| Теория вероятностей и ее применения |
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Краткие сообщения Almost sure results for generalized normal distribution Wen L.a, Zhuang Z.b a School of Mathematics and Big Data, Foshan University, Foshan, P. R. China b Department of Mathematics, Hong Kong University of Science and Technology, Hong Kong, P. R. China
Английская версия:
Theory of Probability and its Applications, 2025, Volume 69, Issue 4, Pages DOI: https://doi.org/10.1137/S0040585X97T992185 
Реферативные базы данных:
  1. Introduction and main resultsA random variable $X$ is said to have the normal distribution with parameters $\mu$ and $\sigma^2$ if its probability distribution function (p.d.f.) is given by
$$
\begin{equation*}
f(x\mid \mu,\sigma)=\frac{1}{\sqrt{2\pi}\,\sigma}\exp\biggl\{-\frac{(x-\mu)^2}{2\sigma^2}\biggr\},
\end{equation*}
\notag
$$
for $-\infty<x<\infty$, $-\infty<\mu<\infty$, $\sigma>0$. We call $X$ obey the standard normal distribution if $\mu=0$ and $\sigma=1$. In probability theory and statistics, the normal distribution is the most significant probability distribution. A large number of random variables are either nearly or exactly represented by the normal distribution, in every physical science and economics. Furthermore, it can be used to approximate other probability distributions. Therefore, many properties, related to the normal distribution, have been studied. For example, let $\{X_n,\, n\geqslant1\}$ be a sequence of independent random variables with the common distribution as $X$, suppose that $X$ obeys the standard normal distribution, then it is well known that
$$
\begin{equation}
\limsup_{n\to\infty}\frac{X_n}{\sqrt{2\log n}}=1\quad \text{a.s.}
\end{equation}
\tag{1.1}
$$
(see [2; Example 2.1.2(a)]) and the cluster set (set of all limit points) of the sequence $\{X_n/\sqrt{2\log n},\, n\geqslant2\}$ is $[-1,1]$ with probability one (see [4]), where and in the following, $\log x=\log_e(\max\{0, x\})$ for any real number $x$, and set $\log 0=-\infty$. Formula (1.1), characterizing the fluctuation of the normal sequence, is very simple and beautiful. And it had been strengthened to
$$
\begin{equation}
\lim_{n\to\infty}\frac{\max_{1\leqslant k\leqslant n}X_k}{\sqrt{2\log n}}=1\quad \text{a.s.},
\end{equation}
\tag{1.2}
$$
see [9; Example viii], or see [2; Example 2.1.2(b)]. Formula (1.2) is a special case of the almost sure asymptotic properties of extremes of independent and identically distributed random variables. For more detail, one can refer to [9] and [5]. Although the normal distribution is the most popular distribution in probability and statistics, many new distributions have been constantly proposed in order to meet the needs. For example, Nadarajah [7] introduced the generalized normal distribution. A random variable $X$ is said to have the generalized normal distribution with parameters $\mu$, $\sigma$, and $s$ if its p.d.f. is given by
$$
\begin{equation}
f(x|\mu,\sigma,s)=\frac{s}{2\sigma \Gamma(1/s)}\exp\biggl\{-\biggl|\frac{x-\mu}{\sigma}\biggr|^s\biggr\},
\end{equation}
\tag{1.3}
$$
for $-\infty<x<\infty$, $-\infty<\mu<\infty$, $\sigma>0$, and $s>0$, where $\Gamma(\,{\cdot}\,)$ is the Gamma function. The generalized normal distribution, also known as the generalized Gaussian distribution, exponential power distribution or generalized error distribution, is the generalization of the normal distribution. The normal distribution belongs to a special case of the generalized normal distribution. The exponential power distribution was first proposed by Subbotin [11], and Vianelli [13] provided an analytical form of the generalized error distribution density function. Box and Tao [1] referred to this distribution as an exponential power distribution. Tadikamala [12] provided a sampling method for exponential power distribution. Nadarajah [7] studied the relevant properties of the generalized normal distribution, such as the extreme value distribution of the failure function, moments, and order statistic, discussed the maximum likelihood estimation, and gave the information matrix. Nadarajah [8] pointed out that the generalized normal distribution proposed by him belongs to exponential power distribution. Li and Li [6] provided high-order expansions for normalizing the maximum density of generalized error distributions and used them to derive high-order expansions for extreme moments. Rui and Nadarajah [10] studied the accurate maximum likelihood estimators of positional and scale parameters for a given generalized Gaussian distribution shape parameter $s=3,4,5$. Due to the importance of the generalized normal distribution, the goal of the paper is to extend (1.1) and (1.2) from the normal distribution to the generalized normal distribution. We now state the main results. The proofs of those results will be detailed in the next section. Theorem 1.1. Let $\{X, X_n,\, n\,{\geqslant}\,1\}$ be a sequence of independent and identically distributed random variables, Suppose that $X$ has the p.d.f. as (1.3), then and the cluster set of the sequence $\{(X_n-\mu)/(\log n)^{1/s},\, n\geqslant2\}$ is $[-\sigma,\sigma]$ with probability one. Theorem 1.2. Under the condition of Theorem 1.1, we have and Theorems 1.1 and 1.2 give the convergence rates of $\limsup_{n\to\infty}(X_n{-}{\kern1pt}\mu){\kern1pt}{=}{\kern1pt}\infty$ a.s. We know that $\liminf_{n\to\infty}|X_n-a|=0$ a.s. for any real number $a$, we also can give the convergence rates of it. Firstly, we give a general result. Theorem 1.3. Let $\{Y,\,Y_n,\, n\geqslant1\}$ be a sequence of independent and identically distributed random variables with Remark 1.1. By (1.6), $\mathbf P\{Y=0\}=0$, hence $\log |Y_n|$ are well defined for all $n\geqslant1$. Remark 1.2. Let $Y$ be a random variable with p.d.f. $g(y)$, suppose that $g(y)$ is continuous in the neighbours of zero and $g(0)\ne 0$, then (1.6) holds with $c_0=2g(0)$. By Theorem 1.3, we have the following corollary. Corollary 1.1. Under the condition of Theorem 1.1, we have, for any $a\in (-\infty, +\infty)$, and the cluster set of the sequence $\{\log|X_n-a|/\log n,\, n\geqslant2\}$ is $[-1,0]$ with probability one. Furthermore, Remark 1.3. Corollary 1.1 is new even in the normal distributed case. Throughout this paper, $C$ always stands for a positive constant whose value is of no importance and may differ from one place to another. 2. Lemmas and proofs of main resultsLet $\{A_n,\, n\,{\geqslant}\,1\}$ be a sequence of events. As usual the abbreviation $\{A_n\ {\rm i.o.}\}$ stands for the case that the events $A_n$ occur infinitely often. That is,
$$
\begin{equation*}
\{A_n\text{ i.o.}\}=\bigcap^\infty_{n=1}\bigcup^\infty_{k=n}A_k.
\end{equation*}
\notag
$$
The following lemmas are needed, the first lemma follows from [2; Example 1.6.25(a)] and the second one is due to [7; Lemma 2.3] can be obtained by the same argument as Lemma 2.2. Lemma 2.1. Let $\{b_n,\, n\geqslant1\}$ be a nondecreasing sequence of positive real numbers such that $\lim_{n\to\infty}b_n=\infty$ and let $\{V_n,\, n\geqslant1\}$ be a sequence of random variables. Then Lemma 2.2. Let $X$ be a random variable with the p.d.f. as (1.3). Then as $x\to+\infty$. Proof of Theorem 1.1. Note that $(X-\mu)/\sigma$ has the p.d.f. as (1.3) with $\mu=0$ and $\sigma=1$, so we can assume that $\mu=0$ and $\sigma=1$. Hence, we should prove that and the cluster set of the sequence $\{X_n/(\log n)^{1/s},\, n\geqslant2\}$ is $[-1,1]$ with probability one.
To prove (2.1), it is enough to show that, for all $\varepsilon\in (0,1)$,
$$
\begin{equation*}
\limsup_{n\to\infty}\frac{X_n}{(\log n)^{1/s}}\leqslant (1+\varepsilon)^{1/s}\quad\text{a.s.}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\limsup_{n\to\infty}\frac{X_n}{(\log n)^{1/s}}\geqslant (1-\varepsilon)^{1/s}\quad\text{a.s.},
\end{equation*}
\notag
$$
which are equivalent to
$$
\begin{equation}
\mathbf P\{X_n> (1+\varepsilon)^{1/s}(\log n)^{1/s}\text{ i.o.}\}=0
\end{equation}
\tag{2.2}
$$
and
$$
\begin{equation}
\mathbf P\{X_n> (1-\varepsilon)^{1/s}(\log n)^{1/s}\text{ i.o.}\}=1,
\end{equation}
\tag{2.3}
$$
respectively. In the view of the Borel–Cantelli lemma, to prove (2.2), it suffices to prove that
$$
\begin{equation}
\sum^\infty_{n=2}\mathbf P\{X> (1+\varepsilon)^{1/s}(\log n)^{1/s}\}=\sum^\infty_{n=2}\mathbf P\{X_n> (1+\varepsilon)^{1/s}(\log n)^{1/s}\}<\infty.
\end{equation}
\tag{2.4}
$$
By Lemma 2.2,
$$
\begin{equation*}
\mathbf P\{X> (1+\varepsilon)^{1/s}(\log n)^{1/s}\}\leqslant C(\log n)^{1/s-1}n^{-(1+\varepsilon)}
\end{equation*}
\notag
$$
and it is obvious that $\sum^\infty_{n=1}(\log n)^{1/s-1}n^{-(1+\varepsilon)}<\infty$. Thus (2.4) holds, and hence (2.2) holds.
In the view of the Borel–Cantelli lemma, to prove (2.3), it suffices to prove that
$$
\begin{equation}
\sum^\infty_{n=2}\mathbf P\{X> (1-\varepsilon)^{1/s}(\log n)^{1/s}\}=\sum^\infty_{n=2}\mathbf P\{X_n> (1-\varepsilon)^{1/s}(\log n)^{1/s}\}=\infty.
\end{equation}
\tag{2.5}
$$
By Lemma 2.2,
$$
\begin{equation*}
\mathbf P\{X> (1-\varepsilon)^{1/s}(\log n)^{1/s}\}> C(\log n)^{1/s-1}n^{-(1-\varepsilon)}
\end{equation*}
\notag
$$
holds, when $n$ is large enough and it is obvious that
$$
\begin{equation*}
\sum^\infty_{n=1}(\log n)^{1/s-1}n^{-(1-\varepsilon)}=\infty.
\end{equation*}
\notag
$$
Thus (2.5) holds, and hence (2.3) holds.
Note that $-X$ has the same distribution as $X$, thus by (2.1),
$$
\begin{equation*}
\limsup_{n\to\infty}\frac{-X_n}{(\log n)^{1/s}}=-1\quad\text{a.s.},
\end{equation*}
\notag
$$
i.e.,
$$
\begin{equation}
\liminf_{n\to\infty}\frac{X_n}{(\log n)^{1/s}}=-1\quad\text{a.s.}
\end{equation}
\tag{2.6}
$$
Therefore, to prove that the cluster set of the sequence $\{X_n/(\log n)^{1/s},\, n\,{\geqslant}\,2\}$ is $[-1,1]$ with probability one, it is enough to prove that, for any $\alpha\in (0,1)$, there exists a subsequence $\{m(n),n\geqslant1\}$ of $\{1,2,\dots\}$, such that
$$
\begin{equation}
\limsup_{n\to\infty}\frac{X_{m(n)}}{(\log m(n))^{1/s}}=\alpha\quad\text{a.s.}
\end{equation}
\tag{2.7}
$$
and
$$
\begin{equation}
\liminf_{n\to\infty}\frac{X_{m(n)}}{(\log m(n))^{1/s}}=-\alpha\quad\text{a.s.}
\end{equation}
\tag{2.8}
$$
In fact, for any subsequence $\{m(n),\, n\geqslant1\}$ of $\{1,2,\dots\}$, $\{X_{m(n)},\, n\geqslant1\}$ is a sequence of independent random variables with common distribution as $X$, by (2.1) and (2.6),
$$
\begin{equation*}
\limsup_{n\to\infty}\frac{X_{m(n)}}{(\log n)^{1/s}}=1\quad\text{a.s.}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\liminf_{n\to\infty}\frac{X_{m(n)}}{(\log n)^{1/s}}=-1\quad\text{a.s.},
\end{equation*}
\notag
$$
which imply (2.7) and (2.8) by setting $m(n)=\lfloor n^{1/\alpha^s}\rfloor$, and the fact that $\alpha^s\log m(n)\sim \log n$, where $\lfloor x\rfloor$ denotes the integer part of the real number $x$. The proof is completed. Proof of Theorem 1.2. We also assume that $\mu=0$ and $\sigma=1$. Hence, we can rewrite (1.4) as
To prove (2.9), it is enough to show that, for all $\varepsilon\in (0,1)$,
$$
\begin{equation*}
\limsup_{n\to\infty}\frac{\max_{1\leqslant k\leqslant n}X_k}{(\log n)^{1/s}}\leqslant (1+\varepsilon)^{1/s} \quad\text{a.s.}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\liminf_{n\to\infty}\frac{\max_{1\leqslant k\leqslant n}X_k}{(\log n)^{1/s}}\geqslant (1-\varepsilon)^{1/s} \quad\text{a.s.},
\end{equation*}
\notag
$$
which are equivalent to
$$
\begin{equation}
\mathbf P\Bigl\{\max_{1\leqslant k\leqslant n}X_k> (1+\varepsilon)^{1/s}(\log n)^{1/s}\text{ i.o.}\Bigr\}=0
\end{equation}
\tag{2.10}
$$
and
$$
\begin{equation}
\mathbf P\Bigl\{\max_{1\leqslant k\leqslant n}X_k<(1-\varepsilon)^{1/s}(\log n)^{1/s}\text{ i.o.}\Bigr\}=0,
\end{equation}
\tag{2.11}
$$
respectively. By Lemma 2.1, (2.10) holds from (2.2). In the view of the Borel–Cantelli lemma, to prove (2.11), it suffices to prove that
$$
\begin{equation}
\sum^\infty_{n=2}\mathbf P\Bigl\{\max_{1\leqslant k\leqslant n}X_k< (1-\varepsilon)^{1/s}(\log n)^{1/s}\Bigr\}<\infty.
\end{equation}
\tag{2.12}
$$
Using the elementary inequality $1-x<e^{-x}$, for $x>0$,
$$
\begin{equation*}
\begin{aligned} \, &\mathbf P\Bigl\{\max_{1\leqslant k\leqslant n}X_k< (1-\varepsilon)^{1/s}(\log n)^{1/s}\Bigr\} =[1-\mathbf P\{X>(1-\varepsilon)^{1/s}(\log n)^{1/s}\}]^n \\ &\qquad<\exp\bigl\{-n\, \mathbf P\{X>(1-\varepsilon)^{1/s}(\log n)^{1/s}\}\bigr\} \end{aligned}
\end{equation*}
\notag
$$
and by Lemma 2.2,
$$
\begin{equation*}
n\, \mathbf P\{X>(1-\varepsilon)^{1/s}(\log n)^{1/s}\}\sim \frac{1}{2\Gamma(s)}\, (1-\varepsilon)^{1-1/s}(\log n)^{1-1/s}n^\varepsilon >n^{\varepsilon/2},
\end{equation*}
\notag
$$
the last inequality holds for large enough $n$. Then (2.12) follows from $\sum^\infty_{n=2}e^{-n^{\varepsilon/2}}<\infty$, and hence (2.11) holds.
$$
\begin{equation*}
\begin{aligned} \, &\lim_{n\to\infty}\frac{\log\max_{1\leqslant k\leqslant n}(X_k-\mu)}{\log\log n} \\ &\qquad=\lim_{n\to\infty}\frac{\log[\max_{1\leqslant k\leqslant n}(X_k-\mu)/(\log n)^{1/s}]+\log(\log n)^{1/s}}{\log\log n}=\frac{1}{s}\quad\text{a.s.}, \end{aligned}
\end{equation*}
\notag
$$
which implies that (1.5) holds. The proof is completed. Proof of Theorem 1.3. In view of the Borel–Cantelli lemma, to prove (1.7), it suffices to prove that, for every $\varepsilon\in (0,1)$,
$$
\begin{equation}
\sum^\infty_{n=2}\mathbf P\{\log|Y_n|<-(1+\varepsilon)\log n\}<\infty
\end{equation}
\tag{2.13}
$$
and
$$
\begin{equation}
\sum^\infty_{n=2}\mathbf P\{\log|Y_n|<-(1-\varepsilon)\log n\}=\infty.
\end{equation}
\tag{2.14}
$$
By (1.6), for $n$ large enough, we have
$$
\begin{equation*}
\mathbf P\{\log|Y_n|<-(1+\varepsilon)\log n\}=\mathbf P\{|Y|<n^{-(1+\varepsilon)}\}\leqslant Cn^{-(1+\varepsilon)}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\mathbf P\{\log|Y_n|<-(1-\varepsilon)\log n\}=\mathbf P\{|Y|<n^{-(1+\varepsilon)}\}\geqslant Cn^{-(1-\varepsilon)}.
\end{equation*}
\notag
$$
Then (2.13) and (2.14) hold from the fact that respectively.
We now prove (1.8). Set $Z=1/|X|$ and $Z_n=1/|Y_n|$, then, by (1.6), which implies that
$$
\begin{equation*}
\lim_{n\to\infty}\frac{\max_{1\leqslant k\leqslant n}\log Z_n}{\log n}= 1\quad\text{a.s.}
\end{equation*}
\notag
$$
by using [3; Theorem 2.3], and hence (1.8) holds.
To prove that $[-1,0]\subset D$, it suffices to show that, for any $\alpha\in (0,1)$, there exists a subsequence $\{m(n),\, n\geqslant1\}$ of $\{1,2,\dots\}$ such that
$$
\begin{equation}
\liminf_{n\to\infty}\frac{\log|Y_{m(n)}|}{\log m(n)}=-\alpha\quad\text{a.s.}
\end{equation}
\tag{2.15}
$$
In fact, for any subsequence $\{m(n),\, n\geqslant1\}$ of $\{1,2,\dots\}$, $\{Y_{m(n)},\, n\geqslant1\}$ is a sequence of independent random variables with the common distribution like $Y$, by (1.7),
$$
\begin{equation*}
\liminf_{n\to\infty}\frac{\log|Y_{m(n)}|}{\log n}=-1\quad\text{a.s.},
\end{equation*}
\notag
$$
which implies (2.15) by setting $m(n)=\lfloor n^{1/\alpha}\rfloor$ and the fact $\alpha \log m(n)\sim \log n$.
Assume that $\mathbf E|Y|^\delta<\infty$, for all $\delta>0$, then we have, for a fixed $\delta>0$,
$$
\begin{equation*}
\sum^\infty_{n=1}\mathbf P\{|Y|>\varepsilon n^{1/\delta}\}<\infty,
\end{equation*}
\notag
$$
which implies that $n^{-1/\delta}|Y_n|\to 0$ a.s. in the view of the Borel–Cantelli lemma, and hence Let $\delta\to\infty$, we have Therefore, $[-1,0]=D$. The proof is completed. Proof of Corollary 1.1. Set $Y=X-a$ and $Y_n=X_n-a$, for all $n\geqslant1$. Then the p.d.f. of $Y$ is $g(y)=f(y+a\,|\,\mu, \sigma, s)$. Obviously, $g(y)$ is continuous for all $y\in(-\infty,\infty)$, and $g(0)=f(a\,|\,\mu, \sigma, s)\ne 0$. Meanwhile, it is easy to show that $\mathbf E|Y|^\delta<\infty$ for all $\delta>0$. Then, by Remark 1.2 and Theorem 1.3, Corollary 1.1 holds. The proof is completed. AcknowledgmentsThe authors would like to thank the referees for the helpful comments. 
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