$, $Var(W_{n})=\sigma _{W}^{2}$ and the characteristic function $\psi _{W}$. Suppose that $\upsilon _{Y_{j}} = 2\sum_{m=-\infty}^{ \infty}P(Y_{j} = m)P(Y_{j} =m+j) > 0$ for all $j=1,2\ldots ,n$. From (3.1)–(3.2) in Theorem 1.1, we have

$$\begin{array}{rl}& |P(W_n=k)-\frac{1}{{\sigma }_W\sqrt{2\pi }}{e}^{-\frac{{(k-{\mu }_W)}^2}{2{\sigma }_W^2}}|\\ & \le \frac{0.0020}{{({\sum }_{j=1}^{n}E|Y_j{|}^{2+\alpha })}^{\frac{1+\alpha }{2}}}+\frac{4.6171\cdot 3^{\frac{1}{\alpha }}}{{\sigma }_W^4}{(\sum _{j=1}^{n}E|Y_j{|}^{2+\alpha })}^{\frac{3-\alpha }{2}}\\ & +\frac{0.3184}{{\sigma }_W^2\tau }{e}^{\frac{-{\sigma }_W^2{\tau }^2}{2}}+|C|.& \end{array}$$

(3.11)

Siripraparat and Neammanee ([1], p. 6) showed that

$$\begin{aligned} \ln \bigl( \bigl\vert \psi _{Y_{j}}(t) \bigr\vert \bigr) & \leq - \sum_{m=-\infty}^{ \infty}\sum _{l=-\infty}^{\infty}P(Y_{j}=m)P(Y_{j}=l) \sin ^{2} \biggl((m-l) \frac{t}{2} \biggr). \end{aligned}$$

From this fact and the fact that

$$ \sum_{j=1}^{n} \sin ^{2} \biggl( \frac{jt}{2} \biggr) \geq \frac{n}{4}\min \biggl( 1, \biggl( \frac{nt}{2\pi} \biggr)^{2} \biggr) $$

for $|t|\leq \pi$ and $n\geq 2$ ([19], p. 399), we have

$$\begin{aligned} \bigl\vert \psi _{W}(t) \bigr\vert & =\prod _{j=1}^{n} \bigl\vert \psi _{ Y_{j}}(t) \bigr\vert \\ & \leq \prod_{j=1}^{n}\exp \Biggl(-2\sum _{m=-\infty}^{ \infty}P(Y_{j} = m)P(Y_{j} =m+j) \sin ^{2} \biggl(\frac{jt}{2} \biggr) \Biggr) \\ & \leq \exp \Biggl(-\sum_{j=1}^{n}\upsilon _{ Y_{j}}\sin ^{2} \biggl(\frac{jt}{2} \biggr) \Biggr) \\ & \leq \exp \Biggl(-\upsilon _{W}\sum_{j=1}^{n} \sin ^{2} \biggl(\frac{jt}{2} \biggr) \Biggr) \\ & \leq \exp \biggl(-\frac{n\upsilon _{W}}{4}\min \biggl( 1, \biggl( \frac{nt}{2\pi} \biggr)^{2} \biggr) \biggr), \end{aligned}$$

where $\upsilon _{W} = \min_{1\leq j \leq n} \upsilon _{Y_{j}}$. Hence,

$$\begin{aligned} \vert C \vert & \leq \frac{1}{2\pi} \int _{\tau \leq \vert t \vert \leq \pi} \bigl\vert \psi _{W}(t) \bigr\vert \,\mathrm{d}t \\ & = \frac{1}{\pi} \int _{\tau}^{\pi} \bigl\vert \psi _{W}(t) \bigr\vert \,\mathrm{d}t \\ & \leq \frac{1}{\pi} \int _{\tau}^{\pi}\exp \biggl(- \frac{n\upsilon _{W}}{4}\min \biggl( 1, \biggl( \frac{nt}{2\pi} \biggr)^{2} \biggr) \biggr) \,\mathrm{d}t \\ & \leq \frac{\pi -\tau}{\pi} \exp \biggl(- \frac{n\upsilon _{W}}{4}\min \biggl( 1, \biggl( \frac{n\tau}{2\pi} \biggr)^{2} \biggr) \biggr) \\ & \leq \exp \biggl(-\frac{n\upsilon _{W}}{4}\min \biggl( 1, \biggl( \frac{n\tau}{2\pi} \biggr)^{2} \biggr) \biggr). \end{aligned}$$

(3.10)

From (3.11) and (3.10),

$$ \begin{aligned} & \biggl\vert P(W_{n} = k) - \frac{1}{\sigma _{W} \sqrt{2\pi}}e^{- \frac{(k-\mu _{W})^{2}}{2\sigma _{W}^{2}}} \biggr\vert \\ &\quad \leq \frac{0.0020}{ (\sum_{j=1}^{n}E \vert Y_{j} \vert ^{2+\alpha} )^{\frac{1+\alpha}{2}}} + \frac{4.6171\cdot 3^{\frac{1}{\alpha}}}{\sigma _{W}^{4}} \Biggl(\sum _{j=1}^{n}E \vert Y_{j} \vert ^{2+\alpha} \Biggr)^{ \frac{3-\alpha}{2}} + \frac{0.3184}{\sigma _{W}^{2}\tau}e^{ \frac{-\sigma _{W}^{2}\tau ^{2}}{2}} \\ &\quad \quad {}+ e^{-\frac{n\upsilon _{W}}{4}\min ( 1, ( \frac{n\tau}{2\pi} )^{2} )}. \end{aligned} $$

Furthermore, if $X_{1},X_{2},\ldots ,X_{n}$ are identically distributed and $\upsilon _{j} > 0$ for all $j=1,2,\ldots ,n$, then

$$ \begin{aligned} &\sup_{k\in \mathbb{Z}} \biggl\vert P(S_{n} = na + kd) - \frac{d}{\sigma _{1} \sqrt{2n\pi}}e^{- \frac{(na + kd-n\mu _{1})^{2}}{2n\sigma _{1}^{2}}} \biggr\vert \\ &\quad \leq \frac{0.0020d^{\frac{(2+\alpha )(1+\alpha )}{2}}}{(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{1+\alpha}{2}}n^{\frac{1+\alpha}{2}}} + \frac{4.6171\cdot 3^{\frac{1}{\alpha}}d^{\frac{\alpha ^{2}-\alpha +2}{2}}(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{3-\alpha}{2}}}{\sigma _{1}^{4}n^{\frac{1+\alpha}{2}}} \\ &\quad \quad {}+ \frac{0.3184\cdot 3^{\frac{1}{\alpha}}\,d(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{1}{2+\alpha}}}{\sigma _{1}^{2}n^{\frac{1+\alpha}{2+\alpha}}} \exp \biggl( \frac{-\sigma _{1}^{2}n^{\frac{\alpha}{2+\alpha}}}{2\cdot 3^{\frac{2}{\alpha}}(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{2}{2+\alpha}}} \biggr) \\ &\quad \quad {}+ \exp \biggl(-\frac{n\upsilon}{4}\min \biggl( 1, \biggl( \frac{n^{\frac{1+\alpha}{2+\alpha}}d}{2\pi \cdot 3^{\frac{1}{\alpha}}( E \vert X_{j}-a \vert ^{2+\alpha})^{\frac{1}{2+\alpha}}} \biggr)^{2} \biggr) \biggr). \end{aligned} $$

From (3.8) and $n\geq ( \frac{2\pi \cdot 3^{\frac{1}{\alpha}}(E|X_{1}-a|^{2+\alpha})^{\frac{1}{2+\alpha}}}{d} )^{\frac{2+\alpha}{1+\alpha}}$, we obtain that

$$ \begin{aligned} \sup_{k\in \mathbb{Z}} \biggl\vert P(S_{n} = na + kd) - \frac{d}{\sigma _{1} \sqrt{2n\pi}}e^{- \frac{(na + kd-n\mu _{1})^{2}}{2n\sigma _{1}^{2}}} \biggr\vert \leq \frac{C_{2}}{n^{\frac{1+\alpha}{2}}} + e^{- \frac{n\upsilon}{4}}, \end{aligned} $$

where

$$\begin{aligned} C_{2} & = \frac{0.0020d^{\frac{(2+\alpha )(1+\alpha )}{2}}}{(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{1+\alpha}{2}}} + \frac{4.6171\cdot 3^{\frac{1}{\alpha}}d^{\frac{\alpha ^{2}-\alpha +2}{2}}(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{3-\alpha}{2}}}{\sigma _{1}^{4}} \\ & \quad {}+ \frac{0.6368\cdot 3^{\frac{3}{\alpha}}\,d(E \vert X_{1}-a \vert ^{2+\alpha})^{\frac{3}{2+\alpha}}}{\sigma _{1}^{4}}. \end{aligned}$$

 □

4 Examples

In our work, we relax the condition third moment to almost the second moment. The following example shows that there is an integer-valued random variable where the third moment does not exist but the aim moment exists.

Example 4.1

For $j=1,2,\ldots ,n$, let

$$ P(X_{j} = 0) = P(X_{j} = 2) = 0.45\quad \text{and}\quad P\bigl(X_{j}=2^{k}\bigr) = \frac{5.6}{2^{3k}}\quad \text{for integer }k\geq 2, $$

and assume that $X_{1},X_{2},\ldots ,X_{n}$ are independent. Note that $X_{1},X_{2},\ldots ,X_{n}$ are maximal lattice random variables with parameter $(0,2)$ and $\mu _{j} = 1.3667$, $\sigma _{j}^{2} = 2.7322$, $\beta _{j}= 0.2025$,

$$ E \vert X_{j} \vert ^{3} = 3.6 + \sum _{k=2}^{\infty}2^{3k}P\bigl(X=2^{k} \bigr) = 3.6 + \sum_{k=2}^{\infty}2^{3k} \biggl( \frac{5.6}{2^{3k}} \biggr) = \infty , $$

and for $\alpha \in (0,1)$,

$$\begin{aligned} E \vert X_{j} \vert ^{2+\alpha} & = 0.45\cdot 2^{2+\alpha} + \sum_{k=2}^{\infty}2^{(2+ \alpha )k}P \bigl(X=2^{k}\bigr) \\ & = 0.45\cdot 2^{2+\alpha} + \frac{5.6}{2^{2(1-\alpha )}-2^{1-\alpha}} < \infty \end{aligned}$$

for all $j=1,2,\ldots ,n$. Let

$$ \Delta _{n} = \sup_{k\in \mathbb{Z}} \biggl\vert P(S_{n} =2k) - \frac{2}{1.6529 \sqrt{2n\pi}}e^{-\frac{(2k -1.3667n)^{2}}{5.4644n}} \biggr\vert . $$

By Theorem 1.1, we have

$$ \Delta _{n} \leq \frac{A_{1}}{n^{\frac{1+\alpha}{2}}} + \frac{A_{2}}{n^{\frac{1+\alpha}{2+\alpha}}}\exp \bigl(-A_{3}n^{ \frac{\alpha}{2+\alpha}} \bigr) + \frac{A_{4}}{n^{\frac{1+\alpha}{2+\alpha}}}\exp \bigl(-A_{5}n^{ \frac{\alpha}{2+\alpha}} \bigr) $$

and Table 1.

Table 1 Explicit constants for Example 4.1

Observe that we cannot apply Theorem 1.2 with Example 4.1 since $\upsilon _{j} = 0$ for some $j \geq 3$.

Example 4.2

Let $X_{1},X_{2},\ldots ,X_{n}$ be independent random variables defined by

$$ P(X_{j} = 0) = \frac{7}{8}-\frac{1}{(2j)^{6}-(2j)^{3}}, \qquad P(X_{j} = 2j) = \frac{1}{8}, \quad \text{and}\quad P \bigl(X_{j}=(2j)^{k}\bigr) = (2j)^{-3k} $$

for integer $k\geq 2$. We see that $X_{1},X_{2},\ldots ,X_{n}$ are common lattice random variables with parameter $(0,2)$ and

$$\begin{aligned} &\mu _{j} = \frac{j}{4}+\frac{1}{16j^{4}-4j^{2}}, \qquad \sigma _{j}^{2} = \frac{j^{2}}{2}+\frac{1}{4j^{2}-2j} - \mu _{j}^{2}, \\ &E \vert X_{j} \vert ^{2+\alpha} = \frac{(2j)^{2+\alpha}}{8}+ \frac{1}{(2j)^{2-2\alpha}-(2j)^{1-\alpha}} \quad \text{and}\quad E \vert X_{j} \vert ^{3} = \infty . \end{aligned}$$

This implies that

$$\begin{aligned} &\frac{n^{3}}{48} \leq \sigma ^{2} \leq n^{3}, \\ &\frac{n^{3}}{6} \leq E \vert X_{j} \vert ^{2+\alpha} \leq \biggl( \frac{2^{2+\alpha}}{8} + \frac{2^{1+2\alpha}}{48(2^{1-\alpha}-1)} \biggr)n^{3+\alpha}. \end{aligned}$$

Moreover, we have that

$$ \upsilon =\min_{1\leq j \leq n} \frac{1}{4} \biggl( \frac{7}{8}- \frac{1}{(2j)^{6}-(2j)^{3}} \biggr) = \frac{3}{14}. $$

Let

$$ \Delta _{n} = \sup_{k\in \mathbb{Z}} \biggl\vert P(S_{n} =2k) - \frac{2}{\sigma \sqrt{2\pi}}e^{-\frac{(2k -\mu )^{2}}{2\sigma ^{2}}} \biggr\vert . $$

By Theorem 1.1, we have

$$ \Delta _{n} \leq \frac{B_{1}}{n^{\frac{3+3\alpha}{2}}} + \frac{B_{2}}{n^{\frac{\alpha ^{2}+3}{2}}} + \frac{B_{3}}{n^{3}}\exp \bigl(-B_{4}n^{\frac{\alpha}{2+\alpha}} \bigr) + \exp \bigl(-B_{5}n^{ \frac{\alpha}{2+\alpha}} \bigr) $$

and Table 2.

Table 2 Explicit constants for Example 4.2

Observe that we cannot apply Theorem 1.1 with Example 4.2 since $\beta _{j} = 0$ for $j \geq 2$.