Switching record and order statistics via random contractions

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Statistics & Probability Letters 73 (2005) 207–217

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Switching record and order statistics via random contractions

Sevgi Yurt Oncela, Mohammad Ahsanullahb, Fazil A. Alievc,�, Funda Ayguna

aDepartment of Statistics, Kirikkale University, Kirikkale, TurkeybDepartment of Management Sciences, Rider University, USA

cFaculty of Science, Department of Statistics, Ankara University, 06100 Tandogan Ankara, Turkey

Received 18 December 2003; received in revised form 11 January 2005

Available online 7 April 2005

Abstract

In this paper, we investigate a random contraction scheme proposed by Wesolowski and Ahsanullah(2004. Austral. NZ J. Statist. 46, 297–303). Random contraction is used in the record and order statisticssettings. Some distributional recurrence relations were found for the probability density functions anddistribution functions of record and order statistics. These recurrence relations lead to new characteriza-tions of some distributions. Our results are extensions of works by Pakes (1992. Austral. J. Statist. 34,323–339), Wesolowski and Ahsanullah (2004. Austral. NZ J. Statist. 46, 297–303) and Alzaid andAhsanullah (2003. Comm. Statist. Theory Methods 32, 2101–2108). Some new characterizations for theWeibull and Pareto distributions and one special distribution are provided in this study.r 2005 Elsevier B.V. All rights reserved.

Keywords: Order statistics; Records statistics; Random contraction; Power distribution; Weibull distribution; Pareto

distribution

1. Introduction

Let X 1;X 2; . . ., be a sequence of independent and identically distributed random variableswith an absolutely continuous distribution function F and a corresponding probability density

see front matter r 2005 Elsevier B.V. All rights reserved.

spl.2005.03.004

ding author.

ress: [email protected] (F.A. Aliev).

www.elsevier.com/locate/stapro

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S.Y. Oncel et al. / Statistics & Probability Letters 73 (2005) 207–217208

function f. For nX1, we denote the order statistics of X 1;X 2; . . . ;X n by X 1:npX 2:np � � �pX n:n.Define

Uð1Þ ¼ 1; Uðn þ 1Þ ¼ minfj : j4UðnÞ; X j4X UðnÞg.

The sequence fX UðnÞg (fUðnÞg) is known as upper record statistics (record times). Record andorder statistics are of importance in many real-life situations involving data relating to sports,economics, weather and life-tests.In this study, we consider random contractions of switching record and order statistics, which

were discussed by Nevzorov (2001) and Wesolowski and Ahsanullah (2004) for order statistics.The general random contraction setting can be described in the following way: Let U be a randomvariable with a distribution concentrated on ð0; 1Þ or ð1;1Þ, and let X be a positive randomvariable which is independent of U. Then the distribution of XU is a random contraction of thedistribution of X. Assume that Y is a random variable such that the distributions of X and Y aresomehow related and consider the equation

Y ¼d

XU ,

where ¼ddenotes equality in distribution. The main question here is the characterization of

distributions when X and/or Y are some order statistics or record values and U is some givencontractor. Interesting results in this context have been obtained by Nevzorov (2001) andWesolowski and Ahsanullah (2004). Similar characterization schemes were considered in theliterature for the cases of length-biasing equality and equality of convolutions (see Gather et al.(1998), Pakes (1992, 1994), Pakes et al. (1996), Kotz and Steutel (1988), Riedel and Rossberg(1994), Rossberg et al. (1997), Yeo and Milne (1991)).For one-sided contraction, Wesolowski and Ahsanullah (2004) have shown that if U has the

Powerð1; aÞ distribution for some a40 and is independent of X 1; . . . ;X n which are independent,identically distributed positive random variables, and if

X k:n ¼d

X k:n1U ðrespectively; X k:n ¼d

X kþ1:nUÞ

for an arbitrary but fixed k 2 f1; 2; . . . ; n 1g, then there exists a40 such that X 1 has thePowerða; a=nÞ (respectively, Powerða; a=kÞÞ distribution. Here and hereafter, the Powerða; aÞdistribution with positive parameters a and a is defined by the probability density function

f ðxÞ ¼ aaaxa1I ð0;aÞðxÞ, (1)

where I is the indicator function.Wesolowski and Ahsanullah (2004) also considered the two-sided power random contraction

X k:n1U1¼d

X kþ1:nU2,

where Ui has a Powerð1; aiÞ distribution for some ai40. In this case, the authors showed that ifa1=n ¼ a2=k ¼ a, then there exists a40 such that X 1 has a Powerða; aÞ distribution, andconversely.Characterizations of the Gumbel distribution using contractions of records and order statistics

are presented in recent work by Alzaid and Ahsanullah (2003).

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S.Y. Oncel et al. / Statistics & Probability Letters 73 (2005) 207–217 209

In this paper, we use the Pareto distribution Paretoða; mÞ with the probability density function

f ðxÞ ¼ amaxðaþ1ÞI ðm;1ÞðxÞ ða40; m40Þ (2)

and the Weibull distribution Weibullðl;bÞ with the probability density function

f ðxÞ ¼ lbxb1 expflxbgI ð0;1ÞðxÞ ð l40; b40Þ. (3)

The following section consists of the main results without proofs, and in the third section wegive some remarks about distributions of order statistics and record values. Proofs of theoremsare presented in the last section.

2. Main results

First, we give a new characterization of the Pareto distribution. The contracting randomvariable also has the Pareto distribution with parameters ða; 1Þ.

Theorem 1. Let U have the Paretoða; 1Þ distribution for some a40, with U being independent of

X 1;X 2; . . . which are positive, independent identically distributed random variables. For any fixedk 2 f1; 2; . . .g, the necessary and sufficient condition for a random variable X 1 to have the

Paretoða;mÞ distribution with some m40 is that

X Uðkþ1Þ ¼d

X UðkÞU . (4)

If U has the Paretoða; 1Þ distribution then W ¼ logðUÞ will have an exponential distributionwith parameter a. Since log is an increasing function, by taking logarithms of both sides of (4) weobtain the following corollary.

Corollary 1. Let W have an exponential distribution with parameter a and be independent ofX 1;X 2; . . . which are positive, independent identically distributed random variables. If

X Uðkþ1Þ ¼d

X UðkÞ þ W

for arbitrary but fixed k 2 f1; 2; . . .g, then there exists m 2 R such that X 1 þ m has an exponentialdistribution with parameter a, and conversely. For k ¼ 1, this result has recently been obtained by

Alzaid and Ahsanullah (2003).

The next theorem gives a characterization of the Weibull distribution with contraction by aPower distribution.

Theorem 2. Let U have a Powerð1; aÞ distribution for some a40, with U being independent of

X 1;X 2; . . . which are non-negative, independent identically distributed random variables. For anyfixed k 2 f1; 2; . . .g, the necessary and sufficient condition for a random variable X 1 to have the

Weibullðl; a=kÞ distribution with some l40 is that

X UðkÞ ¼d

X Uðkþ1ÞU . (5)

Now we give a characterization of one special distribution with contraction by the Paretodistribution using order statistics.

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Theorem 3. Let U have the Paretoða; 1Þ distribution for some a40, with U being independent of

X 1;X 2; . . . which are non-negative, independent identically distributed random variables. For anyfixed k 2 f1; 2; . . . ; n 1g, the necessary and sufficient condition for a random variable X 1 to have adistribution function F ðxÞ ¼ 1 1=ð1þ ðcxÞa=ðnkÞ

Þ with some c40 is that

X k:n1¼d

X k:nU . (6)

As in the proposition given below, by taking logarithms of both sides of (6) we get the followingcorollary.

Corollary 2. Let W have an exponential distribution with parameter a and be independent ofX 1;X 2; . . . which are non-negative, independent identically distributed random variables. If

X k:n1¼d

X k:n þ W

for arbitrary but fixed k 2 f1; 2; . . .g then there exists m 2 R such that X 1 þ m has distribution

function of the form FðxÞ ¼ 1 1=f1þ exp½xa=ðn kÞ�g and conversely.

Remark 1. Note that for the case X kþ1:n ¼d

X k:n þ W , Wesolowski and Ahsanullah (2004) provedthat X 1 þ m will have an exponential distribution with parameter a=ðn kÞ.

Remark 2. Following the proofs of the theorems, one can show that there is no distribution whichsatisfies the two-sided contraction condition

X UðkÞU1¼d

X Uðkþ1ÞU2,

where Ui has a Powerð1; aiÞ distribution for some ai40. This is true also for the Pareto andWeibull distributions.

3. Distributional recurrences for record and order statistics

Let X 1; . . . ;X n be a random sample from an absolutely continuous distribution function F. LetX Uð1Þ;X Uð2Þ; . . . ;X UðnÞ; . . . be record statistics and X 1:n;X 2:n; . . . ;X n:n be order statistics. Denoteby Fk:n and f k:n the distribution function and probability distribution function of the orderstatistic X k:n ðk ¼ 1; . . . ; nÞ, respectively. The distribution function FkðxÞ and probability densityfunction f kðxÞ of k0th upper record value X UðkÞ is given by

FkðxÞ ¼ PðX UðkÞpxÞ ¼

Z x

1

Rk1ðuÞ

ðk 1Þ!dF ðuÞ (7)

and

f kðxÞ ¼1

ðk 1Þ!Rk1ðxÞf ðxÞ, (8)

where RðxÞ ¼ lnð1 F ðxÞÞ and F ðxÞ ¼ 1 FðxÞ (see Ahsanullah (1995)). By changing variableRðuÞ to t in (7), we can write

FkðxÞ ¼

Z RðxÞ

0

tk1

ðk 1Þ!et dt; 1oxo1. (9)

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Ahsanullah (1995) has shown that

FkðxÞ ¼ F ðxÞXk1j¼0

RjðxÞ

j!¼ eRðxÞ

Xk1j¼0

RjðxÞ

j!. (10)

Using this equality it is easy to check that for any kX1,

FkðxÞ Fkþ1ðxÞ ¼ F ðxÞRkðxÞ

k!. (11)

We present the next proposition from the paper by Wesolowski and Ahsanullah (2004), alongwith their proof. Similar distributional recurrences also can be found in Balasubramanian et al.(1992).

Proposition (Wesolowski and Ahsanullah (2004)). For any k ¼ 1; . . . ; n 1 and any n ¼ 2; 3; . . .

n½Fk:n Fk:n1�f ¼ Ff k:n, (12)

k½Fk:n Fkþ1:n�f ¼ Ff k:n, (13)

nð1 F Þf k:n1 ¼ ðn kÞf k:n, (14)

kð1 F Þf kþ1:n ¼ ðn kÞFf k:n. (15)

Proof. Since for any k ¼ 1; 2; . . . ; n

f k:n ¼ kn

k

� �Fk1ð1 FÞ

nkf

then we have

n½Fk:n Fk:n1�f

¼ n

Zk

n

k

!Fk1ð1 FÞ

nkf

Zk

n 1

k

!Fk1ð1 FÞ

nk1f

" #f

¼ knXnk

i¼0

n

k

!n k

i

!ð1Þi

ZFkþi1f

"

Xnk1

i¼0

n 1

k

!n k 1

i

!ð1Þi

ZFkþi1f

#f

¼ knXnk1

i¼0

n

k

!n k

i

!

n 1

k

!n k 1

i

!" #ð1Þi

Fkþi

k þ i

(

þn

k

!ð1Þnk Fn

n

)f .

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Since

n

k

� �n k

i

� �

n 1

k

� �n k 1

i

� �¼

n

k

� �n k

i

� �k þ i

n

then we get

n½Fk:n Fk:n1�f ¼ kn

k

� �Xnk

i¼0

n k

i

� �ð1ÞiFkþif ¼ k

n

k

� �Fk 1 Fð Þ

nkf .

So (12) is proved. We can prove (13) by a similar method. To prove (14), note that

nð1 F Þf k:n1 ¼ nð1 FÞkn 1

k

!Fkð1 FÞ

nk1f

¼ nkðn 1Þ!

k!ðn k 1Þ!Fkð1 F Þ

nkf ¼ ðn kÞkn

k

!Fkð1 FÞ

nkf

¼ ðn kÞf k:n.

So (14) is proved. The proof of (15) is similar.

4. Proofs of Theorems

Proof of Theorem 1. Necessity: Let a; m40, let X 1 has the Paretoða; mÞ distribution and let U hasthe Paretoða; 1Þ distribution. By (2) the probability density function of X 1 is f ðxÞ ¼ amaxa1 andRðxÞ ¼ a lnðm=xÞ ðx4mÞ.From (10) the distribution function of X UðkÞ is FkðxÞ ¼ 1 ðm=xÞa

Pk1j¼0 ða lnðm=xÞÞj=j! and the

distribution function of X UðkÞU is

PðX UðkÞUoxÞ ¼

Z x=m

1

Fkðx=uÞaua1 du

¼

Z x=m

1

½1 ðmu=xÞaXk1j¼0

ða lnðmu=xÞÞj=j!�aua1 du

¼ 1 ðm=xÞa aðm=xÞaXk1j¼0

ðaÞj

j!

Z x=m

1

lnjðmu=xÞu1 du.

Noting that

d

du

lnjþ1ðmu=xÞ

j þ 1¼ lnj

ðmu=xÞu1

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in the above integral we have

PðX UðkÞUoxÞ ¼ 1 ðm=xÞa aðm=xÞaXk1j¼0

ðaÞj

j!

lnjþ1ðmu=xÞ

j þ 1

��x=m

1

¼ 1 ðm=xÞa ðm=xÞaXk1j¼0

ðaÞjþ1

ðj þ 1Þ!lnjþ1

ðm=xÞ

¼ 1 F ðxÞXk

s¼0

RsðxÞ=s! ¼ Fkþ1ðxÞ,

which proves the necessity part.Sufficiency: Let m ¼ supfx40 : F ðxÞo1g and l ¼ supfxX0 : F ðxÞ ¼ 0g. Since U has the

Paretoða; 1Þ distribution, by (2) and (4) we have for any x40 (for both cases l ¼ 0 and l40)

Fkþ1ðxÞ ¼

Z x=l

1

Fkðx=uÞaua1 du. (16)

If mo1 then by definition of m we have Fkðm=2Þo1, and

1 ¼ Fkþ1ðmÞpZ 2

1

Fkðm=uÞaua1 du þ

Z 1

2

Fkðm=uÞaua1 du

pZ 2

1

aua1 du þ

Z 1

2

Fkðm=uÞaua1 du

pZ 2

1

aua1 du þ Fkðm=2Þ

Z 1

2

aua1 du

oZ 2

1

aua1 du þ

Z 1

2

aua1 du ¼ 1,

which is impossible.Thus, we will consider the only possible case m ¼ 1. Now let us show that F ðxÞ is strictly

increasing. From (8) it follows that Fkþ1ðxÞ also is strictly increasing and vice versa. For someaob, let Fkþ1ðaÞ ¼ Fkþ1ðbÞ40. We can assume that

a ¼ inffx40; Fkþ1ðxÞ ¼ Fkþ1ðbÞg. (17)

From (16) one can write

0 ¼ Fkþ1ðbÞ Fkþ1ðaÞ ¼ aZ a=l

1

½Fkðb=uÞ Fkða=uÞ�ua1 du þ

Z b=l

a=l

Fkðb=uÞua1 du.

Since the expressions under the integral signs are non-negative and continuous, both of theseintegrals must be equal to 0. In the case l40 from the second integral we have Fkðb=uÞ ¼ 0 for alla=lpxpb=l, or, equally FkðlÞ ¼ Fkðlb=aÞ ¼ 0. This contradicts the definition of l; becauselolb=a. When l ¼ 0 from the first integral it follows that Fkðb=uÞ ¼ Fkða=uÞ for all u 2 ½1;1Þ.Noting that a40 and taking u ¼ 1 and u ¼ b=a in the last equality we have FkðbÞ ¼ FkðaÞ ¼

Fkðaða=bÞÞ. From (8) we conclude that F ðaÞ ¼ F ðaða=bÞÞ, or equally Fkþ1ðaÞ ¼ Fkþ1ðaða=bÞÞ. This

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contradicts the assumption (17), because aða=bÞoa; therefore, we conclude that FðxÞ is strictlyincreasing.By changing the variable x=u into t in (16) we obtain

Fkþ1ðxÞ ¼ axaZ x

l

FkðtÞta1 dt (18)

After differentiating both sides with respect to x, we have

f kþ1ðxÞ ¼ a2xa1Z x

l

FkðtÞta1 dt þ axaFkðxÞx

a1

¼ ax1 axaZ x

l

FkðtÞta1 dt þ FkðxÞ

�

and by (18) we can write

f kþ1ðxÞ ¼ ax1½FkðxÞ Fkþ1ðxÞ�. (19)

Substituting expressions (8) and (11) into Eq. (19) we have

RkðxÞ

k!f ðxÞ ¼ ax1F ðxÞ

RkðxÞ

k!,

dFðxÞ

1 F ðxÞ¼

adx

x

and

lnð1 F ðxÞÞ ¼ a lnðxcÞ,

where c40 is arbitrary.Consequently, F ðxÞ ¼ 1 ðcxÞa for any x4l, and the condition F ðlÞ ¼ 0 implies that l ¼ 1=c.

Therefore, F ðxÞ corresponds to the Paretoða; mÞ distribution with m ¼ 1=c. &

Proof of Theorem 2. This can be done using a method similar to that in the proof of Theorem 1and hence is omitted. Only note that in this case we will have the identity

xf kðxÞ ¼ a½FkðxÞ Fkþ1ðxÞ�

instead of (19). &

Proof of Theorem 3. Necessity: Let U has the Paretoða; 1Þ distribution for some a40, andlet the distribution function of X 1 be F ðxÞ ¼ 1 1=ð1þ ðcxÞa=ðnkÞ

Þ, ðxX0Þ. By changingthe variable x=u into t and taking the derivative we obtain the probability density function ofX k:nU asZ 1

1

Fk:nðx=uÞaua1 du

� �0

¼ axaZ x

0

Fk:nðtÞta1 dt

� �0

¼ ax1 Fk:nðxÞ axaZ x

0

Fk:nðtÞta1 dt

�¼ ax1½Fk:nðxÞ Fk:n1ðxÞ�. ð20Þ

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On the other hand, we have 1 FðxÞ ¼ 1=ð1þ ðcxÞa=ðnkÞÞ and

f ðxÞ ¼a

ðn kÞx

ðcxÞa=ðnkÞ

ð1þ ðcxÞa=ðnkÞÞ2.

Therefore,

f ðxÞ

1 F ðxÞ¼

aðn kÞx

ðcxÞa=ðnkÞ

1þ ðcxÞa=ðnkÞ¼

aðn kÞx

F ðxÞ,

which is equivalent to

ðn kÞf k:nðxÞ

nð1 FðxÞÞ¼

ax

FðxÞf k:nðxÞ

nf ðxÞ.

Substituting (12) and (14) in the last equation and simplifying, we have

f k:n1ðxÞ ¼ ax1½Fk:nðxÞ Fk:n1ðxÞ�.

This means that the probability density function of X k:nU by (20) and the last expression forf k:n1ðxÞ are the same, which proves the necessity part. &

Sufficiency: Let m ¼ supfx40 : FðxÞo1g and l ¼ supfxX0 : F ðxÞ ¼ 0g. Since U has theParetoða; 1Þ distribution, by (2) and (6) we have for any x40 (for both cases l ¼ 0 and l40)

Fk:n1ðxÞ ¼

Z x=l

1

Fk:nðx=uÞaua1 du. (21)

If mo1 then we have Fk:nðm=2Þo1, and

1 ¼ Fk:n1ðmÞpZ 2

1

Fk:nðm=uÞaua1 du þ

Z 1

2

Fk:nðm=uÞaua1 du

pZ 2

1

aua1 du þ

Z 1

2

Fk:nðm=uÞaua1 du

pZ 2

1

aua1 du þ Fk:nðm=2Þ

Z 1

2

aua1 du

oZ 2

1

aua1 du þ

Z 1

2

aua1 du ¼ 1,

which is impossible.As in the proof of Theorem 1, consider the only possible case m ¼ 1. By changing the variable

x=u in (21) into t we obtain

Fk:n1ðxÞ ¼ axaZ x

l

Fk:nðtÞta1 dt. (22)

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After differentiating both sides with respect to x we have

f k:n1ðxÞ ¼ a2xa1Z x

l

Fk:nðtÞta1 dt þ axaFk:nðxÞx

a1

¼ ax1 Fk:nðxÞ axaZ x

l

Fk:nðtÞta1 dt

�

and combining with (22) we can write

f k:n1ðxÞ ¼ ax1½Fk:nðxÞ Fk:n1ðxÞ�.

Now taking into consideration the expressions for f k:n1ðxÞ and Fk:nðxÞ Fk:n1ðxÞ from (12) and(14) and substituting them in the last equation, we have

ðn kÞf k:nðxÞ

nð1 FðxÞÞ¼

ax

FðxÞf k:nðxÞ

nf ðxÞ,

or simplifying,

f ðxÞ

1 F ðxÞ¼

aðn kÞx

F ðxÞ.

The left-hand side of this equation is R0ðxÞ. Since FðxÞ ¼ 1 eRðxÞ we have

R0ðxÞ ¼a

ðn kÞxð1 eRðxÞÞ;

dR

1 eR¼

aðn kÞ

dx

x,

lnðeRðxÞ 1Þ ¼ lnðcxÞa=ðnkÞ,

where c40 is arbitrary; therefore,

eRðxÞ ¼ 1þ ðcxÞa=ðnkÞ,

FðxÞ ¼ 1 eRðxÞ ¼ 11

1þ ðcxÞa=ðnkÞ; xXl.

The condition F ðlÞ ¼ 0 implies that l ¼ 0. Thus,

FðxÞ ¼ 11

1þ ðcxÞa=ðnkÞ; xX0,

which proves the theorem.

Acknowledgements

The authors would like to thank an anonymous referee for detailed comments andsuggestions.

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Switching record and order statistics via random contractions

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