Characterizations Of The Inverse Weibull Distribution And ...
APPLICATIONES MATHEMATICAE
27,2 (2000), pp. 197–202
P. P A W L A S and D. S Z Y N A L (Lublin)
CHARACTERIZATIONS OF THE INVERSE
WEIBULL DISTRIBUTION AND GENERALIZED
EXTREME VALUE DISTRIBUTIONS BY MOMENTS
OF k TH RECORD VALUES
Abstract. We give characterization conditions for the inverse Weibull
distribution and generalized extreme value distributions by moments of kth
record values.
1. Introduction. We discuss characterization problems for an inverse
Weibull distribution function
(1.1)
F (x) = e−(θ/x)α ,
x > 0, α > 0, θ > 0,
and the standard generalized extreme value distribution function given by
e−{1−γx}1/γ, x < 1/γ when γ > 0,
(1.2)
F (x) =
x > 1/γ when γ < 0,
e−e−x,
x ∈ R when γ = 0.
Note that F (x) given by (1.1) with θ = 1 is a Fr´echet distribution function
(cf. [3]).
We present characterization conditions for distribution functions given
by (1.1) and (1.2) by moments of kth lower record values introduced in [3].
The kth upper record values were discussed in [1]. So first we recall the
concept of kth lower record values (cf. [3]).
Let {Xn, n ≥ 1} be a sequence of i.i.d. random variables with a cumu-
lative distribution function F (x) and a probability density function f (x).
The jth order statistic of a sample (X1, . . . , Xn) is denoted by Xj:n. For a
fixed k ≥ 1 we define the sequence {Lk(n), n ≥ 1} of kth lower record times
2000 Mathematics Subject Classification: Primary 62E10, 60E99.
Key words and phrases: sample, order statistics, record values, kth record values,
inverse Weibull distribution, generalized extreme value distributions.
[197]
198
P. Pawlas and D. Szynal
of {Xn, n ≥ 1} as follows:
Lk(1) = 1,
Lk(n + 1) = min{j > Lk(n) : Xk:Lk(n)+k−1 > Xk:j+k−1}.
For k = 1 we put L(n) := L1(n), n ≥ 1, which are lower record times
of {Xn, n ≥ 1}. The sequence {Z(k)
n , n ≥ 1} with Z (k)
n
= Xk:Lk(n)+k−1,
n = 1, 2, . . . , is called the sequence of kth lower record values of {Xn, n ≥ 1}.
For convenience, we also set Z(k)
0
= 0. Note that for k = 1 we have Z(1)
n
=
XL(n), n ≥ 1, i.e. the record values of {Xn, n ≥ 1}. Moreover, we see that
Z(k)
1
= max(X1, . . . , Xk) := Xk:k.
It is known (cf. [3]) that the pdf of Z(k)
n
and the joint pdf of (Z(k)
m , Z (k)
n )
are given respectively by
kn
(1.3)
f
(x) =
[− ln F (x)]n−1[F (x)]k−1f (x),
n ≥ 1,
Z(k)
n
(n − 1)!
kn
(1.4)
f
(x, y) =
[ln F (x) − ln F (y)]n−m−1
Z(k)
m ,Z(k)
n
(m − 1)!(n − m − 1)!
× [− ln F (x)]m−1 f (x) [F (y)]k−1f (y),
F (x)
x > y, 1 ≤ m < n, n ≥ 2.
Results for kth upper record values can be found in [1].
Section 2 contains characterization conditions for an inverse Weibull dis-
tribution and in Section 3 we give recurrence relations for product moments
of kth lower record values of that distribution. Characterization conditions
for the standard generalized extreme distribution (1.2) are presented in Sec-
tion 4.
2.
Characterization conditions for an inverse Weibull distri-
bution. The characterizations of distributions presented in this paper are
based on the following result by Lin (cf. [2]).
Proposition. Let n0 be any fixed non-negative integer , −∞ < a < b
< ∞, and g(x) > 0 an absolutely continuous function with g′(x) = 0 a.e. on
(a, b). Then the sequence of functions {(g(x))ne−g(x), n ≥ n0} is complete
in L(a, b) iff g(x) is strictly monotone on (a, b).
Let us note that for the inverse Weibull distribution (1.1) we have
(2.1)
xf (x) = αF (x)(− ln F (x)).
We start with recurrence relations for moments of the inverse Weibull
distribution. From them we derive a formula for single moments expressed
in terms of moments of Xk:k.
Characterizations of inverse Weibull distribution
199
Theorem 1. Fix a positive integer k ≥ 1. Then for any positive integer
r, we have
r
(2.2)
E(Z(k)
n
)r =
1 −
E(Z(k)
(n − 1)α
n−1)r
whenever (n − 1)α > r, and consequently,
n−1
r
(2.3)
E(Z(k)
n
)r =
1 −
E(X
iα
k:k)r .
i=1
P r o o f. For n ≥ 1 and r = 1, 2, . . . , from (1.3) we have
αkn
E(Z(k)
n
)r =
xr−1[− ln F (x)]n[F (x)]k dx.
(n − 1)!
Ì
Integrating by parts in the above integral written as xr−1d(. . .), we get
α(n − 1)
E(Z(k)
[E(Z(k)
n−1)r =
r
n−1)r − E(Z (k)
n
)r].
This gives (2.2). Using an induction argument leads to (2.3).
Corollary. Under the assumptions of Theorem 1 with α = 1 we obtain
a recurrence relation for single moments of kth lower record values from the
inverse exponential distribution:
r
E(Z(k)
n
)r =
1 −
E(Z(k)
n − 1
n−1)r .
Now we show that one can have a stronger result.
Theorem 2. Fix a positive integer k ≥ 1 and let r be a positive integer.
A necessary and sufficient condition for a random variable X to be distrib-
uted with pdf given by (1.1) is that
r
E(Z(k)
n
)r =
1 −
E(Z(k)
(n − 1)α
n−1)r
for all positive integers n such that (n − 1)α > r.
P r o o f. The necessity part follows immediately from Theorem 1.
On the other hand if the recurrence relation (2.2) is satisfied, then
xr−1[− ln F (x)]n−1[F (x)]k−1{xf (x) − α(− ln F (x))F (x)} dx = 0.
It now follows from the Proposition that
xf (x) = α(− ln F (x))F (x),
which proves by (2.1) that f (x) has the form (1.1).
200
P. Pawlas and D. Szynal
Corollary. Let α = 1 in Theorem 2. A necessary and sufficient con-
dition for a random variable X to have the inverse exponential distribution
is that
r
E(Z(k)
n
)r =
1 −
E(Z(k)
n − 1
n−1)r
for all positive integers n such that n − 1 > r.
Corollary. Under the assumptions of Theorem 2 with r = 1 the equa-
tions
1
EZ(k)
n
=
1 −
EZ(k)
(n − 1)α
n−1,
(n − 1)α > 1,
characterize an inverse Weibull distribution.
Example. Let
0
(1 − 1/(jα)) := 1 and assume that α > 1. Then
j=1
the equations
θ
1
n−1
1
EZ(k)
n
=
Γ 1 −
1 −
for n = 1, 2, . . .
kα
α
jα
j=1
all hold iff
F (x) = e−(θ/x)α ,
x > 0, α > 1.
Remark. If we let additionally k = 1 then
1
n−1
1
EXL(n) = θΓ 1 −
1 −
for n = 1, 2, . . .
α
iα
j=1
iff F (x) is given by (1.1) with α > 1.
Note that the assumption α > 1 is needed for the existence of EX.
3. Product moments of kth lower record values from an in-
verse Weibull distribution. We complete our considerations by giving
recurrence relations for product moments of kth lower record values from
an inverse Weibull distribution.
Theorem 3. Fix a positive integer k ≥ 1 and let r be a non-negative
integer such that r < mα. Then for m ≥ 1, s = 1, 2, . . . ,
r
(3.1)
E(Z(k)
m+1)r+s =
1 −
E[(Z(k)
mα
m )r (Z (k)
m+1)s]
and for 1 ≤ m ≤ n − 2,
r
(3.2)
E[(Z(k)
m+1)r (Z (k)
n )s] =
1 −
E[(Z(k)
mα
m )r (Z (k)
n )s].
Characterizations of inverse Weibull distribution
201
P r o o f. From (1.1) for 1 ≤ m ≤ n − 1 and r, s = 1, 2, . . . ,
(3.3) E[(Z(k)
m )r (Z (k)
n )s] =
xrysfm,n(x, y) dx dy
kn
=
ys[F (y)]k−1f (y)I(y) dy,
(m − 1)!(n − m − 1)!
where
I(y) = xr[− ln F (x)]m−1 f (x) [ln F (x) − ln F (y)]n−m−1 dx
F (x)
= α xr−1[− ln F (x)]m[ln F (x) − ln F (y)]n−m−1 dx.
But
αm
I(y) =
xr[− ln F (x)]m−1[ln F (x) − ln F (y)]n−m−1 f (x) dx
r
F (x)
α(n − m − 1)
−
xr[− ln F (x)]m[ln F (x) − ln F (y)]n−m−2 f (x) dx .
r
F (x)
Upon substituting the above equation in (3.3) and simplifying, we obtain
αm
E[(Z(k)
m )r (Z (k)
n )s] =
[E[(Z(k)
r
m )r (Z (k)
n )s] − E[(Z (k)
m+1)r (Z (k)
n )s]]
Hence we have (3.2). When n = m + 1 we obtain (3.1).
4.
Characterization conditions for the generalized extreme
value distribution. Recurrence relations for moments of kth lower record
values from a generalized extreme value distribution were presented in [3].
We now only give characterization conditions for the standard generalized
extreme value distribution (1.2) based on those relations.
To characterize the df F given by (1.2) we use an equivalent representa-
tion of (1.2), namely
(4.1)
(1 − γx)f (x) = F (x)(− ln F (x))
for γ = 0,
and f (x) = F (x)(− ln F (x)) for γ = 0.
The main result of this section is as follows.
Theorem 4. A necessary and sufficient condition for a random variable
X to be distributed according to (2.1) is that
r
r
(4.2)
E(Z(k)
n
)r =
1 + γ
E(Z(k)
(EZ(k)
n − 1
n−1)r − n − 1
n−1)r−1
for n = 2, 3, . . .
P r o o f. The necessity part was proved in [3]. Assume now that (4.2) is
satisfied. Then
[− ln F (x)]n−1[F (x)]k−1xr[−(− ln F (x))F (x) − γxf (x) + f (x)] dx = 0.
202
P. Pawlas and D. Szynal
It now follows from the above Proposition that
(1 − γx)f (x) = (− ln F (x))F (x),
which proves that
f (x) = e(1−γx)1/γ (1 − γx)1/γ−1.
Example. Let
n−1(1 + γ/j) := 1. Then for γ > 0,
j=n
1
1
n−1
γ
EZ(k)
n
=
1 −
Γ (γ + 1)
1 +
γ
kγ
j
j=1
n−1 1 n−1
γ
−
1 +
for n = 2, 3, . . .
i
j
i=1
j=i+1
iff F (x) is given by (1.2).
Remark. If we let additionally k = 1, then
1
n−1
γ
XL(n) = (1 − Γ (γ + 1))
1 +
γ
j
j=1
n−1 1 n−1
γ
−
1 +
for n = 2, 3, . . .
i
j
i=1
j=i+1
iff F (x) is given by (1.2).
Acknowledgments. We are grateful to the referee for his comments
which allowed us to give a more general presentation of the results.
References
[1]
W. D z i u b d z i e l a and B. K o p o c i ´
n s k i, Limiting properties of the k-th record value,
Zastos. Mat. 15 (1976), 187–190.
[2]
G. D. L i n, On a moment problem, Tˆ
ohoku Math. J. 38 (1986), 595–598.
[3]
P. P a w l a s and D. S z y n a l, Relations for single and product moments of k-th record
values from exponential and Gumbel distribution, J. Appl. Statist. Sci. 7 (1998),
53–62.
Piotr Pawlas and Dominik Szynal
Department of Mathematics
Maria Curie-Sklodowska University
Pl. M. Curie-Sklodowskiej 1
20-031 Lublin, Poland
E-mail: szynal@golem.umcs.lublin.pl
Received on 14.4.1999;
revised version on 26.7.1999
