# Empirical Bayes Inference for the Parameter of Power Distribution Based on Ranked Set Sampling

Authors: Naiyi Li; Yuan Li; Yongming Li; Yang Liu
Journal: Discrete Dynamics in Nature and Society
 (2015)

---

## Abstract

This research is based on ranked set sampling. Through the analysis and proof, the empirical Bayes test rule and asymptotical property for the parameter of power distribution are obtained.


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## Body

## 1. Introduction

Ranked set sampling (RSS) is now regarded as an effective tool in statistical inference and important alternative to simple random sampling. RSS was first applied in agriculture [1]. In recent years, it has been applied more and more in areas such as environment and ecology. It is also a potential for the method to be successfully applied in industrial statistics and sociology, for which readers can refer to the monograph by Chen et al. [2].Empirical Bayes (EB) approach was originally proposed by Robbins [3, 4], soon after it had been studied in the literature [5–9]. Up to now, EB methods are commonly based on simple random sampling (SRS). As a statistical procedure based on the RSS would perform better than its counterpart based on simple random sampling (SRS), a natural idea is to develop EB methods based on RSS. We will construct empirical Bayes test rule for the parameter of power distribution based on RSS.LetX have a conditional density function for given θ:(1)
f
x
∣
θ
=
θ
x
-
1
exp
⁡
θ
-
1
ln
⁡
x
,where θ is an unknown parameter, Ω
=
{
x
∣
x
>
1
} is sample space, and Θ
=
{
θ
∣
θ
>
0
} is parameter space.In the paper, the main motivation of studying the density of (1) is as follows: (a) it is widely used in the fields of reliability and economy and so on; (b) it is a common distribution in reliability distribution; (c) on the basis of studying it, we intend to further consider empirical Bayes inference for some other distributions based on ranked set sampling. In the paper, we study the following test problem: (2)
H
0
:
θ
≤
θ
0
⟺
H
1
:
θ
>
θ
0
,where θ
0 is a given positive constant.To construct test function, we take loss function:(3)
L
0
θ
,
d
0
=
0
,
θ
≤
θ
0
a
1
-
θ
0
θ
,
θ
>
θ
0
;
L
1
θ
,
d
1
=
a
θ
0
θ
-
1
,
θ
≤
θ
0
0
,
θ
>
θ
0
,where a
>
0, d
=
{
d
0
,
d
1
} is action space, respectively, and d
0 and d
1 imply acceptance and rejection of H
0.Suppose that the prior distributionG
(
θ
) of parameter θ is unknown, we can get randomized decision function:(4)
δ
x
=
P
accept
H
0
∣
X
=
x
.Then, the risk function ofδ
(
x
) is given by(5)
R
δ
x
,
G
θ
=
∫
Θ
∫
Ω
L
0
θ
,
d
0
f
x
∣
θ
δ
x
+
L
1
θ
,
d
1
f
x
∣
θ
1
-
δ
x
d
x
d
G
θ
=
a
∫
Ω
β
x
δ
x
d
x
+
C
G
,where(6)
C
G
=
∫
Θ
L
1
θ
,
d
1
d
G
θ
,
β
x
=
∫
Θ
1
-
θ
0
θ
f
x
∣
θ
d
G
θ
.The marginal density function ofX is shown by (7)
f
G
x
=
∫
Θ
f
x
∣
θ
d
G
θ
=
∫
Θ
θ
x
-
1
exp
⁡
θ
-
1
ln
⁡
x
d
G
θ
.Applying (6), we get (8)
β
x
=
∫
Θ
1
-
θ
0
θ
f
x
∣
θ
d
G
θ
=
∫
Θ
1
-
θ
0
+
θ
0
-
θ
0
θ
f
x
∣
θ
d
G
θ
=
∫
Θ
1
-
θ
0
f
x
∣
θ
d
G
θ
+
∫
Θ
θ
0
-
θ
0
θ
f
x
∣
θ
d
G
θ
=
1
-
θ
0
f
G
x
-
θ
0
x
·
∫
Θ
-
1
+
1
θ
1
x
f
x
∣
θ
d
G
θ
=
1
-
θ
0
f
G
x
-
θ
0
x
·
∫
Θ
-
1
x
f
x
∣
θ
d
G
θ
-
θ
0
x
·
∫
Θ
1
θ
x
f
x
∣
θ
d
G
θ
.Sincef
G
(
x
)
=
∫
Θ
‍
(
θ
x
)
-
1
exp
⁡
{
θ
-
1
ln
⁡
x
}
d
G
(
θ
), we get (9)
f
G
1
x
=
∫
Θ
-
1
x
f
x
∣
θ
d
G
θ
+
1
θ
x
∫
Θ
f
x
∣
θ
d
G
θ
.Thus, we have(10)
β
x
=
1
-
θ
0
f
G
x
-
θ
0
x
·
f
G
1
x
=
u
x
f
G
x
+
v
x
f
G
1
x
,where f
G
(
1
)
(
x
) is derivative of f
G
(
x
), u
(
x
)
=
1
-
θ
0, and v
(
x
)
=
-
θ
0
x.By (5), Bayes test function is obtained as follows: (11)
δ
G
x
=
1
,
if
β
x
≤
0
,
0
,
if
β
x
>
0
.Hence, we can get minimum Bayes risk:(12)
R
G
=
inf
δ
⁡
R
δ
,
G
=
R
δ
G
,
G
=
a
∫
Ω
β
x
δ
G
x
d
x
+
C
G
.If the prior distribution ofG
(
θ
) is known and δ
(
x
)
=
δ
G
(
x
), R
(
G
) is achieved. While G
(
θ
) is unknown, we cannot make use of δ
G
(
x
) and need to introduce EB method.


## 2. Construction of EB Test Based on Ranked Set Sampling

A balanced RSS procedure can be described by Chen et al. [2]. Under the following conditions, we need to construct EB test function. Let X
(
1
)
1
,
X
(
1
)
2
,
…
,
X
(
1
)
m
,
X
(
2
)
1
,
X
(
2
)
2
,
…
,
X
(
2
)
m
,
…
,
X
(
k
)
1
,
X
(
k
)
2
,
…
,
X
(
k
)
m be a balanced ranked set sample from population which has the common marginal density function f
G
(
x
). We assume perfect ranking. Denote by X
(
1
)
1
,
X
(
1
)
2
,
…
,
X
(
1
)
m
,
X
(
2
)
1
,
X
(
2
)
2
,
…
,
X
(
2
)
m
,
…
,
X
(
k
)
1
,
X
(
k
)
2
,
…
,
X
(
k
)
m the historical samples, and X is present sample. Assume f
G
(
x
)
∈
C
s
,
α, x
∈
R
1, where C
s
,
α
=
g
(
x
)
∣
g
x
is a probability density function and has continuous
s
th order derivative
g
(
s
)
(
x
) with  |
g
s
(
x
)
|
≤
α
,
s
≥
3
,
α
>
0. First construct estimator of β
(
x
).LetK
r
(
x
) be a Borel measurable bounded function vanishing off (
0,1
) such that(13)
C
1
:
1
t
!
∫
0
1
y
t
K
r
y
d
y
=
-
1
t
,
when
t
=
r
,
0
,
when
t
≠
r
,
t
=
0,1
,
2
,
…
,
s
-
1
.Kernel estimator off
G
(
x
) is defined by (14)
f
n
r
x
=
1
m
k
h
n
1
+
r
∑
i
=
1
k
∑
j
=
1
m
K
r
x
-
X
i
j
h
n
,where h
n is a positive and smoothing bandwidth and lim
n
→
∞
⁡
h
n
=
0.Denote byf
n
(
0
)
(
x
)
=
f
n
(
x
), f
n
(
r
)
(
x
) the rth order derivative of f
n
(
x
), for r
=
0,1.The estimator ofβ
(
x
) is obtained by (15)
β
n
x
=
u
x
f
n
x
+
v
x
f
n
1
x
.Hence, EB test function is defined by(16)
δ
n
x
=
1
,
β
n
x
≤
0
,
0
,
β
n
x
>
0
.LetE stand for mathematical expectation with respect to the joint distribution of X
(
1
)
1
,
X
(
1
)
2
,
…
,
X
(
1
)
m
,
X
(
2
)
1
,
X
(
2
)
2
,
…
,
X
(
2
)
m
,
…
,
X
(
k
)
1
,
X
(
k
)
2
,
…
,
X
(
k
)
m. Hence, we get the overall Bayes risk of δ
n
(
x
):(17)
R
δ
n
x
,
G
=
a
∫
Ω
β
x
E
δ
n
x
d
x
+
C
G
.Iflim
n
→
∞
⁡
R
(
δ
n
,
G
)
=
R
(
δ
G
,
G
), δ
n
x is called asymptotical optimality of EB test function. If R
(
δ
n
,
G
)
-
R
(
δ
G
,
G
)
=
O
(
n
-
q
), where q
>
0, O
(
n
-
q
) is asymptotically optimal convergence rate of EB test function δ
n
x. Before proving the theorems, we need the following lemmas.Letc, c
1, c
2, c
3 be different constants in different cases even in the same expression.Lemma 1.
LetX
(
1
)
1
,
X
(
1
)
2
,
…
,
X
(
1
)
m
,
X
(
2
)
1
,
X
(
2
)
2
,
…
,
X
(
2
)
m
,
…
,
X
(
k
)
1
,
X
(
k
)
2
,
…
,
X
(
k
)
m be balanced ranked set samples. Suppose that (C1) holds, ∀
x
∈
Ω.(I)
Whenf
G
(
r
)
(
x
) is continuous function, lim
n
→
∞
⁡
h
n
=
0, and lim
n
→
∞
⁡
n
h
n
2
r
+
1
=
∞, one has (18)
lim
n
→
∞
⁡
E
f
n
r
x
-
f
G
r
x
2
=
0
.
(II)
Whenf
G
(
x
)
∈
C
s
,
a, putting h
n
=
n
-
1
/
(
2
+
s
), for 0
<
λ
≤
1, one has (19)
E
f
n
r
x
-
f
G
r
x
2
λ
≤
c
·
n
-
λ
s
-
2
r
+
1
/
2
+
s
.Proof.
Consider the following.
Proof of (I). Using C
r inequation, we obtain (20)
E
f
n
r
x
-
f
G
r
x
2
≤
2
E
f
n
r
x
-
f
G
r
x
2
+
2
Var
⁡
f
n
r
x
≔
2
A
1
2
+
A
2
,where(21)
E
f
n
r
x
=
∑
i
=
1
k
∑
j
=
1
m
n
-
1
h
n
-
r
+
1
E
K
r
x
-
X
i
j
h
n
=
h
n
-
1
+
r
E
K
r
x
-
X
h
n
=
h
n
-
1
+
r
∫
0
∞
K
r
x
-
y
h
n
f
G
y
d
y
=
h
n
-
r
∫
0
1
K
r
u
f
G
x
-
h
n
u
d
u
.
Using Taylor expansion, we get(22)
f
G
x
-
h
n
u
-
f
G
u
=
f
G
′
x
1
!
-
h
n
u
+
f
G
′
′
x
2
!
-
h
n
u
2
+
⋯
+
f
G
s
x
-
ξ
h
n
u
s
!
-
h
n
u
s
.
Becausef
G
(
r
)
(
x
) is continuous in x and condition (C1), it follows that(23)
0
≤
lim
n
→
∞
⁡
E
f
n
r
x
-
f
G
r
x
=
lim
n
→
∞
⁡
1
h
n
r
∫
0
1
K
r
u
f
G
x
-
h
n
u
d
u
-
f
G
r
x
≤
1
r
!
∫
0
1
u
r
K
r
u
·
lim
n
→
∞
⁡
f
G
r
x
-
ξ
h
n
u
-
f
G
r
x
d
u
=
0
;further, we have (24)
lim
n
→
∞
⁡
A
1
2
=
lim
n
→
∞
⁡
E
f
n
r
x
-
f
G
r
x
2
=
0
.
It is easy to see that(25)
A
2
=
2
Var
⁡
f
n
r
x
=
2
∑
i
=
1
k
∑
j
=
1
m
n
-
2
h
n
-
2
r
+
1
Var
⁡
K
r
x
-
X
i
j
h
n
≤
2
n
-
2
h
n
-
2
r
+
1
∑
i
=
1
k
∑
j
=
1
m
E
K
r
x
-
X
i
j
h
n
2
≤
c
·
n
h
n
2
r
+
1
-
1
.
Whenh
n
→
0 and n
h
n
2
r
+
1
→
∞, we get (26)
lim
n
→
∞
⁡
A
2
=
lim
n
→
∞
⁡
Var
⁡
f
G
r
x
=
0
.
Substituting (24) and (26) into (20), proof of (I) is finished.
Proof of (II). Similarly to (20), we can show that(27)
E
f
n
r
x
-
f
G
r
x
2
λ
≤
2
E
f
n
r
x
-
f
G
r
x
2
λ
+
2
Var
⁡
f
n
r
x
λ
≔
2
B
1
2
λ
+
B
2
λ
.
By Taylor expansion, we obtain(28)
f
G
x
-
h
n
v
=
f
G
x
+
f
G
′
x
1
!
-
h
n
v
+
f
G
′
′
x
2
!
-
h
n
v
2
+
⋯
+
f
G
s
x
-
ξ
h
n
v
s
!
-
h
n
v
s
,where 0
<
ξ
<
1, due to condition (C1) and f
G
(
x
)
∈
C
s
,
α, we have (29)
E
f
n
r
x
-
f
G
r
x
≤
∫
0
1
K
r
v
h
n
s
-
r
v
s
f
G
s
x
-
ξ
h
n
v
s
!
d
v
≤
c
·
h
n
s
-
r
.
Therefore, takingh
n
=
n
-
1
/
(
2
+
s
), we get(30)
B
1
2
λ
=
E
f
n
r
x
-
f
G
r
x
2
λ
≤
c
·
n
-
2
λ
s
-
r
/
s
+
2
.
By (26), choosing h
n
=
n
-
1
/
(
2
+
s
), we can get (31)
B
2
λ
≤
c
1
n
h
n
2
r
+
1
-
1
λ
≤
c
·
n
-
λ
s
-
2
r
+
1
/
2
+
s
.
Substituting (30) and (31) into (27), proof of (II) is finished.Lemma 2 (see [8]).
R
(
δ
G
,
G
) and R
(
δ
n
,
G
) are defined by (12) and (17); then (32)
0
≤
R
δ
n
,
G
-
R
δ
G
,
G
≤
a
∫
Ω
β
x
P
β
n
x
-
β
x
≥
β
x
d
x
.


## 3. Asymptotic Optimality and Convergence Rates Based on Ranked Set Sampling

Theorem 3.
Assume (C1) and the following regularity conditions hold:(i)
lim
n
→
∞
⁡
n
h
n
3
=
∞;
(ii)
∫
Θ
‍
θ
-
2
d
G
(
θ
)
<
+
∞;
(iii)
f
G
(
1
)
(
x
) is continuous function, and one has (33)
lim
n
→
∞
⁡
R
δ
n
,
G
=
R
δ
G
,
G
.Proof of Theorem3.
Applying Lemma2, we can get (34)
0
≤
R
δ
n
,
G
-
R
δ
G
,
G
≤
a
∫
Ω
β
x
p
β
n
x
-
β
x
≥
β
x
d
x
.
PutM
n
(
x
)
=
|
β
(
x
)
|
p
(
|
β
n
(
x
)
-
β
(
x
)
|
≥
|
β
(
x
)
|
).
It is easy to see thatM
n
(
x
)
≤
|
β
(
x
)
|.
Again using (10) and Fubini theorem, we obtain (35)
∫
Ω
β
x
d
x
=
∫
Ω
∫
Θ
1
-
θ
0
θ
2
f
x
∣
θ
d
G
θ
d
x
=
1
+
θ
0
2
∫
Ω
∫
Θ
θ
-
2
f
x
∣
θ
d
G
θ
d
x
=
1
+
θ
0
2
∫
Θ
θ
-
2
d
G
θ
<
+
∞
.
By domain convergence theorem, then(36)
0
≤
lim
n
→
∞
⁡
R
δ
n
,
G
-
R
δ
G
,
G
≤
∫
Ω
lim
n
→
∞
⁡
M
n
x
d
x
.
If Theorem3 holds, we only need to prove lim
n
→
∞
⁡
M
n
(
x
)
=
0
a
.
s
.
x.
Applying Markov’s and Jensen’s inequations, we have(37)
M
n
x
≤
E
β
n
x
-
β
x
≤
u
x
E
f
n
x
-
f
G
x
+
v
x
E
f
n
1
x
-
f
G
1
x
≤
u
x
E
f
n
x
-
f
G
x
2
1
/
2
+
v
x
E
f
n
1
x
-
f
G
1
x
2
1
/
2
.
Again using Lemma1(I), for fixed x
∈
Ω and r
=
0,1, we get(38)
0
≤
lim
n
→
∞
⁡
M
n
x
≤
u
x
lim
n
→
∞
⁡
E
f
n
x
-
f
G
x
2
1
/
2
+
v
x
lim
n
→
∞
⁡
E
f
n
1
x
-
f
G
1
x
2
1
/
2
=
0
.
Substituting (38) into (36), proof of Theorem 3 is finished.Theorem 4.
Assume (C1) and the following regularity conditions hold.
(C2): consider∫
Ω
‍
x
-
m
λ
|
β
(
x
)
|
1
-
λ
d
x
<
+
∞, where 0
<
λ
≤
1, m
=
0,1. When h
n
=
n
-
1
/
(
2
+
2
s
), where s
≥
2, one can obtain(39)
R
δ
n
,
G
-
R
δ
G
,
G
=
O
n
-
λ
s
-
1
/
2
s
+
1
,
w
h
e
r
e
s
≥
2
.Proof of Theorem4.
Applying Lemma2 and Markov’s inequations, we have(40)
0
≤
R
δ
n
,
G
-
R
δ
G
,
G
≤
∫
Ω
β
x
1
-
λ
E
β
n
x
-
β
x
λ
d
x
≤
c
1
∫
Ω
β
x
1
-
λ
u
x
E
f
n
x
-
f
G
x
d
x
+
c
2
∫
Ω
β
x
1
-
λ
v
x
E
f
n
1
x
-
f
G
1
x
d
x
=
D
n
+
E
n
.
By Lemma1(II) and condition (C2), we get (41)
D
n
≤
c
1
n
-
λ
s
/
2
s
+
2
∫
Ω
β
x
1
-
λ
u
x
λ
d
x
≤
c
3
n
-
λ
s
/
2
s
+
2
,
E
n
≤
c
2
n
-
λ
s
-
1
/
2
s
+
2
∫
Ω
β
x
1
-
λ
v
x
λ
x
d
x
≤
c
4
n
-
λ
s
-
1
/
2
s
+
2
.
Substituting (41) into (40), we can obtain R
(
δ
n
,
G
)
-
R
(
δ
G
,
G
)
=
O
(
n
-
λ
(
s
-
1
)
/
2
(
s
+
1
)
). Proof of Theorem 4 is finished.Remark 5.
Whenλ
→
1 and s
→
∞, O
(
n
-
λ
(
s
-
1
)
/
2
(
s
+
1
)
) nears O
(
n
-
1
/
2
).


## 4. Conclusions

In this paper, we propose the empirical Bayes test rule for the parameter of power distribution. Based on ranked set sampling, the asymptotical optimality and convergence rates of EB test function are obtained. Another extension of this work would consider empirical Bayes inference for some other distributions based on ranked set sampling.



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*Source: 760768-2015-05-28.xml*