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, 1≤k≤m, n≥1} from the Pareto distribution. We randomly select two adjacent order statistics from each row, Xn(i) andXn(i+1), where 1≤i≤m−1. 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)=px−p−1I(x≥ 1), where p >0. Then, we observe two adjacent order statistics from our sample, that is,X(i)≤X(i+1)for 1≤i≤m−1. Next, we obtain the random variableRi=X(i+1)/X(i), i=1,...,m−1, which is the ratio of our adjacent order statistics. The density ofRiis
f(r)=p(m−i)r−p(m−i)−1I(r≥1). (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
=pmx−1p−1···xm−p−1Ix1≥1···Ixm≥1, (1.2)
hence the density of the corresponding order statisticsX(1),...,X(m)is fx(1),...,x(m)
=pmm!x−(1)p−1···x−(mp)−1I1≤x(1)≤x(2)≤ ··· ≤x(m)
. (1.3)
Next, we obtain the joint density ofX(1),R1,...,Rm−1. 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
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···Rm−1. (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)
∂rm−1
∂x(2)
∂x(1)
∂x(2)
∂r1
∂x(2)
∂r2 ··· ∂x(2)
∂rm−1
∂x(3)
∂x(1)
∂x(3)
∂r1
∂x(3)
∂r2 ··· ∂x(3)
∂rm−1
... ... ... ... ...
∂x(m)
∂x(1)
∂x(m)
∂r1
∂x(m)
∂r2 ··· ∂x(m)
∂rm−1
, (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···rm−1 x(1)r2···rm−1 x(1)r1r3···rm−1 ··· x(1)r1···rm−2
. (1.7)
Thus the Jacobian isxm(1)−1r1m−2r2m−3r3m−4···rm−2. So, the joint density ofX(1),R1,...,Rm−1is fx(1),r1,...,rm−1
=pmm!x−(1)p−1x(1)r1
−p−1 x(1)r1r2
−p−1
···
x(1)r1···rm−1
−p−1
·x(1)m−1r1m−2r2m−3···rm−2
·I1≤x(1)≤x(1)r1≤x(1)r1r2≤ ··· ≤x(1)r1···rm−1
=pmm!x−(1)pm−1r1−p(m−1)−1r−2p(m−2)−1···rm−2−p2−1rm−−p−11
·Ix(1)≥1Ir1≥1Ir2≥1···Irm−1≥1.
(1.8)
This shows that the random variablesX(1),R1,...,Rm−1are independent and that the den- sity of our smallest order statistic is
fX(1)
x(1)
=pmx−(1)pm−1Ix(1)≥1, (1.9)
while the density of the ratio of theith adjacent order statisticRi,i=1,...,m−1 is
fRi(r)=p(m−i)r−p(m−i)−1I(r≥1). (1.10)
We repeat this procedurentimes, assuming independence between sets of data, ob- taining the sequence{Rn=Rni,n≥1}. 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 them−1 ratios of the adjacent order statistics. Finally, we select one of these as our random variableY. Repeating thisntimes, we obtain the sequence {Yn, n≥1}. We do that via our preset prior distribution{Π1,...,Πm−1}, whereΠi≥0 and mi=−11Πi=1. The random variableYnis one of theRni,i=1,...,m−1, chosen via this prior distribution. In other words,P{Yn=Rni} =Πifori=1, 2,...,m−1. 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{k:Πk>0}. We need to do this in caseΠm−1=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(m−ν)=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 andbN−bN−1represents 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(m−ν)=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)β−2/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.
Theorem2.1. Ifp(m−ν)=1, then for allβ >0,
Nlim→∞
Nn=1
(lgn)β−2/nYn
(lgN)β =
Πν
β almost surely. (2.1)
Proof. Letan=(lgn)β−2/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
YnI1≤Yn≤cn
−EYnI1≤Yn≤cn
+ 1 bN
N n=1
anYnIYn> cn + 1
bN
N n=1
anEYnI1≤Yn≤cn .
(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
c2nEYn2I1≤Yn≤cn
=
m−1 i=1
Πi
∞ n=1
1
c2nER2nI1≤Rn≤cn
= ν i=1
Πi
∞ n=1
1 c2n
cn
1 p(m−i)r−p(m−i)+1dr
= ν i=1
Πi
∞ n=1
p(m−i) c2n
cn
1 r−p(m−ν)−p(ν−i)+1dr
= ν i=1
Πi
∞ n=1
p(m−i) c2n
cn
1 r−p(ν−i)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)
The second term vanishes, with probability one, by the Borel-Cantelli lemma since ∞
n=1
PYn> cn
=
m−1 i=1
Πi
∞ n=1
PRn> cn
=ν
i=1
Πi
∞ n=1
∞
cn
p(m−i)r−p(m−i)−1dr
≤C ν i=1
∞ n=1
∞
cn
r−p(m−ν)−p(ν−i)−1dr
=C ν i=1
∞ n=1
∞
cn
r−p(ν−i)−2dr
≤C ν i=1
∞ n=1
∞
cn
r−2dr
≤C ∞ n=1
∞
cn
r−2dr
=C ∞ n=1
1 cn<∞.
(2.4)
The limit of our normalized partial sums is realized via the third term in our partition EYnI1≤Yn≤cn
= ν i=1
ΠiERnI1≤Rn≤cn
=ν
i=1
Πi
cn
1 p(m−i)r−p(m−i)dr
= ν i=1
Πi
cn
1 p(m−i)r−p(m−ν)−p(ν−i)dr
= ν i=1
Πi
cn
1 p(m−i)r−p(ν−i)−1dr
=ν− 1 i=1
Πi
cn
1 p(m−i)r−p(ν−i)−1dr+Πν
cn
1 p(m−ν)r−1dr
∼Πνp(m−ν) lgcn∼Πνlgn
(2.5)
since
ν−1
i=1
Πi
cn
1 p(m−i)r−p(ν−i)−1dr≤C
ν−1 i=1
cn
1 r−p−1dr≤C cn
1 r−p−1dr=O(1). (2.6)
Thus
Nn=1anEYnI1≤Yn≤cn
bN ∼Πν N
n=1(lgn)β−1/n
(lgN)β −→
Πν
β , (2.7)
which completes the proof.
3. Exact weak laws whenp(m−ν)=1
We investigate the behavior of our random variables{Yn,n≥1}, where we slightly in- crease the coefficient ofYn. Instead ofanbeing a power of logarithm timesn−1, 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(m−ν)=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
Yn≥bN
an
=ν
i=1
Πi
N n=1
P
Rn≥bN
an
= ν i=1
Πip(m−i) N n=1
∞
bN/an
r−p(m−i)−1dr
=p ν i=1
Πi(m−i) N n=1
∞
bN/an
r−p(m−ν)−p(ν−i)−1dr
=p ν i=1
Πi(m−i) N n=1
∞
bN/an
r−p(ν−i)−2dr
<
ν i=1
N n=1
∞
bN/an
r−2dr
< C N n=1
an
bN
=C N n=1
nαL(n) Nα+1L(N) lgN
< C lgN −→0.
(3.2)
Similarly, N n=1
Var an
bNYnI
1≤Yn≤bN
an
= ν i=1
Πi
N n=1
Var an
bNRnI
1≤Rn≤bN
an
< C ν i=1
N n=1
a2n b2N
bN/an
1 r−p(m−i)+1dr
=C ν i=1
N n=1
a2n b2N
bN/an
1 r−p(m−ν)−p(ν−i)+1dr
=C ν i=1
N n=1
a2n b2N
bN/an
1 r−p(ν−i)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
1≤Yn≤bN
an
= ν i=1
ΠiERnI
1≤Rn≤bN
an
=ν
i=1
Πip(m−i) bN/an
1 r−p(m−i)dr
=p ν i=1
Πi(m−i) bN/an
1 r−p(m−ν)−p(ν−i)dr
=p ν i=1
Πi(m−i) bN/an
1 r−p(ν−i)−1dr
=p
ν−1 i=1
Πi(m−i) bN/an
1 r−p(ν−i)−1dr+Πν
bN/an
1 r−1dr.
(3.4)
The last term is the dominant term since N
n=1
an
bNp
ν−1 i=1
Πi(m−i) bN/an
1 r−p(ν−i)−1dr < C N n=1
an
bN
bN/an
1 r−p−1dr < C N n=1
an
bN −→0, (3.5)
while N n=1
an
bNΠν
bN/an
1 r−1dr
=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) + lg2N−αlgn−lgL(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=1anEYnI1≤Yn≤bN/an
bN −→Πν
1− α
α+ 1
= Πν
α+ 1, (3.9)
which completes this proof.
4. Further almost sure behavior whenp(m−ν)=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(·).
Theorem4.1. Ifp(m−ν)=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
anYnI1≤Yn≤n
= 1 bN
N n=1
an
YnI1≤Yn≤n−EYnI1≤Yn≤n
+ 1 bN
N n=1
anEYnI1≤Yn≤n.
(4.3)
The first term goes to zero, almost surely, sincebnis essentially increasing and ∞
n=1
cn−2EYn2I1≤Yn≤n≤ν
i=1
∞ n=1
c−n2ER2nI1≤Rn≤n
≤C ν i=1
∞ n=1
cn−2 n
1 r−p(ν−i)dr
≤C ν i=1
∞ n=1
cn−2 n
1 dr
≤C ν i=1
∞ n=1
n c2n
≤C ∞ n=1
1
n(lgn)2 <∞.
(4.4)
As for the second term, we once again focus on the last term, our two largest permis- sible order statistics,
EYnI1≤Yn≤n= ν i=1
ΠiEYnI1≤Yn≤n
= ν i=1
Πi
n
1 p(m−i)r−p(m−i)dr
=p ν i=1
Πi(m−i) n
1 r−p(m−ν)−p(ν−i)dr
=p ν i=1
Πi(m−i) n
1 r−p(ν−i)−1dr
=p
ν−1 i=1
Πi(m−i) n
1 r−p(ν−i)−1dr+Πνp(m−ν)n
1 r−1dr
∼Πνlgn
(4.5)
since p
ν−1 i=1
Πi(m−i) n
1 r−p(ν−i)−1dr < C
ν−1 i=1
n
1 r−p−1dr < C n
1 r−p−1dr=O(1). (4.6) Thus,
lim inf
N→∞
Nn=1anYn
bN ≥lim inf
N→∞
Nn=1anEYnI1≤Yn≤n 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(m−i) ∞
Mcn
r−p(m−i)−1dr
≥ ν i=νΠi
∞ n=1
p(m−i) ∞
Mcn
r−p(m−i)−1dr
=Πν
∞ n=1
p(m−ν) ∞
Mcn
r−p(m−ν)−1dr