asymptotic properties of bayes factor in one way repeated measurements model
TRANSCRIPT
Mathematical Theory and Modeling www.iiste.org
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.4, No.11, 2014
143
Asymptotic Properties of Bayes Factor in One- Way Repeated
Measurements Model
Ameera Jaber Mohaisen and Khawla Abdul Razzaq Swadi
Mathematics Department College of Education for Pure Science AL-Basrah University-Iraq
E-mail: [email protected]
Abstract
In this paper, we consider the linear one- way repeated measurements model which has only one within units
factor and one between units factor incorporating univariate random effects as well as the experimental error
term. In this model, we investigate the consistency property of Bayes factor for testing the fixed effects linear
one- way repeated measurements model against the mixed one- way repeated measurements alternative model.
Under some conditions on the prior and design matrix, we identify the analytic form of the Bayes factor and
show that the Bayes factor is consistent.
Keywords: One- Way Repeated Measurements Model, ANOVA, Mixed model, Prior Distribution, Posterior
Distribution, covariance matrix, Bayes Factor, Consistent.
1. Introduction
Repeated measurements occur frequently in observational studies which are longitudinal in nature, and in
experimental studies incorporating repeated measures designs. For longitudinal studies, the underlying
metameter for the occasions at which measurements are taken is usually time. Here, interest often centers around
modeling the response as linear or nonlinear function of time, Repeated measures designs, on the other hand,
entail one or more response variables being measured repeatedly on each individual over arrange of conditions.
Here the metameter may be time or it may be a set of experimental conditions (e.g. ,dose levels of a drug).
Repeated measurements analysis is widely used in many fields, for example , in the health and life science,
epidemiology, biomedical, agricultural, industrial, psychological, educational researches and so on.[2],[3],[17]
A balanced repeated measurements design means that the p occasions of measurements are the same for all of
the experimental units. A complete repeated measurements design means that measurements are available each
time point for each experimental unit. A typical repeated measurements design consist of experimental units or
subjects randomized to a treatment (a between-units factor) and remaining on the assigned treatment throughout
the course of the experiment. [2],[3],[17]
A one âway repeated measurements analysis of variance refers to the situation with only one within-units
factor and a multi-way repeated measurements analysis of variance to the situation with more than one within
units factor. This is because the number of within factors, and not the number or between factor, dictates the
complexity of repeated measurements analysis of variance.[2],[3],[4],[12],[17]
In this paper, we consider the linear one- way repeated measurements model which has only one within units
factor and one between units factor incorporating univariate random effects as well as the experimental error
term, we can represent repeated measurements model as a mixed model.
The asymptotic properties of the Bayes factor have been studied mainly in nonparametric density estimation
problems related to goodness of fit testing [1],[5],[7],[8],[9],[15],[16]. Choi, Taeryon and et al in (2009) [6]
studied the semiparametric additive regression models as the encompassing model with algebraic smoothing and
obtained the closed form of the Bayes factor and studied the asymptotic behavior of the Bayes factor based on
the closed form. Mohaisen, A. J. and Abdulhussain , A. M. in (2013)[11] investigated large sample properties of
the Bayes factor for testing the pure polynomial component of spline null model whose mean function consists
of only the polynomial component against the fully spline semiparametric alternative model whose mean
function comprises both the pure polynomial and the component spline basis functions. In this paper, we
investigate the consistency property of Bayes factor for testing the fixed effects linear one- way repeated
measurements model against the mixed one- way repeated measurements alternative model. Under some
conditions on the prior and design matrix, we identify the analytic form of the Bayes factor and show that the
Bayes factor is consistent.
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2. One- Way Repeated Measurements Model
Consider the model
yijk = ÎŒ+ Ïj + ÎŽi(j) + γk+ (Ïγ)jk + eijk (1)
Where
i=1,âŠ.,n is an index for experimental unit within group j ,
j=1,âŠ,q is an index for levels of the between-units factor (Group) ,
k=1,âŠ,p is an index for levels of the within-units factor (Time) ,
yijk is the response measurement at time k for unit i within group j,
µ is the overall mean ,
Ïj is the added effect for treatment group j ,
ÎŽ i(j) is the random effect for due to experimental unit i within treatment group j ,
ðŸð is the added effect for time k ,
(Ïγ)jk is the added effect for the group j à time k interaction ,
e ijk is the random error on time k for unit i within group j ,
For the parameterization to be of full rank, we imposed the
following set of conditions
â Ïj=0qj=1 , â γ
k=0
pk=1 , â (Ïγ)jk=0
qj=1 for each k=1,âŠ,p
â (Ïγ)jk=0pk=1 for each j=1,âŠ,q .
And we assumed that the eijk and ÎŽi(j) are indepndent with
eijk ~ i.i.d N (0,Ïe2) , ÎŽij ~ i.i.d N (0,ÏÎŽ2) (2)
The model (1) is rewritten as follows
ð = ððœ + ðð + ð (3)
Where
Y = [
y111y112â®
ynqp
]
nqpÃ1
, β = [
ÎŒ1Ã1ÏqÃ1γpÃ1
(Ïγ)qpÃ1
]
(qp+q+p+1)Ã1
, Z = (1PÃ1 ⯠0PÃ1â® â± â®
0PÃ1 ⯠1PÃ1
)
nqpÃnq
,
b = ÎŽ =
[ ÎŽ1(1)
ÎŽ1(2)â®
ÎŽn(q)]
nqÃ1
, ð = [
e111e112â®enqp
]
nqpÃ1
and ð = (ð1ð , ð2
ð , ð3ð , ð4
ð)ðis an
(ððð Ã (1 + ð + ð + ðð)) design matrix of fixed effects.
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We assume ððð < 1 + ð + ð + ðð + ðð, ððð + ð = 1 + ð + ð + ðð + ðð, (ð is integer number and greater
than 1 ), and we assume that the design matrix as the following way:
ððððð ðððð = ððððŒððð âððð ⥠1 (4)
where:
ðððð = [ð ðð] and ðð be the ððð Ã [ððð â (1 + ð + ð + ðð)] matrix, and let ð = [ðð ðð ].
We would like to choose between a Bayesian mixed one- way repeated measurements model and itâs one- way
repeated measurements model (the model without random effects) by the criterion of the Bayes factor for two
hypotheses,
ð»0: ð = ððœ + ð i . e ð»0: yijk = ÎŒ+ Ïj + γk+ (Ïγ)jk + eijk versus
ð»1: ð = ððœ + ðð + ð i . e ð»1: yijk = ÎŒ+ Ïj + ÎŽi(j) + γk+ (Ïγ)jk + eijk . (5)
We assume ðœ and ð are independent and the prior distribution for the parameters are
ðœ ~ ð(0, â0) , where
â0 = ððœ2ðŒ1+ð+ð+ðð =
[
ðð2 01Ãð 01Ãð
0ðÃ1 ðð2ðŒðÃð 0ðÃð
0ðÃ1 0ðÃð ððŸ2ðŒðÃð
01Ãðð 0ðÃðð 0ðÃðð
0ððÃ1 0ððÃð 0ððÃð ð(ððŸ)2 ðŒððÃðð]
(1+ð+ð+ðð)Ã(1+ð+ð+ðð)
, (6)
and ð ~ ð(0, â1) , where
â1 = ðð2ðŒðð = ðð¿
2ðŒðð =
[ ðð¿112 0 ⯠0
0 ðð¿122 ⯠0
â®0
â®0
â± â®â¯ ðð¿ðð
2]
ððÃðð
. (7)
2.Posterior distribution
From the model (1), we have the following:
ðððð|ðððð ~ ðððð(ðððð, ðð2ðŒððð) , ðððð ~ ðððð(0ððð, ðâ0ð
ð + ðâ1ðð) (8)
where
ðððð = ÎŒ+ Ïj + ÎŽi(j) + γk+ (Ïγ)jk + eijk,for, ð = 1,2, ⊠, ð, ð = 1,⊠, ð ððð ð = 1,⊠, ð.
Then, the posterior distribution ð1(ðððð|ðððð) is the multivariate normal with
ðž(ðððð|ðððð , ðð2) = â2(â2 + ðð
2ðŒð)â1ðððð , where N=1+q+p+qp+nq and
ððð(ðððð|ðððð, ðð2) = (â2
â1 + (ðð2ðŒð)
â1)â1 = â2(â2 + ðð2ðŒð )
â1ðð2ðŒð
where
â2 = ðâ0ðð + ðâ1ð
ð = ð
[
ðð2 01Ãð 01Ãð
0ðÃ1 ðð2ðŒðÃð 0ðÃð
0ðÃ1 0ðÃð ððŸ2ðŒðÃð
01Ãðð 0ðÃðð 0ðÃðð
0ððÃ1 0ððÃð 0ððÃð ð(ððŸ)2 ðŒððÃðð]
ðð + ð
[ ðð¿112 0 ⯠0
0 ðð¿122 ⯠0
â®0
â®0
â± â®â¯ ðð¿ðð
2]
ðð
Then, ðððð~ ððð(0, ðð2ðŒððð + â2). (9)
Hence, the distribution of ðððð under ð»0 and ð»1 are respectively ðððð(0, ðð2ðŒððð + ðâ0ð
ð) and
ðððð(0, ðð2ðŒððð + ðâ0ð
ð + ðâ1ðð). (10)
We investigate the consistency of the Bayes factor, under the one- way repeated measurements model with
design matrix of the random effects (ð) to the first ððð terms. We get
yijk = ÎŒ + Ïj + ÎŽKâ + γk+ (Ïγ)jk + eijk , K
â = 1,⊠. , (ððð â (1 + ð + ð + ðð)).The covariance matrix of the
distribution of ðððð under ð»1 as:
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ðð2ðŒððð + ðâ0ð
ð + ððâ1,ðððð = ðð
2ðŒððð +ððððð·ðððððððð (11)
where â1,ð=ðð2ðŒðððâ(1+ð+ð+ðð) =
[ ðð¿112 0 ⯠0
0 ðð¿122 ⯠0
â®0
â®0
â± â®â¯ ðð¿ðððâ(1+ð+ð+ðð)
2]
ðððâ(1+ð+ð+ðð)Ãðððâ(1+ð+ð+ðð)
= ðð¿2ðŒðððâ(1+ð+ð+ðð) and ð·ððð = [
â0 0ð101ð â1,ð
] , where 0ð1, 01ð are the matrices for zeros element with size
(1 + ð + ð + ðð) Ã (ððð â (1 + ð + ð + ðð)) and (ððð â (1 + ð + ð + ðð)) Ã (1 + ð + ð + ðð) ,
respectively.
Then, we can rewrite (11) by using ð·ððð as:
ðððð [ðð2
ððððŒððð + ð·ððð]ðððð
ð . (12)
Then, the covariance matrix of the distribution of ðððð under ð»0, can be represented as:
ðð2ðŒððð + ðâ0ð
ð + ðð0ððððð = ðð
2ðŒððð +ððððð¶ðððððððð , where:
ð¶ððð = [â0 0ð101ð 0ðð
]
and 0ðð is the matrix for zeros element with size (ððð â (1 + ð + ð + ðð)) Ã (ððð â (1 + ð + ð + ðð)), then
we can rewrite the covariance matrix of the distribution of ðððð under ð»0, the following:
ðððð [ðð2
ððððŒððð + ð¶ððð]ðððð
ð (13)
and the covariance matrix of the distribution of ðððð under ð»1, can be represented as:
ðð2ðŒððð + ðâ0ð
ð + ðâ1ðð = ðð
2ðŒððð + [ð ð]ðºððð[ð ð]ð, where:
ðºððð = [â0 0ð101ð â1
]
then we can rewrite the covariance matrix of the distribution of ðððð under ð»1, the following:
[ð ð] [ðð2
ððððŒððð + ðºððð] [ð ð]
ð (14)
Lemma(1):Let the covariance matrices of the distribution of ðððð under ð»ð and ð»1 be defined as in (12) and
(13) respectively, and suppose ððððsatisfy (4). Then, the following hold.
1 âððððððððð = ðððð
ð ðððð = ððððŒððð
2 â det(ðð2ðŒððð + ðâ0ð
ð) = ððð1+ð+ð+ðð(ðð2
ððð+ ðð
2)(ðð2
ððð+ ðð
2)ð(ðð2
ððð+ ððŸ
2)ð(ðð2
ððð+ ð(ððŸ)
2 )ðð
(ðð2)ðððâ(1+ð+ð+ðð)
3 â det(ðð2ðŒððð + ðâ0ð
ð + ððâ1,ðððð) =
(ððð)ððð (ðð2
ððð+ ðð
2) (ðð2
ððð+ ðð
2)ð
(ðð2
ððð+ ððŸ
2)ð
(ðð2
ððð+ ð(ððŸ)
2 )ðð
(ðð2
ððð+ ðð¿
2)ðððâ(1+ð+ð+ðð)
4 â (ðð2ðŒððð + ððŽ0ð
ð)â1 = 1
ððð2ðððð [
ðð2
ððððŒððð + ð¶ððð]
â1
ððððð
5 â (ðð2ðŒððð + ðâ0ð
ð + ððâ1,ðððð)â1 =
1
ððð2ðððð [
ðð2
ððððŒððð + ð·ððð]
â1
ððððð
Proof:-
1 â By multiplying ðððð and ððððð to equation (4) from right and left side, we have:-
ððððððððð ðððððððð
ð = ððððððððŒðððððððð
= ðððððððððððð
= ðððððððððððð
Multiplying (ððððððððð )â1 to the above equation from left side, we get:-
ððððððððð ðððððððð
ð (ððððððððð )â1 = ððððððððððð
ð (ððððððððð )â1
ððððððððð = ððððŒððð
2 â ðð2ðŒððð + ðâ0ð
ð = ðððð [ðð2
ððððŒððð + ð¶ððð]ðððð
ð , and by using (4), we get
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= [ð , ðð]
[ (
ðð2
ððð+ ðð
2) 01Ãð 01Ãð 01Ãðð 01Ãðð
0ðÃ1 (ðð2ðŒð
ððð+ ðð
2ðŒð) 0ðÃð 0ðÃðð 0ðÃðð
0ðÃ1
0ððÃ1
0ððÃ1
0ðÃð
0ððÃð
0ððÃð
(ðð2ðŒð
ððð+ ððŸ
2ðŒð) 0ðÃðð 0ðÃðð
0ððÃð (ðð2ðŒðð
ððð+ ð(ððŸ)
2 ðŒðð) 0ððÃðð
0ððÃð 0ððÃðððð2
ððððŒðððâ(1+ð+ð+ðð)]
[ð , ðð]ð
=
[ ððð(
ðð2
ððð+ ðð
2) 01Ãð 01Ãð 01Ãðð 01Ãðð
0ðÃ1 ððð(ðð2ðŒð
ððð+ ðð
2ðŒð) 0ðÃð 0ðÃðð 0ðÃðð
0ðÃ1
0ððÃ1
0ððÃ1
0ðÃð
0ððÃð
0ððÃð
ððð(ðð2ðŒð
ððð+ ððŸ
2ðŒð) 0ðÃðð 0ðÃðð
0ððÃð ððð(ðð2ðŒðð
ððð+ ð(ððŸ)
2 ðŒðð) 0ððÃðð
0ððÃð 0ððÃðð ðð2ðŒðððâ(1+ð+ð+ðð)]
Then
det(ðð2ðŒððð + ðâ0ð
ð) = ððð1+ð+ð+ðð(ðð2
ððð+ ðð
2)(ðð2
ððð+ ðð
2)ð(ðð2
ððð+ ððŸ
2)ð(ðð2
ððð+ ð(ððŸ)
2 )ðð
(ðð2)ðððâ(1+ð+ð+ðð)
3 â ðð2ðŒððð + ðâ0ð
ð + ððâ1,ðððð = ðððð [
ðð2
ððððŒððð + ð·ððð]ðððð
ð
= [ð ðð]
[ (ðð2
ððð+ ðð
2) 01Ãð 01Ãð 01Ãðð 01Ãðð
0ðÃ1 (ðð2ðŒðððð
+ ðð2ðŒð) 0ðÃð 0ðÃðð 0ðÃðð
0ðÃ1
0ððÃ1
0ððÃ1
0ðÃð
0ððÃð
0ððÃð
(ðð2ðŒðððð
+ ððŸ2ðŒð) 0ðÃðð 0ðÃðð
0ððÃð (ðð2ðŒððððð
+ ð(ððŸ)2 ðŒðð) 0ððÃðð
0ððÃð 0ððÃðð (ðð2
ððððŒðððâ(1+ð+ð+ðð) + ðð¿
2ðŒðððâ(1+ð+ð+ðð))]
[ð ðð]ð
=
[ ððð(
ðð2
ððð+ ðð
2) 01Ãð 01Ãð 01Ãð 01Ãð
0ðÃ1 ððð(ðð2ðŒðððð
+ ðð2ðŒð) 0ðÃð 0ðÃðð 0ðÃðð
0ðÃ1
0ððÃ1
0ððÃ1
0ðÃð
0ððÃð
0ððÃð
ððð(ðð2ðŒðððð
+ ððŸ2ðŒð) 0ðÃðð 0ðÃðð
0ððÃð ððð(ðð2ðŒððððð
+ ð(ððŸ)2 ðŒðð) 0ððÃðð
0ððÃð 0ððÃðð ððð(ðð2
ððððŒðððâ(1+ð+ð+ðð) + ðð¿
2ðŒðððâ(1+ð+ð+ðð))]
.
Then
det(ðð2ðŒððð + ðâ0ð
ð + ððâ1,ðððð) = (ððð)ððð(
ðð2
ððð+ ðð
2)(ðð2
ððð+ ðð
2)ð(ðð2
ððð+ ððŸ
2)ð(ðð2
ððð+ ð(ððŸ)
2 )ðð(ðð2
ððð+
ðð¿2)ðððâ(1+ð+ð+ðð)
4 â (ðð2ðŒððð + ðâ0ð
ð)â1 = {ðððð [ðð2
ððððŒððð + ð¶ððð]ðððð
ð }â1
= ððððð â1
[ðð2
ððððŒððð + ð¶ððð]
â1
ððððâ1
=ðŒðððâ1
ðððððððŒððððððð
ð â1[ðð2
ððððŒððð + ð¶ððð]
â1
ððððâ1ððððŒððð
ðŒðððâ1
ððð
=1
(ððð)2ðððððððð
ð ððððð â1
[ðð2
ððððŒððð + ð¶ððð]
â1
ððððâ1ðððððððð
ð
= 1
(ððð)2ðððð [
ðð2
ððððŒððð + ð¶ððð]
â1
ððððð (by lemma 1)
5 â (ðð2ðŒððð + ðâ0ð
ð + ððâ1,ðððð)â1 = {ðððð [
ðð2
ððððŒððð + ð·ððð]ðððð
ð }â1
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= ððððð â1
[ðð2
ððððŒððð + ð·ððð]
â1
ððððâ1
=ðŒðððâ1
ðððððððŒððððððð
ð â1[ðð2
ððððŒððð + ð·ððð]
â1
ððððâ1ððððŒððð
ðŒðððâ1
ððð
=1
(ððð)2ðððððððð
ð ððððð â1
[ðð2
ððððŒððð + ð·ððð]
â1
ððððâ1ðððððððð
ð
= 1
(ððð)2ðððð [
ðð2
ððððŒððð + ð·ððð]
â1
ððððð . â
3- The Bayes factor
The Bayes factor for testing problem (5) is given by:
ðµ01(yijk) = m(yijk|ð»ð)
m(yijk|ð»1)
=
1
â2ðððð1+ð+ð+ðð(ðð2
ððð+ðð2 )(
ðð2
ððð+ðð2)ð(
ðð2
ððð+ððŸ2)ð(
ðð2
ððð+ð(ððŸ)2 )ðð(ðð
2)ðððâ(1+ð+ð+ð)
exp{â1
2ððððð (
1
ððð2ðððð[
ðð2
ððððŒððð+ ð¶ððð]
â1
ððððð )ðððð}
1
â2ð(ððð)ððð(ðð2
ððð+ðð2)(
ðð2
ððð+ðð2)
ð
(ðð2
ððð+ððŸ2)
ð
(ðð2
ððð+ð(ððŸ)2 )
ðð
(ðð2
ððð+ðð¿2)ðð
exp{â1
2ððððð (
1
ððð2[ð ð][
ðð2
ððððŒððð+ ðºððð]
â1
[ð ð]ð)ðððð}
=
â(ððð)ððð(ðð2
ððð+ðð
2)(ðð2
ððð+ðð
2)ð
(ðð2
ððð+ððŸ
2)ð
(ðð2
ððð+ð(ððŸ)
2 )ðð
(ðð2
ððð+ðð¿
2)ðð
âððð1+ð+ð+ðð(ðð2
ððð+ðð
2)(ðð2
ððð+ðð
2)ð(ðð2
ððð+ððŸ
2)ð(ðð2
ððð+ð(ððŸ)
2 )ðð(ðð2)ðððâ(1+ð+ð+ðð)
exp{ â1
2ððð2ððððð
(ðððð [ðð2
ððððŒððð + ð¶ððð]
â1
ððððð â[ð ð] [
ðð2
ððððŒððð + ðºððð]
â1
[ð ð]ð)ðððð}. (15)
Now let the approximation of ðµ01 with design matrix of the random effects (ð) to the first ððð terms as
following
ï¿œÌï¿œ01 =
â(ððð)ððð(ðð2
ððð+ðð
2)(ðð2
ððð+ðð
2)ð(ðð2
ððð+ððŸ
2)ð(ðð2
ððð+ð(ððŸ)
2 )ðð(ðð2
ððð+ðð¿
2)ðððâ(1+ð+ð+ðð)
âððð1+ð+ð+ðð(ðð2
ððð+ðð
2)(ðð2
ððð+ðð
2)ð(ðð2
ððð+ððŸ
2)ð(ðð2
ððð+ð(ððŸ)
2 )ðð(ðð2)ðððâ(1+ð+ð+ðð)
exp{â1
2ððð2ððððð
(ðððð [ðð2
ððððŒððð + ð¶ððð]
â1
ððððð âðððð [
ðð2
ððððŒððð + ð·ððð]
â1
ððððð )ðððð}. (16)
Then,
logï¿œÌï¿œ01 = 1
2log
(ððð)ððð(ðð2
ððð+ðð
2)(ðð2
ððð+ðð
2)ð(ðð2
ððð+ððŸ
2)ð(ðð2
ððð+ð(ððŸ)
2 )ðð(ðð2
ððð+ðð¿
2)
ðððâ(1+ð+ð+ðð)
ððð1+ð+ð+ðð(ðð2
ððð+ðð
2)(ðð2
ððð+ðð
2)ð(ðð2
ððð+ððŸ
2)ð(ðð2
ððð+ð(ððŸ)
2 )ðð(ðð2)ðððâ(1+ð+ð+ðð)
â 1
2ððð2ððððð
(ðððð [ðð2
ððððŒððð + ð¶ððð]
â1
ððððð âðððð [
ðð2
ððððŒððð + ð·ððð]
â1
ððððð )ðððð
2 logï¿œÌï¿œ01 = log(ððð)ððð(
ðð2
ððð+ðð¿
2)
ðððâ(1+ð+ð+ðð)
ððð1+ð+ð+ðð(ðð2)ðððâ(1+ð+ð+ðð)
â 1
ððð2ððððð (ðððð [
ðð2
ððððŒððð + ð¶ððð]
â1
ððððð
âðððð [ðð2
ððððŒððð + ð·ððð]
â1
ððððð )ðððð
â 2 logï¿œÌï¿œ01 = log(ððð)ððð(
ðð2
ððð+ðð¿
2)ðððâ(1+ð+ð+ðð)
ððð1+ð+ð+ðð(ðð2)ðððâ(1+ð+ð+ðð)
â1
(ððð)2 {ðððð
ð ðððð {[ðð2
ððððŒððð + ð¶ððð]
â1
â [ðð2
ððððŒððð +
ð·ððð]â1
}ððððð ðððð}
= log(ððð)ðððâ(1+ð+ð+ðð)(
ðð2
ððð+ðð¿
2)
ðððâ(1+ð+ð+ðð)
(ðð2)ðððâ(1+ð+ð+ðð)
â1
(ððð)2{ððððð ðððð
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149
[ 0 01Ãð 01Ãð 01Ãðð 01Ãðð
0ðÃ1 0ðÃð 0ðÃð 0ðÃðð 0ðÃðð
0ðÃ1
0ððÃ1
0ððÃ1
0ðÃð
0ððÃð
0ððÃð
0ðÃð 0ðÃðð 0ðÃðð 0ððÃð 0ððÃðð 0ððÃðð
0ððÃð 0ððÃðð ( (ððð)2ðð¿
2
(ðð2+ððððð¿
2)ðð2 )ðŒðððâ(1+ð+ð+ðð)]
ðððÃððð
ððððð ðððð}
= log(ðð2+ððððð¿
2
ðð2 )ðððâ(1+ð+ð+ðð) â {ðððð
ð ðððððððð}
where
ðððð = 1
(ððð)2ðððð
[ 0 01Ãð 01Ãð 01Ãðð 01Ãðð
0ðÃ1 0ðÃð 0ðÃð 0ðÃðð 0ðÃðð
0ðÃ1
0ððÃ1
0ððÃ1
0ðÃð
0ððÃð
0ððÃð
0ðÃð 0ðÃðð 0ðÃðð 0ððÃð 0ððÃðð 0ððÃðð
0ððÃð 0ððÃðð ( (ððð)2ðð¿
2
(ðð2+ððððð¿
2)ðð2 )ðŒðððâ(1+ð+ð+ðð)]
ðððÃððð
ððððð .
The Bayes factor for testing the model in (5) satisfy is consistent if [6]
ððððððââ ðµ01 = â , ððððð ð¡ ð ð¢ððððŠ ðð ððððððððð¡ðŠ ð0â under ð»0, (17)
ððððððââ ðµ01 = 0 , ððððð ð¡ ð ð¢ððððŠ ðð ððððððððð¡ðŠ ð1â under ð»1, (18)
Lemma(2):let ðððð = (ðŠ111 , ⊠, ðŠððð)ð where ðŠððð â²ð are independent normal random variable
with mean ÎŒ + Ïj + γk+ (Ïγ)jk and variance ðð
2 = 1, then there exist a positive constant ð¶1 such that 1
ðððð,1[â log(1 + ððððð¿
2) â ðððâ(1+ð+ð+ðð)ð=1 ðððð
ð ðððððððð] > ð¶1 , with probability tending to 1, i.e. logï¿œÌï¿œ01
ðððâââ â, in probability, where Snqp,1 = â
nqpÏÎŽ2
(1+nqpÏÎŽ2)
nqpi=1 .
Proof:
Let Xnqp = ( X111, X121, ⊠, Xnqp) are independent standard normal random variables, Note that Ynqp Xnqp=d +
ððœ and
YnqpT QnqpYnqp = (Xnqp + ððœ)
TQnqp (Xnqp + ððœ)
= XnqpT QnqpXnqp + (ððœ)
TQnqp(ððœ) + XnqpT Qnqp(ððœ) + (ððœ)
TQnqpXnqp.
By the property of the design matrix (4) get
(ððœ)TQnqp = (ððœ)T 1
(nqp)2 Mnqp
[ 0 01Ãð 01Ãð 01Ãðð 01Ãðð
0ðÃ1 0ðÃð 0ðÃð 0ðÃðð 0ðÃðð
0ðÃ1
0ððÃ1
0ððÃ1
0ðÃð
0ððÃð
0ððÃð
0ðÃð 0ðÃðð 0ðÃðð 0ððÃð 0ððÃðð 0ððÃðð
0ððÃð 0ððÃðð ( (ððð)2ðð¿
2
(ðð2+ððððð¿
2)ðð2 )ðŒðððâ(1+ð+ð+ðð)]
MnqpT =
1
(ððð)2[ð, ð, ðŸ, (ððŸ)]ððð (ðŒ1+ð+ð+ðð , 0ðððâ(1+ð+ð+ðð)
[ 0 01Ãð 01Ãð 01Ãðð 01Ãðð
0ðÃ1 0ðÃð 0ðÃð 0ðÃðð 0ðÃðð
0ðÃ1
0ððÃ1
0ððÃ1
0ðÃð
0ððÃð
0ððÃð
0ðÃð 0ðÃðð 0ðÃðð 0ððÃð 0ððÃðð 0ððÃðð
0ððÃð 0ððÃðð ( (ððð)
2ðð¿2
(ðð2+ðððð
ð¿
2)ðð2 )ðŒðððâ(1+ð+ð+ðð)
]
ððð(ðŒ1+ð+ð+ðð , 0ððâ(1+ð+ð+ðð) [ð, ð, ðŸ, (ððŸ)]ð = 0.
Then YnqpT QnqpYnqp = Xnqp
T QnqpXnqp.
Now by using the expectation formula of the quadratic form[4],[10],[12],[14], get
0 < ðž(XnqpT QnqpXnqp) = tr(Qnqp) = â
nqpÏÎŽ2
(1+nqpÏÎŽ2)
nqpi=1 = Snqp,1
Note that
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Qnqp2 =
1
(nqp)4Mnqp
[ 0 01Ãð 01Ãð 01Ãðð 01Ãðð
0ðÃ1 0ðÃð 0ðÃð 0ðÃðð 0ðÃðð
0ðÃ1
0ððÃ1
0ððÃ1
0ðÃð
0ððÃð
0ððÃð
0ðÃð 0ðÃðð 0ðÃðð 0ððÃð 0ððÃðð 0ððÃðð
0ððÃð 0ððÃðð ( (ððð)2ðð¿
2
(ðð2+ððððð¿
2)ðð2 )ðŒðððâ(1+ð+ð+ðð)]
MnqpT Mnqp
[ 0 01Ãð 01Ãð 01Ãðð 01Ãðð
0ðÃ1 0ðÃð 0ðÃð 0ðÃðð 0ðÃðð
0ðÃ1
0ððÃ1
0ððÃ1
0ðÃð
0ððÃð
0ððÃð
0ðÃð 0ðÃðð 0ðÃðð 0ððÃð 0ððÃðð 0ððÃðð
0ððÃð 0ððÃðð ( (ððð)
2ðð¿2
(ðð2+ðððð
ð¿
2)ðð2 )ðŒðððâ(1+ð+ð+ðð)
]
MnqpT
= 1
(nqp)3Mnqp
[ 0 01Ãð 01Ãð 01Ãðð 01Ãðð
0ðÃ1 0ðÃð 0ðÃð 0ðÃðð 0ðÃðð
0ðÃ1
0ððÃ1
0ððÃ1
0ðÃð
0ððÃð
0ððÃð
0ðÃð 0ðÃðð 0ðÃðð 0ððÃð 0ððÃðð 0ððÃðð
0ððÃð 0ððÃðð ( (ððð)2ðð¿
2
(ðð2+ððððð¿
2)ðð2 )
2ðŒðððâ(1+ð+ð+ðð) ]
MnqpT
and
tr(Qnqp2 ) =
1
(nqp)2â (
(nqp)2ÏÎŽ2
1+nqpÏÎŽ2)2nqp
i=1 = â (nqpÏÎŽ
2
1+nqpÏÎŽ2)2 = nqp3(
ÏÎŽ2
1+nqpÏÎŽ2)2 = Snqp,2
nqpi=1 .
Using the variance formula of the quadratic form[4],[10],[12],[14], get
var(XnqpT QnqpXnqp) = 2tr(Qnqp
2 ) = 2Snqp,2 = 2â (nqpÏÎŽ
2
1+nqpÏÎŽ2)2nqp
i=1 .
Let cnqp = Snqp,3âSnqp,1
2Snqp,1 , where Snqp,3 = â log(1 + ððððð¿
2)ðððð=1 = ððð log(1 + ððððð¿
2) . Then,
Pr { 1
Snqp,1[â log(1 + nqpÏÎŽ
2)2 â nqpâ(1+q+p+qp)i=1 Ynqp
T QnqpYnqp] †cnqp } = Pr [ {YnqpT QnqpYnqp}
Snqp,1 â¥
Snqp,3
Snqp,1â
cnqp ]
= Pr [ {YnqpT QnqpYnqp}
Snqp,1â
E(YnqpT QnqpYnqp)
Snqp,1â¥Snqp,3
Snqp,1â cnqp â
E(YnqpT QnqpYnqp)
Snqp,1 ]
= Pr [ 1
Snqp,1{(Ynqp
T QnqpYnqp) â E(YnqpT QnqpYnqp)} â¥
Snqp,3âE(YnqpT QnqpYnqp)
Snqp,1â cnqp ]
â€1
Snqp,12
var(YnqpT QnqpYnqp)
(Snqp,2âSnqp,1
2Snqp,1)2
= 8Snqp,2
(Snqp,3âSnqp,1)2
nqpâââ 0, from the Chebyshev inequality[13] .
Since Snqp,3 â Snqp,1 ⥠c1Snqp,1 for some c1 > 0 and let cnqp ⥠c1/2.
Then, there exists a positive constant C1 = c1/2 such as 1
Snqp,1[â log(1 + nqpÏÎŽ
2)2 â nqpâ(1+q+p+qp)i=1 Ynqp
T QnqpYnqp] > C1, with probability tending to 1.
Then,
2logBÌ01 = â log(1 + nqpÏÎŽ2)2 â
nqpâ(1+q+p+qp)i=1 Ynqp
T QnqpYnqpnqpâââ â in probability under H0 i.e.
logBÌ01nqpâââ â
Lemma 3: Suppose we have the one- way repeated measurements model, then
1
ðððð,1ð¡ð (
1
ððð2ðððð [
ðð2
ððððŒððð + ð·ððð]
â1
ððððð â [ð ð] [
ðð2
ððððŒððð + ðºððð]
â1
[ð ð]ð)ðððâââ 0
Proof:
Since[ð ð] [ðð2
ððððŒððð + ðºððð] [ð ð]
ð âðððð [ðð2
ððððŒððð + ð·ððð]ðððð
ð is the covariance matrix of
ðâ = ð¿ðŸâ = ðð ð, ðŸâ = ððð â (1 + ð + ð + ðð), ⊠, ððð + ð , it is a positive definite matrix. Thus
1
ððð2ðððð [
ðð2
ððððŒððð + ð·ððð]
â1
ððððð â
1
ððð2[ð ð] [
ðð2
ððððŒððð + ðºððð]
â1
[ð ð]ðis also a positive definite matrix.
Thus, ð¡ð (1
ððð2ðððð [
ðð2
ððððŒððð + ð·ððð]
â1
ððððð ) ⥠ð¡ð(
1
ððð2[ð ð] [
ðð2
ððððŒððð + ðºððð]
â1
[ð ð]ð ), also
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ð¡ð (ðð ð 1
ððð2ðððð [
ðð2
ððððŒððð + ð·ððð]
â1
ððððð ) †âð¡ð(ðð
ððð )ð¡ð (1
ððð2ðððð [
ðð2
ððððŒððð + ð·ððð]
â1
ððððð )
= âðð â â1
ððð2ðððð [
ðð2
ððððŒððð + ð·ððð]
â1
ððððð â by the Cauchy-Schwarz inequality, where
â1
ððð2ðððð [
ðð2
ððððŒððð + ð·ððð]
â1
ððððð â = { ð¡ð ((
1
ððð2ðððð [
ðð2
ððððŒððð + ð·ððð]ðððð
ð )â2)}1
2
= [1
ððð2{ ((
ðð2
ððð+ ðð
2) + â (ðð2
ððð+ ðð
2) + â (ðð2
ððð+ ððŸ
2) + â (ðð2
ððð+ ð(ððŸ)
2 )ððð=1
ðð=1
ðð=1 )
â2
+
â (ðð2
ððð+ ðð¿
2)â2
ðððâ(1+ð=ð+ðð)ð=1 }]
1
2
.
Since
((ðð2
ððð+ ðð
2) + â (ðð2
ððð+ ðð
2) + â (ðð2
ððð+ ððŸ
2) + â (ðð2
ððð+ ð(ððŸ)
2 )ððð=1
ðð=1
ðð=1 )
â2
= 1
((ðð2
ððð+ðð
2)+ð(ðð2
ððð+ðð
2)+ð(ðð2
ððð+ððŸ
2)+ðð(ðð2
ððð+ð(ððŸ)
2 ))
2 =
ððð2
((ðð2+ððððð
2)+ð(ðð2+ððððð
2)+ð(ðð2+ðððððŸ
2)+ðð(ðð2+ðððð(ððŸ)
2 ))2 â€
ððð2
((ððððð2)+ð(ððððð
2)+ð(ðððððŸ2)+ðð(ðððð(ððŸ)
2 ))2
= 1
((ðð2)+ð(ðð
2)+ð(ððŸ2)+ðð(ð(ððŸ)
2 ))2 = ð(1), and
â (ðð2
ððð+ ðð¿
2)â2
ðððâ(1+ð+ð+ðð)ð=1 =
ðððâ(1+ð+ð+ðð)
(ðð2
ððð+ðð¿
2)2 â€
ððð
(ðð2
ððð+ðð¿
2)2 =
(ððð)3
(ðð2+ððððð¿
2)2 â€
(ððð)3
(ððððð¿2)2 = ð(ððð).
Then, â1
ððð2ðððð [
ðð2
ððððŒððð + ð·ððð]
â1
ððððð â = ð{(ððð)
1
2 }.
Since ð ð¢ðâðð â = 1 from the distances between the elements of design matrix ð, then âðð â2 †ððð.
Thus , 1
ðððð,1ð¡ð (
1
ððð2ðððð [
ðð2
ððððŒððð + ð·ððð]
â1
ððððð â
1
ððð2[ð ð] [
ðð2
ððððŒððð + ðºððð]
â1
[ð ð]ð) =
1
ðððð,1ð¡ð (
1
ððð2ðððð [
ðð2
ððððŒððð + ð·ððð]
â1
ððððð {
1
ððð2[ð ð] [
ðð2
ððððŒððð + ðºððð] [ð ð]
ð â1
ððð2ðððð [
ðð2
ððððŒððð +
ð·ððð]ððððð } [ð ð] [
ðð2
ððððŒððð + ðºððð]
â1
[ð ð]ð )
= 1
ðððð,1â ðð¿
2(ððŸâð 1
ððð2ðððð [
ðð2
ððððŒððð + ð·ððð]
â1
ððððð (ððŸâ
ð 1
ððð2[ð ð] [
ðð2
ððððŒððð +
ððð+ð ðŸâ=ðððâ(1+ð+ð+ðð)
ðºððð]â1
[ð ð]ð)ð) †ðð¿2
ðððð,1â
1
ððð2ðððð [
ðð2
ððððŒððð + ð·ððð]
â1
ððððð â
2
â âð§ðŸââ2ððð+ð
ðŸâ=ðððâ(1+ð+ð+ðð)
â€ððððð¿
2
ððððððð,1â 1ððð+ð ðŸâ=ðððâ(1+ð+ð+ðð) â€
ðŸâðð¿2
ððð ðððð,1=
ðŸâðð¿2
(ððð)2ðð¿2
(1+ððððð¿2)
=ðŸâ(1+ððððð¿
2)
(ððð)2
Thus
1
ðððð,1ð¡ð (
1
ððð2ðððð [
ðð2
ððððŒððð + ð·ððð]
â1
ððððð â
1
ððð2[ð ð] [
ðð2
ððððŒððð + ðºððð]
â1
[ð ð]ð) †ðŸâ(1+ððððð¿
2)
(ððð)2
ðððâââ 0
Theorem(1): Suppose that the true model is the one- way repeated measurements model with fixed effect
only (the model under ð»0 in 5), and let ð0ððð
denote the true distribution of the data, with probability density
function ð0( yijk|ÎŒ, Ïj, γk, (Ïγ)jk) = ð·{(ðŠððð â ÎŒ â Ïj â γkâ (Ïγ)jk)/ ðð} , where ð·(â) is the standard normal
density, then, the Bayes factor is consistent under the null hypothesis ð»0 i.e
limðððââ ðµ01 = â in probability ð0ððð
.
Proof:
log ðµ01 = 1
2log
det (ðð2ðŒððð+ðâ0ð
ð+ðâ1ðð)
det (ðð2ðŒððð+ðâ0ð
ð)â1
2ððððð ( (ðð
2ðŒððð + ðâ0ðð)â1â(ðð
2ðŒððð + ðâ0ðð + ðâ1ð
ð)â1)ðððð
=1
2log
det (ðð2ðŒððð+ðâ0ð
ð+ðâ1ðð)
det (ðð2ðŒððð+ðâ0ð
ð+ððâ1,ðððð)+
1
2log
det (ðð2ðŒððð+ðâ0ð
ð+ððâ1,ðððð)
det (ðð2ðŒððð+ðâ0ð
ð)â1
2ððððð (ðð
2ðŒððð + ðâ0ðð) â1
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152
â(ðð2ðŒððð + ðâ0ð
ð + ððâ1,ðððð)â1)ðððð â
1
2ððððð ((ðð
2ðŒððð + ðâ0ðð + ððâ1,ððð
ð)â1â(ðð2ðŒððð + ðâ0ð
ð +
ðâ1ðð)â1)ðððð
= logï¿œÌï¿œ01 +1
2log
det (ðð2ðŒððð+ðâ0ð
ð+ðâ1ðð)
det (ðð2ðŒððð+ðâ0ð
ð+ððâ1,ðððð) â
1
2ððððð ((ðð
2ðŒððð + ðâ0ðð + ððâ1,ððð
ð)â1â(ðð2ðŒððð + ðâ0ð
ð +
ðâ1ðð)â1)ðððð
Since (ðð2ðŒððð + ðâ0ð
ð + ðâ1ðð) â (ðð
2ðŒððð + ðâ0ðð + ððâ1,ððð
ð) is a positive definite matrix,
ððð¡ (ðð2ðŒððð + ðâ0ð
ð + ðâ1ðð) ⥠ððð¡ (ðð
2ðŒððð + ðâ0ðð + ððâ1,ððð
ð), thus
1
2log
det (ðð2ðŒððð+ðâ0ð
ð+ðâ1ðð)
det (ðð2ðŒððð+ðâ0ð
ð+ððâ1,ðððð) ⥠0
ðž( ððððð ((ðð
2ðŒððð + ðâ0ðð + ððâ1,ððð
ð)â1â(ðð2ðŒððð + ðâ0ð
ð + ðâ1ðð)â1)ðððð) = ðð
2 ð¡ð((ðð2ðŒððð +
ðâ0ðð + ððâ1,ððð
ð)â1 â (ðð2ðŒððð + ðâ0ð
ð + ðâ1ðð)â1) + (ððœ)ð((ðð
2ðŒððð + ðâ0ðð + ððâ1,ððð
ð)â1 â
(ðð2ðŒððð + ðâ0ð
ð + ðâ1ðð)â1)(ððœ) ,
since (ðð2ðŒððð + ðâ0ð
ð + ððâ1,ðððð)â1 â (ðð
2ðŒððð + ðâ0ðð + ðâ1ð
ð)â1 , (ðð2ðŒððð + ðâ0ð
ð + ððâ1,ðððð)â1
and (ðð2ðŒððð + ðâ0ð
ð + ðâ1ðð)â1 are all nonnegative definite matrices, it follows that
0 †(ððœ)ð((ðð2ðŒððð + ðâ0ð
ð + ððâ1,ðððð)â1 â (ðð
2ðŒððð + (ðâ0ðð + ðâ1ð
ð)â1)(ððœ) †(ððœ)ð
(ðð2ðŒððð + ðâ0ð
ð + ððâ1,ðððð)â1 (ððœ).
By the property of design matrix (4), we get
ðððððð = ððððð ð = ððð (ðŒ1+ð+ð+ðð , 0ðððâ(1+ð+ð+ðð).
Now as the proof of lemma (3), we get:
(ððœ)ð(ðð2ðŒððð + ðâ0ð
ð + ððâ1,ðððð)â1(ððœ) =
1
(ððð)2(ððœ)ððððð [
ðð2
ððððŒððð + ð·ððð]
â1
ððððð (ððœ)
=1
(ððð)2[ð, ð, ðŸ, (ððŸ)]ððð (ðŒ1+ð+ð+ðð , 0ððâ(1+ð+ð+ðð)
[ (ðð2
ððð+ ðð
2) 01Ãð 01Ãð 01Ãðð 01Ãðð
0ðÃ1 (ðð2
ððð+ ðð
2)ðŒð 0ðÃð 0ðÃðð 0ðÃðð
0ðÃ1
0ððÃ1
0ððÃ1
0ðÃð
0ððÃð
0ððÃð
(ðð2
ððð+ ððŸ
2)ðŒð 0ðÃðð 0ðÃðð
0ððÃð (ðð2
ððð+ ð(ððŸ)
2 )ðŒðð 0ððÃðð
0ððÃð 0ððÃðð (ðð2
ððð+ ðð¿
2)ðŒðððâ(1+ð+ð+ðð) ] â1
ððð
(ðŒ1+ð+ð+ðð , 0ððâ(1+ð+ð+ðð) [ð, ð, ðŸ, (ððŸ)]ð
= ð2 [ðð2
ððð+ ðð
2]â1
+ â ðð2ð
ð=1 [ðð2
ððð+ ðð
2]â1
+ â ðŸð2ð
ð=1 [ðð2
ððð+ ððŸ
2]â1
+ â (ððŸ)ð2ðð
ð=1 [ðð2
ððð+ ð(ððŸ)
2 ]â1
= ð(1).
Since ðð { 1
ðððð,1ððððð ((ðð
2ðŒððð + ðâ0ðð + ððâ1,ððð
ð)â1â(ðð2ðŒððð + ðâ0ð
ð + ðâ1ðð)â1)ðððð > í } â€
1
ðððð,1 {
ðž( ððððð ((ðð
2ðŒððð+ðâ0ðð+ððâ1,ððð
ð)â1â(ðð2ðŒððð+ðâ0ð
ð+ðâ1ðð)â1)ðððð)
}
†1
ðððð,1
ð¡ð((ðð2ðŒððð+ðâ0ð
ð+ððâ1,ðððð)â1â(ðð
2ðŒððð+ðâ0ðð+ðâ1ð
ð)â1
+ 1
ðððð,1
(ððœ)ð(ðð2ðŒððð+ðâ0ð
ð+ððâ1,ðððð)â1 (ððœ)
ðððâââ 0 (by the Markov inequality[13], where í > 0) .
Since (ðð2ðŒððð + ðâ0ð
ð + ððâ1,ðððð)â1 â (ðð
2ðŒððð + ðâ0ðð + ðâ1ð
ð)â1 is nonnegative definite â
ððððð ((ðð
2ðŒððð + ðâ0ðð + ððâ1,ððð
ð)â1â(ðð2ðŒððð + ðâ0ð
ð + ðâ1ðð)â1)ðððð is nonnegative .
Therefore,
â1
2ððððð ((ðð
2ðŒððð + ðâ0ðð + ððâ1,ððð
ð)â1â(ðð2ðŒððð + ðâ0ð
ð + ðâ1ðð)â1)ðððð = ð(ðððð,1)
â â
1
2ððððð ((ðð
2ðŒððð + ðâ0ðð + ððâ1,ððð
ð)â1â(ðð2ðŒððð + ðâ0ð
ð + ðâ1ðð)â1)ðððð
ðððð,1
= ð(1)
Then, as in lemma (2) there exists a constant ð¶1 such that
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1
ðððð,1log ðµ01 â¥
1
ðððð,1log ï¿œÌï¿œ01 + ð(1) > ð¶1 + ð(1).
That is,
ðµ01 ⥠exp( ðððð,1{ ð¶1 + ð(1)}) ðððâââ â in probability ð0
ððð.â
Lemma(4):
Assuming that the true model is the mixed one-way repeated measurements model (the model under ð»1in 5). Let
ðððð = (ðŠ1 , ⊠, ðŠððð) where ðŠððð â²ð are independent normal random variables with mean ÎŒ+ Ïj + ÎŽi(j) + γk+
(Ïγ)jk and variance ðð2 = 1 and let ð€ððð = ðž(ðððð)
ððððð ðž(ðððð).
Then, limðððââ1
ð€ðððððððð ðððððððð = 1, in probability ð1
â.
Proof:
From the moment formula of the quadratic form of normal random variables, the expectation and variance of
quadratic form ððððð ðððððððð are given by:
ðž(ððððð ðððððððð ) = ð¡ð (ðððð) + ðž(ðððð)
ððððð ðž(ðððð)
= âððððð¿
2
(1+ððððð¿2)
ðððâ(1+ð+ð+ðð)ð=1 + ðž(ðððð)
ððððð ðž(ðððð)
= ðððð,1 + ð€ððð , and
ð£ðð(ððððð ðððððððð ) = 2ð¡ð (ðððð
2) + 4ðž(ðððð)ððððð
2 ðž(ðððð).
Note
ð€ððð = ðž(ðððð)ððððð ðž(ðððð)
= (ð, ðð , ðŸð , (ððŸ)ð , ð¿ðððâ(1+ð+ð+ðð)ð )ðððð
ð1
ððð2ðððð
[ 0 01Ãð 01Ãð 01Ãðð 01Ãðð
0ðÃ1 0ðÃð 0ðÃð 0ðÃðð 0ðÃðð
0ðÃ1
0ððÃ1
0ððÃ1
0ðÃð
0ððÃð
0ððÃð
0ðÃð 0ðÃðð 0ðÃðð 0ððÃð 0ððÃðð 0ððÃðð
0ððÃð 0ððÃðð ( (ððð)
2ðð¿2
(ðð2+ðððð
ð¿
2)ðð2 )ðŒðððâ(1+ð+ð+ðð)
]
ððððð ðððð (ð, ðð, ðŸð, (ððŸ)
ð, ð¿ðððâ(1+ð+ð+ðð)ð
)ð
= (ð, ðð , ðŸð , (ððŸ)ð , ð¿ðððâ(1+ð+ð+ðð)ð )
[ 0 01Ãð 01Ãð 01Ãðð 01Ãðð
0ðÃ1 0ðÃð 0ðÃð 0ðÃðð 0ðÃðð
0ðÃ1
0ððÃ1
0ððÃ1
0ðÃð
0ððÃð
0ððÃð
0ðÃð 0ðÃðð 0ðÃðð
0ððÃð 0ððÃðð 0ððÃðð
0ððÃð 0ððÃðð ( (ððð)2ðð¿
2
(ðð2+ððððð¿
2)ðð2 ) ðŒðððâ(1+ð+ð+ðð)]
(ð, ðð , ðŸð , (ððŸ)ð , ð¿ðððâ(1+ð+ð+ðð)ð )
ð
= ð¿ðððâ(1+ð+ð+ðð)ð (
(ððð)2ðð¿2
(ðð2+ððððð¿
2)ðð2 ) ðŒðððâ(1+ð+ð+ðð)ð¿ðððâ(1+ð+ð+ðð) = â
(ððð)2ðð¿2ð¿ð2
1+ððððð¿2
ðððâ(1+ð+ð+ðð)ð=1 ,
since ð ð¢ðð ð¿ð < â , we have
ð€ððð †ð ð¢ðð ð¿ð2 ðððâ
ððððð¿2
1+ððððð¿2
ðððð=1 = ð(ððð ðððð,1), then ð€ððð = ð(ððð
2).
Now we can consider ð¿ð2 > 0 , where ð ⥠1 is a fixed positive integer and a sufficiently large ððð with
ððððð¿2
1+ððððð¿2 â¥
1
2 . Then for such ððð and ð.
ð€ððð = âððð2ðð¿
2ð¿ð2
1+ððððð¿2
ðððâ(1+ð+ð+ðð)ð=1 â¥
ððð2ðð¿2
1+ððððð¿2 ð¿ð
2
= ððð ððððð¿
2
1+ððððð¿2 ð¿ð
2 ⥠ðððð¿ð2
2.
Similarly to the previous calculation, we have
ðž(ðððð)ððððð2 ðž(ðððð) = (ð, ð
ð , ðŸð , (ððŸ)ð , ð¿ðððâ(1+ð+ð+ðð)ð )ðððð
ð1
(ððð)3ðððð
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[ 0 01Ãð 01Ãð 01Ãðð 01Ãðð
0ðÃ1 0ðÃð 0ðÃð 0ðÃðð 0ðÃðð
0ðÃ1
0ððÃ1
0ððÃ1
0ðÃð
0ððÃð
0ððÃð
0ðÃð 0ðÃðð 0ðÃðð 0ððÃð 0ððÃðð 0ððÃðð
0ððÃð 0ððÃðð ( (ððð)
2ðð¿2
(ðð2+ðððð
ð¿
2)ðð2 )2
ðŒðððâ(1+ð+ð+ðð)]
ððððð ðððð (ð, ðð, ðŸð, (ððŸ)
ð, ð¿ðððâ(1+ð+ð+ðð)ð
)ð
=1
ðððð¿ðððâ((1+ð+ð+ðð)ð (
(ððð)2ðð¿2
(ðð2+ððððð¿
2)ðð2 )
2ðŒðððâ(1+ð+ð+ðð)ð¿ðððâ(1+ð+ð+ðð)
=1
ðððâ
(ððð)4ðð¿4ð¿ð2
(1+ððððð¿2)2
ðððâ(1+ð+ð+ðð)ð=1 .
Since ðž(ðððð)ððððð2 ðž(ðððð) †â
(ððð)3ðð¿4ð¿ð2
(1+ððððð¿2)2
ðððð=1 †ððð ð ð¢ðð ð¿ð
2â(ððð)2ðð¿
4
(1+ððððð¿2)2
ðððð=1 = ð( ððððððð,2) (as in case
of ðž(ðððð)ððððð ðž(ðððð)).
Now for previous ððð and ð we can find
ðž(ðððð)ððððð2 ðž(ðððð) = â
(ððð)3ðð¿4ð¿ð2
(1+ððððð¿2)2
ðððâ(1+ð+ð+ðð)ð=1 ⥠ððð ð¿ð
2 (ððððð¿
2
1+ððððð¿2)2
⥠ð¿ð2
4 ððð.
By combining value of ðððð,1 and the previous result, get
ð£ðð(ððððð ðððððððð ) †ð(ððð ðððð,1).
Furthermore, note that ðž(ðððð)ððððð ðž(ðððð) ⥠ðž(ðððð)
ððððð
2 ðž(ðððð).
By the Chebshev inequality[13] get
ðð {|1
ð€ðððððððð ðððððððð â
ðž(ððððð ðððððððð)
ð€ððð| > í } â€
ð£ðð(ððððð ðððððððð)
ð€ððð2 2 = ð(ðððâ2)
ðððâââ 0, where í > 0 .
Since limðððââðž(ðððð
ð ðððððððð)
ð€ððð= limðððââ (
ðððð,1
ð€ððð+ð€ððð
ð€ððð) = 1
⎠limðððââ1
ð€ðððððððð ðððððððð = 1 , in probability â
Lemma(5):
Suppose we have the mixed one- way repeated measurements model, then
limðððââ1
ð€ðððlog
det(ðð2ðŒððð+ðâ0ð
ð+ðâ1ðð)
det(ðð2ðŒððð+ðâ0ð
ð+ððâ1,ðððð)= 0
Proof:
Since det(ðð2ðŒððð + ðâ0ð
ð + ðâ1ðð) †â (ðð
2 + ðð2ð¥ð2 + â ðð
2ð¥ðð2 + â ððŸ
2ð¥ðð2 + â ð(ððŸ)
2ð¥ðð2ðð
ð=1
ð
ð=1
ð
ð=1 +ðððð=1
â ðð¿2ðð
ð=1 ð¥ðð2 ) (from the property of the determinant of the positive definite matrix)
= (ððð)ðððâ (ðð2
ððð+(ðð2ð¥ð2+â ðð
2ð¥ðð2 +â ððŸ
2ð¥ðð2 +â ð(ððŸ)
2 ð¥ðð2ðð
ð=1ðð=1
ðð=1 +â ðð¿
2ððð=1 ð¥ðð
2
ððð )
ðððð=1 .
And we have
det(ðð2ðŒððð + ðâ0ð
ð + ððâ1,ðððð) = (ððð)ððð(
ðð2
ððð+ ðð
2)(ðð2
ððð+ ðð
2)ð(ðð2
ððð+ ððŸ
2)ð(ðð2
ððð+ ð(ððŸ)
2 )ðð(ðð2
ððð+
ðð¿2)ðððâ(1+ð+ð+ðð) (from lemma 1).
â logdet(ðð
2ðŒððð+ðâ0ðð+ðâ1ð
ð)
det(ðð2ðŒððð+ðâ0ð
ð+ððâ1,ðððð)= logdet(ðð
2ðŒððð + ðâ0ðð + ðâ1ð
ð) â log det(ðð2ðŒððð + ðâ0ð
ð +
ððâ1,ðððð)
†{(ððð)log(ððð) + â log (ðð2
ððð+(ðð2ð¥ð2+â ðð
2ð¥ðð2 +â ððŸ
2ð¥ðð2 +â ð(ððŸ)
2 ð¥ðð2ðð
ð=1ðð=1
ðð=1 +â ðð¿
2ððð=1 ð¥ðð
2
ððð )
ðððð=1 } â {ðððlog(ððð) +
log (ðð2
ððð+ ðð
2) + ðlog (ðð2
ððð+ ðð
2) + ðlog (ðð2
ððð+ ððŸ
2) + ððlog (ðð2
ððð+ ð(ððŸ)
2 ) + ððð â (1 + ð + ð +
ðð)log (ðð2
ððð+ ðð¿
2)} = ð(1) â ð(1) = ð(1).
Then
1
ð€ðððlog
det(ðð2ðŒððð+ðâ0ð
ð+ðâ1ðð)
det(ðð2ðŒððð+ðâ0ð
ð+ððâ1,ðððð)
ðððâââ 0
Theorem(2):
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Suppose that the true model is the mixed one- way repeated measurements model (the model under ð»1in 5),
and let ð1ððð
denote the true distribution of the data, with probability density function
ð1(yijk|ÎŒÏj, ÎŽi(j), γk, (Ïγ)jk) = ð·{(yijk â (ÎŒ + Ïj + ÎŽi(j) + γk+ (Ïγ)jk))/ðð}, where ð·(â) is the standard normal
density. Then the Bayes factor is consistent under the alternative hypothesis ð»1 i.e
limðððââ ðµ01 = 0 in probability ð1ððð
.
Proof:
log ðµ01 = 1
2log
det(ðð2ðŒððð+ðâ0ð
ð+ðâ1ðð)
det (ðð2ðŒððð+ðâ0ð
ð)â1
2ððððð ((ðð
2ðŒððð + ðâ0ðð) â1â(ðð
2ðŒððð + ðâ0ðð + ðâ1ð
ð)â1)ðððð
=
1
2log
det (ðð2ðŒððð+ðâ0ð
ð+ðâ1ðð)
det(ðð2ðŒððð+ðâ0ð
ð+ððâ1,ðððð)+
1
2log
det(ðð2ðŒððð+ðâ0ð
ð+ððâ1,ðððð)
det (ðð2ðŒððð+ðâ0ð
ð)â1
2ððððð ((ðð
2ðŒððð + ðâ0ðð) â1â(ðð
2ðŒððð +
ðâ0ðð + ððâ1,ððð
ð)â1)ðððð â
1
2ððððð ((ðð
2ðŒððð + ðâ0ðð + ððâ1,ððð
ð)â1
â(ðð2ðŒððð + ðâ0ð
ð + ðâ1ðð)â1)ðððð
= logï¿œÌï¿œ01 â1
2ððððð ((ðð
2ðŒððð + ðâ0ðð) â1â(ðð
2ðŒððð + ðâ0ðð + ððâ1,ððð
ð)â1)ðððð â
1
2ððððð
((ðð2ðŒððð + ðâ0ð
ð + ððâ1,ðððð)â1â(ðð
2ðŒððð + ðâ0ðð + ðâ1ð
ð)â1)ðððð , where
logï¿œÌï¿œ01 =
1
2log
det (ðð2ðŒððð+ðâ0ð
ð+ððâ1,ðððð)
det (ðð2ðŒððð+ðâ0ð
ð)â1
2ððððð ((ðð
2ðŒððð + ðâ0ðð) â1â(ðð
2ðŒððð + ðâ0ðð + ððâ1,ððð
ð)â1)ðððð .
From lemma (1), we have
det(ðð
2ðŒððð+ðâ0ðð+ððâ1,ððð
ð)
det(ðð2ðŒððð+ðâ0ð
ð)=
(ððð)ððð(ðð2
ððð+ðð
2)(ðð2
ððð+ðð
2)ð(ðð2
ððð+ððŸ
2)ð(ðð2
ððð+ð(ððŸ)
2 )ðð(ðð2
ððð+ðð¿
2)ðððâ(1+ð+ð+ðð)
ððð1+ð+ð+ðð(ðð2
ððð+ðð
2)(ðð2
ððð+ðð
2)ð(ðð2
ððð+ððŸ
2)ð(ðð2
ððð+ð(ððŸ)
2 )ðð(ðð2)ðððâ(1+ð+ð+ðð)
=ððððððâ(1+ð+ð+ðð)(
ðð2
ððð+ðð¿
2)
ðððâ(1+ð+ð+ðð)
ðð2ðððâ(1+ð+ð+ðð)
= (ðð2+ððððð¿
2
ðð2 )
ðððâ(1+ð+ð+ðð)
=(1 +ððððð¿
2
ðð2 )
ðððâ(1+ð+ð+ðð)
= ð(ððð)
â¹ log (1 +ððððð¿
2
ðð2 )
ðððâ(1+ð+ð+ðð)
= ð(ððð).
And from lemma (4) ð€ððð = ð(ððð2). Then
1
ð€ðððlog
det(ðð2ðŒððð+ðâ0ð
ð+ððâ1,ðððð)
det (ðð2ðŒððð+ðâ0ð
ð)= ð(ðððâ1)
ðððâââ 0.
Then,
1
ð€ðððlogï¿œÌï¿œ01 =
1
2(
1
ð€ððð log
det(ð1,ððð)
det(ðð2ðŒððð+ðâ0ð
ð)â
1
ð€ðððððððð ((ðð
2ðŒððð + ðâ0ðð) â1â(ðð
2ðŒððð + ðâ0ðð +
ððâ1,ðððð)â1)ðððð).
Since (ðð2ðŒððð + ðâ0ð
ð) â1 â (ðð2ðŒððð + ðâ0ð
ð + ððâ1,ðððð)â1= ðððð, then
1
ð€ðððððððð ((ðð
2ðŒððð + ðâ0ðð) â1â(ðð
2ðŒððð + ðâ0ðð + ððâ1,ððð
ð)â1)ðððð =
1
ð€ðððððððð ðððððððð. from lemma
(4), limðððââ1
ð€ðððððððð ðððððððð = 1 , then
1
ð€ðððlogï¿œÌï¿œ01
ðððâââ =
1
2(
1
ð€ððð log
det(ðð2ðŒððð+ðâ0ð
ð+ððâ1,ðððð)
det(ðð2ðŒððð+ðâ0ð
ð)â
1
ð€ðððððððð ((ðð
2ðŒððð + ðâ0ðð) â1â(ðð
2ðŒððð + ðâ0ðð +
ððâ1,ðððð)â1)ðððð)
ðððâââ =
1
2(0 â 1) =
â1
2
âŽ1
ð€ðððlogï¿œÌï¿œ01
ðððâââ
â1
2 , in probability ð1
ððð.
Now from lemma (5),
1
ð€ððð 1
2log
det (ðð2ðŒððð+ðâ0ð
ð+ðâ1ðð)
det (ðð2ðŒððð+ðâ0ð
ð+ððâ1,ðððð) ðððâââ 0, and â
1
2ððððð ((ðð
2ðŒððð + ðâ0ðð + ððâ1,ððð
ð)â1â(ðð
2ðŒððð +
ðâ0ðð + ðâ1ð
ð)â1)ðððð is nonpositive random variable since (ðð
2ðŒððð + ðâ0ðð + ððâ1,ððð
ð)â1â
(ðð2ðŒððð + ðâ0ð
ð + ðâ1ðð)â1
is a nonnegative definite matrix.
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Therefore, by combining the above results, there exist ð¶2 > 0 such that
limðððââ sup1
ð€ðððlog ðµ01 â€
âð¶2
2, with probability tending to 1.
Then , ðµ01ðððâââ 0, in probability ð1
ððð. â
4.Conclusions
1- The posterior distribution ð1(ðððð|ðððð) is the multivariate normal with
ðž(ðððð|ðððð , ðð2) = â2(â2 + ðð
2ðŒð)â1ðððð , where N=1+q+p+qp+nq and
ððð(ðððð|ðððð, ðð2) = (â2
â1 + (ðð2ðŒð)
â1)â1 = â2(â2 + ðð2ðŒð )
â1ðð2ðŒð, where
ðððð = ÎŒ+ Ïj + ÎŽi(j) + γk+ (Ïγ)jk + eijk,for, ð = 1,2, ⊠, ð, ð = 1,⊠, ð ððð ð = 1,⊠, ð, and
â2 = ðâ0ðð + ðâ1ð
ð = ð
[
ðð2 01Ãð 01Ãð
0ðÃ1 ðð2ðŒðÃð 0ðÃð
0ðÃ1 0ðÃð ððŸ2ðŒðÃð
01Ãðð 0ðÃðð 0ðÃðð
0ððÃ1 0ððÃð 0ððÃð ð(ððŸ)2 ðŒððÃðð]
ðð + ð
[ ðð¿112 0 ⯠0
0 ðð¿122 ⯠0
â®0
â®0
â± â®â¯ ðð¿ðð
2]
ðð
2-The Bayes factor for testing problem ð»0: yijk = ÎŒ + Ïj + γk + (Ïγ)jk + eijk versus
ð»1: yijk= ÎŒ + Ïj + ÎŽi(j) + γk + (Ïγ)jk + eijk is given by:
ðµ01 =
â(ððð)ððð(ðð2
ððð+ðð
2)(ðð2
ððð+ðð
2)ð
(ðð2
ððð+ððŸ
2)ð
(ðð2
ððð+ð(ððŸ)
2 )ðð
(ðð2
ððð+ðð¿
2)ðð
âððð1+ð+ð+ðð(ðð2
ððð+ðð
2)(ðð2
ððð+ðð
2)ð(ðð2
ððð+ððŸ
2)ð(ðð2
ððð+ð(ððŸ)
2 )ðð(ðð2)ðððâ(1+ð+ð+ðð)
exp{ â1
2ððð2ððððð
(ðððð [ðð2
ððððŒððð + ð¶ððð]
â1
ððððð âðððð [
ðð2
ððððŒððð + ðºððð]
â1
ððððð )ðððð}.
3-The approximation of ðµ01 with design matrix of the random effects (ð) to the first ððð terms as following
ï¿œÌï¿œ01 =
â(ððð)ððð(ðð2
ððð+ðð
2)(ðð2
ððð+ðð
2)ð(ðð2
ððð+ððŸ
2)ð(ðð2
ððð+ð(ððŸ)
2 )ðð(ðð2
ððð+ðð¿
2)ðððâ(1+ð+ð+ðð)
âððð1+ð+ð+ðð(ðð2
ððð+ðð
2)(ðð2
ððð+ðð
2)ð(ðð2
ððð+ððŸ
2)ð(ðð2
ððð+ð(ððŸ)
2 )ðð(ðð2)ðððâ(1+ð+ð+ðð)
exp{â1
2ððð2ððððð
(ðððð [ðð2
ððððŒððð + ð¶ððð]
â1
ððððð âðððð [
ðð2
ððððŒððð + ð·ððð]
â1
ððððð )ðððð}.
4- If we have the one- way repeated measurements model, then
1
ðððð,1ð¡ð (
1
ððð2ðððð [
ðð2
ððððŒððð + ð·ððð]
â1
ððððð â [ð ð] [
ðð2
ððððŒððð + ðºððð]
â1
[ð ð]ð)ðððâââ 0
5-If the true model is the one-way repeated measurement model with fixed effects only then the Bayes factor is
consistent under the null hypothesis ð»0 . This implies that, limðððââ ðµ01 = â in probability ð0ððð
.
6- If we have the mixed one- way repeated measurements model, then
limðððââ1
ð€ðððlog
det(ðð2ðŒððð+ðâ0ð
ð+ðâ1ðð)
det(ðð2ðŒððð+ðâ0ð
ð+ððâ1,ðððð)= 0.
7- If the true model is the mixed one- way repeated measurements model, then the Bayes factor is consistent
under the alternative hypothesis ð»1.This implies that, limðððââ ðµ01 = 0 in probability ð1ððð
.
5.References
[1] Aerts, M. , Claeskens, G. and Hart, J.D., (2004). Bayesian âmotivated tests of function fit and their
asymptotic frequentist properties, Ann. Statist. 32 (6), 2580-2615.
[2] Al-Mouel, A.H.S.,(2004) . Multivariate Repeated Measures Models and Comparison of Estimators, h. D.
Thesis, East China Normal University, China.
[3] Al-Mouel , A.H.S., and Wang, J.L,(2004). One âWay Multivariate Repeated Measurements analysis of
Variance Model, Applied Mathematics a Journal of Chinese Universities, 19,4,pp435- 448.
Mathematical Theory and Modeling www.iiste.org
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.4, No.11, 2014
157
[4] Arnold, Steven F., (1981). The theory of linear models and multivariate analysis. John Wiley and
Sons, New York. Chichester. Brisbane. Toronto .
[5] Berger, J. O., (1985). Statistical Decision Theory and Bayesian Analysis. Springer, New York.
[6] Choi, T., Lee, J. and Roy, A., (2008). A note on the Bayes factor in a semiparametric regression
model, Journal of Multivariate Analysis. 1316-1327.
[7] Dass, S.C. and Lee, J., (2004). A note on the consistency of Bayes factors for testing point null versus
non-parametric alternatives, J. Statist. plan. Inference 119 (1) 143-152.
[8] Ghosh, J.K. , Delampady, M. and Samanta, T., (2006). Introduction to Bayesian Analysis: Theory and
Methods, Springer, New York.
[9] Gosal, G., Lember, J. and Van der Vaart, A., (2008). Nonparametric Bayesian model selection and
averaging, Electron. J. Stat. 2 63-89.
[10] Graybill, F.A., (1976). Theory and Application of the linear Model, Duxbury press, North Scituate, Mass,
xiv+704.
[11] Mohaisen, A. J. and Abdulhussain , A. M., (2013). Large Sample Property of The Bayes Factor in a Spline
Semiparametric Regression Model , Mathematical Theory and Modeling www.iiste.org ISSN
(Paper)2224-5804 ISSN (Online)2225-0522 Vol 3, No. 11, 2013.
[12] Neter, J. and Wasserman, W. , (1974). Applied linear statistical models, regression analysis of variance
and experimental designs, Richard. D. IRWIN, INC .
[13] Papoulis,A.,(1997). Markoff Sequences in Probability, Random Variables, and Stochastic Processes,
2nd ed. New York . McGraw-Hill, PP. 528-535, 1984.
[14] Tong, Y.L., (1989). The Multivariate Normal Distribution, Springer series in statistics, Springer.
[15] Verdinelli, I. and Wasserman, L., (1998). Bayesian goodness âof-fit testing using infinite- dimensional
exponential families, Ann. Stat.26 (4) 1215-1241.
[16] Vinish, R. Mc , Rousseau, J. and Mengerson, K., (2008). Bayesian goodness âof-fit testing with
mixtures of triangular distributions, Scand. J, statist. in press (doi:10.1111/j.1467- 9469.2008.00620.x).
[17] Vonesh, E.F. and Chinchilli, V.M., (1997). Linear and Nonlinear Models for the Analysis of Repeated
Measurements, Marcel Dakker, Inc., New York.
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