Journal of Statistical Theory and Applications

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Volume 17, Issue 3, September 2018, Pages 408 - 418

Characterizations of Exponential Distribution Based on Two-Sided Random Shifts

Authors
Santanu Chakraborty, George P. Yanev
School of Mathematical and Statistical Sciences, The University of Texas Rio Grande Valley
Received 25 May 2017, Accepted 28 February 2018, Available Online 30 September 2018.
DOI
10.2991/jsta.2018.17.3.2How to use a DOI?
Keywords
characterization; exponential distribution; order statistics; random shifts
Abstract

A new characterization of the exponential distribution is obtained. It is based on an equation involving randomly shifted (translated) order statistics. No specific distribution is assumed for the shift random variables. The proof uses a recently developed technique including the Maclaurin series expansion of the probability density of the parent variable.

Copyright
© 2018, the Authors. Published by Atlantis Press.
Open Access
This is an open access article under the CC BY-NC license (http://creativecommons.org/licences/by-nc/4.0/).

1. Introduction and main result

Let X1, X2, …, Xn be a simple random sample from a continuously distributed parent X. Denote by Xj:n for 1 ≤ j ≤ n the jth order statistic (OS). Let n1, n2, k1, and k2 be fixed integers, such that 0 ≤ ki ≤ ni − 1 for i = 1, 2. Consider the distributional equation

Xn1−k1:n1+c1ξ1=dXn2−k2:n2+c2ξ2,
where c1 and c2 are certain constants and the “shift” (or “translation”) variables ξ1 and ξ2 are independent from X1, X2, . . . , Xn. A variety of particular cases of (1) have appeared in a number of papers devoted to characterizations of certain classes of continuous distributions. Recent surveys can be found in [4] and [6].

If c2 = 0, then (1) is called one-sided random shift equation. Let k and n be fixed integers, such that 1 ≤ k ≤ n − 1. Under some regularity conditions, it is proven in [3], [12] and [7] that each one of the following two equations characterizes the exponential distribution:

  Xn−k:n+1kξ=dXn−k+1:n (consecutive OS),Xn−k:n−1+1nξ=dXn−k+1:n (consecutive OS and sample sizes)
where ξ is unit exponential.

If both c1 > 0 and c2 > 0, then we have equation with to two-sided random shifts. It is established in [12] that for unit exponential ξ1 and ξ2, the equation

Xn−k:n−1+1nξ1=dXn−k:n+1kξ2,
where k and n are fixed integers, such that 1 ≤ k ≤ n − 1, characterizes the exponential distribution.

In all equations above the random shifts are exponentially distributed and therefore (as the characterization implies) are also identically distributed with the parent variable X. In these notes the condition shifts variables to be exponential is dropped. Instead the weaker assumption that the random shifts are identically distributed with X is made. The one-sided shift case without specifying the distribution of the shift variable is studied in [5] for n = 2. Namely, it is proven that the equation

X1+12X2=dX2:2
characterizes the exponential distribution. This is generalized in [8] and [13] where it is shown that for fixed n ≥ 2 the equation
Xn−1:n−1+1nXn=dXn:n,
characterizes the exponential distribution. It is worth mentioning that a similar characterization in [1] for a sample of size n = 2 is based on the equation
X1:2+X=dX2:2.

In our main result below we obtain an analog of the two-sided characterization (2) dropping the assumption on the shifts to be exponential. In the proof of the theorem a recently developed technique based on the Maclaurin series expansion of the density function f(x) is used. This method of proof grew out of an argument given first in [5].

Theorem

Let k and n be fixed integers, such that 1 ≤ k ≤ n − 1. Let X1, X2, . . . , Xn+1 be a simple random sample from a distribution with cdf F(x) (F(0) = 0) and pdf f(x) (f(0) > 0). Assume that f(x) is analytic for x > 0. Then for some λ > 0

F(x)=1−e−λx, x≥0
if and only if
Xn−k:n−1+1nXn=dXn−k:n+1kXn+1.

In the next section, we present several lemmas needed to prove the theorem. In Section 3, we prove our main theorem.

2. Preliminaries

The first lemma plays a key role in the proof of the theorem. Originally it appeared in a similar form as an argument in [5]. In the form presented below, it was proven in [10].

Lemma 1

Let f(x) be analytic for all x > 0 and f(0) > 0. If for all non-negative m

f(m)(0)=(−1)mfm+1(0),
then
f(x)=f(0)e−f(0)x.

Next, define for all non-negative integers n and i, and any real x the numbers

Hn,i(x):=∑j=0n(−1)j(nj)(x−j)i.
It is known (see [11]) that Hn,n(·) = n! and Hn,i(·) = 0 for 0 ≤ i ≤ n − 1.

In the rest of the section we prove three more lemmas needed in the proof of the theorem.

Lemma 2

For m, n and k positive integers with 1 ≤ k ≤ n,

∑l=0m(ml)Hn−k+l,i(n−k+l+1)=Hn−k,i(n−k+m+1)

Proof.

Let s and r be positive integers. We shall prove that

Hs,r(s+1)+Hs+1,r(s+2)=Hs,r(s+2).
Using the identity
(s+1j)−(sj−1)=(sj)
and the definition of Hi,j(x), we obtain for r ≥ s + 1
Hs,r(s+1)+Hs+1,r(s+2) =∑j=0s(−1)j(sj)(s+1−j)r+∑j=0s+1(−1)j(s+1j)(s+2−j)r =∑j=1s+1[(−1)j−1(sj−1)+(−1)j(s+1j)](s+2−j)t+(s+2)r =∑j=0s+1(−1)j(sj)(s+2−j)r =Hs,r(s+2).
It follows from (8) that
Hn−k,i(n−k+1)+Hn−k+1,i(n−k+2)=Hn−k,i(n−k+2)
Proceeding one more step, we have,
Hn−k,i(n−k+1)+2Hn−k+1,i(n−k+2)+Hn−k+2,i(n−k+3)=Hn−k,i(n−k+3)
Suppose the lemma is true for m ≤ r. We shall prove it for m = r + 1. Indeed,
∑l=0r+1(r+1l)Hn−k,l(n−k+l+1) =∑l=0r(rl)Hn−k+l,i(n−k+l+1)+∑l=0r(rl)Hn−k+l+1,i(n−k+l+2) =Hn−k,i(n−k+r+1)+Hn−k+1,i(n−k+r+2)=Hn−k,i(n−k+r+2).
The proof is complete.

Lemma 3

Let k, n, and r be positive integers such that 1 ≤ k ≤ n − 1. Denote t = n − k − 1. The following two identities are true:

∑i=0rkiHt,t+r+1−i(n)=1t+1Ht+1,t+r+2(n)−kr+1t!
and
∑i=0rniHt,t+r+1−i(n−1)=1t+1Ht+1,t+r+2(n)−nr+1t!.

Proof.

We shall prove (9). The proof of (10) is similar. We have

∑i=0rkiHt,t+r+1−i(n)=∑i=0rki∑j=0t(−1)j(tj)(n−j)t+r+1−i     =∑j=0t(−1)j(tj)∑i=0rki(n−j)t+r+1−i     =∑j=0t(−1)j(tj)(n−j)t+r+1∑i=0r(kn−j)i     =∑j=0t(−1)j(tj)(n−j)t+r+1[1−(kn−j)r+11−kn−j]     =∑j=0t(−1)j(tj)(n−j)t+r+2t+1−j[1−(kn−j)r+1]     =1t+1∑j=0t(−1)j(t+1j)(n−j)t+r+2[1−(kn−j)r+1]     =1t+1∑j=0t(−1)j(t+1j)(n−j)t+r+2−kr+1t+1∑j=0t(−1)j(t+1j)(n−j)t+1     =1t+1∑j=0t+1(−1)j(t+1j)(n−j)t+r+2−kr+1t+1∑j=0t+1(−1)j(t+1j)(n−j)t+1     =1t+1Ht+1,t+r+2(n)−kr+1t+1Ht+1,t+1(n)     =1t+1Ht+1,t+r+2(n)−kr+1t!.

Lemma 4

Let j ≥ 1 and d be integers, such that j + d ≥ 0. Assume F (0) = 0 and for d ≥ 1

f(m)(0)=(−1)mfm+1(0), m=1,2,…,d.
Then for j = 1, 2, . . .
Gj(j+d)(0)={Hj,j+d(j+1)fj+1−d(0)(f′(0))d if   d≥0;0 if −j≤d<0.
where Gj(x) := Fj(x)f(x).

Proof.

  1. (i)

    If −j ≤ d < 0, then Gj(j+d)(0)=0 because all the terms in the expansion of Gj(j+d)(0) have a factor F (0) = 0.

  2. (ii)

    Let d = 0. We shall prove (13) by induction on j. One can verify directly the case j = 1. Assuming (13) for j = k, we shall prove it for j = k + 1. Since Gk+1(x) = F(x)Gk(x), applying (i), we see that

    Gk+1(k+1)(0)=∑i=0k+1(k+1i)F(i)(0)Gk(k+1−i)(0)       =F(0)Gk(k+1)(0)+(k+1)F′(0)Gk(k)(0)+∑i=2k+1(k+1i)F(i)(0)Gk(k+1−i)(0)       =(k+1)!fk+2(0),
    which completes the proof of (ii).

  3. (iii)

    Let d > 0 and j be any positive integer. For simplicity, we will write f(i) := f(i)(0) below.

    1. (a)

      Let j = 1. If d = 1, then we have G1(2)(0)=3f′f=f′fH1,2(2) since H1,2(2) = 3. Thus, (13) is true for d = 1. Next, assuming (13) for G1(k)(0), we shall prove it for G1(k+1)(0). Since G1(x) = F(x)f(x), using (12) we obtain

      G1(k+1)(0)=∑i=1k+1(k+1i)f(i-1)f(k+1-i)=∑i=1k+1(k+1i)(-1)i-1fi(-1)k+1-ifk+2-i=(-1)fk+2∑j=1k+1(k+1j)=(-1)fk+2H1,1+k(2).
      This completes the proof for the case (a) j = 1 and any d > 0.

    2. (b)

      Assuming (13) for j = 1, 2, . . . k and any d > 0 we shall prove it for j = k + 1 and any d > 0. Since Gk+1(x) = F (x)Gk(x), by (12) and the induction assumption, we obtain

      Gk+1(k+1+d)(0)=∑i=1k+1+d(k+1+di)f(i-1)Gk(k+1+d-i)(0)=∑i=1d+1(k+1+di)f(i-1)Gk(k+1+d-i)(0)=∑i=1d+1(k+1+di)(-1)i-1fifk-d+i(f′)1+d-iHk,k+1+d-j(k+1)=fk+2-d(f′)d∑i=1k+1+d(k+1+di)Hk,k+1+d-i(k+1)=fk+2-d(f′)d∑l=0k+d(k+1+dl)Hk,l(k+1),
      where in the last equality we have made the index change l = k + 1 + d − j. Therefore, to finish the proof of the induction step (b), we need to show that
      ∑l=0k+d(k+1+dl)Hk,l(k+1)=Hk+1,k+1+d(k+2).
      For brevity denote r = k + 1 + d. Using the definition of Hk,l(k + 1), we obtain
      ∑l=0r−1(rl)Hk,l(k+1)=∑i=0k(−1)i(ki)∑l=0r−1(rl)(k+1−i)l    =∑i=0k(−1)i(ki)[(k−i+2)r−(k−i+1)r]    =(k+2)r−[(k+1)r+(k1)(k+1)r]+…       +(−1)k[(kk−1)2r+2r]+(−1)k+1    =(k+2)r−(k+11)(k+1)r+…+(−1)k(k+1k)2r+(−1)k+1    =∑j=0k+1(−1)j(k+1j)(k+2−j)r    =Hk+1,k+1+d(k+2).
      This proves the induction step (b). Now (iii) follows from (a) and (b). The proof of the lemma is complete.

3. Proof of the Theorem

It is not difficult to see that (4) is equivalent to (for brevity t := n − k − 1)

∫0xFt(u)(1−F(u))k−1f(u)f(n(x−u))du=∫0xFt(u)(1−F(u))kf(u)f(k(x−u))du.
Recalling from Lemma 4 that Gj(x) := Fj(x)f(x) for j = 1, 2, . . . and using the binomial formula, we write last equation as
∫0x∑l=0k−1(k−1l)(−1)lGt+l(u)f(n(x−u))du    =∫0x∑l=0k(kl)(−1)lGt+l(u)f(k(x−u))du.
Differentiating with respect to x as many as t + r + 2 times for r ≥ 0 and substituting x = 0, we obtain
∑l=0k(−1)l(k−1l)∑i=0t+r+1niGt+l(t+r+1−i)(0)f(i)(0)      =∑l=0k(−1)l(kl)∑i=0t+r+1kiGt+l(t+r+1−i)(0)f(i)(0).
In view of Lemma 1, to prove that (15) implies (6), it is sufficient to show that it implies (5). We shall prove (5) by induction with respect to m. Let us first verify (5) for m = 1, i.e., f′(0) = −f2(0). If r = 0 then (15) becomes
∑l=0k(−1)l(k−1l)∑i=0t+1niGt+l(t+1−i)(0)f(i)(0)     =∑l=0k(−1)l(kl)∑i=0t+1kiGt+l(t+1−i)(0)f(i)(0).
Since, by Lemma 4, Gt+l(t+1−i)=0 when t + 1 − i < t + l, omitting the zero terms in the sums and simplifying we obtain
(t+1)Gt(t)(0)f′(0)=−Gt+1(t+1)(0)f(0),
which, in view of Lemma 4, is equivalent to
(t+1)ft+1(0)t!f′(0)=−ft+2(0)(t+1)!f(0)
and thus f′(0) = −f2(0). Thus, (5) is true for m = 1.

To prove the induction step suppose (5) holds for m = 1, 2, . . . , r, i.e.,

f(m)(0)=(−1)mfm+1(0), m=1,2,…,r.
We shall prove it for m = r + 1. Under the induction hypothesis (17), Lemma 2 implies Gt+l(t+r+1−i)(0)>0 only if i ≤ r − l + 1 and thus, omitting the zero terms in the sums of (15), we have
∑l=0k(−1)l(k−1l)∑i=0r−l+1niGt+l(t+r+1−i)(0)f(i)(0)      =∑l=0k(−1)l(kl)∑i=0r−l+1kiGt+l(t+r+1−i)(0)f(i)(0)
and interchanging the sums, we write it as
∑i=0r+1nif(i)(0)∑l=0r−i+1(−1)l(k−1l)Gt+l(t+r+1−i)(0)     =∑i=0r+1kif(i)(0)∑l=0r−i+1(−1)l(kl)Gt+l(t+r+1−i)(0).
Collecting in the left-hand side the terms with i = r + 1, results in
(nr+1−kr+1)f(r+1)(0)Gt(t)(0)    =∑i=0rf(i)(0)∑l=0r−i+1[ki(kl)−ni(k−1l)](−1)lGt+l(t+r+1−i)(0).
Applying Lemma 4 to both sides of (18) and taking into account the induction hypothesis (17) in the right-hand side, it is not difficult to see that
(nr+1−kr+1)f(r+1)(0)t!ft+1(0) =(−1)r+1ft+r+3(0)∑i=0r∑l=0k[ki(kl)−ni(k−1l)]Ht+l,t+r+1−i(t+1+l),
where (p−1k)=0. Now, it is clear that proving (5) for m = r +1 is equivalent to proving
(nr+1−kr+1)t!=∑i=0r∑l=0k[ki(kl)−ni(k−1l)]Ht+l,t+r+1−i(t+1+l).
Applying Lemma 2 we write last equation as
(nr+1−kr+1)t!=∑i=0rkiHt,t+r+1−i(n)−∑i=0rniHt,t+r+1−i(n−1)
Finally, it is easily to see that (19) with m = r + 1 follows from (9) and (10) in Lemma 3. This proves (5) for m = r + 1, which in turn completes the induction argument and theorem’s proof.

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Journal
Journal of Statistical Theory and Applications
Volume-Issue
17 - 3
Pages
408 - 418
Publication Date
2018/09/30
ISSN (Online)
2214-1766
ISSN (Print)
1538-7887
DOI
10.2991/jsta.2018.17.3.2How to use a DOI?
Copyright
© 2018, the Authors. Published by Atlantis Press.
Open Access
This is an open access article under the CC BY-NC license (http://creativecommons.org/licences/by-nc/4.0/).

Cite this article

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TY  - JOUR
AU  - Santanu Chakraborty
AU  - George P. Yanev
PY  - 2018
DA  - 2018/09/30
TI  - Characterizations of Exponential Distribution Based on Two-Sided Random Shifts
JO  - Journal of Statistical Theory and Applications
SP  - 408
EP  - 418
VL  - 17
IS  - 3
SN  - 2214-1766
UR  - https://doi.org/10.2991/jsta.2018.17.3.2
DO  - 10.2991/jsta.2018.17.3.2
ID  - Chakraborty2018
ER  -