Characterizations Of Power Distributions Via Moments Of Order ...
APPLICATIONES MATHEMATICAE
26,4 (1999), pp. 467–475
Z. G R U D Z I E ´
N and D. S Z Y N A L (Lublin)
CHARACTERIZATIONS OF POWER DISTRIBUTIONS VIA
MOMENTS OF ORDER STATISTICS
AND RECORD VALUES
Abstract. Power distributions can be characterized by equalities involv-
ing three moments of order statistics. Similar equalities involving three
moments of k-record values can also be used for such a characterization.
The case of samples with random sizes is also considered.
1. Introduction. Too and Lin [8] have given a characterization of the
uniform distribution by an equality involving only two moments of order
statistics. We extend that result to power distributions. Moreover, we give
a characterization of power distributions in terms of moments of k-record
values. In Sections 4 and 5 we treat the characterization problem when
sample sizes are random (cf. [1], [6], [9]).
2. A characterization of power distributions. Let Xk:n be the
kth smallest order statistic of a random sample (X1, . . . , Xn) from a dis-
tribution F . Let m be a negative integer. We start with the problem of
characterizing the power distribution function F defined as follows (cf. [1]):
(2.1)
F (x) = 1 − (1 + mx)−1/m,
x ∈ (0, −1/m).
Theorem 1.
With the above notation suppose that EX2
< ∞ for
k:n
some pair (k, n). Then the equality
2
n
(2.2)
EX2
[k]
k:n −
EX
m (n − m)
k:n−m − EXk:n
[k]
1
n
2n
+
[k]
−
[k]
+ 1 = 0,
m2 (n − 2m)[k]
(n − m)[k]
where n[k] = n(n − 1) . . . (n − k + 1), holds iff F is given by (2.1).
1991 Mathematics Subject Classification: Primary 62E10, 60E99.
Key words and phrases: sample, record values, power, uniform distributions.
[467]
468
Z. Grudzie ´
n and D. Szynal
P r o o f. Let F −1(t) = inf{x : F (x) ≥ t, t ∈ (0, 1)}. Taking into account
that
1
n!
(2.3)
EXlk:n =
(F −1(t))ltk−1(1 − t)n−kdt,
l ≥ 1,
(k − 1)!(n − k)! 0
we see that E|Xk:n| < ∞, and E|Xk:n−m| < ∞. Furthermore, when F is
given by (2.1), we find that
k
s
(2.4)
EXk:s =
[B(k, s − m − k + 1)
m k
− B(k, n − k + 1)],
n ≤ s ≤ n − m.
and
k
n
(2.5) EX2k:n =
[B(k, n − k − 2m + 1) − 2B(k, n − k − m + 1)
m2 k
+ B(k, n − k + 1)],
where B(a, b) is the Beta function, and so (2.2) holds true.
Conversely, assume that (2.2) holds. Applying (2.3) we see that (2.2)
can be written as
1
(1 − t)−m − 1 2
F −1(t) −
tk−1(1 − t)n−k dt = 0,
m
0
which implies that F (x) is given by (2.1).
When m = −1 Theorem 1 reduces to the following characterization of
the uniform distribution.
Corollary 1 (cf. [8]). Let EX2
< ∞ for some pair (k, n). Then
k:n
2k
k(k + 1)
(2.2′)
EX2k:n −
EX
= 0
n + 1
k+1:n+1 + (n + 1)(n + 2)
iff F (x) = x on (0, 1).
In proving (2.2′) we use the equality
(n − k)EXk:n + kEXk+1:n = nEXk:n−1
with k = 1 (cf. [2]).
Using (2.4) and (2.5) we obtain the following characterizing conditions.
Theorem 1′.
Under the assumptions of Theorem 1 the distribution
function F is given by (2.1) iff
1
s
EX
[k]
k:s =
− 1 ,
s = n, n − m,
m (s − m)[k]
1
n
n
EX2
[k]
[k]
k:n =
− 2
+ 1 .
m2 (n − 2m)[k]
(n − m)[k]
Characterizations of power distributions
469
The same results can be derived for the distribution
(2.6)
F (x) = 1 − (1 + mx)−1/m,
x > 0,
where m is a positive integer (cf. [1]). In this case (2.2) holds with n − 2m −
k > 0.
3. Characterizations in terms of moments of k-record values.
Let {Xn, n ≥ 1} be a sequence of i.i.d. random variables with a common
distribution function F . For a fixed integer k ≥ 1 we define (cf. [3]) the
sequence of k-record values as follows:
Y (k)
n
= XLk(n):Lk(n)+k−1,
n ∈ N,
where the sequence {Lk(n), n ≥ 1} of k-record times is given by Lk(1) = 1,
Lk(n + 1) = min{j : j > Lk(n), Xj:j+k−1 > XLk(n):Lk(n)+k−1}, n ∈ N.
A characterization of F in (2.1) is contained in the following theorem.
Theorem 2. Let {Xn, n ≥ 1} be a sequence of i.i.d. random variables
with a common distribution function F such that E|min(X1, . . . , Xk)|2p < ∞
for a fixed k ≥ 1 and some p > 1. Then F is given by (2.1) iff
2
k
n
(3.1)
E(Y (k)
n
)2 −
EY (k−m)
m
k − m
n
− EY (k)
n
1
k
n
k
n
+
1 − 2
+
= 0
m2
k − m
k − 2m
for n = 1, 2, . . .
P r o o f. Suppose that F is given by (2.1). Then we have
1
kn
(3.2)
EY (k)
n
=
F −1(t)[− log(1 − t)]n−1(1 − t)k−1 dt
(n − 1)! 0
1
k
m
=
− 1
m
k − m
and
1
kn
(3.3)
E(Y (k)
n
)2 =
(F −1(t))2[− log(1 − t)]n−1(1 − t)k−1 dt
(n − 1)! 0
1
kn
=
[(1 − t)−m − 1]2[− log(1 − t)]n−1(1 − t)k−1 dt
(n − 1)! 0
kn
Γ (n)
2Γ (n)
1
=
−
+
Γ (n)
(n − 1)!m2 (k − 2m)n
(k − m)n
kn
1
k
n
k
n
=
− 2
+ 1 ,
m2
k − 2m
k − m
470
Z. Grudzie ´
n and D. Szynal
k
n
kn
1
1
(3.4)
EY (k−m)
−
,
k − m
n
= m (k − 2m)n (k − m)n
which establishes (3.1).
Conversely, assuming that (3.1) is satisfied we see that
1
(1 − t)−m − 1 2
F −1(t) −
[− log(1 − t)]n−1(1 − t)k−1 dt = 0.
m
0
Since the sequence {(− log(1 − t))n, n ≥ 1}, is complete in L(0, 1) (cf. [7])
we conclude that F (x) is of the form (2.1).
Theorem 2′.
Under the assumptions of Theorem 2 the distribution
function F (x) is given by (2.1) for k > 2m iff the following relations hold:
1
s
n
EY (s)
n
=
− 1 ,
s = k, k − m,
m
s − m
1
k
n
k
n
E(Y (k)
n
)2 =
− 2
+ 1
m2
k − 2m
k − m
for n = 1, 2, . . .
Putting m = −1 we obtain the characterization results given in [6]. For
n = 1, m = −1 we obtain the result of Too and Lin [8].
Similar considerations lead to the analogous characterizations for the
distribution (2.6). Namely, we have the following results.
Theorem 3. Let {Xn, n ≥ 1} be a sequence of i.i.d. random variables
with a common distribution function F such that E|min(X1, . . . , Xk)|2p < ∞
for a fixed k ≥ 1 and some p > 1. Then F (x) has the form (2.6) iff for
k − 2m > 0, where m is a positive integer ,
2
k
n
E(Y (k)
n
)2 −
EY (k−m)
m
k − m
n
− EY (k−m−1)
n
1
k
n
k
n
+
1 − 2
+
= 0
m2
k − m
k − 2m
for n = 1, 2, . . .
Theorem 3′.
Under the assumptions of Theorem 3, the distribution
function F is given by (2.6) iff for k − 2m > 0,
1
s
n
EY (s)
n
=
− 1 ,
s = k, k − m,
m
s − m
1
k
n
k
n
E(Y (k)
n
)2 =
− 2
+ 1
m2
k − 2m
k − m
for n = 1, 2, . . .
Characterizations of power distributions
471
Letting m = 1 we obtain the following characterization result.
Corollary 2. F (x) = 1 − (1 + x)−1, x > 0, iff
k
n
k
n
k
n
E(Y (k)
n
)2−2
EY (k−1)
+
= 0.
k − 1
n
−EY (k)
n
+1−2 k − 1
k − 2
4. Characterizations by moments of randomly indexed order
statistics. Let Xk:N be the kth smallest order statistics of a random sample
(X1, . . . , XN ) with common distribution function F , where N is a random
variable independent of {Xn, n ≥ 1} with a probability function p(k) =
P [N = k], k = 1, 2, . . . We write Pk = P [N ≥ k].
In this section we give a characterization for the distribution (2.1) in
terms of moments of order statistics with a random index.
Theorem 4. With the above notation, suppose that E(X2
| N ≥ k)
k:N
< ∞ for some k and a given probability function p(·) of N . Then
(4.1)
E(X2k:N | N ≥ k)
2
N
−
E
[k]
X
m
(N − m)
k:N −m N ≥ k
− E(Xk:N | N ≥ k)
[k]
1
N
N
+
E
[k]
N ≥ k − 2E
[k]
N ≥ k + 1 = 0,
m2
(N − 2m)[k]
(N − m)[k]
where N[k] = N (N − 1) . . . (N − k + 1), iff F is given by (2.1).
P r o o f. Let F −1(t) = inf{x : F (x) ≥ t, t ∈ (0, 1)}. We have
1
n!
(4.2)
EXlk:n =
(F −1(t))ltk−1(1 − t)n−k dt,
l ≥ 1.
(k − 1)!(n − k)! 0
Suppose that F is given by (2.1). Since N is independent of {Xn, n ≥ 1},
from (2.3) we have
∞
1
n!
(4.3)
E(Xk:N | N ≥ k) = mPk
(k − 1)!(n − k)!
n=k
1
× [(1 − t)−m − 1]tk−1(1 − t)n−k dt P [N = n]
0
1
N
=
E
[k]
N ≥ k − 1 ,
m
(N − m)[k]
472
Z. Grudzie ´
n and D. Szynal
N
(4.4)
E
[k]
X
(N − m)
k:N −m N ≥ k
[k]
∞
1
(n − m)!
= mPk
(k − 1)!(n − k − m)!
n=k
n(n − 1) . . . (n − k + 1)
× (n − m)(n − m − 1)... (n − m − k + 1)
1
× ((1 − t)−m − 1)tk−1(1 − t)n−m−k dt P [N = n]
0
1
N
N
=
E
[k]
N ≥ k − E
[k]
N ≥ k
m
(N − 2m)[k]
(N − m)[k]
and
(4.5)
E(X2k:N | N ≥ k)
1
N
N
=
E
[k]
N ≥ k − 2E
[k]
N ≥ k + 1 .
m2
(N − 2m)[k]
(N − m)[k]
We see that (4.1) holds true.
Conversely, assume that (4.1) holds. It can be written as
∞
1
(1 − t)−m − 1 2
F −1(t) −
tk−1(1 − t)n−k dt P [N = n] = 0,
m
n=k 0
which implies that F is given by (2.1).
Using (4.3)–(4.5) we have the following characterization conditions in
terms of conditional moments of order statistics.
Theorem 4′.
Under the assumptions of Theorem 4 the distribution
function F is given by (2.1) iff
1
N
E(X
[k]
k:N | N ≥ k) =
E
N ≥ k − 1 ,
m
(N − m)[k]
N
E
[k]
X
(N − m)
k:N −m N ≥ k
[k]
1
N
N
=
E
[k]
N ≥ k − E
[k]
m
(N − 2m)[k]
(N − m)[k]
and
1
N
E(X2
[k]
k:N | N ≥ k) =
E
N ≥ k
m2
(N − 2m)[k]
N
− 2E
[k]
N ≥ k + 1 .
(N − m)[k]
Characterizations of power distributions
473
Corollary 3. Let N be a random variable with probability function
αθn
1
(4.7)
P [N = n] =
,
n = 1, 2, . . . ; α = −
, θ ∈ (0, 1).
n
ln(1 − θ)
Then X has the distribution (2.1) iff
−
1
m θn
EX1:N =
θm 1 − α
− 1 ,
m
n
n=1
−
−
N
1
2m θn
m θn
E
X
θ2m − θm − θ2mα
+ θmα
,
N − m 1:N−m = m
n
n
n=1
n=1
−
−
1
2m θn
m θn
EX21:N =
θ2m − 2θm − αθ2m
+ 2θmα
+ 1 .
m2
n
n
n=1
n=1
Remark. Putting m = −1 we obtain a characterization of the uniform
distribution in terms of X1:N , which after using the equality
N
1
EX21:N − E
X
EX
N + 1 1:N = αEX1 + 1 − θ
1:N
leads to the result of [9], i.e.
1
3
1
1 2
EX21:N − 2 αEX + 1 −
EX
−
− 1 −
ln(1 − θ) .
θ
1:N
= −α 2 θ
θ
5. Characterizations via moments of randomly indexed record
statistics
Theorem 5. Let Y (k) be the kth record value, where N is a positive
N
integer-valued random variable independent of {Xn, n ≥ 1}, and suppose
that E(Y (k))2 < ∞. Then F is given by (2.1) iff
N
2
k
N
(5.1)
E(Y (k))2 −
E
Y (k−m) − EY (k)
N
m
k − m
N
N
1
k
N
k
N
+
1 − 2E
+ E
= 0.
m2
k − m
k − 2m
P r o o f. Suppose that F is given by (2.1). Since
1
kn
E(Y (k)
n
)l =
(F −1(t))l[− log(1 − t)]n−1(1 − t)k−1 dt
(n − 1)! 0
and N and {Xn, n ≥ 1} are independent, it follows that
1
k
N
EY (k) =
E
− 1 ,
N
m
k − m
k
N
1
k
N
k
N
E
Y (k−m) =
E
− E
,
k − m
N
m
k − 2m
k − m
474
Z. Grudzie ´
n and D. Szynal
1
k
N
k
N
E(Y (k))2 =
E
− 2E
+ 1 ,
N
m2
k − 2m
k − m
which establishes (5.1).
Assuming now that (2.1) is satisfied we see that
∞
1
kn
(1 − t)−m − 1 2
F −1(t) −
(n − 1)!
m
n=1
0
× [− log(1 − t)]n−1(1 − t)k−1 dt P [N = n] = 0.
Since the sequence {(− log(1 − t))n, n ≥ 1} is complete in L(0, 1) (cf. [7])
it follows that F (x) has the form (2.1).
Putting m = −1 we have the following characterization.
Corollary 4. F (x) = x, x ∈ (0, 1), iff
k
N
E(Y (k))2 + 2 E
Y (k+1) − EY (k)
N
k + 1
N
N
k
N
k
N
+ E
− 2E
+ 1 = 0
k + 2
k + 1
(cf. [5] with m = 1).
Corollary 5. Let N be a random variable with the probability function
(4.7). Then X has the distribution (2.1) iff
2
k
N
E(Y (k))2 −
E
Y (k−m) − EY (k)
N
m
k − m
N
N
1
k(1 − θ) − m
k(1 − θ) − 2m
+
1 + 2α log
− α log
.
m2
k − m
k − 2m
Remark. F (x) = x, x ∈ (0, 1), iff
k
N
E(Y (k))2 + 2 E
Y (k+1) − EY (k)
N
k + 1
N
N
k(1 − θ) + 1
k(1 − θ) + 2
+ 1 + 2α log
− α log
= 0.
k + 1
k + 2
Acknowledgements. The authors are grateful to the referee for useful
comments which improved the presentation of the paper.
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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 7.5.1999 ;
revised version on 12.7.1999