A covariance matrix test for high-dimensional data

Abstract

For the multivariate normally distributed data with the dimension larger than or equal to the number of observations,
or the sample size, called high-dimensional normal data, we proposed a test for testing the null hypothesis that the covariance
matrix of a normal population is proportional to a given matrix on some conditions when the dimension goes to infinity.
We showed that this test statistic is consistent. The asymptotic null and non-null distribution of the test statistic is also
given. The performance of the proposed test is evaluated via simulation study and its application.

Full text

Original Article
A covariance matrix test for high-dimensional data
Saowapha Chaipitak and Samruam Chongcharoen*
School of Applied Statistics, National Institute of Development Administration (NIDA),
Bang Kapi, Bangkok, 10240 Thailand.
Received: 25 October 2015; Accepted: 25 January 2016
Abstract
For the multivariate normally distributed data with the dimension larger than or equal to the number of observations,
or the sample size, called high-dimensional normal data, we proposed a test for testing the null hypothesis that the covariance
matrix of a normal population is proportional to a given matrix on some conditions when the dimension goes to infinity.
We showed that this test statistic is consistent. The asymptotic null and non-null distribution of the test statistic is also
given. The performance of the proposed test is evaluated via simulation study and its application.
Keywords: asymptotic distribution, high-dimensional data, null distribution, non-null distribution, multivariate normal,
hypothesis testing
Songklanakarin J. Sci. Technol.
38 (5), 521-535, Sep. - Oct. 2016
1. Introduction
Let 1
,...,
N
X X
be a set of independent observations
from a multivariate normal distribution
( , )
p
N


where both
the mean vector

and covariance matrix ,  is a positive
definite matrix, are unknown. In this paper, we are interested
in the problem of testing the hypothesis that the covariance
matrix of a normal population is proportional to a given
matrix, that is,
0 0
:
H t
  
against
1 0
:
H t
  
where
both
0
0 ,
t
   
are known. The likelihood ratio test
(LRT),which is based on the sample covariance matrix, is the
traditional technique to handle this hypothesis and requires
n p

. But many applications in modern science and eco-
nomics, e.g. the analysis of DNA microarrays, the dimension
is usually in thousands of gene expressions whereas the
sample size is small, which makes
n p

, called high-dimen-
sional data. For such data, the LRT is not applicable because
the sample covariance matrix, S, is singular when
n p

(see,
for examples, Muirhead, 1982, Sections 8.3 and 8.4; Anderson,
1984, Sections 10.7 and 10.8).
Recently, several authors have proposed methods for
testing the related problems. Some of them are: John (1971);
Nagao (1973); Ledoit and Wolf (2002); and Srivastava (2005),
and Fisher et al. (2010). Those are given as follows.
John (1971) proposed a test statistic for testing that
the covariance matrix of a normal population is proportional
to an identity matrix, that is, 0:
H tI

 
,
0
t
  
a known
value which is the locally most powerful invariant test as
2
1
(1 / ) ( )
S
U tr I
p p tr S
 
 
 
 
 
 
 
 
and Nagao (1973) proposed a test statistic for testing
0
:
H

I
 
as
 
2
1
V tr S I
p
 
 
 
Both U and V test statistics are consistent and have been
studied under assuming that n goes to infinity while p
remains fixed. So, Ledoit and Wolf (2002) demonstrated that
the test statistic for testing
0
H

based on U statistic is still
consistent if n goes to infinity with p that is as ( , )n p
 
and
/ ,
p n c

(0, )
c
 
. The null hypothesis
0
H

is rejected
if
* Corresponding author.
Email address: samruam@as.nida.ac.th
http: //www.sjst.ps u.ac. th

S. Chaipitak & S. Chongcharoen / Songklanakarin J. Sci. Technol. 38 (5), 521-535, 2016
522
2
j
npU
U(1.1)
exceeds the appropriate quantile from the
2


distribution
with
( 1) / 2 1
p p
 
degree of freedom. For testing
0
:
H

I
 
if goes to infinity with p as n < p, Ledoit and Wolf
(2002) showed that the statistic V is not consistent against
every alternative and its n

limiting distribution differs from
its
( , )
n p

limiting distribution under the null hypothesis.
Then they modified the statistic V as
 
2
1 1
2
.
p p
W tr S I trS
p n p n
   
 
   
   
They have shown that the statistic W is consisten t as
( , )n p
 
, including the case n < p. The test statistic based
on W rejects the null hypothesis
0
H

if
/ 2
npW exceeds the
appropriate quantile from the 2


distribution with
( 1) / 2
p p

degrees of freedom. Srivastava (2005) proposed
a test statistic when ( , )n p
 
,
( )
n O p


0 1,

 
to
reject the null hypothesis
2
0:
H I

  , 2
0


unknown,
with a test statistic which did not relate with unknown
2
0


.
Then we applied his test statistic for testing
0
H

and reject
0
H

if
1
2
2 1
ˆ ˆ
( / 1)
2
S
n
T h h
 
(1.2)
exceeds the appropriate quantile of the standard normal dis-
tribution, where 1
ˆ
(1 / ) ( )
h p tr S
 and
2
2
ˆ
( 1)( 2)
n
h
n n

 
2 2
1 1
( )
trS trS
p n

 
 
 
. The statistics
1
ˆ
h
and
2
ˆ
h
are
( , )
n p

consistent estimators of
(1 / )
p tr

and
2
(1/ )
p tr

respec-
tively. Also he proposed a test to reject the null hypothesis
0
H

if
2 1 2
ˆ ˆ
( 2 1)
2
S
n
T h h
  
(1.3)
exceeds the appropriate quantile of the standard normal distri-
bution. Motivated by the result in Srivastava (2005), which
requires, /
p n c

,
(0, )
c
 
Fisher et al. (2010)
proposed the test for testing
0
H

based on unbiased and
consistent estimators of the second and fourth arithmetic
means of the sample eigenvalues. With the constants:
2
4 2 3 6 2(5 6)
*
, , ,
2 2
( 2) ( 2)
n n n
b c d
nn n n n n n
  
    
   
5 6
2 2
( 2)
n
e
n n n

 
 
and
5 2
( 2)
( 1)( 2)( 4)( 6)( 1)( 2)( 3)
n n n
n n n n n n n

 

      
,
they proposed the test statistic to reject the null hypothesis
0
H

if
2
4 2
2
ˆ ˆ
( / 1)
8(8 12 )
F
n h h
T
c c


 
exceeds the appropriate quantile of the standard normal distri-
bution, where
4 3 * 2 2 2 2 4
4
ˆ
( ) ( ) ( )
h trS btrS trS c trS dtrS trS e trS
p

    
 
 
 
is
( , )
n p

consistent estimator of
4
(1/ )
p tr

.
The remainder of this paper is organized as follows.
Section 2 provides the proposed test statistic and its asymp-
totic distribution under both the null and alternative hypo-
theses as (n, p) go to infinity even if n < p. Section 3 shows
the performance of the proposed test statistic through
simulation technique. Section 4 applies the test statistic to
real data. Section 5 contains the conclusions. The theoretical
derivations are given in the Appendix.
2. Description of the Proposed Test
Suppose


1 1
,..., ~ ,
n p
X X N



and we are
interested in testing that the covariance matrix of a normal
population is proportional to a given matrix, that is,
0
:
H
0
t
  
against
1 0
:
H t
  
where 0t
  
is known
value and
0

is a given known positive definite matrix. Wee
proposed the test statistic by considering a measure of a
distance between the two matrices
1 1 2 1 2
0 0 0
2
1 1 2
( ( ) ( ) ( )
t
tr tI tr tr t
p p p

  
          
(2.1)
where tr denotes the trace of matrix and if and
0


only if
the null hypothesis holds. Thus, we may consider testing
0
: 0
H


against 1
: 0
H


.
We shall make the following assumptions:
(A) 0 0
lim , (0, ), 1,..., 8
i i i
p
a a a i
 
   
(B)
( , )
lim / , (0, )
n p
p n c c
 
  
where 1
0
1
(1 / ) ( ) (1/ ) ( / )
p
i i
i j j
j
a p tr p d



    . The
j

’s are the eigenvalues of the covariance matrix  and dj’ss
are the eigenvalues of a known positive definite matrix 0.
We need estimators of a1 and a2 to be consistent estimators

523
S. Chaipitak & S. Chongcharoen / Songklanakarin J. Sci. Technol. 38 (5), 521-535, 2016
for large p and n even if n < p. The following theorem
provides these consistent estimators.
Theorem 2.1 The unbiased and consistent estimators of a1 =
1
1 0
(1 / ) ( )
a p tr

  
and
1 2
2 0
(1 / ) ( )
a p tr 
  
are respectively
given by
1
1 0
ˆ
(1 / ) ( )
a p tr S

  (2.2)
and
2
1 2 1 2
2 0 0
1 1
ˆ
( ) ( ( ))
( 1)( 2)
n
a tr S tr S
n n p n
 
   
 
 
 
 
(2.3)
Thus we use estimators in Theorem 2.1 to define the unbiased
and consistent estimator of

in (2.1) as
2
2 1
ˆ ˆ ˆ
2
a ta t

  
(2.4)
The following theorem gives the asymptotic distribution of
the estimators
1
ˆ
a
and
2
ˆ
a
in (2.4).
Theorem 2.2 Under the assumption (A), and (B), as
( , )n p
 
2
2
2 3
1 1
2
2 2 3 4
2 / 4 /
ˆ
,
ˆ4 / 4(2 ) /
a np a np
a a
DN
a a
a np a ca np

 
 
   
 
 

   
   
 
 
   
 
 
where D
x y

denotes x converges in distribution to y.
The following theorem and corollary provide the
asymptotic distribution of
ˆ

under the alternative and null
hypothesis by applying the delta method of a function of
two random variables.
Theorem 2.3 Under the assumption (A), and (B), as
( , )n p
 


2
ˆ0,
DN
  
 (2.5)
with
2 2
2 3 4 2
4
(2 4 2 )
ta ta a ca
np

    .
Corollary 2.1 Under the null hypothesis
0 0
:
H t
  
then
0


and under the assumption (A), and (B), as ( , )n p
 
2
4
ˆˆ
(0,1)
2
2
np nD
T N
t
ct

   (2.6)
Remark If t = 1 and 0
I
 
where I is identity matrix, then
the proposed statistic T is the test statistic
2
S
T
in (1.3) given
by Srivastava (2005).
3. Simulation Study
For studying the performance of the proposed test
statistic T, we compute the attained significance level (ASL)
of the proposed test by simulation technique. Based on 10,000
replications of the data set simulated under the null hypo-
thesis
0 0
:
H t
  
, test statistic T is computed and then we
obtain the attained significance level (ASL) of the test by
recording the proportion rejection of test statistic for the
null hypothesis with the nominal significance level at 0.05.
We simulate the ASL for different four null hypotheses as
1) 1
0 0 01
:
H t C
    wh e r e 01 ,
( )
i j p p
C c 
 
( )
i j p p
c
 
is a Toeplitz matrix with elements 0 1
1, c c
 
1
0.5
c
 
and the rest elements are equal to zero
2) 2
0 0 02
: 0.5 0.51 1
p p p
H t C I

      , where
p
I
denotes the
p p

identity matrix, and
1
p
denotes the
1
p

vector having each element equal to 1
3) 3
0 0 03
:
H t C
    where 03 ,
( )
i j p p
C c 
 
,
( )
j i p p
c

with ,
( 1) ( / 2 ) 1,..,
i j
i j
c i j i j p

    
and ,
1.0 1,...,
i i
c i p
  
4) 4
0 0 04
:
H t C
    where 04 ,
( )
i j p p
C c 
 
,
( )
j i p p
c

with
1 / 5
,
| |
( 1) 0.9 , 1, ..,
i j i j
i j
c i j p
 
   
For each null hypothesis, we simulate the empirical
power of the proposed test T under the alternative hypothesis
for each of four null hypotheses as
1) 1
0 01
:
H C
  against 1
1 1
:
H C
 
where
1
C

,
( ) ( )
i j p p i j p p
c c
  
 is a Toeplitz matrix with elements
0 1 1
1, 0.49
c c c

    and the rest elements are equal to
zero
2) 2
0 02
:
H C
  against 2
1 2
: 0.9
p
H C I
   
(0.1)1 1
p p

3) 3
0 03
:
H C
  against 3
1 3
:
H C
 
where 3
C

, ,
( ) ( )
i j p p j i p p
c c
 
 with ,
( 1) ( / 4 )
i j
i j
c i j i

  
1,..,
j p
 
and ,
1.0 1,...,
i i
c i p
  
4) 4
0 04
:
H C
  against 4
1 4
:
H C
 
where 4
C

, ,
( ) ( )
i j p p j i p p
c c
 
 with
2 / 5
,
| |
( 1) 0.9
i j i j
i j
c 
 
, 1,..,
i j p
 
3.1 Simulation results
The ASL is provided in Table 1 corresponding to the
null hypotheses. As expected, the ASL of the test statistic T

S. Chaipitak & S. Chongcharoen / Songklanakarin J. Sci. Technol. 38 (5), 521-535, 2016
524
is reasonably close to the nominal significant level 0.05 and
gets better when p and n get large. We found that four sets
of the ASL are almost the same that means the consistency
of our test statistic is not affected by varying the null co-
variance matrices.
The empirical powers are shown in Table 2. It shows
that four sets of the empirical power of test statistic T rapidly
converge to one and stay high as n and p get large for n < p.
We also compute the ASL in a special case of the null
covariance matrix with setting t = 2 and 0
I
 
, that is, the
test with the null hypothesis as 0
: 2
H I

  (spherecity).
We compare the performance of the proposed test statistic
T with the test statistics defined in Ledoit and Wolf (2002),
denoted Uj in (1.1) and Srivastava (2005), denoted TS1 in
(1.2). We compare them under the alternative hypothesis
1
: 2
H D

  where 1
( ,..., ); (0,1),
p i
D diag d d d Unif

1, 2, ...,
i p

. The ASL and the empirical powers are provided
in Table 3. Table 3 reports that the ASL of the proposed test
statistic T is similar to those provided in Table 1 and closed
to those from the test statistic TS1 and Uj. But the test statistic
T gives the best performance for all of the setting (n,p) and
has substantially higher powers than those of Uj and TS1 for
almost every n and p considered. These results suggest that
the proposed test may more appropriate to use than Uj test
and TS1 test, especially when is small.
4. A Real Example
In this section, the microarray dataset is collected from
Notterman et al. (2001) is available at http://genomics-
pubs.p rincet on .ed u/oncol og y/Data/ Car cin omaNo rmal
datasetCancerResearch.xls (last accessed: 9 October 2011).
There are 18 colon adenocarcinomas and their paired normal
colon tissues and they are obtained on oligonucleotide
arrays. The expression levels of 6500 human genes are
measured on each. For simplicity, we will restrict attention to
18 colon adenocarcinomas with only first 256 measurements
each. We examine whether the covariance matrix is the
sphericity. The data gives the observed test statistic values
as
T = 8 .500,
284.567
j
U and 1
270.582
S
T with p-
value
0
p value
 
each, thus the hypothesis of being sphericity is
rejected at any reasonable significance level.
5. Conclusions
For testing the covariance matrix in high-dimensional
data, our test statistic is proposed under normality assump-
tion. The test statistic is approximated by normal distribution.
Numerical simulations indicate that our test statistic T in
(2.6) constructed from the consistent estimators with accu-
rately control size of test and their powers get better when
(n,p) get large for n < p. Moreover, the test statistic gives
higher power than, for testing being sphericity of the co-
variance matrix, those of the tests in Ledoit and Wolf (2002)
and Srivastava (2005).
Acknowledgements
The authors would like to express their gratitude to
the Commission on Higher Education (CHE) of Thailand for
their financial support.
Table 1. The ASL of test statistic T under four null hypotheses at Nominal Significance
Level
0.05


.
The ASL of T
p
1
n N
 
1:
0 01
H C
  2:
0 02
H C
  3:
0 03
H C
  4:
0 04
H C
 
10 9 0.059 0.058 0.059 0.059
40 9 0.055 0.055 0.055 0.055
39 0.056 0.056 0.056 0.057
80 9 0.057 0.056 0.057 0.057
39 0.052 0.052 0.052 0.052
79 0.053 0.052 0.052 0.052
160 9 0.053 0.053 0.054 0.054
39 0.056 0.055 0.055 0.056
79 0.056 0.056 0.056 0.055
159 0.053 0.053 0.053 0.053
320 9 0.052 0.052 0.052 0.052
39 0.052 0.051 0.052 0.052
79 0.051 0.051 0.050 0.050
159 0.051 0.050 0.051 0.051
319 0.053 0.051 0.053 0.053
as

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S. Chaipitak & S. Chongcharoen / Songklanakarin J. Sci. Technol. 38 (5), 521-535, 2016
Table 2. The empirical power of T under four alternative hypotheses.
The empirical power of T
p
1
n N
 
1
1 1
H C
  
2:
1 2
H C
  3:
1 3
H C
  4:
1 4
H C
 
10 9 0.480 0.560 0.174 0.159
40 9 0.996 0.617 0.265 0.286
39 1.000 1.000 0.772 0.877
80 9 1.000 0.624 0.300 0.330
39 1.000 1.000 0.837 0.939
79 1.000 1.000 0.998 1.000
160 9 1.000 0.625 0.319 0.346
39 1.000 1.000 0.866 0.966
79 1.000 1.000 0.999 1.000
159 1.000 1.000 1.000 1.000
320 9 1.000 0.629 0.342 0.361
39 1.000 1.000 0.891 0.977
79 1.000 1.000 1.000 1.000
159 1.000 1.000 1.000 1.000
319 1.000 1.000 1.000 1.000
Table 3. The ASL (under
: 2
0
H I
  ) and the empirical power (under
: 2
1
H D
  )
of
,
T U
j
and
1
T
S
at Nominal Significance Level
0.05


.
ASL Empirical Power
p
1
n N
 
T
U
j
1
T
S
T
U
j
1
T
S
10 9 0.059 0.049 0.048 1.000 0.412 0.405
40 9 0.055 0.054 0.051 1.000 0.368 0.360
39 0.056 0.056 0.053 1.000 0.999 0.999
80 9 0.057 0.057 0.053 1.000 0.356 0.348
39 0.052 0.052 0.050 1.000 0.999 0.999
79 0.052 0.051 0.050 1.000 1.000 1.000
160 9 0.053 0.056 0.054 1.000 0.354 0.346
39 0.055 0.056 0.055 1.000 0.999 0.999
79 0.055 0.057 0.055 1.000 1.000 1.000
159 0.053 0.052 0.052 1.000 1.000 1.000
320 9 0.052 0.055 0.052 1.000 0.352 0.343
39 0.052 0.054 0.052 1.000 0.999 0.999
79 0.050 0.050 0.050 1.000 1.000 1.000
159 0.051 0.050 0.050 1.000 1.000 1.000
319 0.053 0.053 0.053 1.000 1.000 1.000
References
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John, S. 1971. Some optimal multivariate tests. Biometrika.
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Appendix
Before proving Theorem 2.1, we need the following information and lemma:
For positive symmetric definite matrix  and by spectral decomposition, we have

  
where 1 2
( , ..., )
p
diag
  
 
with
i

being the ith eigenvalue of and  is an orthogonal matrix with each column as normalized corresponding
eigenvectors 1 2
, ,...,
p
γ γ γ
. Similarly, we also can write
0

as 0
RDR

  where 1 2
( , ..., )
p
D diag d d d
 with di being the
ith eigenvalue of
0

and R is an orthogonal matrix with each column as normalized corresponding eigenvectors 1 2
, ,...,
p
r r r
(Rencher, 2003).
Let
~ ( , )
p
nS YY W n

  where 1 2
( , , ... )
n
Y y y y
 and each
~ ( , )
j p
y N

0 and independent (Anderson (1984),
Section 3.3; Srivastava (2005); Fisher et al. (2010)). Let 1 2
( , ,... )
n
U u u u
 where
j
u
is independently and identically
distributed (iid.)
( , )
p
N I
0
and we can write
1
2
Y U
  where
1 1
2 2
   
. Define 1 2
( , ,..., )
p
W U w w w
  
   and each wi
are iid.
( , ).
n
N I
0 Thus, define
v w w
ii i i

 are iid chi-squared random variables with n degree of freedom.
Lemma A.1. For
ii i i
v w w

 and
ij i j
v w w

 for any
i j

.
( ) ( 2)...( 2 2), 1, 2, ..
r
ii
E v n n n r r    
( ) 2 ,
ii
Var v n

2
( ) 8 ( 2)( 3),
ii
Var v n n n
  
3
( ) 8 ,
ii
E v n n
 
4
( ) 12 ( 4),
ii
E v n n n
  
2 4 4 3
( ( 2)) 3 ( 2)[272 ( )],
ii
E v n n n n n O n    
2
( ) ,
ij
E v n

4
( ) 3 ( 2),
ij
E v n n
 
2
( ) ( 2),
ii ij
E v v n n  2 2
( ) ( 2)( 4),
ii ij
E v v n n n
  
2 2
( ) ( 2) .
ij ii jj
E v v v n n 
Proof. The first 6 results can be found in Srivastava (2005) and the last 5 results can be found in Fisher et al. (2010).
As in similar proofs of Srivastava (2005), we can write 1
0
(1 / ) ( )
p tr S

 and
1 2
0
(1/ ) ( )
p tr S

 in terms of chi-squared
random variables.
.
1 1
1 0
1
1 1 1 1
ˆ( ) ( )
pi
ii
ii
a tr S tr RDR YY v
p p n np d

 

 
    
 
 
  (A.1)
Similarly, we also have

527
S. Chaipitak & S. Chongcharoen / Songklanakarin J. Sci. Technol. 38 (5), 521-535, 2016
2
1 2 2 2
02 2 2
1
1 1 2
( ) ,
p p i j
i
ii ij
i i j i j
i
tr S v v
p d d
n p d n p
 


 
  
  (A.2)
where
.
ij i j
v w w

 We let
2
2
1 2 1 2
0 0
1 1
ˆ
( ) ( ( )) .
( 1)( 2)
n
a tr S tr S
n n p n
 
   
 
 
 
 
(A.3)
Thus
[
2
2
2
2 2
22 2 2
1 1
2
2
2 2
3 2 2
1
2
1 2
1 1 2 1
ˆ( 1)( 2)
1 2 1
( )
( 1)( 2)
],
( 1)( 2)
p p p
i j
i i
ii ij ii
i i j i
i j i
i
p p i j
i
ii ij ii jj
i i j i j
i
n
a v v v
n n p d d np d
n p d n p
n n v v v v
n n d d n
n p d n p
nb b
n n
 
 
 

  
 
  
 
  
 

  
 
 
 
 
 
 
 
 
 
 
 
 
 
  (A.4)
where
2
2 2
1 2
3 2 2
1
1 2 1
, ( ).
p p i j
i
ii ij ii jj
i i j i j
i
n
b v b v v v
d d n
n p d n p
 

 

  
 
Proof of Theorem 2.1.
Since
1
1 0 1 1 1
1
0 1
1
1 1 1 1
ˆ
( ) ( ) ( )
1 1 ( ) .
p p p
i i i
ii ii
i i i
i i i
pi
ii
E a E tr S E v E v n
p np d np d np d
tr a
p d p
  


  


  

    
    
 
   
 
   
And from Lemma A.1, we easily find that 21
( ) 0
ij ii jj
E v v v
n
 
then 2
( ) 0
E b

. Thus
2 2
2 2
2 2
23 2 3 2
1 1
2 2
2
1 2
0 2
3 2 2
1 1
1 1
ˆ
( ) ( )
( 1)( 2) ( 1)( 2)
1 1 1
( 2) ( )
( 1)( 2)
p p
i i
ii ii
i i
i i
p p
i i
i i
i i
n n n n
E a E v E v
n n n n
n p d n p d
n n n n tr a
n n p p
n p d d
 
 
 

 
 
 
 
 
   

      
 
   
   
   
   
This is shown that both
1
ˆ
a
and
2
ˆ
a
are unbiased estimators of
1
a
and
2
a
respectively. To show that
1
ˆ
a
and
2
ˆ
a
are consistent
estimators considered by
2 2
2
2 2 2
1 1
11
1 1 2 1 2
ˆ
( ) ( ) .
2
p p i i
i i
i i
p
i
ii ii
i
i
Var a Var v Var v a
np d np p np
n p d d
  
 
   
  
 
   
 
   
   
(A.5)

S. Chaipitak & S. Chongcharoen / Songklanakarin J. Sci. Technol. 38 (5), 521-535, 2016
528
And since
2
2 1 2
ˆ
( 1)( 2)
n
a b b
n n
 
 
 
 
thus
 
22
2 1 2 1 2
ˆ
( ) ( ) ( ) 2 ( , .
( 1)( 2)
n
Var a Var b Var b COV b b
n n
  
 
 
 
 
 
2 4
2 2
2 2
16 2 2 6 2 4
1 1
2
4
5
( 1) ( 1)
( ) ( )
8( 1) ( 2)( 3) .
p p
i i
ii ii
i i
i i
n n
Var b Var v Var v
n p d n p d
n n n a
n p
 
 
 
 
 
  

 
 
 
 
(A.6)
2
24 2
2
2
2 2
2 2
4 2 2 2
2
2 4
4
4 1
( )
4 1 1
4( 1)( 2) 1 .
pi j
ij ii jj
i j i j
pi i
ij ii jj ij ii jj
i j i i
Var b Var v v v
d d n
n p
E v v v E v v v
n n
n p d d
n n a a
p
n
 
 




 
   
 
 
 
 
 
 
 
 
 
 
 
   
 
   
 
 
   
 
 
 
 
 
(A.7)
And since 2
( ) 0
E b

then
1 2 1 2
2
2 2
5 2
1
2 2
2 2 2 2
1 2
11 22
2 2
1 2
5
( , ) ( )
2( 1) 1
1 1
2( 1)
..
p p j k
i
ii jk jj kk
i i j i j
i
p p
j k j k
jk jj kk jk jj kk
i j i j
i j i j
COV b b E b b
nE v v v v
d d n
n p d
E v v v v E v v v v
d d n d d n
d d
n
n p
 

   
 
 
 
 
 


 
  



 
 
 
 
 
 
 
   
   
   
   
   
   
   
2
2 2
2
1
.
0
p
p j k
pp jk jj kk
i j i j
p
E v v v v
d d n
d
  

 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
(A.8)
becau se

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S. Chaipitak & S. Chongcharoen / Songklanakarin J. Sci. Technol. 38 (5), 521-535, 2016
2 2
2 2 2
2 2 3
1
1 1
( 2)( ) ( 2)( )( ) 0, ,
1( 2)(
pj k
ii jk jj kk
i j i j
p p
j k j k
ii jk ii jj kk
i j i j
i j i j
pj k j k
ii ik ii kk
i j i j i j
E v v v v
d d n
E v v v v v n n n n n n n for i j k
d d n d d n
E v v v v n n n
d d n d d
 
   
   

 


 


        
    
 
 
 
 
 
 
 
   
   
   
 
 
 
2 2 3
1
4) ( 2)( 4)( ) 0, ,
1 1
( 2)( 4) ( 2)( 4)( ) 0, .
p
i j
p p
j k j k
ii ij ii jj
i j i j
i j i j
n n n n for i j k
n
E v v v v n n n n n n n for i j k
d d n d d n
   

 

 
     
          
 
 
 
   
   
   
(A.9)
By (A.6) – (A.8), then we have
 
4
2 1 2 1 2
2 2
4 2
2
4 2 4
2 2 5 4
2
2
4 2
ˆ
( ) ( ) ( ) 2 ( , )
( 1) ( 2)
8( 1) ( 2)( 3) 4( 1)( 2) 1
( 1) ( 2)
4(2 3 6) 4 .
( 1)( 2) ( 1)( 2)
n
Var a Var b Var b COV b b
n n
n n n n n n
a a a
p
n n n p n
n n a a
n n n p n n
  
 
    
  
 
 
 
   
 
 
 
  
 
 
 
 
 
 
 
  
(A.10)
Since
1
ˆ
a
and
2
ˆ
a
are unbiased estimators of
1
a
and
2
a
, respectively and from (A.5), (A.9), and by applying the Chebyshev’ss
inequality, for any
0


as ( , )n p
 
,
2
1 1 1
2 2
2
1 1
ˆ ˆ
( ) 0
a
P a a Var a np

 
    
 
  and
2
2 2
2 2 2 4 2 4 2
2 2 2
1 1 4(2 3 6) 4 8 4
ˆ ˆ
( ) 0.
( 1)( 2) ( 1)( 2)
n n
P a a Var a a a a a
n n n p n n np n

 
 
       
   
   
 
     
   
 
 
Hence
1
ˆ
a
and
2
ˆ
a
are unbiased and consistent estimators of
1
a
and
2
a
, respectively. The proof is completed.
Proof of Theorem 2.2.
From Theorem 2.1, we have
1 1 2 2
ˆ ˆ
( ) , ( )
E a a E a a
 
(A.11)
By Lemma A.1., with simple calculations and in similar proofs of Srivastava (2005) under assumption (A), and (B), and as
( , )n p
 
, we obtain
1 2
ˆ
( ) 2 / ,
Var a a np
(A.12)
2
1 4 4
5
8( 1) ( 2)( 3)
( ) 8 / ,
n n n
Var b a a np
n p
  
  (A.13)
2 2
2 2 4 2 4
4
4( 1)( 2) 1
( ) 4 ( / ) / ,
n n
Var b a a c a a p np
p
n
 
   
 
 
  (A.14)

S. Chaipitak & S. Chongcharoen / Songklanakarin J. Sci. Technol. 38 (5), 521-535, 2016
530
2
2 2
2 4 2 4 2
4(2 3 6) 4
ˆ
( ) 4(2 ) / ,
( 1)( 2) ( 1)( 2)
n n
Var a a a a ca np
n n n p n n
 
   
    (A.15)
1 1 1 1 1 1
2 2
2 2
4 2 3 2
1 1 1 1
2
3
3 2
4 3 2 2
1
ˆ ˆ ˆ
( , ) ( ) ( ) ( )
1 1 1
1 ( 1)( 2)
p p p p
i i i i
ii ii ii ii
i i i i
i i
i i
p p i j
i
ii ii jj
i i j
i i j
COV a b E a b E a E b
n n
E v v E v E v
d np d
n p d n p d
n n n
E v v v
n p d d d n p
   
 

   
 
 
 
 
   
  
  
 
   
 
   
 
   
 
 
   
 
 
 
 
 
2
2 2
1 1
p p
i i
i i
ii
dd
 
 
 
2
3
2
4 2 3 2
1
2
3
2 2 3 2
1
2
32
4 2 3 4 2 2
1
3
2 2 3
1( 2)( 4) ( 2)
( 1)( 2)
( 1)( )( 2)( 4) ( 1) ( 2)
( 1)( 2)
p p i j
i
i i j
i i j
p p i j
i
i i j
i i j
p p
i j
i
i i j
i i j
i
ii
nn n n n n
n p d d d
n n
n p d d d
n n n n n n n
n p d n p d d
n n
n p d
 

 

 


 
 
 

 
 
 

    
 
 
    
 
 

 
 
 
 
 
 
 
 
2
2 2 2
1
3 3
3 2 3 2 2 3
1 1
3 3
3 2
( 1)( 2)
( 1)( 2)( 4) ( 1)( 2)
4( 1)( 2) 4 .
p p i j
i j i j
p p
i i
i i
i i
n n
n p d d
n n n n n
n p d n p d
n n a a
np
n p
 
 

 
 
 
 

    
 
 
 
(A.16)
From the fact that 2
( ) 0
E b

and similar to the proof for
1 2
( )
E b b
2 1
2
ˆ ˆ
( , ) ( ) 0.
1 2 1 2 3 2 1
p p i j
i
COV a b E a b E v v v v
ii ij ii jj
d d d n
n p i i j
i i j
 

   
 
 
 
 
   
  
   
  
 
 
  
  (A.17)
Note that

531
S. Chaipitak & S. Chongcharoen / Songklanakarin J. Sci. Technol. 38 (5), 521-535, 2016
2
2
1 1 2 1
2 2
ˆ ˆ
( )
1 2 3 2 2
( 1)( 2)
1 1
2
1 2 1
2 2
2 2 2 2
( 2) ( 1)( 2)
1 1 1
p p p
n n i j
i i
E a a E v v v v v
ii ii ij ii jj
np d n n d d n
n p d n p
i i i j
i i j
i
p p p i j
i i i
E v v v v v v
ii ii ii ij ii jj
d d d d n
n n p d n n np
i i i
i i i j
i
 
 
 
  

  
  
 
  
  
  
  
  
 
 
 
  
 
 
 
 
 
 

p
i j


 

 
 
 
 
 
(A.18)
By similar proof to
1 2
( )
E b b
we have 1
2
0
1
p p i j
i
E v v v v
ii ij ii jj
d d d n
i i j
i i j
 

 
 
 
 
 
 
 
 
 
 
 
then the expectation of the second
term in (A.18) equals to zero. Thus, we obtain that
2
3
3 2
1 2 2 2 3 2
1
2
3
2
2 2 3 2
1
2
3
2 3 2 2
1
3 2
2 3 2 2
1
1
ˆ ˆ
( )
( 2)
1
( 2)( 4) ( 2)
( 2)
( 4) 1
( 4) 1
p p i j
i
ii ii jj
i i j
i i j
p p i j
i
i i j
i i j
p p i j
i
i i j
i i j
pi i
ii i
E a a E v v v
n n p d d d
n n n n n
n n p d d d
n
np d p d d
n
np d p d
 

 

 

 
 
 
 

 
 
 

 

    


 

 
 
 
 
 
 
 
 
 
3
3
1 1 1
3 2
2 3 2 2
1 1 1
3 2 1
4 1
4 / .
p p p
i i
i i i
ii
p p p
i i i
i i i i
i i
dd
d
np d p d
a np a a
 
  
  
  
  
  

 
 
 
 
 
(A.19)
By (A.11) and (A.19) as ( , )n p
 
, we obtain
ˆ ˆ ˆ ˆ ˆ ˆ
( , ) ( ) ( ) ( ) 4 /
1 2 1 2 1 2 3
COV a a E a a E a E a a np
   (A.20)
To find the distribution of
1
ˆ
a
and
2
ˆ
a
, we used Multivariate central limit theorem (Rao,1973,p.147) and Lindebergg
Central Limit Theorem (Billingsley, 1995, p.359)
Since
2
ˆ
[ ]
2 1 2
( 1)( 2)
n
a b b
n n
 
  , so we need to find the distribution of
1 1
ˆ
,
a b
and
2
b
which will distribute as
Normal distribution, respectively. First, we find the distribution of
1 1
ˆ
,
a b
because both are functions of
ii
v
and the second
is of
2
b
because it is a function of ,
ij
v i j

. Finally, the distribution of
2
ˆ
a
which is a distribution of a linear function of two
normal random variables is obtained.
First, in order to find the distribution of
1
ˆ
a
and
1
b
. Under
i

and
i
d
as before, we let

S. Chaipitak & S. Chongcharoen / Songklanakarin J. Sci. Technol. 38 (5), 521-535, 2016
532
( )
1
v n
i ii
ui
d n
i


 and
2 2
( ( 2))
22
( 2)( 3)
v n n
i ii
uid n n n
i

 

 
where
2 2 4 4 3 3
( ) 0, ( ) 0, ( ) 2 / , ( ) 8 / , ( , ) 4 /
1 2 1 2 1 2
E u E u Var u d Var u d COV u u e d
i i i i i i i i i i i n i
  
     an d
2 / 3 1
e n n
n
   
as
n
 
Since
ii
v
s are independent, thus
( , )
1 2
u u
i i i

u are independently distributed
random vectors,
1,...,
i p

with ( )Ei

u 0
and covariance matrices
in

given by
2 2 3 3
2 / 4 /
, 1, ...,
3 3 4 4
4 / 8 /
d e d
i i i n i
i p
in
e d d
i n i i i
 
 
  
 
 
 
 
.
For any n as
p
 
( ... ) /
1
2 3
4
2
2 3 2 41 1 2 3 0
3 4 4 8
4 3 4
8
3 4
1 1
p
n n pn
p p
e
i n i
p p
d d a e ai i
i i n
n
e a a
p p
e n
n i i
p P
d di i
i i
 
 
     
 
 
    
 
 
 
 
   
   
   
 
 
 
 
0
where
0 0
2 3
0
0 0
3 4
2 4
.
4 8
n
n
n
a e a
e a a
 
 
 
 
 
If
i
F
is the distribution function of
i
u
then
(
2 2 2 2 4 4
1 2 1 2
2 2 2 2 2
1 1 1 1
1 1 1 1 2
) ( ) ( ),
i i
p p p p
i i i i i i i i i i
i i i i
p
dF dF E u u E u u
p p p p p
  
 
   

   
 
    u u u u
u u
from
r
C

inequality in Rao (1973, p.149). Since as
p
 
and from Lemma A.1.,
4 4
2 2 2
4 4
( ) ( ) 12 ( 4) 0
1
2 2 2 2 4 2 2 2 4 2
1 1 1
p p p
i i
E u E v n n n
i ii
p p d n p d ni i i
i i
 
  
    
  
  
and by an analogous derivation as
p
 
,
Hence 24 4
( ) 0
1 2
2 2 1
p
E u u
i i
p i

 


as
p
 
. By applying the multivariate central limit theorem, as
p
 
for any n

533
S. Chaipitak & S. Chongcharoen / Songklanakarin J. Sci. Technol. 38 (5), 521-535, 2016
( )
1
1
1
0
( ... ) ( , )
2 2
1 2 2
( ( 2))
1
2( 2)( 3)
1
pv n
i ii
d
np iiDN
p n
pv n np i ii
np d n n
i
i





    
 

 

 
 
   
 
 
 
 
u u u 0
Note that as
,
n
 
,
0 0
2 4
2 3
0 0
1, ,
0 0
4 8
3 4
a e a
n
en n
e a a
n
    
 
 
 
 
where
0 0
2 4
2 3
0
.
0 0
4 8
3 4
a a
a a
 
 
 
 
 
Thus, it follows that as
( , )n p
 
,
( )
1
1
0
( , ).
2 2 2
( ( 2))
1
2( 2)( 3)
1
pv n
i ii
d
np iiDN
pv n n
i ii
np d n n
i
i






 

 

 
 
   
 
 
 
 
0
And under assumption (A) which leads to assuming that
0
  
, where
2 4
2 3
4 8
3 4
a a
a a
 
 
 
 
 
then we have that
( )
1
1
( , ).
2 2 2
( ( 2))
1
2( 2)( 3)
1
pv n
i ii
d
np iiDN
pv n n
i ii
np d n n
i
i






 

 

 
 
   
 
 
 
 
0
For the first element in the previous random vector, since
1 1 1 1 2
1 1 1
( )1 1 1 ˆ ˆ
( ) ( ) (0, 2 ),
p p p D
i ii i ii i
i i i
i i i
v n v n
npa npa np a a N a
d d d
np np np
  
  
  
     
  
 
 
then
ˆ
( , 2 / ).
1 1 2
D
a N a a np
 (A.21)
For the second element, we have that
2 2 2 2 2
2 2 2
1 1 1
3
1 2
4
( ( 2)) ( 2)
1 1
( 2)( 3) ( 2)( 3) ( 2)( 3)
( 2)1
(0, 8 ).
( 1) ( 2)( 3) ( 2)( 3)
p p p
i ii i ii i
i i i
i i i
D
v n n v n n
np np
d n n d n n d n n
n pb n n pa
N a
np n n n n n
  
  
  
  
 
     

 
    
 
 
 
  
 
 
Since as
n
 
,

S. Chaipitak & S. Chongcharoen / Songklanakarin J. Sci. Technol. 38 (5), 521-535, 2016
534
3( 2)
1 1
1 2
( ) ( )
1 2 1 2
( 1) ( 2)( 3) ( 2)( 3)
n pb n n pa
npb npa np b a
np n n n n n np

    
    
 
 
 
 
then
( ) (0,8 )
1 2 4
D
np b a N a
 also, and with a linear transformation we have the result that
( ,8 / ).
1 2 4
D
b N a a np
 (A.22)
The next is to find the distribution of
2
b
. Srivastava (2005) gave the important results, which are used for the next
proof, that
/ ~ (0,1)
v n N
ij as
n
 
and
2 2
/ ~
1
v n
ij

which are asymptotically independently distributed for all distinct
i and j.
Note that
2
b
defined in (A.4), now we let 2
2
21
( )
i j
ij ij ii ij
i j
v v v
n
n pd d
 

  . From Lemma A.1., we have
( ) 0
Eij


and
let
 
2 2 1
2
21
4 2
2
2 2
4 4
2 2
4 2
2
( ) ( )
4( )
( )
4( 1)( 2) 4
p p i j
p ij ij ii jj
n
i j i j i j
pi j
ij ii jj
n
i j i j
S Var Var v v v
n pd d
Var v v v
d d
n p
Var b
a an n c
a a
p p
n n
 

 
 

 

  
 

 
   
 
 
 
 
 
   
   
   
as ( , )n p
 
.
Let 2
2
2
2 1
( )
p p i j
p ij ij ii ij
i j i j i j
M v v v b
d d n
n p
 

 
 
   
. If
ij
P
is the distribution function of
ij

. Since, for
0


2 2
2 2 2
2
2 2
2
2
4 2 2 2
2 2
4 2 2 2 2 2
2 2
2 2 2 2 2 2
1 1
1( )
4 1
8( 1)( 2)
80
ij p
p p
ij ij ij ij
i j i j
S
p p
p
ij
i j p
pi j
ij ii jj
i j i j
p
pi j
i j p i j
pi j
i j p i j
dP dP
S S
E
S
E v v v
d d n
n p S
n n
n p S d d
n p S d d
 
 



 

 

 

 





 
 






 
 

 
 
 
 
 
 
 
 

535
S. Chaipitak & S. Chongcharoen / Songklanakarin J. Sci. Technol. 38 (5), 521-535, 2016
as
.
p
 
Then, it follows from the Lindeberg Central Limit Theorem in Lemma A.3.,
2
(0,1).
2
2 ( / )
2 4
Mnpb D
pN
Spc a a p



Then we have 42
0, ( / ) .
2 2 4
c
D
b N a a p
np 
 

 
 
(A.23)
By (A.8), then
1
b
and
2
b
are asymptotically independent. Note that
2
ˆ
a
is a linear function of two random variables
1
b
and
2
b
that is,
2
ˆ
[ ]
2 1 2
( 1)( 2)
n
a b b
n n
  
 
1 2
b b

as
.
n
 
By (A.5), (A.15.), (A.22.), and (A.23.), then we have


2
ˆ
, 4(2 ) / .
2 2 4 2
D
a N a a ca np
 (A.24)
From (A.20), ˆ ˆ
( , ) 4 /
1 2 3
COV a a a np
, (A.21), and (A.24), we have
2 / 4 /
ˆ2 3
1 1
, .
22
ˆ4 / 4( 2 ) /
2 2 3 4 2
a np a np
a a
DN
a a a np a ca np
  
   
 
 

   
   
 
 
   
  
The proof is completed. 
Proof of Theorem 2.3. Note that our test statistic is
2
ˆˆ ˆ
2
2 1
a ta t

  
and we have
ˆ
2
ˆ
1
t
a


 
 and ˆ
1
ˆ
2
a



.
By applying the delta method (Lehmann and Romano, 2005, p.436), thus,
2
ˆ
(0, )
DN
  
 where
 
2
2 / 4 /
2 3 2 4
2 2
2 1 (2 4 2 )
2 3 4 2
21
4 / 4(2 ) /
3 4 2
a np a np t
t t a ta a ca
np
a np a ca np


     

  
  
  
 
The proof is completed. 
Proof of Corollary 2.1. Under
2 3
, ,
0 2 3
H a t a t
 
and
4
4
a t

Thus, 2 4
4 / .
ct np

 It follows from Theorem 2.3. that
the null asymptotic distribution of T is
(0,1)
N. The proof is completed. 