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# 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 [36], and [7] for examples) and in the case of lattice random variables (see [79], 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

### 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}

□

### 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

$\begin{array}{rl}& |P\left({W}_{n}=k\right)-\frac{1}{{\sigma }_{W}\sqrt{2\pi }}{e}^{-\frac{{\left(k-{\mu }_{W}\right)}^{2}}{2{\sigma }_{W}^{2}}}|\\ & \phantom{\rule{1em}{0ex}}\le \frac{0.0020}{{\left({\sum }_{j=1}^{n}E|{Y}_{j}{|}^{2+\alpha }\right)}^{\frac{1+\alpha }{2}}}+\frac{4.6171\cdot {3}^{\frac{1}{\alpha }}}{{\sigma }_{W}^{4}}{\left(\sum _{j=1}^{n}E|{Y}_{j}{|}^{2+\alpha }\right)}^{\frac{3-\alpha }{2}}\\ & \phantom{\rule{2em}{0ex}}+\frac{0.3184}{{\sigma }_{W}^{2}\tau }{e}^{\frac{-{\sigma }_{W}^{2}{\tau }^{2}}{2}}+|C|.& \end{array}$
(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.

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.

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

Not applicable.

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The authors would like to thank the reviewers for their valuable comments and suggestions.

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Kammoo, P., Laipaporn, K. & Neammanee, K. Local limit theorems without assuming finite third moment. J Inequal Appl 2023, 21 (2023). https://doi.org/10.1186/s13660-023-02928-y

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### Keywords

• Local limit theorem
• Normal density function
• Lattice random variable
• Rate of convergence
• Characteristic function