INTRODUCTION

Let independent and identically distributed random variables \(X_{1},...,X_{n}\) have a common distribution function (DF) \(F(x)=\mathbf{P}\{X\leqslant x\}\) and distribution density \(f(x)=F^{\prime}(x).\) We denote by \(X_{k}^{(n)}\) the \(k\)th order statistic in variational series \(X_{1}^{(n)}\leqslant...\leqslant X_{n}^{(n)}\) constructed using the variables \(X_{1},...,X_{n}\), \(\bar{X}=\sum\limits_{i=1}^{n}X_{i}/n\), and \(S^{2}=\sum\limits_{i=1}^{n}(X_{i}-\bar{X})^{2}/n.\) For some \(c_{n}>0\) and \(d_{n}\). Below, we study the asymptotic distribution (as \(n\to\infty\)) of the quantity

$$((X_{k}^{(n)}-\bar{X})/S-d_{n})/c_{n},$$
(1)

where

$$k=k(n)\to\infty,\quad\lambda_{k,n}=k/n\to 0,\quad n\to\infty.$$
(2)

In [1], a similar problem was considered as applied to a limiting distribution of extremal order statistics.

1. MAIN RESULT

Theorem. Let

$$\mathbf{E}X_{1}=0,\quad\mathbf{E}X_{1}^{2}=1,\quad 0<\mathbf{E}X_{1}^{4}<\infty.$$
(3)

If, for some \(c_{n}>0\) and \(d_{n}\), the quantity

$$(X_{k}^{(n)}-d_{n})/c_{n}$$
(4)

has a limiting DF \(H(x)\) as \(n\to\infty\) and the limit relation

$$d_{n}/(c_{n}\sqrt{n})\to 0$$
(5)

holds, quantity (1) has the same limiting DF \(H(x).\)

Proof. We write (1) in the form

$$\left(\frac{X_{k}^{(n)}}{S}-d_{n}\right)\frac{1}{c_{n}}-\frac{\bar{X}}{c_{n}S}\,.$$

It follows from conditions (3) that, as \(n\to\infty\),

$$S\stackrel{{\scriptstyle P}}{{\longrightarrow}}1,\quad\frac{1}{S}=\frac{1}{1-(1-S)}=1+(1-S)+o_{p}(1-S),$$

expression (1) is equivalent to the expression

$$\frac{X_{k}^{(n)}-d_{n}}{c_{n}}+\frac{X_{k}^{(n)}}{c_{n}}(1-S)(1+o_{p}(1))-\frac{\bar{X}}{c_{n}S}.$$
(6)

Let us show that quantity (6) has the same limit as quantity (4). It follows from conditions (3) that quantities \(\sqrt{n}\bar{X}\) and \(\sqrt{n}(1-S)\) as \(n\to\infty\) are asymptotically normal.

We consider the expression

$$\frac{X_{k}^{(n)}}{c_{n}\sqrt{n}}=\frac{X_{k}^{(n)}-d_{n}}{c_{n}\sqrt{n}}+\frac{d_{n}}{c_{n}\sqrt{n}}.$$
(7)

Since both terms on the right-hand side of relation (7) converge in probability to zero as \(n\to\infty\) by the hypotheses of the theorem, we have

$$\frac{X_{k}^{(n)}}{c_{n}\sqrt{n}}\stackrel{{\scriptstyle P}}{{\longrightarrow}}0.$$
(8)

It thus follows that the second term in expression (6) converges to zero in probability as \(n\to\infty\). Let us show that as \(n\to\infty\),

$$1/(c_{n}\sqrt{n})\to 0.$$
(9)

Let us consider two cases.

1. Dstribution \(F\) is unbound from the left; i.e., \(F(x)>0\) for all \(x.\) As \(n\to\infty\), we then have \(X_{k}^{(n)}\stackrel{{\scriptstyle P}}{{\longrightarrow}}-\infty,\) and condition (8) implies condition (9).

2. There exists a finite number \(x_{0},\) such that \(d_{n}\to x_{0}\) and \(X_{k}^{(n)}\stackrel{{\scriptstyle P}}{{\longrightarrow}}x_{0}\) as \(n\to\infty,\) and the condition \(\mathbf{E}X_{1}=0\) implies that \(x_{0}<0.\) Again, condition (9) follows from condition (8); consequently, as \(n\to\infty\), the third term in expression (6) converges in probability to zero. The theorem is proved.

2. APPLICATIONS

Suppose that condition (2) holds. A necessary and sufficient condition for asymptotic normality as \(n\to\infty\) of statistics \(T_{n}=(X_{k}^{(n)}-d_{n})/c_{n}\) for some \(c_{n}>0\) and \(d_{n}\) is that the following relation is satisfied for any \(x\) [2, 3]:

$$\lim_{n\to\infty}(F(c_{n}x+d_{n})-\lambda_{k,n})\sqrt{k}/\lambda_{k,n}=x.$$
(10)

For absolutely continuous distributions, quantities \(c_{n}\) and \(d_{n}\) are determined using relations

$$F(d_{n})=\lambda_{k,n},\quad c_{n}=\sqrt{k}/(nf(d_{n})).$$

In [4, 5], it was shown that under condition (2) and as \(n\to\infty\), probable limiting distributions of statistics \(T_{n}\) are normal and lognormal distributions. The joint asymptotic distribution of intermediate order statistics was studied in [6, 7].

We assume below that \(z_{F}=\inf\{x:F(x)>0\}\), \(k=[n^{\alpha}]\), \(0<\alpha<1\), and \([x]\) denotes the integral part of number \(x,\) and we consider typical classes of distributions widely used in statistical applications.

Class \(B_{1}:\)

$$F(x)\sim a|x|^{\gamma}\exp(-b|x|^{\Delta}),\quad f(x)\sim ab\Delta|x|^{\gamma+\Delta-1}\!\exp(-b|x|^{\Delta})\quad\textrm{for \, }x\to-\infty,\quad a,b,\Delta>0,$$
$$d_{n}=-\left(\frac{(1-\alpha)\ln n}{b}\right)^{1/\Delta}\left(1+\frac{\gamma\ln\ln n+\ln c}{\Delta^{2}(1-\alpha)\ln n}\right),\quad c=\left(\frac{1-\alpha}{b}\right)^{\gamma}a^{\Delta},$$
$$c_{n}=\left(\frac{(1-\alpha)\ln n}{b}\right)^{(1-\Delta)/\Delta}\frac{1}{\Delta bn^{\alpha/2}}.$$

Class \(B_{2}:\)

$$F(x)\sim a|x|^{-\Delta},\quad f(x)\sim a\Delta|x|^{-\Delta-1}\quad\textrm{for \, }x\to-\infty,\quad a,\Delta>0,$$
$$d_{n}=-a^{1/\Delta}n^{(1-\alpha)/\Delta},\quad c_{n}=a^{1/\Delta}/(\Delta n^{-(1-\alpha)/\Delta+\alpha/2}).$$

Class \(B_{3}:\)

$$F(x)\sim a(x-z_{F})^{\Delta},\quad f(x)\sim a\Delta(x-z_{F})^{\Delta-1}\quad\textrm{for \, }x\to z_{F},\quad-\infty<z_{F}<\infty,\quad a,\Delta>0,$$
$$d_{n}=z_{F}+(an^{1-\alpha})^{-1/\Delta},\quad c_{n}=1/(\Delta a^{1/\Delta}n^{(1-\alpha)/\Delta+\alpha/2}).$$

It is easy to show that relation (10) holds for all three classes \(B_{1}\), \(B_{2}\), and \(B_{3}\), and the limiting distribution of the statistics \(T_{n}\) as \(n\to\infty\) is the standard normal distribution. For classes \(B_{1}\) and \(B_{2}\), condition (5) is satisfied,while for class \(B_{3}\) condition (5) is satisfied under the constraint \(\Delta>2.\) Conditions (3) require additional constraints on the parameters for all three classes. For class \(B_{3}\), it follows from (3) that \(z_{F}<0.\)

Examples of the distributions from class \(B_{1}\) that satisfy the hypotheses of the theorem are the standard normal distribution and the Laplace distribution with density \(f(x)=\exp(-\sqrt{2}|x|)/\sqrt{2}\), \(|x|<\infty.\)

3. EXAMPLE OF THE DISTRUBUTION FROM CLASS \(B_{2}\) SATISFYING THE HYPOTHESES OF THE THEOREM

Let us consider a distribution with density \(f(x)=b_{1}/(x^{6}+b_{2}^{6})\),\(|x|<\infty.\) Positive numbers \(b_{1}\) and \(b_{2}\) are determined later by using condition (3). We have

$$F(x)\sim\frac{b_{1}}{5|x|^{5}},\quad f(x)\sim\frac{b_{1}}{x^{6}}\quad\textrm{ as}\quad x\to-\infty,$$
$$1=b_{1}\int\limits_{-\infty}^{\infty}\frac{u^{2}du}{u^{6}+b_{2}^{6}}=\frac{b_{1}}{3b_{2}^{3}}\int\limits_{-\infty}^{\infty}\frac{dt}{t^{2}+1}=\frac{b_{1}\pi}{3b_{2}^{3}};$$

hence, \(b_{2}^{3}=b_{1}\pi/3.\) Further,

$$1=b_{1}\int\limits_{-\infty}^{\infty}\frac{du}{u^{6}+(b_{1}\pi/3)^{2}}=b_{1}(b_{1}\pi/3)^{-5/3}\int\limits_{-\infty}^{\infty}\frac{dt}{t^{6}+1}.$$

Since

$$\int\limits_{0}^{\infty}\frac{dt}{t^{6}+1}=\frac{\pi}{3}\quad\text{(see [8, p. 165 of Russian translation])},$$

it follows that \(b_{1}=6\sqrt{2}/\pi\) and \(b_{2}^{6}=8.\) We have

$$d_{n}=-\left(\frac{b_{1}}{5}\right)^{1/5}n^{(1-\alpha)/5},\quad\ c_{n}=\left(\frac{b_{1}}{5}\right)^{1/5}\frac{1}{5}n^{(1-\alpha)/5-\alpha/2},$$
$$\frac{d_{n}}{c_{n}\sqrt{n}}=-5n^{-(1-\alpha)/2}\to 0\quad\textrm{ as }\quad n\to\infty.$$

The conditions of the theorem are met.

4. AN EXAMPLE OF THE DISTRUBUTION FROM CLASS \(B_{3}\) SATISFYING THE HYPOTHESES OF THE THEOREM

Let \(F(x)=a(x-z_{F})^{\Delta}\), \(f(x)=\Delta a(x-z_{F})^{\Delta-1}\), \(x\in(z_{F},b)\), and \(a,\Delta>0.\) In light of condition \(F(b)=1\) and relations (3), we obtain

$$z_{F}=-\sqrt{\Delta(\Delta+2)},\quad b=\sqrt{\Delta(\Delta+2)},\quad a=\Delta^{\Delta/2}/((\Delta+2)^{\Delta/2}(\Delta+1)^{\Delta}).$$

Under constraint \(\Delta>2\), the conditions of the theorem hold.

5. EXAMPLE OF A DISTRUBUTION FROM CLASS \(B_{3}\) NOT SATISFYING THE HYPOTHESES OF THE THEOREM

Let \(F(x)=1-\exp(-(x+1)),\quad f(x)=\exp(-(x+1))\), \(x>-1.\) We assume that conditions (3) holds,

$$c_{n}=\frac{1}{n^{1-\alpha/2}},\quad d_{n}=-1+\frac{1}{n^{1-\alpha}},\quad\frac{d_{n}}{c_{n}\sqrt{n}}\sim-n^{(1-\alpha)/2}\to-\infty,$$

and condition (5) is not satisfied. We have

$$(X_{[n^{\alpha}]}^{(n)}-d_{n})/c_{n}\stackrel{{\scriptstyle d}}{{\longrightarrow}}\mathcal{N}(0,1)\quad\textrm{ as }\quad n\to\infty,$$

but since \(X_{[n^{\alpha}]}^{(n)}\stackrel{{\scriptstyle P}}{{\longrightarrow}}-1\), \(\sqrt{n}\bar{X}\) and \(\sqrt{n}(1-S)\) are asymptotically normal,

$$\frac{1}{c_{n}\sqrt{n}}\to\infty\quad\textrm{ as }\quad n\to\infty,$$

the second and third terms in representation (6) grow infinitely in absolute magnitude.