# Journal of Statistical Theory and Applications

Volume 17, Issue 2, June 2018, Pages 206 - 212

# Record Ranges for Samples From Asymmetrical Laplace Distributions

Authors
I.V. BELKOV
Department of Mathematics and Mechanics, St. Petersburg State University, St. Petersburg, Russia
M. AHSANULLAH
Department of Management Sciences, Rider University, Lawrenceville, New Jersey, United States of America
V. B. NEVZOROV
Department of Mathematics and Mechanics, St. Petersburg State University, St. Petersburg, Russia
Received 2 November 2017, Accepted 11 February 2018, Available Online 30 June 2018.
DOI
10.2991/jsta.2018.17.2.2How to use a DOI?
Keywords
Record values; Exponential distribution; Negative exponential distribution; Laplace distribution; Sample ranges
Abstract

The representations of record ranges via sums of independent identically distributed exponential random variables are obtained for asymmetrical Laplace distributions. This result generalizes the corresponding relations for record values in the cases of exponential and negative exponential distributions

Open Access
This is an open access article under the CC BY-NC license (http://creativecommons.org/licences/by-nc/4.0/).

## Introduction

Let X1, X2,… be a sequence of independent identically distributed random variables (r.v.’s) with an absolutely continuous distribution function (d.f.) F(x). For any n=1,2,… let us introduce r.v.’s

mn=min{X1,X2,,Xn},Mn=max{X1,X2,,Xn}
and the corresponding sample ranges
Wn=Mnmn

Upper record times L(n) and upper record values X(n), n=1,2,…, are defined as follows:

L(1)=1,L(n)=min{j>L(n1):Xj>XL(n1)},n=2,3,,
and
X(n)=XL(n),n=1,2,.

Analogously one can define the so-called lower record times l (n) and lower record values x(n), n=1,2,… :

l(1)=1,l(n)=min{j>l(n1):Xj<Xl(n1)},n=2,3,,
and
x(n)=Xl(n),n=1,2,.

Below the notation X=dY means the equality in distribution of random variables or vectors X and Y.

In the record theory (see, for example,  – ) results for the initial sequences of exponentially distributed random variables are very important and popular. We give here two of the corresponding statements.

Let Z1, Z2,… be a sequence of independent E(1)-distributed r.v.’s with d.f.

H1(x)=max{0,1exp(x)}
and Z(1) = Z1 < Z(2) <… be the corresponding upper record values. Let ξ1, ξ2,… also be independent E(1)-exponential r.v.’s and Sk = ξ12+…+ ξk, k=1,2,…. The following result is valid.

### •  Representation 1.

For any n=1,2,…

(Z(1),Z(2),,Z(n))=d(S1,S2,,Sn)

Now let us consider r.v.’s Vk= − Zk, k=1,2,…, with the negative exponential d.f. H2(x), where

H2(x)=1H1(x)=min{ex,1},
and the corresponding lower record values v(1) = V1 > v(2)> ….

In this situation the relation, analogous to (5), can be written.

### •  Representation 2.

For any n=1,2,… the following equality holds:

(v(1),v(2),,v(n))=d(S1,S2,,Sn)

Equalities (5) and (6) together with Smirnov’s transformation allow to obtain some useful results for X’s with any continuous d.f. F(x). For example, one can immediately write that

F(X(k))=d1exp(Z(k))=d1exp(Sk)=1exp((ξ1+ξ2++ξk))=d1(1U1)(1U2)(1Ur)=d1U1U2Uk,k=1,2,,
where U1, U2,… are independent uniformly U([0,1))-distributed random variables. Hence in this case the following relations are valid for record values:
(X(1),X(2),,X(n))=d(G(1U1),G(1U1U2),,G(1U1U2Un)),n=1,2,,
where G(x) is the inverse function for d.f. F(x).

Analogously,

F(x(k))=dexp(v(k))=dexp((ξ1+ξ2++ξk))=d(1U1)(1U2)(1Uk)=dU1U2Uk
and
(x(1),x(2),,x(n))=d(G(U1),G(U1U2),,G(U1U2Un)),n=1,2,.
The aim of the given paper is to present some new relations which generalize presentations 1 and 2.

## Record Ranges and Laplace Distributions

Above the exponential and negative exponential random variables with d.f.’s H1(x)=max{0,1−exp(−x)} and H2(x)= min{ex,1} were under our consideration. Let us consider now the distribution which is the mixture with weights q and p = (1−q), 0≤q≤1, of these two distributions. The corresponding d.f. H(x) is taken in this case as H(x) = qH1(x)+pH2(x). Thus,

H(x)=pex,ifx0,
and
H(x)=1qex,ifx>0
The probability density function (pdf) h(x) is given as
h(x)=pex,ifx0,andh(x)=qex,x>0
If p = 0, one gets E(1)-exponential distribution. The value p =1 corresponds to the negative exponential distribution. If p = q =1/2 one deals with the classical Laplace pdf h(x) = exp(−|x|)/2. Thus, we can say that relations (10) and (11) present asymmetrical Laplace distributions.

Let now independent r.v.’s X1, X2,… have the common distribution function (10) and let us add the degenerate r.v. X0 = 0 in the beginning of this sequence. For X0, X1, X2, … we consider maximal and minimal values

Mn=max{X0,X1,,Xn}andmn=min{X0,X1,,Xn},n=1,2,,
as well as ranges
Wn=Mnmn,n=1,2,.

Record values W(1) = W1 < W(2) <…< W(n) <… in the sequence of ranges W1,W2,… are the subjects of our interest. It appears that one can express these records via r.v.’s ξ1, ξ2,…, which are defined above. The following result is valid.

### •  Representation 3.

For any n=1,2,… and 0 ≤ p = 1−q ≤ 1 the relation

(W(1),W(2),,W(n))=d(S1,S2,,Sn)
holds, where Sk= ξ12+…+ ξk, k=1,2,….
1. 1.

Remark 1. It appears that the RHS of (12) doesn’t depend on p and q.

2. 2.

Remark 2. It is easy to see that Representation 1 is the partial (under p = 0) case of (12). Analogously, the result of Representation 2 immediately follows from (12) if to take p =1 and q=0.

3. 3.

Remark 3. We discuss here record ranges in the sequences of random variables which have different forms of Laplace distributions, which are rather close in some sense to the exponential distributions. Let us note that some analogous results for record ranges were obtained in  for the initial sequences of the uniformly distributed r.v.’s.

### Proof of Representation 3

It is evidently that W(1) = W1 = max{0,X1} − min{0,X1}= |X1| has the exponential E(1)-distribution. Hence one can write that

W(1)=dξ1.
Let us find now distributions of differences
T(n)=W(n)W(n1),n=2,3,.
The following notation is used below:
M(n)=ML(n)andm(n)=mL(n),n=1,2,,
where
1=L(1)<L(2)<<L(n1)<L(n)<
represent the corresponding record moments.

We examine the behavior of conditional probabilities

R(x,y,v)=P{T(n)>x| m(n1)=y,M(n1)=v}, x>0, y0, v0.

Note that the condition {m(n−1) = −y, M(n−1) = v} corresponds to the situation, when W(n−1) = y+v.

Let us denote events

{ymin{XL(n1)+1,,XL(n1)+k}<max{XL(n1)+1,,XL(n1)+k}v}
as A(y,v,k,n).

One can see that

P(A(y,v,k,n))=(H(v)H(y))k=(1qevpey)k,n=2,3,,k=0,1,2,.

Now it is possible to write that

R(x,y,v)=k=0P(A(y,v,k,n))P{XL(n1)+k+1<yx}+P{XL(n1)+k+1>v+x}=k=0(1qevpey)k(H(yx)+1H(v+x))=(pe(y+x)+qe(v+x))/(qev+pey)=ex,
and
P{T(n)<x|m(n1)=y,M(n1)=v}=1ex=P{ξn<x}
for any x>0.

It appears that conditional probabilities R(x,y,v) don’t depend on y and v. Thus, it follows that r.v. T(n) doesn’t depend on W(n−1) and this difference of two neighbouring record values W(n) −W(n−1) has the standard E(1)-exponential distribution. Recalling relation (13) one can write now that for any record value W(n), n=1,2,…, the following presentation is valid:

W(n)=dξ1+ξ2++ξn.
Moreover, taking into account that r.v.’s W(1), W(2)-W(1), W(3)-W(2),… are independent we get finally that representation 3 holds for any n=1,2,….

## The Number of Record Ranges

Let Z1,Z2,… be a sequence of independent E(1)-distributed r.v.’s with d.f. H1(x)=max{0,1−exp(−x)} and N1(n) be the number of the upper records among Z1,Z2,…,Zn.

It is known that

EN1(n)=k=1n1k~lnn,n.

If independent r.v.’s V1,V2,… have d.f. H2(x)=min(1,e x) and N2(n) is the number of the lower record values among V1,V2,…,Vn, then also we come to the relation

EN2(n)=k=1n1k~lnn,n.

Let us consider now the number N(n) of record values in the set of ranges W1,W2,…,Wn. We denote

Yk,1=max{0,Xk}andYk,2=min{0,Xk},k=1,2,
It is evident that
P{Yk,1=0}=p,P{Yk,1<x}=1qexp(x),x>0,
and
P{Yk,2<x}=pex,x<0,P{Yk,2=0}=1p=q.

The number N(n) of records in the set W1,W2,…,Wn is equal to the sum N (1)(n) + N(2)(n), where

N(1)(n)=η1+η2++ηn
is the number of the positive upper records among r.v.’s Y1,1,Y2,1,…,Yn,1 and
N(2)(n)=γ1+γ2++γn
is the number of the negative lower records among r.v.’s Y1,2,Y2,2,…,Yn,2. One can find that
P{η1=1}=P{Y1,1>0}=1P{Y1,1=0}=1p,
and
P{ηk=1}=1P{ηk=0}=P{Yk,1>max{0,Y1,1,Y2,1,,Yk,1,1}}=(1pk)/k,k=2,3,.
Analogously,
P{γk=1}=1P{γk=0}=(1qk)/k,k=1,2,,n.
Thus, we get that
EN(n)=k=1n(2pkqk)/k,n=1,2,.
The maximal value of EN(n) is reached if p=q=1/2.