## Abstract

In this paper, we consider the ratios of order statistics in samples from uniform distribution and establish strong and weak laws for these ratios.

Show Summary Details# Various limit theorems for ratios from the uniform distribution

#### Open Access

## Abstract

## 1 Introduction

## 2 Main results

## 2.1 Properties of *R*_{n12}

## 2.2 Properties of*R*_{n23}

## 2.3 Properties of *R*_{n1j}

## Acknowledgement

## References

## About the article

## Citing Articles

*Journal of Inequalities and Applications*, 2017, Volume 2017, Number 1

More options …# Open Mathematics

### formerly Central European Journal of Mathematics

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Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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In this paper, we consider the ratios of order statistics in samples from uniform distribution and establish strong and weak laws for these ratios.

Keywords: Uniform distribution; Order statistics; Almost sure convergence; Strong laws of large numbers

If the random variables *X*_{1}, …, *X _{n}* are arranged in order of magnitude and then written

$$P({X}_{n,k}\le x)={\displaystyle \sum _{i=k}^{n}\left(\begin{array}{c}n\\ i\end{array}\right)}{[F(x)]}^{i}{[1-F(x)]}^{n-i},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\infty <x<\infty .$$

In the theory of order statistics, the uniform distribution plays an important role. For instance, let us introduce independent identically distributed random variables *η*_{1}, ···, *η _{n}* and

$$({\eta}_{1},\cdots ,{\eta}_{n})\stackrel{d}{=}\left(F({\xi}_{1}),\cdots ,F({\xi}_{n})\right)$$

if *F* is continuous and

$$\left({\xi}_{1},\cdots ,{\xi}_{n}\right)\stackrel{d}{=}\left({F}^{-1}({\eta}_{1}),\cdots ,{F}^{-1}({\eta}_{n})\right)$$

where *F*^{–1} is quantile function. Let *U _{n}*

$$({U}_{n,1},\cdots ,{U}_{n,n})\stackrel{d}{=}\left(F({X}_{n,1}),\cdots ,F({X}_{n,n})\right)$$

and

$$({X}_{n,1},\cdots ,{X}_{n,n})\stackrel{d}{=}\left(F({U}_{n,1}),\cdots ,F({U}_{n,n})\right).$$

In general, spacings of uniform random variables cannot be independent. However it was shown by Malmquist [3] that certain ratios of order statistics *U _{ni}* of uniform random variables are independent, namely,

$$1-{U}_{n,1},\frac{1-{U}_{n,2}}{1-{U}_{n,1}},\cdots ,\frac{1-{U}_{n,n}}{1-{U}_{n,n-1}}$$

are independent random variables and

$$\frac{1-{U}_{n,r}}{1-{U}_{n,r-1}}\stackrel{d}{=}{U}_{n-r+1,n-r+1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}r=1,\cdots ,n,$$

(with the convention that *U _{n}*

In this paper, we consider the following ratios of order statistics from uniform distribution.

$${R}_{nij}=\frac{{X}_{n}(j)}{{X}_{n}(i)},1\le i<j\le {m}_{n},$$

where *X _{n}*

Assume that {*X _{ni}*,

$${R}_{nij}=\frac{{X}_{n}(j)}{{X}_{n}(i)},1\le i<j\le {m}_{n}$$

denote the ratio of the order statistics {*X _{n}*

*Let* 0 = *a _{n}* <

$$f(r)=\frac{1}{{r}^{2}}I(r>1)$$

*which is independent of n*.

*Proof*. It is easy to see

$$f({x}_{1},{x}_{2})=\frac{{m}_{n}({m}_{n}-1)}{{b}_{n}^{{m}_{n}}}{({b}_{n}-{x}_{2})}^{{m}_{n}-2}I(0\le {x}_{1}\le {x}_{2}\le {b}_{n}).$$

Let *w* =*x*_{1}, $r=\frac{{x}_{2}}{{x}_{1}}$, then the Jacobian is *w* and it is not difficult to get the joint probability density function of *w* and *r* is

$$f(w,r)=\frac{{m}_{n}({m}_{n}-1)}{{b}_{n}^{{m}_{n}}}w{({b}_{n}-rw)}^{{m}_{n}-2}I(w\ge 0,r>1,rw\le {b}_{n}),$$

then we have

$$\begin{array}{ll}f(r)\hfill & =\frac{{m}_{n}({m}_{n}-1)}{{b}_{n}^{{m}_{n}}}{\displaystyle \underset{0}{\overset{\frac{{b}_{n}}{r}}{\int}}w{({b}_{n}-rw)}^{m{}_{n}-2}dw}\hfill \\ \hfill & =\frac{{m}_{n}({m}_{n}-1)}{{b}_{n}^{{m}_{n}}{r}^{2}}{\displaystyle \underset{{b}_{n}}{\overset{0}{\int}}({u}^{{m}_{n}-1}-{b}_{n}{u}^{{m}_{n}-2})du}\hfill \\ \hfill & =\frac{1}{{r}^{2}}.\hfill \end{array}$$

□*Obviously*, *the expectation of R _{n}*

*Let* 0 = *a _{n}* <

$$\underset{N\to \infty}{\mathrm{lim}}\frac{{\displaystyle \sum _{n=1}^{N}\frac{{(\mathrm{ln}n)}^{\alpha}{X}_{n(2)}}{n{X}_{n(1)}}}}{{(\mathrm{ln}N)}^{\alpha +2}}=\frac{1}{\alpha +2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}almost\text{\hspace{0.17em}}\text{\hspace{0.17em}}surely.$$

*Theorem 2.3 extends the result in Adler [6*, *Theorem 3.1]*, *which proved the same result for the sample from the uniform distribution on* [0, *p*]. *Since the proof is similar to Adler*, *we omit it*.

*Let* 0 < *a _{n}* <

$$f(r)=\frac{{({b}_{n}-{a}_{n}r)}^{{m}_{n}-1}[({m}_{n}-1){a}_{n}r+b}{{({b}_{n}-{a}_{n})}^{{m}_{n}}{r}^{2}}I\left(1<r<\frac{{b}_{n}}{{a}_{n}}\right).$$

*Proof*. By a straightforward computation we have

$$f({x}_{1},{x}_{2})=\frac{{m}_{n}({m}_{n}-1)}{{\left({b}_{n}-{a}_{n}\right)}^{{m}_{n}}}{({b}_{n}-{x}_{2})}^{{m}_{n}-2}I({a}_{n}\le {x}_{1}\le {x}_{2}\le {b}_{n}).$$

Let *w* = *x*_{1}, $r=\frac{{x}_{2}}{{x}_{1}}$, then the Jacobian is *w*, it is easy to get the joint probability density function of *w*and *r* is

$$f(w,r)=\frac{{m}_{n}({m}_{n}-1)}{{({b}_{n}-{a}_{n})}^{{m}_{n}}}w{({b}_{n}-rw)}^{{m}_{n}-2}I(w\ge {a}_{n},r>1,rw\le {b}_{n}),$$

then we have

$$\begin{array}{ll}f(r)\hfill & =\frac{{m}_{n}({m}_{n}-1)}{{({b}_{n}-{a}_{n})}^{{m}_{n}}}{\displaystyle \underset{{a}_{n}}{\overset{\frac{{b}_{n}}{r}}{\int}}w{({b}_{n}-rw)}^{m{}_{n}-2}dw}\hfill \\ \hfill & =\frac{{m}_{n}({m}_{n}-1)}{{({b}_{n}-{a}_{n})}^{{m}_{n}}{r}^{2}}{\displaystyle \underset{{b}_{n}-{a}_{n}r}{\overset{0}{\int}}({b}_{n}{u}^{{m}_{n}-2}-{u}^{{m}_{n}-1})du}\hfill \\ \hfill & =\frac{{({b}_{n}-{a}_{n}r)}^{{m}_{n}-1}[({m}_{n}-1){a}_{n}r+{b}_{n}]}{{({b}_{n}-{a}_{n})}^{{m}_{n}}{r}^{2}},\hfill \end{array}$$

which completes the proof of the theorem. □

*Let* 0 < *a _{n}* <

$$E({R}_{n12})=1+\frac{1}{{({b}_{n}-{a}_{n})}^{{m}_{n}}}{\displaystyle \underset{1}{\overset{\frac{{b}_{n}}{{a}_{n}}}{\int}}\frac{{({b}_{n}-{a}_{n}r)}^{{m}_{n}}}{r}dr=:1+{\Delta}_{n},}$$

*where*

$$\frac{{b}_{n}-a{}_{n}}{{b}_{n}({m}_{n}+1)}\le {\Delta}_{n}\le \frac{{b}_{n}-{a}_{n}}{{a}_{n}({m}_{n}+1)}$$

*In particular, if as n* → ∞,

$$\frac{{b}_{n}-{a}_{n}}{{a}_{n}{m}_{n}}\to 0,$$

*then we have*

$$E({R}_{n12})\to 1.$$

*Proof*. By Theorem 2.5 we have

$$\begin{array}{ll}E({R}_{n12})\hfill & =\frac{1}{{({b}_{n}-{a}_{n})}^{{m}_{n}}}{\displaystyle \underset{1}{\overset{\frac{{b}_{n}}{{a}_{n}}}{\int}}\frac{{({b}_{n}-{a}_{n}r)}^{{m}_{n}-1}[({m}_{n}-1){a}_{n}r+{b}_{n}]}{r}dr}\hfill \\ \hfill & =\frac{1}{{({b}_{n}-{a}_{n})}^{{m}_{n}}}{\displaystyle \underset{1}{\overset{\frac{{b}_{n}}{{a}_{n}}}{\int}}{a}_{n}{m}_{n}{({b}_{n}-{a}_{n}r)}^{{m}_{n}-1}+\frac{({b}_{n}-{a}_{n}r)}{r}dr}\hfill \\ \hfill & =1+\frac{1}{{({b}_{n}-{a}_{n})}^{{m}_{n}}}{\displaystyle \underset{1}{\overset{\frac{{b}_{n}}{{a}_{n}}}{\int}}\frac{{({b}_{n}-{a}_{n}r)}^{{m}_{n}}}{r}dr,}\hfill \end{array}$$

which yields the theorem. □

*Let* 0 < *a _{n}* <

$$E\left({R}_{n12}^{2}\right)=1+\frac{2({b}_{n}-{a}_{n})}{{a}_{n}({m}_{n}+1)}$$(1)

*and the variance of R _{n}*

$$Var({R}_{n12})\le \frac{{({b}_{n}-{a}_{n})}^{2}}{({m}_{n}+1){b}_{n}}\left(\frac{2}{{a}_{n}}-\frac{1}{{b}_{n}({m}_{n}+1)}\right).$$

*Proof*. From Theorem 2.5 and Theorem 2.6, it is easy to obtain the desired results.

*From Theorem 2.7*, *when* $\frac{{b}_{n}-{a}_{n}}{{a}_{n}{m}_{n}}\to 0$, *then Var*(*R _{n}*

Based on the above theorems for the cases 0 < *a _{n}* <

*Let* 0 < *a _{n}* <

$$\frac{1}{\sqrt{N}}{\displaystyle \sum _{n=1}^{N}\frac{{R}_{n12}-\mathbb{E}{R}_{n12}}{\sqrt{Var}({R}_{n12})}\stackrel{d}{\to}N(0,1).}$$

*Proof*. By the Liapounov’s condition, the theorem holds. □

In this subsection, we discuss the properties of the ratio of the second and third order statistics from an independent identically distributed sample of uniform distribution.

Let 0 = *a _{n}* <

$$f(r)=\frac{2}{{r}^{3}}I(r>1)$$

*which is independent of n*.

*Proof*. It is not difficult to obtain

$$f({x}_{2},{x}_{3})=\frac{{m}_{n}!}{({m}_{n}-3)!{b}_{n}^{{m}_{n}}}{x}_{2}{({b}_{n}-{x}_{3})}^{{m}_{n}-3}I(0\le {x}_{2}\le {x}_{3}\le {b}_{n}).$$

Let *w* = *x _{2}*, $r=\frac{{x}_{3}}{{x}_{2}}$, then the Jacobian is

$$f(w,r)=\frac{{m}_{n}!}{({m}_{n}-3)!{b}_{n}^{{m}_{n}}}{w}^{2}{({b}_{n}-rw)}^{{m}_{n}-3}I(w\ge 0,r>1,rw\le {b}_{n}),$$

then we have

$$\begin{array}{ll}f(r)\hfill & =\frac{{m}_{n}!}{({m}_{n}-3)!{b}_{n}^{{m}_{n}}}{\displaystyle \underset{0}{\overset{\frac{{b}_{n}}{r}}{\int}}{w}^{2}{({b}_{n}-rw)}^{{m}_{n}-3}dw}\hfill \\ \hfill & =\frac{{m}_{n}!}{({m}_{n}-3)!{b}_{n}^{{m}_{n}}}{\displaystyle \underset{{b}_{n}}{\overset{0}{\int}}\left(\frac{{b}_{n}-u}{r}\right){u}^{{m}_{n}-3}\frac{-du}{r}}\hfill \\ \hfill & =\frac{{m}_{n}!(-1)}{({m}_{n}-3)!{b}_{n}^{{m}_{n}}{r}^{3}}{\displaystyle \underset{{b}_{n}}{\overset{0}{\int}}\left({b}_{n}^{2}{u}^{{m}_{n}-3}+{u}^{{m}_{n}-1}+2{b}_{n}{u}^{{m}_{n}-2}\right)du}\hfill \\ \hfill & =\frac{2}{{r}^{3}}.\hfill \end{array}$$

which completes the proof. □

*It is not difficult to check that for any* 0 < *δ* < 2, *the δ-order moment of R _{n}*

*Let* 0 = *a _{n}* <

$$\frac{1}{\sqrt{N}}{\displaystyle \sum _{n=1}^{N}({R}_{n23}-E{R}_{n23})\stackrel{d}{\to}N(0,1).}$$

*Proof*. Let us define *L*(*x*) := *E*(_{ξ}^{2}1_{{|ξ| ≤ x}} and

$${c}_{N}=1\vee \mathrm{sup}\{x\ge 0;NL(N)\ge {x}^{2}\},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}N\in \mathbb{N}.$$

It is easy to see that ${c}_{N}=\sqrt{N}$, so by Theorem 4.17 in [7], the desired result can be obtained. □

*Let* 0 < *a _{n}* <

$$f(r)=\frac{{({b}_{n}-{a}_{n}r)}^{{m}_{n}-1}[({m}_{n}-2){a}_{n}r+2{b}_{n}]}{{({b}_{n}-{a}_{n})}^{{m}_{n}}{r}^{3}}I\left(1<r<\frac{{b}_{n}}{{a}_{n}}\right).$$

*Proof*. It is easy to get

$$f({x}_{2},{x}_{3})=\frac{{m}_{n}!}{({m}_{n}-3)!{({b}_{n}-{a}_{n})}^{{m}_{n}}}({x}_{2}-{a}_{n}){({b}_{n}-{x}_{3})}^{m{}_{n}-3}I({a}_{n}\le {x}_{2}\le {x}_{3}\le {b}_{n}).$$

Let *w* = *x*_{2}, $r=\frac{{x}_{3}}{{x}_{2}}$, then the Jacobian is *w* and it is not difficult to get the joint probability density function of *w* and *r* is

$$f(w,r)=\frac{{m}_{n}!}{({m}_{n}-3)!{({b}_{n}-{a}_{n})}^{{m}_{n}}}w(w-{a}_{n}){({b}_{n}-rw)}^{m{}_{n}-3}I(w\ge {a}_{n},r>1,rw\le {b}_{n}),$$

then we have

$$\begin{array}{ll}f(r)\hfill & =\frac{{m}_{n}!}{({m}_{n}-3)!{({b}_{n}-{a}_{n})}^{{m}_{n}}}{\displaystyle \underset{{a}_{n}}{\overset{\frac{{b}_{n}}{r}}{\int}}w(w-{a}_{n}){({b}_{n}-rw)}^{{m}_{n}-3}dw}\hfill \\ \hfill & =\frac{{m}_{n}!(-1)}{({m}_{n}-3)!{({b}_{n}-{a}_{n})}^{{m}_{n}}{r}^{3}}{\displaystyle \underset{{b}_{n}-{a}_{n}r}{\overset{0}{\int}}\left[{u}^{{m}_{n}-1}+({a}_{n}r-2{b}_{n}){u}^{{m}_{n}-2}+({b}_{n}-{a}_{n}r){b}_{n}{u}^{{m}_{n}-3}\right]du}\hfill \\ \hfill & =\frac{{({b}_{n}-{a}_{n}r)}^{m{}_{n}-1}[({m}_{n}-2){a}_{n}r+2{b}_{n}]}{{({b}_{n}-{a}_{n})}^{{m}_{n}}{r}^{3}},\hfill \end{array}$$

which completes the proof. □

From Theorem 2.13, we know that for all *δ >* 0, the *δ*-order moments of *R _{n}*

*Let* 0 < *a _{n}* <

$$E({R}_{n23})=1+\frac{{b}_{n}-{a}_{n}}{{a}_{n}}{\displaystyle \underset{0}{\overset{1}{\int}}\frac{{(1-t)}^{{m}_{n}}}{{\left(\frac{{b}_{n}-{a}_{n}}{{a}_{n}}t+1\right)}^{2}}dt.}$$

*In particular*, *if as n* → ∞,

$$\frac{{b}_{n}-{a}_{n}}{{a}_{n}{m}_{n}}\to 0,$$

*then we have*

$$E({R}_{n23})\to 1.$$

*Proof*. By using Theorem 2.13 we obtain

$$\begin{array}{ll}E({R}_{n23})\hfill & =\frac{1}{{({b}_{n}-{a}_{n})}^{{m}_{n}}}{\displaystyle \underset{1}{\overset{\frac{{b}_{n}}{{a}_{n}}}{\int}}\frac{{({b}_{n}-{a}_{n}r)}^{{m}_{n}-1}[({m}_{n}-2){a}_{n}r+2{b}_{n}]}{{r}^{2}}}\hfill \\ \hfill & =\frac{1}{({b}_{n}-{a}_{n})}{\displaystyle \underset{1}{\overset{\frac{{b}_{n}}{{a}_{n}}}{\int}}\left[\frac{{a}_{n}{m}_{n}{({b}_{n}-{a}_{n}r)}^{{m}_{n}-1}}{r}+\frac{2{({b}_{n}-{a}_{n}r)}^{{m}_{n}}}{{r}^{2}}\right]dr}\hfill \\ \hfill & =\frac{{a}_{n}}{{({b}_{n}-{a}_{n})}^{{m}_{n}}}{\displaystyle \underset{{a}_{n}}{\overset{{b}_{n}}{\int}}\left[\frac{{m}_{n}{({b}_{n}-u)}^{m{}_{n}-1}}{u}+\frac{2{({b}_{n}-u)}^{{m}_{n}}}{{u}^{2}}\right]du}\hfill \\ \hfill & ={\displaystyle \underset{0}{\overset{1}{\int}}\frac{{m}_{n}{(1-t)}^{{m}_{n}-1}}{\frac{{b}_{n}-{a}_{n}}{{a}_{n}}t+1}dt+{\displaystyle \underset{0}{\overset{t}{\int}}\frac{2{(1-t)}^{{m}_{n}}}{\frac{{b}_{n}-{a}_{n}}{{a}_{n}}{\left(t+\frac{{a}_{n}}{{b}_{n}-{a}_{n}}\right)}^{2}}dt}}\hfill \\ \hfill & =2-{\displaystyle \underset{0}{\overset{1}{\int}}\frac{{m}_{n}{(1-t)}^{{m}_{n}-1}}{\frac{{b}_{n}-{a}_{n}}{{a}_{n}}t+1}dt}\hfill \\ \hfill & =1+\frac{{b}_{n}-{a}_{n}}{{a}_{n}}{\displaystyle \underset{0}{\overset{1}{\int}}\frac{{(1-t)}^{{m}_{n}}}{{\left(\frac{{b}_{n}-{a}_{n}}{{a}_{n}}t+1\right)}^{2}}dt}\hfill \end{array}$$

where we use the variable replacement $t:=\frac{u-{a}_{n}}{{b}_{n}-{a}_{n}}$ in the fourth equality. □

*Let* 0 < *a _{n}* <

$$E\left({R}_{n23}^{2}\right)=1+2{\displaystyle \underset{0}{\overset{1}{\int}}\frac{{(1-t)}^{{m}_{n}}}{t+{a}_{n}/({b}_{n}-{a}_{n})}dt.}$$

*In particular*, *if as n* → ∞,

$$\frac{{b}_{n}-{a}_{n}}{{a}_{n}{m}_{n}}\to 0,$$

*then we have*

$$E\left({R}_{n23}^{2}\right)\to 1.$$

*Proof*. By Theorem 2.13 we have

$$\begin{array}{ll}E\left({R}_{n23}^{2}\right)\hfill & =\frac{1}{{({b}_{n}-{a}_{n})}^{{m}_{n}}}{\displaystyle \underset{1}{\overset{\frac{{b}_{n}}{{a}_{n}}}{\int}}\frac{{({b}_{n}-{a}_{n}r)}^{{m}_{n}-1}[({m}_{n}-2){a}_{n}r+1{b}_{n}]}{r}dr}\hfill \\ \hfill & =\frac{1}{{({b}_{n}-{a}_{n})}^{{m}_{n}}}{\displaystyle \underset{1}{\overset{\frac{{b}_{n}}{{a}_{n}}}{\int}}\left[{a}_{n}{m}_{n}{({b}_{n}-{a}_{n}r)}^{{m}_{n}-1}+\frac{2{({b}_{n}-{a}_{n}r)}^{{m}_{n}}}{r}\right]}dr\hfill \\ \hfill & =1+\frac{2}{{({b}_{n}-{a}_{n})}^{{m}_{n}}}{\displaystyle \underset{1}{\overset{\frac{{b}_{n}}{{a}_{n}}}{\int}}\frac{{({b}_{n}-{a}_{n}r)}^{{m}_{n}}}{r}dr}\hfill \\ \hfill & =1+\frac{2}{{({b}_{n}-{a}_{n})}^{{m}_{n}}}{\displaystyle \underset{{a}_{n}}{\overset{{b}_{n}}{\int}}\frac{{({b}_{n}-u)}^{{m}_{n}}}{u}}du\hfill \\ \hfill & =1+2{\displaystyle \underset{0}{\overset{1}{\int}}\frac{{(1-t)}^{{m}_{n}}}{t+{a}_{n}/({b}_{n}-{a}_{n})}dt\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(t:=\frac{u-{a}_{n}}{{b}_{n}-{a}_{n}}\right).}\hfill \end{array}$$

□*From Theorem 2.14 and Theorem 2.15*, *we know that*

$$Var\left({R}_{n23}\right)\to 0.$$

In this subsection, we establish the properties of the ratio of the first and j ^{th} order statistics from an independent identically distributed sample of uniform distribution with parameters *a _{n}* and

Let 0 = *a _{n}* <

$$f(r)=\frac{m!{(r-1)}^{j-2}}{(j-2)!(m-j)!{r}^{j}}{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{l}m-j\hfill \\ k\hfill \end{array}\right)}\frac{{(-1)}^{k}}{j+k}I(r>1).$$

*Proof*. It is easy to see

$$f({x}_{1},{x}_{j})=\frac{m!}{(j-2)!(m-j)!{b}_{n}^{m}}{({x}_{j}-{x}_{1})}^{j-2}{({b}_{n}-{x}_{j})}^{m-j}I(0\le {x}_{1}\le {x}_{j}\le {b}_{n}).$$

Let *w* = *x*_{1}, $r=\frac{{x}_{j}}{{x}_{1}}$, then the Jacobian is *w* and it is not difficult to get the joint probability density function of *w* and *r* is

$$f(w,r)=\frac{m!}{(j-2)!(m-j)!{b}_{n}^{m}}{w}^{j-1}{(r-1)}^{j-2}{({b}_{n}-rw)}^{m-j}I(w\ge 0,r>1,rw\le b),$$

then we have

$$\begin{array}{ll}f(r)\hfill & =\frac{m!{(r-1)}^{j-2}}{(j-2)!(m-j)!{b}_{n}^{m}}{\displaystyle \underset{0}{\overset{\frac{{b}_{n}}{r}}{\int}}{w}^{j-1}{({b}_{n}-rw)}^{m-j}dw}\hfill \\ \hfill & =\frac{m!{(r-1)}^{j-2}}{(j-2)!(m-j)!{b}^{m}}{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{l}m-j\hfill \\ k\hfill \end{array}\right){b}_{n}^{m-j-k}{(-1)}^{k}{r}^{k}{\displaystyle \underset{0}{\overset{\frac{{b}_{n}}{r}}{\int}}{w}^{j-1+k}dw}}\hfill \\ \hfill & =\frac{m!{(r-1)}^{j-2}}{(j-2)!(m-j)!{r}^{j}}{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{l}m-j\hfill \\ k\hfill \end{array}\right)\frac{{(-1)}^{k}}{j+k},}\hfill \end{array}$$

which completes this proof. □

*Let* 0 = *a _{n}* <

$$\underset{N\to \infty}{\mathrm{lim}}\frac{{\displaystyle \sum _{n=1}^{N}\frac{{(\mathrm{ln}n)}^{\alpha}{X}_{n(j)}}{n{X}_{n(1)}}}}{{(\mathrm{ln}N)}^{\alpha +2}}=\frac{m!}{(j-2)!(m-j)!(\alpha +2)}{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{c}m-j\\ k\end{array}\right)\frac{{(-1)}^{k}}{j+k}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}almost\text{\hspace{0.17em}}surely.}$$

*Proof*. Let *z _{n}* = (ln

$$f(r)=\frac{m!{(r-1)}^{j-2}}{(j-2)!(m-j)!{r}^{j}}{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{c}m-j\\ k\end{array}\right)\frac{{(-1)}^{k}}{j+k}I(r>1).}$$

Next we use the partition

$$\begin{array}{ll}\frac{1}{{d}_{N}}{\displaystyle \sum _{n=1}^{N}{z}_{n}{R}_{n1j}}\hfill & =\frac{1}{{d}_{N}}{\displaystyle \sum _{n=1}^{N}{z}_{n}[{R}_{n1j}I(1\le {R}_{n1j}\le {c}_{n})-E{R}_{n1j}I(1\le {R}_{n1j}\le {c}_{n})]}\hfill \\ \hfill & +\frac{1}{{d}_{N}}{\displaystyle \sum _{n=1}^{N}{z}_{n}{R}_{n1j}I\left({R}_{n1j}>{c}_{n}\right)}\hfill \\ \hfill & +\frac{1}{{d}_{N}}{\displaystyle \sum _{n=1}^{N}{z}_{n}E{R}_{n1j}I\left(1\le {R}_{n1j}\le {c}_{n}\right).}\hfill \end{array}$$

Combining with the the Khintchine-Kolmogorov convergence theorem (see Chow and Teicher [8, Theorem 1 in Page 113]), Kronecker’s lemma (see Chow and Teicher [8, Lemma 2 in Page 114]) and the following analysis

$$\begin{array}{ll}{\displaystyle \sum _{n=1}^{\infty}\frac{1}{{c}_{n}^{2}}E{R}_{n1j}^{2}I\left(1\le {R}_{n1j}\le {c}_{n}\right)}\hfill & ={\displaystyle \sum _{n=1}^{\infty}\frac{1}{{c}_{n}^{2}}\frac{m!}{(j-2)!(m-j)!}\left[{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{c}m-j\\ k\end{array}\right)\frac{{(-1)}^{k}}{j+k}}\right]{\displaystyle \underset{1}{\overset{{c}_{n}}{\int}}{\left(\frac{r-1}{r}\right)}^{j-2}dr}}\hfill \\ \hfill & \le C{\displaystyle \sum _{n=1}^{\infty}\frac{1}{{c}_{n}^{2}}{\displaystyle \underset{1}{\overset{{c}_{n}}{\int}}dr\le C{\displaystyle \sum _{n=1}^{\infty}\frac{1}{{c}_{n}}=C{\displaystyle \sum _{n=1}^{\infty}\frac{1}{n{(\mathrm{ln}n)}^{2}}<\infty ,}}}}\hfill \end{array}$$

we find that the first term vanishes almost surely. By the Borel-Cantelli lemma and the following analysis

$$\begin{array}{ll}{\displaystyle \sum _{n=1}^{\infty}P\{{R}_{n1j}>{c}_{n}\}}\hfill & ={\displaystyle \sum _{n=1}^{\infty}\frac{m!}{(j-2)!(m-j)!}\left[{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{c}m-j\\ k\end{array}\right)\frac{{(-1)}^{k}}{j+k}}\right]{\displaystyle \underset{{c}_{n}}{\overset{\infty}{\int}}{\left(\frac{r-1}{r}\right)}^{j-2}dr}}\hfill \\ \hfill & \le C{\displaystyle \sum _{n=1}^{\infty}{\displaystyle \underset{{c}_{n}}{\overset{\infty}{\int}}\frac{{(r-1)}^{j-2}}{{r}^{j}}}dr=C{\displaystyle \sum _{n=1}^{\infty}{\displaystyle \underset{{c}_{n}}{\overset{\infty}{\int}}{\left(\frac{r-1}{r}\right)}^{j}\frac{1}{{(r-1)}^{2}}dr}}}\hfill \\ \hfill & \le C{\displaystyle \sum _{n=1}^{\infty}{\displaystyle \underset{1}{\overset{{c}_{n}}{\int}}\frac{dr}{{(r-1)}^{2}}=C{\displaystyle \sum _{n=1}^{\infty}\frac{1}{{c}_{n}-1}<\infty ,}}}\hfill \end{array}$$

the second term vanishes. For the third term in the partition, we have

$$\begin{array}{l}E{R}_{n1j}I(1\le {R}_{n1j}\le {c}_{n})\hfill \\ =\frac{m!}{(j-2)!(m-j)!}\left[{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{c}m-j\\ k\end{array}\right)\frac{{(-1)}^{k}}{j+k}}\right]{\displaystyle \underset{1}{\overset{{c}_{n}}{\int}}\frac{{(r-1)}^{j-2}}{{r}^{j-1}}dr}\hfill \\ =\frac{m!}{(j-2)!(m-j)!}\left[{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{c}m-j\\ k\end{array}\right)\frac{{(-1)}^{k}}{j+k}}\right]{\displaystyle \sum _{t=0}^{j-2}\left(\begin{array}{c}j-2\\ t\end{array}\right){(-1)}^{j-2-t}{\displaystyle \underset{1}{\overset{{c}_{n}}{\int}}{r}^{t+1-j}}dr}\hfill \\ =\frac{m!}{(j-2)!(m-j)!}\left[{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{c}m-j\\ k\end{array}\right)\frac{{(-1)}^{k}}{j+k}}\right]\left[{\displaystyle \sum _{t=0}^{j-2}\left(\begin{array}{c}j-2\\ t\end{array}\right){(-1)}^{j-2-t}\frac{{c}_{n}^{t+2-j}-1}{t+2-j}+\mathrm{ln}{c}_{n}}\right]\hfill \\ ~\frac{m!}{(j-2)!(m-j)!}\left[{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{c}m-j\\ k\end{array}\right)\frac{{(-1)}^{k}}{j+k}}\right]\mathrm{ln}{c}_{n}\hfill \\ ~\frac{m!}{(j-2)!(m-j)!}\left[{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{c}m-j\\ k\end{array}\right)\frac{{(-1)}^{k}}{j+k}}\right]\mathrm{ln}n,\hfill \end{array}$$

then we obtain

$$\begin{array}{l}\begin{array}{l}\frac{1}{{d}_{N}}{\displaystyle \sum _{n=1}^{N}{z}_{n}E{R}_{n1j}I(1\le {R}_{n1j}\le {c}_{n})~\frac{\frac{m!}{(j-2)!(m-j)!}\left[{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{c}m-j\\ k\end{array}\right)\frac{{(-1)}^{k}}{j+k}}\right]{\displaystyle \sum _{n=1}^{N}\frac{{(\mathrm{ln}n)}^{\alpha +1}}{n}}}{{(\mathrm{ln}N)}^{\alpha +2}}}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\to \frac{m!}{(j-2)!(m-j)!(\alpha +2)}{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{c}m-j\\ k\end{array}\right)\frac{{(-1)}^{k}}{j+k},}\hfill \end{array}\hfill \end{array}$$

so, the theorem holds. □

*Let* 0 < *a _{n}* <

$$f(r)={C}_{n}(r){\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{c}m-j\\ k\end{array}\right)\frac{{(-1)}^{k}{b}_{n}^{m-j-k}}{k+j}}\left({b}_{n}^{k+j}-{a}_{{}_{n}}^{k+j}{r}^{k+j}\right)I\left(1<r<\frac{{b}_{n}}{{a}_{n}}\right).$$

*where*

$${C}_{n}(r):={C}_{m,{a}_{n},b{}_{n},j}(r):=\frac{m!{(r-1)}^{j-2}}{(j-2)!(m-j)!{({b}_{n}-{a}_{n})}^{{m}_{r}j}}.$$

*Proof*. It is not difficult to obtain

$$f({x}_{1},{x}_{j})=\frac{m!}{(j-2)!(m-j)!{({b}_{n}-{a}_{n})}^{m}}{({x}_{j}-{x}_{1})}^{j-2}{({b}_{n}-{x}_{j})}^{m-j}I({a}_{n}\le {x}_{1}\le {x}_{j}\le {b}_{n}).$$

Let *w* = *x*_{1}, $r=\frac{{x}_{j}}{{x}_{1}}$, then the Jacobian is *w* and it is easy to get the joint probability density function of *w* and *r* is

$$f(w,r)=\frac{m!}{(j-2)!(m-j)!{({b}_{n}-{a}_{n})}^{m}}{w}^{j-1}{(r-1)}^{j-2}{({b}_{n}-rw)}^{m-j}I(w\ge {a}_{n},r>1,rw\le {b}_{n}),$$

then we have

$$\begin{array}{ll}f(r)\hfill & =\frac{m!{(r-1)}^{j-2}}{(j-2)!(m-j)!{({b}_{n}-{a}_{n})}^{m}}{\displaystyle \underset{{a}_{n}}{\overset{\frac{{b}_{n}}{r}}{\int}}{w}^{j-1}{({b}_{n}-rw)}^{m-j}dw}\hfill \\ \hfill & =\frac{m!{(r-1)}^{j-2}}{(j-2)!(m-j)!{({b}_{n}-{a}_{n})}^{m}}{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{c}m-j\\ k\end{array}\right){b}_{n}^{m-j-k}{(-r)}^{k}{\displaystyle \underset{{a}_{n}}{\overset{\frac{{b}_{n}}{r}}{\int}}{w}^{k+j-1}dw}}\hfill \\ \hfill & =\frac{m!{(r-1)}^{j-2}}{(j-2)!(m-j)!{({b}_{n}-{a}_{n})}^{m}{r}^{j}}{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{c}m-j\\ k\end{array}\right)\frac{{(-1)}^{k}{b}_{n}^{m-j-k}}{k+j}}\left({b}_{n}^{k+j}-{a}_{n}^{k+j}{r}^{k+j}\right),\hfill \end{array}$$

which completes this proof. □

*Let* 0 < *a _{n}* <

$$\begin{array}{l}E\left({R}_{n1j}\right)=\frac{m!}{(j-2)!(m-j)!{({b}_{n}-{a}_{n})}^{m}}{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{c}m-j\\ k\end{array}\right)\frac{{(-1)}^{k}{b}_{n}^{m-j-k}}{k+j}}[{b}_{n}^{k+j}\mathrm{ln}\frac{{b}_{n}}{{a}_{n}}\hfill \\ +\frac{{a}_{n}^{k+j}-{b}_{n}^{k+j}}{k+j}+{\displaystyle \sum _{t=0}^{j-3}\left(\begin{array}{c}j-2\\ t\end{array}\right){(-1)}^{j-2-t}\left(\frac{{a}_{n}^{j-t-2}{b}_{n}^{t+k+2}-{b}_{n}^{k+j}}{t-j+2}+\frac{{a}_{n}^{k+j}-{a}_{n}^{j-t-2}{b}_{n}^{t+k+2}}{t+k+2}\right)]}.\hfill \end{array}$$

*Proof*. By Theorem 2.19 we have

$$\begin{array}{ll}E\left({R}_{n1j}\right)\hfill & =\frac{m!}{(j-2)!(m-j)!{({b}_{n}-{a}_{n})}^{m}}{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{c}m-j\\ k\end{array}\right)\frac{{(-1)}^{k}{b}_{n}^{m-j-k}}{k+j}}{\displaystyle \underset{1}{\overset{\frac{{b}_{n}}{{a}_{n}}}{\int}}{(r-1)}^{j-2}}\hfill \\ \hfill & \left(\frac{{b}_{n}^{k+j}}{{r}^{j-1}}-{a}_{n}^{k+j}{r}^{k+1}\right)dr\hfill \\ \hfill & =\frac{m!}{(j-2)!(m-j)!{({b}_{n}-{a}_{n})}^{m}}{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{c}m-j\\ k\end{array}\right)\frac{{(-1)}^{k}{b}_{n}^{m-j-k}}{k+j}}{\displaystyle \sum _{t=0}^{j-2}\left(\begin{array}{c}j-2\\ t\end{array}\right){(-1)}^{j-2-t}}\hfill \\ \hfill & {\displaystyle \underset{1}{\overset{\frac{{b}_{n}}{{a}_{n}}}{\int}}\left({b}_{n}^{k+j}{r}^{t-j+1}-{a}_{n}^{k+j}{r}^{t+k+1}\right)dr}\hfill \\ \hfill & =\frac{m!}{(j-2)!(m-j)!{({b}_{n}-{a}_{n})}^{m}}{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{c}m-j\\ k\end{array}\right)\frac{{(-1)}^{k}{b}_{n}^{m-j-k}}{k+j}}[{\displaystyle \underset{1}{\overset{\frac{{b}_{n}}{{a}_{n}}}{\int}}\left(\frac{{b}_{n}^{k+j}}{r}-{a}_{n}^{k+j}{r}^{j+k-1}\right)dr}\hfill \\ \hfill & +{\displaystyle \sum _{t=0}^{j-3}\left(\begin{array}{c}j-2\\ t\end{array}\right){(-1)}^{j-2-t}{\displaystyle \underset{1}{\overset{\frac{{b}_{n}}{{a}_{n}}}{\int}}\left({b}_{n}^{k+j}{r}^{t-j+1}-{a}_{n}^{k+j}{r}^{t-j+1}\right)dr}]}\hfill \\ \hfill & \frac{m!}{(j-2)!(m-j)!{({b}_{n}-{a}_{n})}^{m}}{\displaystyle \sum _{k=0}^{m-j}\left(\begin{array}{c}m-j\\ k\end{array}\right)\frac{{(-1)}^{k}{b}_{n}^{m-j-k}}{k+j}}[{b}_{n}^{k+j}\mathrm{ln}\frac{{b}_{n}}{{a}_{n}}+\frac{{a}_{n}^{k+j}-{b}_{n}^{k+j}}{k+j}\hfill \\ \hfill & +{\displaystyle \sum _{t=0}^{j-3}\left(\begin{array}{c}j-2\\ t\end{array}\right){(-1)}^{j-2-t}\left(\frac{{a}_{n}^{j-t-2}{b}_{n}^{t+k+2}-{b}_{n}^{k+j}}{t-j+2}+\frac{{a}_{n}^{k+j}-{a}_{n}^{j-t-2}{b}_{n}^{t+k+2}}{t+k+2}\right)].}\hfill \end{array}$$

so, the theorem holds. □

This work is supported by IRTSTHN (14IRTSTHN023), NSFC (11471104, 71501016), Science and Technology Project of Beijing Municipal Education Commission (71E1610975).

- [1]
Arnold B. C., Balakrishnan N., Nagaraja H. N., A first course in order statistics, Philadelphia, PA, 2008. Google Scholar

- [2]
Reiss R. D., Approximate distributions of order statistics, Springer-Verlag, New York, 1989. Google Scholar

- [3]
Malmquist S., On a property of order statistics from a rectangular distribution, Skand. Aktuarietidskr., 33, 1950, 214-222. Google Scholar

- [4]
Adler A., Strong laws for ratios of order statistics from exponentials, Bull. Inst. Math. Acad. Sin. (N.S.), 2015, 10(1), 101-111. Google Scholar

- [5]
Miao, Y., Wang, R. J., Adler, A., Limit theorems for order statistics from exponentials. Statist. Probab. Lett., 2016, 110, 51-57. Google Scholar

- [6]
Adler, A., Laws of large numbers for ratios of uniform random variables. Open Math., 2015, 13, 571-576. Google Scholar

- [7]
Kallenberg O., Foundations of modern probability, Springer-Verlag, New York, 1997. Google Scholar

- [8]
Chow Y. S., Teicher H., Probability Theory: Independence, Interchangeability, Martingales, 3rd ed., Springer-Verlag, New York, 1997. Google Scholar

**Received**: 2016-02-17

**Accepted**: 2016-05-27

**Published Online**: 2016-06-11

**Published in Print**: 2016-01-01

**Citation Information: **Open Mathematics, Volume 14, Issue 1, Pages 393–403, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2016-0037.

© 2016 Miao *etal*., published by De Gruyter Open. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License. BY-NC-ND 3.0

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Shou-Fang Xu and Yu Miao

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