Q·IARACTERI ZATIONS OF THE EXPONENfIAL DISTRIBtlfION
BY CONDITIONAL r·{)MENTS
by
Steel T. Huang
Institute of Statistics tUrneo Series 1#1214
February, 1979
DEPARTMENT OF STATISTICS
Chapel Hill, North Carolina
CHARACTERIZATIONS OF THE EXPONENTIAL DISTRIBUTION
BY CONDITIONAL MOMENTS
by Steel T. Huang*
Department of Mathematics, University of Cincinnati and
Department of Statistics, University of North Carolina.
ABSTRACT
In this note, two characterizations of the exponential distribution are
established.
Theorem 1: Let a > -1, a
~
O.
A positive random variable X has
an exponential distribution if and only if E[CX-y)aI X > y] is a finite constant
for all y
~
O.
Theorem 2: Let Xl ,X2, ..• ,Xn be i.i.d. random variables having
a continuous distribution FCx), and XCl)'XC2)' ... 'XCn) their order statistics.
Let a > 0 and m = 1,2, ..• ,n-l be given.
Then FCx) is an exponential distribu-
tion up to a location parameter if and only if E[CX Cm+ l ) - XCm))aIXCm)
a finite constant for almost all y with respect to FCy).
= y] is
*This research was supported by the Air Force Office of Scientific Research
under contract AFOSR-75-2796 and the Office of Naval Research under contract
NOOOI4-75-C-0809.
CHARACTERIZATIONS OF THE EXPONENTIAL DISTRIBUTION
BY CONDITIONAL MOMENTS
1.
INTRODUCTION
In recent years there has been much interest and development in the theory
of characterizations of the exponential distribution.
A comprehensive account
of this theory and its significance in model building is given by Galambos and
Kotz [3].
In this note, two known characterization theorems are generalized.
A functional equation is solved first (Theorem 3) using Mellin transforms.
As
a consequence, the following characterization theorems are established.
THEOREM 1:
Let a > -1, a 1 0 be given.
A positive random variable X has an
exponential distribution if and only if
E[(X - y)al x > y]
=c
for all y > 0,
(1.1)
where c is a (finite) constant.
If P(X > y)
= 0, the conditional expectation in (1.1) is chosen to satisfy
the equality.
THEOREM 2:
Let X ,X 2 , ... ,Xn be i.i.d. random variables having a continuous
l
distribution F(x), and X(1)'X(2)' ... 'X(n) their order statistics.
and m = 1,2, ... ,n-l be given.
location parameter a (i.e. F(x)
Let a > 0
Then F(x) is an exponential distribution with some
= 1 - e -~(x-a) , x
~
a, for some
~ >
.
0) 1f and
only if
(1.2)
where c is a (finite) constant.
2
REMARK:
If F(x) is absolutely continuous and has a bounded density. then
Theorem 2 holds for a > -1. a
F o.
Shanbhag [5] showed that (1.1) is a characteristic property of the exponential distribution for the case a
abitrary a
=
Theorem 2 (a
= 1.
Later. this result was generalized to
1.2 •... by Sahobov and Geshev (see [3, p.33]).
= 1)
A special case of
was obtained by Ferguson [2] (also see [3. p. 61]).
In the context of applied probability, X represents the life time of a
component and X - Y given X > Y represents its residual life time; Xl .X 2 •... ,Xn
represent the life times of n identical components working independently.
Then
Theorem 1 asserts that the life time is exponentially distributed if and only if
the a-th moment (a > -1. a
F 0)
of the residual life time is a finite constant.
and Theorem 2 asserts that the life time has an exponential distribution up to
a location parameter if and only if the conditional cl.-th moment (a > 0) of the
waiting time between the m-th and (m+l)-st failures ,given the waiting time to
the m-th failure is a finite constant.
2.
DERIVATION OF CHARACTERIZATION THEOREMS
Some preliminaries are needed for the proof of Theorem 3.
transfoPms of a distribution function F(x) with F(O+)
=0
The MelUn
and a nonnegative
measurable function ¢(x). 0 < x < 00, are given by
roo xs-l dF(x) and n(s) = 10
roo xs-l ¢(x)dx
= 10
m(s)
respectively. where s is a complex number.
converges is either the single line Re s
(_00
9 cl
< Re s < c
2
(~oo);
The set of all s for which m(s)
=1
or an open strip
the set of all s for which n(s) converges is empty
or a single vertical line or an open strip.
When a Mellin transform converges
3
for all s in an open strip, it is analytic in that region.
It follows from the
uniqueness property of the characteristic functions that if two Mellin transforms
=1
coincide on the line Re s
identical.
LEMMA 1:
then their corresponding distribution functions are
For detailed properties of Mellin transforms, see [4].
Suppose F(x) is a distribution function with F(O+)
c l < Re s < c ' its Mellin transform.
2
convex) on the interval c
Proof.
l
=0
and m(s),
Then m(s) is log convex (i.e. log m(s) is
< s < c .
2
The lemma is a slight generalization of Theorem 1.9 in [1]; the proof is
essentially the same.
Consider the Riemann-Stieltjes sum of the integral
c
l
<
fba+
x s-l dF(x) for
s < c ' where 0 < a < b < 00:
2
·e
n-l
S (s)
n
where h
=
=
(b-a)jn.
function.
r
. 0
J=
(a + jh)s-l [F(a + (j+l)h) - F(a + jh)] ,
Each summand is either identically zero or a log convex
For large b, S (s) is positive.
n
Being the sum of log convex functions,
S (s) is itself log convex [I, Theorem 1.8].
As n
n
b
+
00, S (s) converges to
n
fa+ xs-l dF(x) for c l < s < c 2 by the Bounded convergence theorem.
Hence this
integral, as the limit of log convex functions, is also log convex [I, Theorem 1.6].
Since m(s), c
log convex.
LEMMA 2:
<
l
s l
s < c ' is the limit of f~+ x - dF(x) as a
2
+
0 and b
+
00, it is
0
If a real-valued function m(s), s
~
I, satisfies the following conditions,
then it is identical with the Gamma function res) for s > 1:
4
i)
For some constant a
m(s
ii)
iii)
Proof.
+
>
0,
a) = res + a) m(s) ,
res)
s > 1.
m(s) is log convex.
mel) = 1.
Suppose m(s) is a function satisfying these three conditions.
Applying
(i) recursively, we have
m(s
+
ka) = res + ka) m(s),
res)
k=0,1,2, ...
(2.1)
Setting s = 1, (2.1) becomes by (iii)
m(l + ka) = r(l + ka),
(2.2)
k = 0,1,2, ...
It suffices to show that m(s) agrees with res) on the interval 1
because of (i).
Fix s between 1 and l+a.
log m(l + ka) - log m(l + ka - a)
(1 + ka) - (1 + ka - a)
2
<
s
<
It follows from (ii) that for k
<
e·
l+a
= 1,2, ...
log m(s + ka) - log m(l + ka)
(s + ka) - (1 + ka)
log m(l + ka + a) - log: m(l + ka)
(1 + ka + a) - (1 + ka)
which reduces to
s-l
[
j
m(l + ka)
m(l + ka - a)
a
<
m(s
ro
+
+
ka)
ka)
~O
<
s-l
+
[ m(l
substituting (2.1) and (2.2) into (2.3), we get
s-l
r (l + ka)
r 0 + ka)
a < m(s)
res)
res + kal ~(l + ka - a~
r
ka + a~a
ka )
+
J
(2.3)
J
r(l + ka)
2 res + kal
[
ro
-t.
ro
s-l
ka + a)Ja
+ kal
J
(2.4)
5
A trite calculation employing Stirling's formula shows that. for any a
the ratio of f(t + a)/f(t) and t a tends to 1 as t
+
00.
THEOREM 3:
= f(s) for
1 < s < l+a
r:y (x_y)a dF a (x)
Then Fa(x)
Proof.
F(O+)
-e
= 1 - e - A(x-a) for x
Let F(x)
=
O.
~
=c
(1 -
Fa(y)).
=0
+
00.
0
as desired.
Let a. a > -1 and a ; O. c > 0 be given (finite) constants.
Fa(x) is a distribution function satisfying Fa(a+)
O.
Applying this result. we
deduce that the quantities at both ends of the inequality (2.4) tend 1 as k
Thus it follows that m(s)
~
Suppose
and
y ~ a
a
a. where A is given by A c
(2.5)
=
f(l + a).
x
= Fa(a + I)'
Then F(x) is a distribution function satisfyifig
Also it follows from (2.5) by a change of variable that F(x) satisfies
J;
(x_y)a dF(x)
=1
We shall show that F(x)
- e
definiteness let us assume a
>
= f(l + a)(l - F(y)). Y ~ O.
-Ax for x
= 1
O. and the Theorem then follows.
For
O. the case -1 < a < 0 being similar.
roo x a dF(x) >
Since (2.6) implies JO
of F(x) converges at s
~
(2.6)
and 1 + a.
00.
the Mellin transform m(s)
= JOroo
xs-l dF(x)
Therefore m(s) converges on a strip
c
< Re s < c where c < 1 < 1 + a < c . Now consider the function $(y). y > O.
2
2
l
l
defined by either side of (2.6). and its Mellin transform n(s). On the one hand
n(s)
roo Ys-l $(y)dy
= JO
=~
J; yS-l
(x_y)a dF(x)dy
= J~ [f~ ys-l (x_y)a dy]dF(x)
= ~ [f~ zS-l (l_z)a dz]x s +a dF(x)
= f(s)f(l+a) m(s+l+a) •
f(s+l+a)
(2.7)
6
where the set of s for which n(s) converges is seen to be
{siRe s > O} n {slc
= {siO
- 1 - a
l
< s < c
Re s
<
c 2 - 1 - a}
<
(2.8)
1 - a} ;
2
on the other hand
n(s)
= r(l+a)
~ 1;+
= r(l+a)
r;
=
r(l+a)
s
y s-l dF(x)dy
s-l dy dF(x)
f~
m(s+l)
(2.9)
Y
,
where the set of s for which n(s) converges is seen to be
{siRe s > O} n {c 1 - 1
<
Re s
c 2 - I}
<
=
{sio
<
Re s
<
c 2 - I}.
(2.10)
e-
The interchange of the order of integration above is justified by Fubini's
theorem because the integrands are nonnegative.
in (2.8) and (2.10), we have that c
2
=
ao
Corrparing the convergence regions
and n(s) converges for all s, Re s > O.
Thus it follows from (2.7) and (2.9) that
m(s
+
1
+
a)
res + 1 + a)
= --'-:r="'('-s-+-:1;-;:)~
m(s
+
1) ,
Re s >
o.
Note that m(s) restricted to s > 1 is a real-valued function which clearly satisfies conditions (i) and (ii) in Lemma 2 and which is log convex by Lemma 1.
Therefore, by Lemma 2, m(s)
= res)
for s 2:. 1.
and res) is analytic in Re s > 0, we have m(s)
particular m(s)
=
res) on Re s
=
1.
Since m(s) is analytic in Re s > c 1
= res)
on Re s > max(0,c ).
1
Finally notice that res), Re s > 0, is the
Mellin transform of the exponential distribution 1 .. e-x, x 2:. 0,
uniqueness property of the Mellin transform, we conclude F(x)
as was to be proved.
n
In
=1
Thus by the
- e -x , x > 0,
7
Proof of Theorem 1.
•
If X has an exponential distribution then it has the lack of
memory property (i.e. X given X
hence (1.1) holds.
Y (y
>
~
0) has the same distribution as X) and
Conversely, suppose (1.1) holds.
It is easy to show that
(1.1) is equivalent to
J;
= c(l - F(y)), Y ~ 0 ,
(x_y)a dF(x)
where F(x) is the distribution function of X.
Thus, by Theorem 3, F(x)
complete.
=1
.4It
-Ax
x
~
0, for some A.
The proof is
0
Proof of Theorem 2 "only if" part.
assume a
- e
Since X is positive, we have
Suppose F(x)
=1
-
e - A(x-a) , x
~
a.
We may
= 0, for otherwise we can argue with Xl-a, ... ,Xn-a. Then X(m+l) - X(m)
is independent of X(m) and it has an exponential distribution (see for example,
[3, Theorem 3.1.1]).
"if" part.
Hence (1.2) holds.
From Theorem 3.1. 3 in [3], i t follows that \m+l) given X(m)
=y
has the distribution function
F*(x)
y
=1
- (1 _ F(x) - F(f) )n-m
1 - F(y
,
for almost all y with respect to F(y).
.J; (x_y)a dF;(X)
x~y
Therefore, (1.2) implies
= c a.e. [dF(y)] ,
or equivalently
r:: (x_y)a dG(x)
y
= c(l - G(y)) a.e. [dF(y)]
(2.11)
where
G(x)
=1
- (l_F(x))n-m .
(2.12)
8
It is easily seen that G(x) is a continuous distribution function and
= O} = sup{xIF(x) = OJ = a (say), inf {xIG(x) = I} = inf
= I} = b (say). We shall show that (2.11) in fact holds for
sup{xIG(x)
{xIF(x)
and a is finite.
all y > a
•
Then, by Theorem 3, G(x) is an exponential distribution with
location parameter a.
By (2.12), F(x) is also an exponential distribution with
location parameter a as asserted.
Denote by A(y) and B(y),
_00
< Y <
the functions defined by the left hand
00,
side and the right hand side of (2.11) respectively.
G(y) is.
Also A(y) can be shown to be continuous by the Monotone convergence
theorem (here the assumption a >
increasing on a < y < b.
F(Yl)
= F(Y2)
(Y2' Y2
B(y) is continuous since
+
E)
0 is used).
We claim that F(y) is strictly
Otherwise, there exist a < Yl < Y2 < b such that
and F(y) is strictly increasing on the intervals (Yl for sufficiently small
>
E
O.
Now we have B(Yl)
E,
= B(Y2)'
Yl) and
and
e·
since the integrand is strictly positive on the set (Y2'oo) which has positive
dF-measure.
(Yl -
E,
Thus, by the continuity of A(y) and B(y), A(y)
Yl) or for every y in (Y2' Y2
for every y in
for sufficiently small
+ E)
contradicts (2.11) since both (Yl - E, Yl) and (Y2' Y2
dF-measure.
F B(y)
+
E.
This
E) have positive
Therefore F(y) is strictly increasing on a < y < b.
This implies
(2.11) holds for a dense subset of (a, b).
By the continuity of A(y) and B(y)
again, (2.11) holds for all y, a < y < b.
A further inspection reveals that
(2.11) in fact holds for all y
the contrary.
~
a.
Finally, we need to show that a >
Then there exists a sequence a
n
< b decreasing to
the distribution function
F(x) - F(a )
n
Fa (x) = ---,l=-----=F-;-(a---::-)n
n
x > a
-
n
_00
•
_00.
Assume
Consider
9
+
=0
Note that F (a)
a
n
and, by the improved version of (2.11),
n
r:: (x-y) cx dF
•
y
~
(x)
=
=
r;
= -=-l---=~~(a--':-)
n
(x_y)cx dF(x)
c
1 - F (a ) (l - F (y) )
n
c(l - F
Now Theorem 3 implies that F (x)
a
~
=
(y)), Y > a .
1 - e
-
n
-A(x-a )
n
x > a where ACXC
- n
n
=
r(l
+
cx).
Hence
F(x)
Letting a
n
+ _00,
a must be finite.
=
-A(x-a )
(1 - F(a ))(1 - e n )
n
we have F(x)
=1
for all x.
+
F(a),
n
x > a
- n
This is a contradiction, and thus
0
The proof is now complete.
,"
·e
To substantiate the remark after Theorem 2, we note that if F(x) has a
bounded density then the function A(y) can be shown to be continuous for cx > -1
and the same proof of Theorem 3 will carryover.
REFERENCES
..
'-
[1]
Artin, E., The Gamma Function, New York:
Holt, Rinehart and Winston, 1964.
[2]
Ferguson, T.S., "A Characterization of the Exponential Distribution,"
Ann. Math. Statist., 35 (1964), 1199-1207.
[3]
Galambos, J. and Kotz, S., Characterizations of Probability Distributions,
Berlin, Springer-Verlag, 1978.
[4]
Kawata, T., Fourier Analysis in Probability Theory, New York: Academic Press,
1972.
[5]
Shanbhag, D.N., "The Characterizations for Exponential and Geometric Distributions,"
J. ArneI'. Statist. Assoc., 65 (1970), 1256-1259 .