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ORDER STATISTICS FROM A PARETO DISTRIBUTION

ANDR ´E ADLER

Received 28 May 2005 and in revised form 23 September 2005

Consider independent and identically distributed random variables {Xnk, 1km, n1} from the Pareto distribution. We randomly select two adjacent order statistics from each row, Xn(i) andXn(i+1), where 1im1. Then, we test to see whether or not strong and weak laws of large numbers with nonzero limits for weighted sums of the random variablesXn(i+1)/Xn(i)exist, where we place a prior distribution on the selection of each of these possible pairs of order statistics.

1. Introduction

In this paper, we observe weighted sums of ratios of order statistics taken from small sam- ples. We look atmobservations from the Pareto distribution, that is, f(x)=pxp1I(x 1), where p >0. Then, we observe two adjacent order statistics from our sample, that is,X(i)X(i+1)for 1im1. Next, we obtain the random variableRi=X(i+1)/X(i), i=1,...,m1, which is the ratio of our adjacent order statistics. The density ofRiis

f(r)=p(mi)rp(mi)1I(r1). (1.1) We will derive this and show how the distributions of these random variables are related.

The joint density of the original i.i.d. Pareto random variablesX1,...,Xmis fx1,...,xm

=pmx1p1···xmp1Ix11···Ixm1, (1.2)

hence the density of the corresponding order statisticsX(1),...,X(m)is fx(1),...,x(m)

=pmm!x(1)p1···x(mp)1I1x(1)x(2)≤ ··· ≤x(m)

. (1.3)

Next, we obtain the joint density ofX(1),R1,...,Rm1. In order to do that, we need the

Copyright©2005 Hindawi Publishing Corporation

International Journal of Mathematics and Mathematical Sciences 2005:21 (2005) 3427–3441 DOI:10.1155/IJMMS.2005.3427

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inverse transformation, which is

X(1)=X(1), X(2)=X(1)R1, X(3)=X(1)R1R2,

(1.4)

through

X(m)=X(1)R1R2···Rm1. (1.5) So, in order to obtain this density, we need the Jacobian, which is the determinant of the matrix

∂x(1)

∂x(1)

∂x(1)

∂r1

∂x(1)

∂r2 ··· ∂x(1)

∂rm1

∂x(2)

∂x(1)

∂x(2)

∂r1

∂x(2)

∂r2 ··· ∂x(2)

∂rm1

∂x(3)

∂x(1)

∂x(3)

∂r1

∂x(3)

∂r2 ··· ∂x(3)

∂rm1

... ... ... ... ...

∂x(m)

∂x(1)

∂x(m)

∂r1

∂x(m)

∂r2 ··· ∂x(m)

∂rm1

, (1.6)

which is the lower triangular matrix

1 0 0 ··· 0

r1 x(1) 0 ··· 0

r1r2 x(1)r2 x(1)r1 ··· 0

... ... ... ... ...

r1···rm1 x(1)r2···rm1 x(1)r1r3···rm1 ··· x(1)r1···rm2

. (1.7)

Thus the Jacobian isxm(1)1r1m2r2m3r3m4···rm2. So, the joint density ofX(1),R1,...,Rm1is fx(1),r1,...,rm1

=pmm!x(1)p1x(1)r1

p1 x(1)r1r2

p1

···

x(1)r1···rm1

p1

·x(1)m1r1m2r2m3···rm2

·I1x(1)x(1)r1x(1)r1r2≤ ··· ≤x(1)r1···rm1

=pmm!x(1)pm1r1p(m1)1r2p(m2)1···rm2p21rmp11

·Ix(1)1Ir11Ir21···Irm11.

(1.8)

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This shows that the random variablesX(1),R1,...,Rm1are independent and that the den- sity of our smallest order statistic is

fX(1)

x(1)

=pmx(1)pm1Ix(1)1, (1.9)

while the density of the ratio of theith adjacent order statisticRi,i=1,...,m1 is

fRi(r)=p(mi)rp(mi)1I(r1). (1.10)

We repeat this procedurentimes, assuming independence between sets of data, ob- taining the sequence{Rn=Rni,n1}. Notice that we have dropped the subscripti, but the density ofRnidoes depend oni. Hence, we first start out withnindependent sets of mi.i.d. Pareto random variables. We then order thesemPareto random variables within each set. Next, we obtain them1 ratios of the adjacent order statistics. Finally, we select one of these as our random variableY. Repeating thisntimes, we obtain the sequence {Yn, n1}. We do that via our preset prior distribution{Π1,...,Πm1}, whereΠi0 and mi=11Πi=1. The random variableYnis one of theRni,i=1,...,m1, chosen via this prior distribution. In other words,P{Yn=Rni} =Πifori=1, 2,...,m1. It is very im- portant to identify which is our largest acceptable pair of order statistics since the largest order statistic does dominate the partial sums. Hence, we defineν=max{kk>0}. We need to do this in caseΠm1=0.

Our goal is to determine whether or not there exist positive constants an and bN

such that Nn=1anYn/bNconverges to a nonzero constant in some sense, where{Yn, n 1} are i.i.d. copies of Y. Another important observation is that when p(mν)=1, we haveEY= ∞. These are called exact laws of large numbers since they create a fair game situation, where theanYnrepresents the amount a player wins on thenth play of some game andbNbN1represents the corresponding fair entrance fee for the partici- pant.

In Adler [1], just one order statistic from the Pareto was observed, while in Adler [2], ratios of order statistics were examined. Here we look at the case of randomly selecting one of these adjacent ratios. As usual, we define lgx=log (max{e,x}) and lg2x=lg(lgx).

We use throughout the paper the constantCas a generic real number that is not neces- sarily the same in each appearance.

2. Exact strong laws whenp(mν)=1

In this situation, we can get an exact strong law, but only if we select our coefficients and norming sequences properly. We use as our weights an=(lgn)β2/n, but we could set an=S(n)/n, whereS(·) is any slowly varying function. Note that if we do changean, then we must also revisebn, and consequentlycn=bn/an.

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Theorem2.1. Ifp(mν)=1, then for allβ >0,

Nlim→∞

Nn=1

(lgn)β2/nYn

(lgN)β =

Πν

β almost surely. (2.1)

Proof. Letan=(lgn)β2/n,bn=(lgn)β, andcn=bn/an=n(lgn)2. We use the usual par- tition

1 bN

N n=1

anYn= 1 bN

N n=1

an

YnI1Yncn

EYnI1Yncn

+ 1 bN

N n=1

anYnIYn> cn + 1

bN

N n=1

anEYnI1Yncn .

(2.2)

The first term vanishes almost surely by the Khintchine-Kolmogorov convergence the- orem, see [3, page 113], and Kronecker’s lemma since

n=1

1

c2nEYn2I1Yncn

=

m1 i=1

Πi

n=1

1

c2nER2nI1Rncn

= ν i=1

Πi

n=1

1 c2n

cn

1 p(mi)rp(mi)+1dr

= ν i=1

Πi

n=1

p(mi) c2n

cn

1 rp(mν)p(νi)+1dr

= ν i=1

Πi

n=1

p(mi) c2n

cn

1 rpi)dr

C ν i=1

n=1

1 c2n

cn

1 dr

C n=1

1 c2n

cn

1 dr

C n=1

1 cn

=C n=1

1

n(lgn)2 <.

(2.3)

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The second term vanishes, with probability one, by the Borel-Cantelli lemma since

n=1

PYn> cn

=

m1 i=1

Πi

n=1

PRn> cn

=ν

i=1

Πi

n=1

cn

p(mi)rp(mi)1dr

C ν i=1

n=1

cn

rp(mν)p(νi)1dr

=C ν i=1

n=1

cn

rpi)2dr

C ν i=1

n=1

cn

r2dr

C n=1

cn

r2dr

=C n=1

1 cn<.

(2.4)

The limit of our normalized partial sums is realized via the third term in our partition EYnI1Yncn

= ν i=1

ΠiERnI1Rncn

=ν

i=1

Πi

cn

1 p(mi)rp(mi)dr

= ν i=1

Πi

cn

1 p(mi)rp(mν)p(νi)dr

= ν i=1

Πi

cn

1 p(mi)rpi)1dr

=ν 1 i=1

Πi

cn

1 p(mi)rp(νi)1drν

cn

1 p(mν)r1dr

Πνp(mν) lgcnΠνlgn

(2.5)

since

ν1

i=1

Πi

cn

1 p(mi)rp(νi)1drC

ν1 i=1

cn

1 rp1drC cn

1 rp1dr=O(1). (2.6)

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Thus

Nn=1anEYnI1Yncn

bN Πν N

n=1(lgn)β1/n

(lgN)β −→

Πν

β , (2.7)

which completes the proof.

3. Exact weak laws whenp(mν)=1

We investigate the behavior of our random variables{Yn,n1}, where we slightly in- crease the coefficient ofYn. Instead ofanbeing a power of logarithm timesn1, we now allowanto bento any power larger than negative one. In this case, there is no way to obtain an exact strong law (seeSection 4), but we are able to obtain exact weak laws.

Theorem3.1. Ifp(mν)=1andα >1, then

Nn=1nαL(n)Yn Nα+1L(N) lgN

−−→P Πν

α+ 1 (3.1)

for any slowly varying functionL(·).

Proof. This proof is a consequence of the degenerate convergence theorem, see [3, page 356]. Here, we setan=nαL(n) andbN=Nα+1L(N) lgN. Thus, for all>0, we have

N n=1

P

YnbN

an

=ν

i=1

Πi

N n=1

P

RnbN

an

= ν i=1

Πip(mi) N n=1

bN/an

rp(mi)1dr

=p ν i=1

Πi(mi) N n=1

bN/an

rp(mν)p(νi)1dr

=p ν i=1

Πi(mi) N n=1

bN/an

rpi)2dr

<

ν i=1

N n=1

bN/an

r2dr

< C N n=1

an

bN

=C N n=1

nαL(n) Nα+1L(N) lgN

< C lgN −→0.

(3.2)

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Similarly, N n=1

Var an

bNYnI

1YnbN

an

= ν i=1

Πi

N n=1

Var an

bNRnI

1RnbN

an

< C ν i=1

N n=1

a2n b2N

bN/an

1 rp(mi)+1dr

=C ν i=1

N n=1

a2n b2N

bN/an

1 rp(mν)p(νi)+1dr

=C ν i=1

N n=1

a2n b2N

bN/an

1 rpi)dr

< C N n=1

a2n b2N

bN/an

1 dr < C N n=1

an

bN

=C N n=1

nαL(n) Nα+1L(N) lgN

C lgN −→0.

(3.3)

As for our truncated expectation, we have

EYnI

1YnbN

an

= ν i=1

ΠiERnI

1RnbN

an

=ν

i=1

Πip(mi) bN/an

1 rp(mi)dr

=p ν i=1

Πi(mi) bN/an

1 rp(mν)p(νi)dr

=p ν i=1

Πi(mi) bN/an

1 rpi)1dr

=p

ν1 i=1

Πi(mi) bN/an

1 rp(νi)1drν

bN/an

1 r1dr.

(3.4)

The last term is the dominant term since N

n=1

an

bNp

ν1 i=1

Πi(mi) bN/an

1 rpi)1dr < C N n=1

an

bN

bN/an

1 rp1dr < C N n=1

an

bN −→0, (3.5)

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while N n=1

an

bNΠν

bN/an

1 r1dr

=Πν

N n=1

an bNlg

bN an

=Πν N

n=1nαL(n) lgNα+1L(N) lgN/nαL(n) Nα+1L(N) lgN

=Πν N

n=1nαL(n)(α+ 1) lgN+ lgL(N) + lg2NαlgnlgL(n)

Nα+1L(N) lgN .

(3.6)

The important terms are

Nn=1nαL(n)(α+ 1) lgN Nα+1L(N) lgN =

(α+ 1) Nn=1nαL(n) Nα+1L(N) −→1,

Nn=1nαL(n)(αlgn) Nα+1L(N) lgN = −

α Nn=1nαL(n) lgn Nα+1L(N) lgN −→ −

α α+ 1,

(3.7)

while the other three terms vanish as N→ ∞. For completeness, we will verify these claims:

Nn=1nαL(n) lgL(N)

Nα+1L(N) lgN <ClgL(N) lgN −→0,

Nn=1nαL(n) lg2N

Nα+1L(N) lgN <Clg2N lgN −→0,

Nn=1nαL(n) lgL(n)

Nα+1L(N) lgN <CNα+1L(N) lgL(N) Nα+1L(N) lgN =

ClgL(N) lgN −→0.

(3.8)

Therefore,

Nn=1anEYnI1YnbN/an

bN −→Πν

1 α

α+ 1

= Πν

α+ 1, (3.9)

which completes this proof.

4. Further almost sure behavior whenp(mν)=1

Using our exact weak law, we are able to obtain a generalized law of the iterated logarithm.

This shows that under the hypotheses of Theorem 4.1, exact strong laws do not exist whenan=nαL(n),α >1, whereL(·) is a slowly varying function. Hence, the coefficients selected inTheorem 2.1are the only permissible ones that will allow us to obtain an exact strong law, that is, an=S(n)/nfor some slowly varying function S(·), where we used logarithms as our functionS(·).

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Theorem4.1. Ifp(mν)=1andα >1, then

lim inf

N→∞

Nn=1nαL(n)Yn

Nα+1L(N) lgN = Πν

α+ 1 almost surely, lim sup

N→∞

Nn=1nαL(n)Yn

Nα+1L(N) lgN = ∞ almost surely,

(4.1)

for any slowly varying functionL(·).

Proof. FromTheorem 3.1, we have

lim inf

N→∞

Nn=1nαL(n)Yn Nα+1L(N) lgN

Πν

α+ 1 almost surely. (4.2)

Setan=nαL(n),bn=nα+1L(n) lgn, andcn=bn/an=nlgn. In order to obtain the oppo- site inequality, we use the following partition:

1 bN

N n=1

anYn 1 bN

N n=1

anYnI1Ynn

= 1 bN

N n=1

an

YnI1YnnEYnI1Ynn

+ 1 bN

N n=1

anEYnI1Ynn.

(4.3)

The first term goes to zero, almost surely, sincebnis essentially increasing and

n=1

cn2EYn2I1Ynnν

i=1

n=1

cn2ER2nI1Rnn

C ν i=1

n=1

cn2 n

1 rp(νi)dr

C ν i=1

n=1

cn2 n

1 dr

C ν i=1

n=1

n c2n

C n=1

1

n(lgn)2 <.

(4.4)

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As for the second term, we once again focus on the last term, our two largest permis- sible order statistics,

EYnI1Ynn= ν i=1

ΠiEYnI1Ynn

= ν i=1

Πi

n

1 p(mi)rp(mi)dr

=p ν i=1

Πi(mi) n

1 rp(mν)pi)dr

=p ν i=1

Πi(mi) n

1 rp(νi)1dr

=p

ν1 i=1

Πi(mi) n

1 rp(νi)1drνp(mν)n

1 r1dr

Πνlgn

(4.5)

since p

ν1 i=1

Πi(mi) n

1 rp(νi)1dr < C

ν1 i=1

n

1 rp1dr < C n

1 rp1dr=O(1). (4.6) Thus,

lim inf

N→∞

Nn=1anYn

bN lim inf

N→∞

Nn=1anEYnI1Ynn bN

= lim

N→∞

Πν N

n=1nαL(n) lgn Nα+1L(N) lgN

= Πν

α+ 1,

(4.7)

establishing our almost sure lower limit.

As for the upper limit, letM >0, then

n=1

PYn> Mcn

= ν i=1

Πi

n=1

PRn> Mcn

=ν

i=1

Πi

n=1

p(mi)

Mcn

rp(mi)1dr

ν i=νΠi

n=1

p(mi)

Mcn

rp(mi)1dr

=Πν

n=1

p(mν)

Mcn

rp(mν)1dr

参照

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