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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 –way 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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Vol.4, No.11, 2014

144

2. One- Way Repeated Measurements Model

Consider the model

yijk = μ+ τj + δi(j) + γk+ (τγ)jk + eijk (1)

Where

i=1,….,n is an index for experimental unit within group j ,

j=1,…,q is an index for levels of the between-units factor (Group) ,

k=1,…,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 ,

τj is the added effect for treatment group j ,

δ i(j) is the random effect for due to experimental unit i within treatment group j ,

𝛾𝑘 is the added effect for time k ,

(τγ)jk is the added effect for the group j × 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

∑ τj=0qj=1 , ∑ γ

k=0

pk=1 , ∑ (τγ)jk=0

qj=1 for each k=1,…,p

∑ (τγ)jk=0pk=1 for each j=1,…,q .

And we assumed that the eijk and δi(j) are indepndent with

eijk ~ i.i.d N (0,σe2) , δij ~ i.i.d N (0,σδ2) (2)

The model (1) is rewritten as follows

𝑌 = 𝑋𝛽 + 𝑍𝑏 + 𝜖 (3)

Where

Y = [

y111y112⋮

ynqp

]

nqp×1

, β = [

μ1×1τq×1γp×1

(τγ)qp×1

]

(qp+q+p+1)×1

, Z = (1P×1 ⋯ 0P×1⋮ ⋱ ⋮

0P×1 ⋯ 1P×1

)

nqp×nq

,

b = δ =

[ δ1(1)

δ1(2)⋮

δn(q)]

nq×1

, 𝜖 = [

e111e112⋮enqp

]

nqp×1

and 𝑋 = (𝑋1𝑇 , 𝑋2

𝑇 , 𝑋3𝑇 , 𝑋4

𝑇)𝑇is an

(𝑛𝑞𝑝 × (1 + 𝑞 + 𝑝 + 𝑞𝑝)) design matrix of fixed effects.




Vol.4, No.11, 2014

145

We assume 𝑛𝑞𝑝 < 1 + 𝑞 + 𝑝 + 𝑞𝑝 + 𝑛𝑞, 𝑛𝑞𝑝 + 𝑠 = 1 + 𝑞 + 𝑝 + 𝑞𝑝 + 𝑛𝑞, (𝑠 is integer number and greater

than 1 ), and we assume that the design matrix as the following way:

𝑀𝑛𝑞𝑝𝑇 𝑀𝑛𝑞𝑝 = 𝑛𝑞𝑝𝐼𝑛𝑞𝑝 ∀𝑛𝑞𝑝 ≥ 1 (4)

where:

𝑀𝑛𝑞𝑝 = [𝑋 𝑍𝑟] and 𝑍𝑟 be the 𝑛𝑞𝑝 × [𝑛𝑞𝑝 − (1 + 𝑞 + 𝑝 + 𝑞𝑝)] matrix, and let 𝑍 = [𝑍𝑟 𝑍𝑠].

We would like to choose between a Bayesian mixed one- way repeated measurements model and it’s one- way

repeated measurements model (the model without random effects) by the criterion of the Bayes factor for two

hypotheses,

𝐻0: 𝑌 = 𝑋𝛽 + 𝜖 i . e 𝐻0: yijk = μ+ τj + γk+ (τγ)jk + eijk versus

𝐻1: 𝑌 = 𝑋𝛽 + 𝑍𝑏 + 𝜖 i . e 𝐻1: yijk = μ+ τj + δi(j) + γk+ (τγ)jk + eijk . (5)

We assume 𝛽 and 𝑏 are independent and the prior distribution for the parameters are

𝛽 ~ 𝑁(0, ∑0) , where

∑0 = 𝜎𝛽2𝐼1+𝑞+𝑝+𝑞𝑝 =

[

𝜎𝜇2 01×𝑞 01×𝑝

0𝑞×1 𝜎𝜏2𝐼𝑞×𝑞 0𝑞×𝑝

0𝑝×1 0𝑞×𝑝 𝜎𝛾2𝐼𝑝×𝑝

01×𝑞𝑝 0𝑞×𝑞𝑝 0𝑝×𝑞𝑝

0𝑞𝑝×1 0𝑞𝑝×𝑞 0𝑞𝑝×𝑝 𝜎(𝜏𝛾)2 𝐼𝑞𝑝×𝑞𝑝]

(1+𝑞+𝑝+𝑞𝑝)×(1+𝑞+𝑝+𝑞𝑝)

, (6)

and 𝑏 ~ 𝑁(0, ∑1) , where

∑1 = 𝜎𝑏2𝐼𝑛𝑞 = 𝜎𝛿

2𝐼𝑛𝑞 =

[ 𝜎𝛿112 0 ⋯ 0

0 𝜎𝛿122 ⋯ 0

⋮0

⋮0

⋱ ⋮⋯ 𝜎𝛿𝑛𝑞

2]

𝑛𝑞×𝑛𝑞

. (7)

2.Posterior distribution

From the model (1), we have the following:

𝑌𝑛𝑞𝑝|𝜉𝑛𝑞𝑝 ~ 𝑁𝑛𝑞𝑝(𝜉𝑛𝑞𝑝, 𝜎𝜖2𝐼𝑛𝑞𝑝) , 𝜉𝑛𝑞𝑝 ~ 𝑁𝑛𝑞𝑝(0𝑛𝑞𝑝, 𝑋∑0𝑋

𝑇 + 𝑍∑1𝑍𝑇) (8)

where

𝜉𝑖𝑗𝑘 = μ+ τj + δi(j) + γk+ (τγ)jk + eijk,for, 𝑖 = 1,2, … , 𝑛, 𝑗 = 1,… , 𝑞 𝑎𝑛𝑑 𝑘 = 1,… , 𝑝.

Then, the posterior distribution 𝜋1(𝜉𝑛𝑞𝑝|𝑌𝑛𝑞𝑝) is the multivariate normal with

𝐸(𝜉𝑛𝑞𝑝|𝑌𝑛𝑞𝑝 , 𝜎𝜖2) = ∑2(∑2 + 𝜎𝜖

2𝐼𝑁)−1𝑌𝑛𝑞𝑝 , where N=1+q+p+qp+nq and

𝑉𝑎𝑟(𝜉𝑛𝑞𝑝|𝑌𝑛𝑞𝑝, 𝜎𝜖2) = (∑2

−1 + (𝜎𝜖2𝐼𝑁)

−1)−1 = ∑2(∑2 + 𝜎𝜖2𝐼𝑁 )

−1𝜎𝜖2𝐼𝑁

where

∑2 = 𝑋∑0𝑋𝑇 + 𝑍∑1𝑍

𝑇 = 𝑋

[

𝜎𝜇2 01×𝑞 01×𝑝


0𝑝×1 0𝑝×𝑞 𝜎𝛾2𝐼𝑝×𝑝



𝑋𝑇 + 𝑍

[ 𝜎𝛿112 0 ⋯ 0

0 𝜎𝛿122 ⋯ 0

⋮0

⋮0


2]

𝑍𝑇

Then, 𝑌𝑛𝑞𝑝~ 𝑀𝑉𝑁(0, 𝜎𝜖2𝐼𝑛𝑞𝑝 + ∑2). (9)

Hence, the distribution of 𝑌𝑛𝑞𝑝 under 𝐻0 and 𝐻1 are respectively 𝑁𝑛𝑞𝑝(0, 𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇) and

𝑁𝑛𝑞𝑝(0, 𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 + 𝑍∑1𝑍𝑇). (10)

We investigate the consistency of the Bayes factor, under the one- way repeated measurements model with

design matrix of the random effects (𝑍) to the first 𝑛𝑞𝑝 terms. We get

yijk = μ + τj + δK∗ + γk+ (τγ)jk + eijk , K

∗ = 1,… . , (𝑛𝑞𝑝 − (1 + 𝑞 + 𝑝 + 𝑞𝑝)).The covariance matrix of the

distribution of 𝑌𝑛𝑞𝑝 under 𝐻1 as:




Vol.4, No.11, 2014

146

𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 + 𝑍𝑟∑1,𝑟𝑍𝑟𝑇 = 𝜎𝜖

2𝐼𝑛𝑞𝑝 +𝑀𝑛𝑞𝑝𝐷𝑛𝑞𝑝𝑀𝑛𝑞𝑝𝑇 (11)

where ∑1,𝑟=𝜎𝑏2𝐼𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝) =

[ 𝜎𝛿112 0 ⋯ 0

0 𝜎𝛿122 ⋯ 0

⋮0

⋮0

⋱ ⋮⋯ 𝜎𝛿𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)

2]

𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)×𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)

= 𝜎𝛿2𝐼𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝) and 𝐷𝑛𝑞𝑝 = [

∑0 0𝑟101𝑟 ∑1,𝑟

] , where 0𝑟1, 01𝑟 are the matrices for zeros element with size

(1 + 𝑞 + 𝑝 + 𝑞𝑝) × (𝑛𝑞𝑝 − (1 + 𝑞 + 𝑝 + 𝑞𝑝)) and (𝑛𝑞𝑝 − (1 + 𝑞 + 𝑝 + 𝑞𝑝)) × (1 + 𝑞 + 𝑝 + 𝑞𝑝) ,

respectively.

Then, we can rewrite (11) by using 𝐷𝑛𝑞𝑝 as:

𝑀𝑛𝑞𝑝 [𝜎𝜖2

𝑛𝑞𝑝𝐼𝑛𝑞𝑝 + 𝐷𝑛𝑞𝑝]𝑀𝑛𝑞𝑝

𝑇 . (12)

Then, the covariance matrix of the distribution of 𝑌𝑛𝑞𝑝 under 𝐻0, can be represented as:


𝑇 + 𝑍𝑟0𝑟𝑟𝑍𝑟𝑇 = 𝜎𝜖

2𝐼𝑛𝑞𝑝 +𝑀𝑛𝑞𝑝𝐶𝑛𝑞𝑝𝑀𝑛𝑞𝑝𝑇 , where:

𝐶𝑛𝑞𝑝 = [∑0 0𝑟101𝑟 0𝑟𝑟

]

and 0𝑟𝑟 is the matrix for zeros element with size (𝑛𝑞𝑝 − (1 + 𝑞 + 𝑝 + 𝑞𝑝)) × (𝑛𝑞𝑝 − (1 + 𝑞 + 𝑝 + 𝑞𝑝)), then

we can rewrite the covariance matrix of the distribution of 𝑌𝑛𝑞𝑝 under 𝐻0, the following:

𝑀𝑛𝑞𝑝 [𝜎𝜖2

𝑛𝑞𝑝𝐼𝑛𝑞𝑝 + 𝐶𝑛𝑞𝑝]𝑀𝑛𝑞𝑝

𝑇 (13)

and the covariance matrix of the distribution of 𝑌𝑛𝑞𝑝 under 𝐻1, can be represented as:


𝑇 + 𝑍∑1𝑍𝑇 = 𝜎𝜖

2𝐼𝑛𝑞𝑝 + [𝑋 𝑍]𝐺𝑛𝑞𝑝[𝑋 𝑍]𝑇, where:

𝐺𝑛𝑞𝑝 = [∑0 0𝑟101𝑟 ∑1

]

then we can rewrite the covariance matrix of the distribution of 𝑌𝑛𝑞𝑝 under 𝐻1, the following:

[𝑋 𝑍] [𝜎𝜖2

𝑛𝑞𝑝𝐼𝑛𝑞𝑝 + 𝐺𝑛𝑞𝑝] [𝑋 𝑍]

𝑇 (14)

Lemma(1):Let the covariance matrices of the distribution of 𝑌𝑛𝑞𝑝 under 𝐻𝑂 and 𝐻1 be defined as in (12) and

(13) respectively, and suppose 𝑀𝑛𝑞𝑝satisfy (4). Then, the following hold.

1 −𝑀𝑛𝑞𝑝𝑀𝑛𝑞𝑝𝑇 = 𝑀𝑛𝑞𝑝

𝑇 𝑀𝑛𝑞𝑝 = 𝑛𝑞𝑝𝐼𝑛𝑞𝑝

2 − det(𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇) = 𝑛𝑞𝑝1+𝑞+𝑝+𝑞𝑝(𝜎𝜖2

𝑛𝑞𝑝+ 𝜎𝜇

2)(𝜎𝜖2

𝑛𝑞𝑝+ 𝜎𝜏

2)𝑞(𝜎𝜖2

𝑛𝑞𝑝+ 𝜎𝛾

2)𝑝(𝜎𝜖2

𝑛𝑞𝑝+ 𝜎(𝜏𝛾)

2 )𝑞𝑝

(𝜎𝜖2)𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)

3 − det(𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 + 𝑍𝑟∑1,𝑟𝑍𝑟𝑇) =

(𝑛𝑞𝑝)𝑛𝑞𝑝 (𝜎𝜖2


2) (𝜎𝜖2


2)𝑞

(𝜎𝜖2


2)𝑝

(𝜎𝜖2


2 )𝑞𝑝

(𝜎𝜖2

𝑛𝑞𝑝+ 𝜎𝛿

2)𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)

4 − (𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋𝛴0𝑋

𝑇)−1 = 1

𝑛𝑞𝑝2𝑀𝑛𝑞𝑝 [

𝜎𝜖2

𝑛𝑞𝑝𝐼𝑛𝑞𝑝 + 𝐶𝑛𝑞𝑝]

−1

𝑀𝑛𝑞𝑝𝑇

5 − (𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 + 𝑍𝑟∑1,𝑟𝑍𝑟𝑇)−1 =

1


𝜎𝜖2

𝑛𝑞𝑝𝐼𝑛𝑞𝑝 + 𝐷𝑛𝑞𝑝]

−1


Proof:-

1 − By multiplying 𝑀𝑛𝑞𝑝 and 𝑀𝑛𝑞𝑝𝑇 to equation (4) from right and left side, we have:-

𝑀𝑛𝑞𝑝𝑀𝑛𝑞𝑝𝑇 𝑀𝑛𝑞𝑝𝑀𝑛𝑞𝑝

𝑇 = 𝑀𝑛𝑞𝑝𝑛𝑞𝑝𝐼𝑛𝑞𝑝𝑀𝑛𝑞𝑝𝑇

= 𝑀𝑛𝑞𝑝𝑛𝑞𝑝𝑀𝑛𝑞𝑝𝑇

= 𝑛𝑞𝑝𝑀𝑛𝑞𝑝𝑀𝑛𝑞𝑝𝑇

Multiplying (𝑀𝑛𝑞𝑝𝑀𝑛𝑞𝑝𝑇 )−1 to the above equation from left side, we get:-

𝑀𝑛𝑞𝑝𝑀𝑛𝑞𝑝𝑇 𝑀𝑛𝑞𝑝𝑀𝑛𝑞𝑝

𝑇 (𝑀𝑛𝑞𝑝𝑀𝑛𝑞𝑝𝑇 )−1 = 𝑛𝑞𝑝𝑀𝑛𝑞𝑝𝑀𝑛𝑞𝑝

𝑇 (𝑀𝑛𝑞𝑝𝑀𝑛𝑞𝑝𝑇 )−1

𝑀𝑛𝑞𝑝𝑀𝑛𝑞𝑝𝑇 = 𝑛𝑞𝑝𝐼𝑛𝑞𝑝

2 − 𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 = 𝑀𝑛𝑞𝑝 [𝜎𝜖2


𝑇 , and by using (4), we get




Vol.4, No.11, 2014

147

= [𝑋 , 𝑍𝑟]

[ (

𝜎𝜖2


2) 01×𝑞 01×𝑝 01×𝑞𝑝 01×𝑛𝑞

0𝑞×1 (𝜎𝜖2𝐼𝑞


2𝐼𝑞) 0𝑞×𝑝 0𝑞×𝑞𝑝 0𝑞×𝑛𝑞

0𝑝×1

0𝑞𝑝×1

0𝑛𝑞×1

0𝑝×𝑞

0𝑞𝑝×𝑞

0𝑛𝑞×𝑞

(𝜎𝜖2𝐼𝑝


2𝐼𝑝) 0𝑝×𝑞𝑝 0𝑝×𝑛𝑞

0𝑞𝑝×𝑝 (𝜎𝜖2𝐼𝑞𝑝


2 𝐼𝑞𝑝) 0𝑞𝑝×𝑛𝑞

0𝑛𝑞×𝑝 0𝑛𝑞×𝑞𝑝𝜎𝜖2

𝑛𝑞𝑝𝐼𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)]

[𝑋 , 𝑍𝑟]𝑇

=

[ 𝑛𝑞𝑝(

𝜎𝜖2


2) 01×𝑞 01×𝑝 01×𝑞𝑝 01×𝑛𝑞

0𝑞×1 𝑛𝑞𝑝(𝜎𝜖2𝐼𝑞


2𝐼𝑞) 0𝑞×𝑝 0𝑞×𝑞𝑝 0𝑞×𝑛𝑞

0𝑝×1

0𝑞𝑝×1

0𝑛𝑞×1

0𝑝×𝑞

0𝑞𝑝×𝑞

0𝑛𝑞×𝑞

𝑛𝑞𝑝(𝜎𝜖2𝐼𝑝


2𝐼𝑝) 0𝑝×𝑞𝑝 0𝑝×𝑛𝑞

0𝑞𝑝×𝑝 𝑛𝑞𝑝(𝜎𝜖2𝐼𝑞𝑝


2 𝐼𝑞𝑝) 0𝑞𝑝×𝑛𝑞

0𝑛𝑞×𝑝 0𝑛𝑞×𝑞𝑝 𝜎𝜖2𝐼𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)]

Then

det(𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇) = 𝑛𝑞𝑝1+𝑞+𝑝+𝑞𝑝(𝜎𝜖2


2)(𝜎𝜖2


2)𝑞(𝜎𝜖2


2)𝑝(𝜎𝜖2


2 )𝑞𝑝


3 − 𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 + 𝑍𝑟∑1,𝑟𝑍𝑟𝑇 = 𝑀𝑛𝑞𝑝 [

𝜎𝜖2


𝑇

= [𝑋 𝑍𝑟]

[ (𝜎𝜖2


2) 01×𝑞 01×𝑝 01×𝑞𝑝 01×𝑛𝑞

0𝑞×1 (𝜎𝜖2𝐼𝑞𝑛𝑞𝑝

+ 𝜎𝜏2𝐼𝑞) 0𝑞×𝑝 0𝑞×𝑞𝑝 0𝑞×𝑛𝑞

0𝑝×1

0𝑞𝑝×1

0𝑛𝑞×1

0𝑝×𝑞

0𝑞𝑝×𝑞

0𝑛𝑞×𝑞

(𝜎𝜖2𝐼𝑝𝑛𝑞𝑝

+ 𝜎𝛾2𝐼𝑝) 0𝑝×𝑞𝑝 0𝑝×𝑛𝑞

0𝑞𝑝×𝑝 (𝜎𝜖2𝐼𝑞𝑝𝑛𝑞𝑝

+ 𝜎(𝜏𝛾)2 𝐼𝑞𝑝) 0𝑞𝑝×𝑛𝑞

0𝑛𝑞×𝑝 0𝑛𝑞×𝑞𝑝 (𝜎𝜖2

𝑛𝑞𝑝𝐼𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝) + 𝜎𝛿

2𝐼𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝))]

[𝑋 𝑍𝑟]𝑇

=

[ 𝑛𝑞𝑝(

𝜎𝜖2


2) 01×𝑞 01×𝑝 01×𝑞 01×𝑞

0𝑞×1 𝑛𝑞𝑝(𝜎𝜖2𝐼𝑞𝑛𝑞𝑝

+ 𝜎𝜏2𝐼𝑞) 0𝑞×𝑝 0𝑞×𝑞𝑝 0𝑞×𝑛𝑞

0𝑝×1

0𝑞𝑝×1

0𝑛𝑞×1

0𝑝×𝑞

0𝑞𝑝×𝑞

0𝑛𝑞×𝑞

𝑛𝑞𝑝(𝜎𝜖2𝐼𝑝𝑛𝑞𝑝

+ 𝜎𝛾2𝐼𝑝) 0𝑝×𝑞𝑝 0𝑝×𝑛𝑞

0𝑞𝑝×𝑝 𝑛𝑞𝑝(𝜎𝜖2𝐼𝑞𝑝𝑛𝑞𝑝

+ 𝜎(𝜏𝛾)2 𝐼𝑞𝑝) 0𝑞𝑝×𝑛𝑞

0𝑛𝑞×𝑝 0𝑛𝑞×𝑞𝑝 𝑛𝑞𝑝(𝜎𝜖2

𝑛𝑞𝑝𝐼𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝) + 𝜎𝛿

2𝐼𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝))]

.

Then


𝑇 + 𝑍𝑟∑1,𝑟𝑍𝑟𝑇) = (𝑛𝑞𝑝)𝑛𝑞𝑝(

𝜎𝜖2


2)(𝜎𝜖2


2)𝑞(𝜎𝜖2


2)𝑝(𝜎𝜖2


2 )𝑞𝑝(𝜎𝜖2

𝑛𝑞𝑝+

𝜎𝛿2)𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)


𝑇)−1 = {𝑀𝑛𝑞𝑝 [𝜎𝜖2


𝑇 }−1

= 𝑀𝑛𝑞𝑝𝑇 −1

[𝜎𝜖2


−1

𝑀𝑛𝑞𝑝−1

=𝐼𝑛𝑞𝑝−1

𝑛𝑞𝑝𝑛𝑞𝑝𝐼𝑛𝑞𝑝𝑀𝑛𝑞𝑝

𝑇 −1[𝜎𝜖2


−1

𝑀𝑛𝑞𝑝−1𝑛𝑞𝑝𝐼𝑛𝑞𝑝

𝐼𝑛𝑞𝑝−1

𝑛𝑞𝑝

=1

(𝑛𝑞𝑝)2𝑀𝑛𝑞𝑝𝑀𝑛𝑞𝑝

𝑇 𝑀𝑛𝑞𝑝𝑇 −1

[𝜎𝜖2


−1

𝑀𝑛𝑞𝑝−1𝑀𝑛𝑞𝑝𝑀𝑛𝑞𝑝

𝑇

= 1

(𝑛𝑞𝑝)2𝑀𝑛𝑞𝑝 [

𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 (by lemma 1)


𝑇 + 𝑍𝑟∑1,𝑟𝑍𝑟𝑇)−1 = {𝑀𝑛𝑞𝑝 [

𝜎𝜖2


𝑇 }−1




Vol.4, No.11, 2014

148

= 𝑀𝑛𝑞𝑝𝑇 −1

[𝜎𝜖2


−1

𝑀𝑛𝑞𝑝−1

=𝐼𝑛𝑞𝑝−1

𝑛𝑞𝑝𝑛𝑞𝑝𝐼𝑛𝑞𝑝𝑀𝑛𝑞𝑝

𝑇 −1[𝜎𝜖2


−1

𝑀𝑛𝑞𝑝−1𝑛𝑞𝑝𝐼𝑛𝑞𝑝

𝐼𝑛𝑞𝑝−1

𝑛𝑞𝑝

=1

(𝑛𝑞𝑝)2𝑀𝑛𝑞𝑝𝑀𝑛𝑞𝑝

𝑇 𝑀𝑛𝑞𝑝𝑇 −1

[𝜎𝜖2


−1

𝑀𝑛𝑞𝑝−1𝑀𝑛𝑞𝑝𝑀𝑛𝑞𝑝

𝑇

= 1

(𝑛𝑞𝑝)2𝑀𝑛𝑞𝑝 [

𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 . ∎

3- The Bayes factor

The Bayes factor for testing problem (5) is given by:

𝐵01(yijk) = m(yijk|𝐻𝑜)

m(yijk|𝐻1)

=

1

√2𝜋𝑛𝑞𝑝1+𝑞+𝑝+𝑞𝑝(𝜎𝜖2

𝑛𝑞𝑝+𝜎𝜇2 )(

𝜎𝜖2

𝑛𝑞𝑝+𝜎𝜏2)𝑞(

𝜎𝜖2

𝑛𝑞𝑝+𝜎𝛾2)𝑝(

𝜎𝜖2

𝑛𝑞𝑝+𝜎(𝜏𝛾)2 )𝑞𝑝(𝜎𝜖

2)𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞)

exp{−1

2𝑌𝑛𝑞𝑝𝑇 (

1

𝑛𝑞𝑝2𝑀𝑛𝑞𝑝[

𝜎𝜖2

𝑛𝑞𝑝𝐼𝑛𝑞𝑝+ 𝐶𝑛𝑞𝑝]

−1

𝑀𝑛𝑞𝑝𝑇 )𝑌𝑛𝑞𝑝}

1

√2𝜋(𝑛𝑞𝑝)𝑛𝑞𝑝(𝜎𝜖2

𝑛𝑞𝑝+𝜎𝜇2)(

𝜎𝜖2

𝑛𝑞𝑝+𝜎𝜏2)

𝑞

(𝜎𝜖2

𝑛𝑞𝑝+𝜎𝛾2)

𝑝

(𝜎𝜖2

𝑛𝑞𝑝+𝜎(𝜏𝛾)2 )

𝑞𝑝

(𝜎𝜖2

𝑛𝑞𝑝+𝜎𝛿2)𝑛𝑞

exp{−1

2𝑌𝑛𝑞𝑝𝑇 (

1

𝑛𝑞𝑝2[𝑋 𝑍][

𝜎𝜖2

𝑛𝑞𝑝𝐼𝑛𝑞𝑝+ 𝐺𝑛𝑞𝑝]

−1

[𝑋 𝑍]𝑇)𝑌𝑛𝑞𝑝}

=

√(𝑛𝑞𝑝)𝑛𝑞𝑝(𝜎𝜖2

𝑛𝑞𝑝+𝜎𝜇

2)(𝜎𝜖2

𝑛𝑞𝑝+𝜎𝜏

2)𝑞

(𝜎𝜖2

𝑛𝑞𝑝+𝜎𝛾

2)𝑝

(𝜎𝜖2

𝑛𝑞𝑝+𝜎(𝜏𝛾)

2 )𝑞𝑝

(𝜎𝜖2

𝑛𝑞𝑝+𝜎𝛿

2)𝑛𝑞

√𝑛𝑞𝑝1+𝑞+𝑝+𝑞𝑝(𝜎𝜖2


2)(𝜎𝜖2


2)𝑞(𝜎𝜖2


2)𝑝(𝜎𝜖2


2 )𝑞𝑝(𝜎𝜖2)𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)

exp{ −1

2𝑛𝑞𝑝2𝑌𝑛𝑞𝑝𝑇

(𝑀𝑛𝑞𝑝 [𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 −[𝑋 𝑍] [

𝜎𝜖2

𝑛𝑞𝑝𝐼𝑛𝑞𝑝 + 𝐺𝑛𝑞𝑝]

−1

[𝑋 𝑍]𝑇)𝑌𝑛𝑞𝑝}. (15)

Now let the approximation of 𝐵01 with design matrix of the random effects (𝑍) to the first 𝑛𝑞𝑝 terms as

following

�̃�01 =



2)(𝜎𝜖2


2)𝑞(𝜎𝜖2


2)𝑝(𝜎𝜖2







2)(𝜎𝜖2


2)𝑞(𝜎𝜖2


2)𝑝(𝜎𝜖2



exp{−1




−1

𝑀𝑛𝑞𝑝𝑇 −𝑀𝑛𝑞𝑝 [

𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 )𝑌𝑛𝑞𝑝}. (16)

Then,

log�̃�01 = 1

2log

(𝑛𝑞𝑝)𝑛𝑞𝑝(𝜎𝜖2


2)(𝜎𝜖2


2)𝑞(𝜎𝜖2


2)𝑝(𝜎𝜖2




2)

𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)

𝑛𝑞𝑝1+𝑞+𝑝+𝑞𝑝(𝜎𝜖2


2)(𝜎𝜖2


2)𝑞(𝜎𝜖2


2)𝑝(𝜎𝜖2



− 1




−1


𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 )𝑌𝑛𝑞𝑝

2 log�̃�01 = log(𝑛𝑞𝑝)𝑛𝑞𝑝(

𝜎𝜖2


2)


𝑛𝑞𝑝1+𝑞+𝑝+𝑞𝑝(𝜎𝜖2)𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)

− 1

𝑛𝑞𝑝2𝑌𝑛𝑞𝑝𝑇 (𝑀𝑛𝑞𝑝 [

𝜎𝜖2


−1


−𝑀𝑛𝑞𝑝 [𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 )𝑌𝑛𝑞𝑝

→ 2 log�̃�01 = log(𝑛𝑞𝑝)𝑛𝑞𝑝(

𝜎𝜖2



𝑛𝑞𝑝1+𝑞+𝑝+𝑞𝑝(𝜎𝜖2)𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)

−1

(𝑛𝑞𝑝)2 {𝑌𝑛𝑞𝑝

𝑇 𝑀𝑛𝑞𝑝 {[𝜎𝜖2


−1

− [𝜎𝜖2

𝑛𝑞𝑝𝐼𝑛𝑞𝑝 +

𝐷𝑛𝑞𝑝]−1

}𝑀𝑛𝑞𝑝𝑇 𝑌𝑛𝑞𝑝}

= log(𝑛𝑞𝑝)𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)(

𝜎𝜖2


2)



−1

(𝑛𝑞𝑝)2{𝑌𝑛𝑞𝑝𝑇 𝑀𝑛𝑞𝑝




Vol.4, No.11, 2014

149

[ 0 01×𝑞 01×𝑝 01×𝑞𝑝 01×𝑛𝑞

0𝑞×1 0𝑞×𝑞 0𝑞×𝑝 0𝑞×𝑞𝑝 0𝑞×𝑛𝑞

0𝑝×1

0𝑞𝑝×1

0𝑛𝑞×1

0𝑝×𝑞

0𝑞𝑝×𝑞

0𝑛𝑞×𝑞

0𝑝×𝑝 0𝑝×𝑞𝑝 0𝑞×𝑛𝑞 0𝑞𝑝×𝑝 0𝑞𝑝×𝑞𝑝 0𝑞𝑝×𝑛𝑞

0𝑛𝑞×𝑝 0𝑛𝑞×𝑞𝑝 ( (𝑛𝑞𝑝)2𝜎𝛿

2

(𝜎𝜖2+𝑛𝑞𝑝𝜎𝛿

2)𝜎𝜖2 )𝐼𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)]

𝑛𝑞𝑝×𝑛𝑞𝑝

𝑀𝑛𝑞𝑝𝑇 𝑌𝑛𝑞𝑝}

= log(𝜎𝜖2+𝑛𝑞𝑝𝜎𝛿

2

𝜎𝜖2 )𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝) − {𝑌𝑛𝑞𝑝

𝑇 𝑄𝑛𝑞𝑝𝑌𝑛𝑞𝑝}

where

𝑄𝑛𝑞𝑝 = 1

(𝑛𝑞𝑝)2𝑀𝑛𝑞𝑝

[ 0 01×𝑞 01×𝑝 01×𝑞𝑝 01×𝑛𝑞


0𝑝×1

0𝑞𝑝×1

0𝑛𝑞×1

0𝑝×𝑞

0𝑞𝑝×𝑞

0𝑛𝑞×𝑞



2



𝑛𝑞𝑝×𝑛𝑞𝑝

𝑀𝑛𝑞𝑝𝑇 .

The Bayes factor for testing the model in (5) satisfy is consistent if [6]

𝑙𝑖𝑚𝑛𝑞𝑝→∞ 𝐵01 = ∞ , 𝑎𝑙𝑚𝑜𝑠𝑡 𝑠𝑢𝑟𝑒𝑙𝑦 𝑖𝑛 𝑝𝑟𝑜𝑏𝑎𝑏𝑙𝑖𝑡𝑦 𝑃0∞ under 𝐻0, (17)

𝑙𝑖𝑚𝑛𝑞𝑝→∞ 𝐵01 = 0 , 𝑎𝑙𝑚𝑜𝑠𝑡 𝑠𝑢𝑟𝑒𝑙𝑦 𝑖𝑛 𝑝𝑟𝑜𝑏𝑎𝑏𝑙𝑖𝑡𝑦 𝑃1∞ under 𝐻1, (18)

Lemma(2):let 𝑌𝑛𝑞𝑝 = (𝑦111 , … , 𝑦𝑛𝑞𝑝)𝑇 where 𝑦𝑖𝑗𝑘 ′𝑠 are independent normal random variable

with mean μ + τj + γk+ (τγ)jk and variance 𝜎𝜖

2 = 1, then there exist a positive constant 𝐶1 such that 1

𝑆𝑛𝑞𝑝,1[∑ log(1 + 𝑛𝑞𝑝𝜎𝛿

2) − 𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)𝑖=1 𝑌𝑛𝑞𝑝

𝑇 𝑄𝑛𝑞𝑝𝑌𝑛𝑞𝑝] > 𝐶1 , with probability tending to 1, i.e. log�̃�01

𝑛𝑞𝑝→∞→ ∞, in probability, where Snqp,1 = ∑

nqpσδ2

(1+nqpσδ2)

nqpi=1 .

Proof:

Let Xnqp = ( X111, X121, … , Xnqp) are independent standard normal random variables, Note that Ynqp Xnqp=d +

𝑋𝛽 and

YnqpT QnqpYnqp = (Xnqp + 𝑋𝛽)

TQnqp (Xnqp + 𝑋𝛽)

= XnqpT QnqpXnqp + (𝑋𝛽)

TQnqp(𝑋𝛽) + XnqpT Qnqp(𝑋𝛽) + (𝑋𝛽)

TQnqpXnqp.

By the property of the design matrix (4) get

(𝑋𝛽)TQnqp = (𝑋𝛽)T 1

(nqp)2 Mnqp

[ 0 01×𝑞 01×𝑝 01×𝑞𝑝 01×𝑛𝑞


0𝑝×1

0𝑞𝑝×1

0𝑛𝑞×1

0𝑝×𝑞

0𝑞𝑝×𝑞

0𝑛𝑞×𝑞



2



MnqpT =

1

(𝑛𝑞𝑝)2[𝜇, 𝜏, 𝛾, (𝜏𝛾)]𝑛𝑞𝑝 (𝐼1+𝑞+𝑝+𝑞𝑝 , 0𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)

[ 0 01×𝑞 01×𝑝 01×𝑞𝑝 01×𝑛𝑞


0𝑝×1

0𝑞𝑝×1

0𝑛𝑞×1

0𝑝×𝑞

0𝑞𝑝×𝑞

0𝑛𝑞×𝑞


0𝑛𝑞×𝑝 0𝑛𝑞×𝑞𝑝 ( (𝑛𝑞𝑝)

2𝜎𝛿2

(𝜎𝜖2+𝑛𝑞𝑝𝜎

𝛿

2)𝜎𝜖2 )𝐼𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)

]

𝑛𝑞𝑝(𝐼1+𝑞+𝑝+𝑞𝑝 , 0𝑛𝑞−(1+𝑞+𝑝+𝑞𝑝) [𝜇, 𝜏, 𝛾, (𝜏𝛾)]𝑇 = 0.

Then YnqpT QnqpYnqp = Xnqp

T QnqpXnqp.

Now by using the expectation formula of the quadratic form[4],[10],[12],[14], get

0 < 𝐸(XnqpT QnqpXnqp) = tr(Qnqp) = ∑

nqpσδ2

(1+nqpσδ2)

nqpi=1 = Snqp,1

Note that




Vol.4, No.11, 2014

150

Qnqp2 =

1

(nqp)4Mnqp

[ 0 01×𝑞 01×𝑝 01×𝑞𝑝 01×𝑛𝑞


0𝑝×1

0𝑞𝑝×1

0𝑛𝑞×1

0𝑝×𝑞

0𝑞𝑝×𝑞

0𝑛𝑞×𝑞



2



MnqpT Mnqp

[ 0 01×𝑞 01×𝑝 01×𝑞𝑝 01×𝑛𝑞


0𝑝×1

0𝑞𝑝×1

0𝑛𝑞×1

0𝑝×𝑞

0𝑞𝑝×𝑞

0𝑛𝑞×𝑞



2𝜎𝛿2


𝛿


]

MnqpT

= 1

(nqp)3Mnqp

[ 0 01×𝑞 01×𝑝 01×𝑞𝑝 01×𝑛𝑞


0𝑝×1

0𝑞𝑝×1

0𝑛𝑞×1

0𝑝×𝑞

0𝑞𝑝×𝑞

0𝑛𝑞×𝑞



2


2)𝜎𝜖2 )

2𝐼𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝) ]

MnqpT

and

tr(Qnqp2 ) =

1

(nqp)2∑ (

(nqp)2σδ2

1+nqpσδ2)2nqp

i=1 = ∑ (nqpσδ

2

1+nqpσδ2)2 = nqp3(

σδ2

1+nqpσδ2)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 = 2∑ (nqpσδ

2

1+nqpσδ2)2nqp

i=1 .

Let cnqp = Snqp,3−Snqp,1

2Snqp,1 , where Snqp,3 = ∑ log(1 + 𝑛𝑞𝑝𝜎𝛿

2)𝑛𝑞𝑝𝑖=1 = 𝑛𝑞𝑝 log(1 + 𝑛𝑞𝑝𝜎𝛿

2) . Then,

Pr { 1

Snqp,1[∑ log(1 + nqpσδ

2)2 − nqp−(1+q+p+qp)i=1 Ynqp

T QnqpYnqp] ≤ cnqp } = Pr [ {YnqpT QnqpYnqp}

Snqp,1 ≥

Snqp,3

Snqp,1−

cnqp ]

= Pr [ {YnqpT QnqpYnqp}

Snqp,1−

E(YnqpT QnqpYnqp)

Snqp,1≥Snqp,3

Snqp,1− cnqp −

E(YnqpT QnqpYnqp)

Snqp,1 ]

= Pr [ 1

Snqp,1{(Ynqp

T QnqpYnqp) − E(YnqpT QnqpYnqp)} ≥

Snqp,3−E(YnqpT QnqpYnqp)

Snqp,1− cnqp ]

≤1

Snqp,12

var(YnqpT QnqpYnqp)

(Snqp,2−Snqp,1

2Snqp,1)2

= 8Snqp,2

(Snqp,3−Snqp,1)2

nqp→∞→ 0, from the Chebyshev inequality[13] .

Since Snqp,3 − 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[∑ log(1 + nqpσδ

2)2 − nqp−(1+q+p+qp)i=1 Ynqp

T QnqpYnqp] > C1, with probability tending to 1.

Then,

2logB̃01 = ∑ log(1 + nqpσδ2)2 −

nqp−(1+q+p+qp)i=1 Ynqp

T QnqpYnqpnqp→∞→ ∞ in probability under H0 i.e.

logB̃01nqp→∞→ ∞

Lemma 3: Suppose we have the one- way repeated measurements model, then

1

𝑆𝑛𝑞𝑝,1𝑡𝑟 (

1


𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 − [𝑋 𝑍] [

𝜎𝜖2


−1

[𝑋 𝑍]𝑇)𝑛𝑞𝑝→∞→ 0

Proof:

Since[𝑋 𝑍] [𝜎𝜖2


𝑇 −𝑀𝑛𝑞𝑝 [𝜎𝜖2


𝑇 is the covariance matrix of

𝜉∗ = 𝛿𝐾∗ = 𝑍𝑠𝑏, 𝐾∗ = 𝑛𝑞𝑝 − (1 + 𝑞 + 𝑝 + 𝑞𝑝), … , 𝑛𝑞𝑝 + 𝑠, it is a positive definite matrix. Thus

1


𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 −

1

𝑛𝑞𝑝2[𝑋 𝑍] [

𝜎𝜖2


−1

[𝑋 𝑍]𝑇is also a positive definite matrix.

Thus, 𝑡𝑟 (1


𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 ) ≥ 𝑡𝑟(

1


𝜎𝜖2


−1

[𝑋 𝑍]𝑇 ), also




Vol.4, No.11, 2014

151

𝑡𝑟 (𝑍𝑠𝑇 1


𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 ) ≤ √𝑡𝑟(𝑍𝑠

𝑇𝑍𝑠)𝑡𝑟 (1


𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 )

= ‖𝑍𝑠‖ ‖1


𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 ‖ by the Cauchy-Schwarz inequality, where

‖1


𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 ‖ = { 𝑡𝑟 ((

1


𝜎𝜖2


𝑇 )−2)}1

2

= [1

𝑛𝑞𝑝2{ ((

𝜎𝜖2


2) + ∑ (𝜎𝜖2


2) + ∑ (𝜎𝜖2


2) + ∑ (𝜎𝜖2


2 )𝑞𝑝𝑖=1

𝑝𝑖=1

𝑞𝑖=1 )

−2

+

∑ (𝜎𝜖2


2)−2

𝑛𝑞𝑝−(1+𝑞=𝑝+𝑞𝑝)𝑖=1 }]

1

2

.

Since

((𝜎𝜖2


2) + ∑ (𝜎𝜖2


2) + ∑ (𝜎𝜖2


2) + ∑ (𝜎𝜖2


2 )𝑞𝑝𝑖=1

𝑝𝑖=1

𝑞𝑖=1 )

−2

= 1

((𝜎𝜖2


2)+𝑞(𝜎𝜖2


2)+𝑝(𝜎𝜖2


2)+𝑞𝑝(𝜎𝜖2


2 ))

2 =

𝑛𝑞𝑝2

((𝜎𝜖2+𝑛𝑞𝑝𝜎𝜇

2)+𝑞(𝜎𝜖2+𝑛𝑞𝑝𝜎𝜏

2)+𝑝(𝜎𝜖2+𝑛𝑞𝑝𝜎𝛾

2)+𝑞𝑝(𝜎𝜖2+𝑛𝑞𝑝𝜎(𝜏𝛾)

2 ))2 ≤

𝑛𝑞𝑝2

((𝑛𝑞𝑝𝜎𝜇2)+𝑞(𝑛𝑞𝑝𝜎𝜏

2)+𝑝(𝑛𝑞𝑝𝜎𝛾2)+𝑞𝑝(𝑛𝑞𝑝𝜎(𝜏𝛾)

2 ))2

= 1

((𝜎𝜇2)+𝑞(𝜎𝜏

2)+𝑝(𝜎𝛾2)+𝑞𝑝(𝜎(𝜏𝛾)

2 ))2 = 𝑂(1), and

∑ (𝜎𝜖2


2)−2

𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)𝑖=1 =


(𝜎𝜖2


2)2 ≤

𝑛𝑞𝑝

(𝜎𝜖2


2)2 =

(𝑛𝑞𝑝)3


2)2 ≤

(𝑛𝑞𝑝)3

(𝑛𝑞𝑝𝜎𝛿2)2 = 𝑂(𝑛𝑞𝑝).

Then, ‖1


𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 ‖ = 𝑂{(𝑛𝑞𝑝)

1

2 }.

Since 𝑠𝑢𝑝‖𝑍𝑠‖ = 1 from the distances between the elements of design matrix 𝑍, then ‖𝑍𝑠‖2 ≤ 𝑛𝑞𝑝.

Thus , 1


1


𝜎𝜖2


−1


1


𝜎𝜖2


−1

[𝑋 𝑍]𝑇) =

1


1


𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 {

1


𝜎𝜖2


𝑇 −1


𝜎𝜖2


𝐷𝑛𝑞𝑝]𝑀𝑛𝑞𝑝𝑇 } [𝑋 𝑍] [

𝜎𝜖2


−1

[𝑋 𝑍]𝑇 )

= 1

𝑆𝑛𝑞𝑝,1∑ 𝜎𝛿

2(𝑍𝐾∗𝑇 1


𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 (𝑍𝐾∗

𝑇 1


𝜎𝜖2


𝑛𝑞𝑝+𝑠𝐾∗=𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)

𝐺𝑛𝑞𝑝]−1

[𝑋 𝑍]𝑇)𝑇) ≤ 𝜎𝛿2

𝑆𝑛𝑞𝑝,1‖

1


𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 ‖

2

∑ ‖𝑧𝐾∗‖2𝑛𝑞𝑝+𝑠

𝐾∗=𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)

≤𝑛𝑞𝑝𝜎𝛿

2

𝑛𝑞𝑝𝑆𝑛𝑞𝑝,1∑ 1𝑛𝑞𝑝+𝑠𝐾∗=𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝) ≤

𝐾∗𝜎𝛿2

𝑛𝑞𝑝 𝑆𝑛𝑞𝑝,1=

𝐾∗𝜎𝛿2

(𝑛𝑞𝑝)2𝜎𝛿2

(1+𝑛𝑞𝑝𝜎𝛿2)

=𝐾∗(1+𝑛𝑞𝑝𝜎𝛿

2)

(𝑛𝑞𝑝)2

Thus

1


1


𝜎𝜖2


−1


1


𝜎𝜖2


−1

[𝑋 𝑍]𝑇) ≤ 𝐾∗(1+𝑛𝑎𝑝𝜎𝛿

2)

(𝑛𝑞𝑝)2

𝑛𝑞𝑝→∞→ 0

Theorem(1): Suppose that the true model is the one- way repeated measurements model with fixed effect

only (the model under 𝐻0 in 5), and let 𝑃0𝑛𝑞𝑝

denote the true distribution of the data, with probability density

function 𝑃0( yijk|μ, τj, γk, (τγ)jk) = 𝛷{(𝑦𝑖𝑗𝑘 − μ − τj − γk− (τγ)jk)/ 𝜎𝜖} , where 𝛷(∙) is the standard normal

density, then, the Bayes factor is consistent under the null hypothesis 𝐻0 i.e

lim𝑛𝑞𝑝→∞ 𝐵01 = ∞ in probability 𝑃0𝑛𝑞𝑝

.

Proof:

log 𝐵01 = 1

2log

det (𝜎𝜖2𝐼𝑛𝑞𝑝+𝑋∑0𝑋

𝑇+𝑍∑1𝑍𝑇)


𝑇)−1

2𝑌𝑛𝑞𝑝𝑇 ( (𝜎𝜖

2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋𝑇)−1−(𝜎𝜖

2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋𝑇 + 𝑍∑1𝑍

𝑇)−1)𝑌𝑛𝑞𝑝

=1

2log




𝑇+𝑍𝑟∑1,𝑟𝑍𝑟𝑇)+

1

2log


𝑇+𝑍𝑟∑1,𝑟𝑍𝑟𝑇)


𝑇)−1

2𝑌𝑛𝑞𝑝𝑇 (𝜎𝜖

2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋𝑇) −1




Vol.4, No.11, 2014

152

−(𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 + 𝑍𝑟∑1,𝑟𝑍𝑟𝑇)−1)𝑌𝑛𝑞𝑝 −

1

2𝑌𝑛𝑞𝑝𝑇 ((𝜎𝜖

2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋𝑇 + 𝑍𝑟∑1,𝑟𝑍𝑟

𝑇)−1−(𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 +

𝑍∑1𝑍𝑇)−1)𝑌𝑛𝑞𝑝

= log�̃�01 +1

2log




𝑇+𝑍𝑟∑1,𝑟𝑍𝑟𝑇) −

1




𝑇 +

𝑍∑1𝑍𝑇)−1)𝑌𝑛𝑞𝑝

Since (𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 + 𝑍∑1𝑍𝑇) − (𝜎𝜖


𝑇) is a positive definite matrix,

𝑑𝑒𝑡 (𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 + 𝑍∑1𝑍𝑇) ≥ 𝑑𝑒𝑡 (𝜎𝜖


𝑇), thus

1

2log




𝑇+𝑍𝑟∑1,𝑟𝑍𝑟𝑇) ≥ 0

𝐸( 𝑌𝑛𝑞𝑝𝑇 ((𝜎𝜖



𝑇 + 𝑍∑1𝑍𝑇)−1)𝑌𝑛𝑞𝑝) = 𝜎𝜖

2 𝑡𝑟((𝜎𝜖2𝐼𝑛𝑞𝑝 +

𝑋∑0𝑋𝑇 + 𝑍𝑟∑1,𝑟𝑍𝑟

𝑇)−1 − (𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 + 𝑍∑1𝑍𝑇)−1) + (𝑋𝛽)𝑇((𝜎𝜖


𝑇)−1 −

(𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 + 𝑍∑1𝑍𝑇)−1)(𝑋𝛽) ,

since (𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 + 𝑍𝑟∑1,𝑟𝑍𝑟𝑇)−1 − (𝜎𝜖


𝑇)−1 , (𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 + 𝑍𝑟∑1,𝑟𝑍𝑟𝑇)−1

and (𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 + 𝑍∑1𝑍𝑇)−1 are all nonnegative definite matrices, it follows that

0 ≤ (𝑋𝛽)𝑇((𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋


2𝐼𝑛𝑞𝑝 + (𝑋∑0𝑋𝑇 + 𝑍∑1𝑍

𝑇)−1)(𝑋𝛽) ≤ (𝑋𝛽)𝑇


𝑇 + 𝑍𝑟∑1,𝑟𝑍𝑟𝑇)−1 (𝑋𝛽).

By the property of design matrix (4), we get

𝑋𝑇𝑀𝑛𝑞𝑝 = 𝑀𝑛𝑞𝑝𝑇 𝑋 = 𝑛𝑞𝑝 (𝐼1+𝑞+𝑝+𝑞𝑝 , 0𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝).

Now as the proof of lemma (3), we get:

(𝑋𝛽)𝑇(𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 + 𝑍𝑟∑1,𝑟𝑍𝑟𝑇)−1(𝑋𝛽) =

1

(𝑛𝑞𝑝)2(𝑋𝛽)𝑇𝑀𝑛𝑞𝑝 [

𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 (𝑋𝛽)

=1

(𝑛𝑞𝑝)2[𝜇, 𝜏, 𝛾, (𝜏𝛾)]𝑛𝑞𝑝 (𝐼1+𝑞+𝑝+𝑞𝑝 , 0𝑛𝑞−(1+𝑞+𝑝+𝑞𝑝)

[ (𝜎𝜖2


2) 01×𝑞 01×𝑝 01×𝑞𝑝 01×𝑛𝑞

0𝑞×1 (𝜎𝜖2


2)𝐼𝑞 0𝑞×𝑝 0𝑞×𝑞𝑝 0𝑞×𝑛𝑞

0𝑝×1

0𝑞𝑝×1

0𝑛𝑞×1

0𝑝×𝑞

0𝑞𝑝×𝑞

0𝑛𝑞×𝑞

(𝜎𝜖2


2)𝐼𝑝 0𝑝×𝑞𝑝 0𝑝×𝑛𝑞

0𝑞𝑝×𝑝 (𝜎𝜖2


2 )𝐼𝑞𝑝 0𝑞𝑝×𝑛𝑞

0𝑛𝑞×𝑝 0𝑛𝑞×𝑞𝑝 (𝜎𝜖2


2)𝐼𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝) ] −1

𝑛𝑞𝑝

(𝐼1+𝑞+𝑝+𝑞𝑝 , 0𝑛𝑞−(1+𝑞+𝑝+𝑞𝑝) [𝜇, 𝜏, 𝛾, (𝜏𝛾)]𝑇

= 𝜇2 [𝜎𝜖2


2]−1

+ ∑ 𝜏𝑖2𝑞

𝑖=1 [𝜎𝜖2


2]−1

+ ∑ 𝛾𝑖2𝑝

𝑖=1 [𝜎𝜖2


2]−1

+ ∑ (𝜏𝛾)𝑖2𝑞𝑝

𝑖=1 [𝜎𝜖2


2 ]−1

= 𝑂(1).

Since 𝑝𝑟 { 1

𝑆𝑛𝑞𝑝,1𝑌𝑛𝑞𝑝𝑇 ((𝜎𝜖



𝑇 + 𝑍∑1𝑍𝑇)−1)𝑌𝑛𝑞𝑝 > 휀 } ≤

1

𝑆𝑛𝑞𝑝,1 {

𝐸( 𝑌𝑛𝑞𝑝𝑇 ((𝜎𝜖

2𝐼𝑛𝑞𝑝+𝑋∑0𝑋𝑇+𝑍𝑟∑1,𝑟𝑍𝑟

𝑇)−1−(𝜎𝜖2𝐼𝑛𝑞𝑝+𝑋∑0𝑋

𝑇+𝑍∑1𝑍𝑇)−1)𝑌𝑛𝑞𝑝)

}

≤ 1

𝑆𝑛𝑞𝑝,1

𝑡𝑟((𝜎𝜖2𝐼𝑛𝑞𝑝+𝑋∑0𝑋

𝑇+𝑍𝑟∑1,𝑟𝑍𝑟𝑇)−1−(𝜎𝜖

2𝐼𝑛𝑞𝑝+𝑋∑0𝑋𝑇+𝑍∑1𝑍

𝑇)−1

+ 1

𝑆𝑛𝑞𝑝,1

(𝑋𝛽)𝑇(𝜎𝜖2𝐼𝑛𝑞𝑝+𝑋∑0𝑋

𝑇+𝑍𝑟∑1,𝑟𝑍𝑟𝑇)−1 (𝑋𝛽)

𝑛𝑞𝑝→∞→ 0 (by the Markov inequality[13], where 휀 > 0) .




𝑇)−1 is nonnegative definite →

𝑌𝑛𝑞𝑝𝑇 ((𝜎𝜖



𝑇 + 𝑍∑1𝑍𝑇)−1)𝑌𝑛𝑞𝑝 is nonnegative .

Therefore,

−1




𝑇 + 𝑍∑1𝑍𝑇)−1)𝑌𝑛𝑞𝑝 = 𝑜(𝑆𝑛𝑞𝑝,1)

⇒ −

1




𝑇 + 𝑍∑1𝑍𝑇)−1)𝑌𝑛𝑞𝑝

𝑆𝑛𝑞𝑝,1

= 𝑜(1)

Then, as in lemma (2) there exists a constant 𝐶1 such that




Vol.4, No.11, 2014

153

1

𝑆𝑛𝑞𝑝,1log 𝐵01 ≥

1

𝑆𝑛𝑞𝑝,1log �̃�01 + 𝑜(1) > 𝐶1 + 𝑜(1).

That is,

𝐵01 ≥ exp( 𝑆𝑛𝑞𝑝,1{ 𝐶1 + 𝑜(1)}) 𝑛𝑞𝑝→∞→ ∞ in probability 𝑃0

𝑛𝑞𝑝.∎

Lemma(4):

Assuming that the true model is the mixed one-way repeated measurements model (the model under 𝐻1in 5). Let

𝑌𝑛𝑞𝑝 = (𝑦1 , … , 𝑦𝑛𝑞𝑝) where 𝑦𝑖𝑗𝑘 ′𝑠 are independent normal random variables with mean μ+ τj + δi(j) + γk+

(τγ)jk and variance 𝜎𝜖2 = 1 and let 𝑤𝑛𝑞𝑝 = 𝐸(𝑌𝑛𝑞𝑝)

𝑇𝑄𝑛𝑞𝑝 𝐸(𝑌𝑛𝑞𝑝).

Then, lim𝑛𝑞𝑝→∞1

𝑤𝑛𝑞𝑝𝑌𝑛𝑞𝑝𝑇 𝑄𝑛𝑞𝑝𝑌𝑛𝑞𝑝 = 1, in probability 𝑃1

∞.

Proof:

From the moment formula of the quadratic form of normal random variables, the expectation and variance of

quadratic form 𝑌𝑛𝑞𝑝𝑇 𝑄𝑛𝑞𝑝𝑌𝑛𝑞𝑝 are given by:

𝐸(𝑌𝑛𝑞𝑝𝑇 𝑄𝑛𝑞𝑝𝑌𝑛𝑞𝑝 ) = 𝑡𝑟 (𝑄𝑛𝑞𝑝) + 𝐸(𝑌𝑛𝑞𝑝)

𝑇𝑄𝑛𝑞𝑝 𝐸(𝑌𝑛𝑞𝑝)

= ∑𝑛𝑞𝑝𝜎𝛿

2

(1+𝑛𝑞𝑝𝜎𝛿2)

𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)𝑖=1 + 𝐸(𝑌𝑛𝑞𝑝)

𝑇𝑄𝑛𝑞𝑝 𝐸(𝑌𝑛𝑞𝑝)

= 𝑆𝑛𝑞𝑝,1 + 𝑤𝑛𝑞𝑝 , and

𝑣𝑎𝑟(𝑌𝑛𝑞𝑝𝑇 𝑄𝑛𝑞𝑝𝑌𝑛𝑞𝑝 ) = 2𝑡𝑟 (𝑄𝑛𝑞𝑝

2) + 4𝐸(𝑌𝑛𝑞𝑝)𝑇𝑄𝑛𝑞𝑝

2 𝐸(𝑌𝑛𝑞𝑝).

Note

𝑤𝑛𝑞𝑝 = 𝐸(𝑌𝑛𝑞𝑝)𝑇𝑄𝑛𝑞𝑝 𝐸(𝑌𝑛𝑞𝑝)

= (𝜇, 𝜏𝑇 , 𝛾𝑇 , (𝜏𝛾)𝑇 , 𝛿𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)𝑇 )𝑀𝑛𝑞𝑝

𝑇1

𝑛𝑞𝑝2𝑀𝑛𝑞𝑝

[ 0 01×𝑞 01×𝑝 01×𝑞𝑝 01×𝑛𝑞


0𝑝×1

0𝑞𝑝×1

0𝑛𝑞×1

0𝑝×𝑞

0𝑞𝑝×𝑞

0𝑛𝑞×𝑞



2𝜎𝛿2


𝛿


]

𝑀𝑛𝑞𝑝𝑇 𝑀𝑛𝑞𝑝 (𝜇, 𝜏𝑇, 𝛾𝑇, (𝜏𝛾)

𝑇, 𝛿𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)𝑇

)𝑇

= (𝜇, 𝜏𝑇 , 𝛾𝑇 , (𝜏𝛾)𝑇 , 𝛿𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)𝑇 )

[ 0 01×𝑞 01×𝑝 01×𝑞𝑝 01×𝑛𝑞


0𝑝×1

0𝑞𝑝×1

0𝑛𝑞×1

0𝑝×𝑞

0𝑞𝑝×𝑞

0𝑛𝑞×𝑞

0𝑝×𝑝 0𝑝×𝑞𝑝 0𝑞×𝑛𝑞

0𝑞𝑝×𝑝 0𝑞𝑝×𝑞𝑝 0𝑞𝑝×𝑛𝑞


2


2)𝜎𝜖2 ) 𝐼𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)]

(𝜇, 𝜏𝑇 , 𝛾𝑇 , (𝜏𝛾)𝑇 , 𝛿𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)𝑇 )

𝑇

= 𝛿𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)𝑇 (



2)𝜎𝜖2 ) 𝐼𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)𝛿𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝) = ∑

(𝑛𝑞𝑝)2𝜎𝛿2𝛿𝑖2

1+𝑛𝑞𝑝𝜎𝛿2

𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)𝑖=1 ,

since 𝑠𝑢𝑝𝑖 𝛿𝑖 < ∞ , we have

𝑤𝑛𝑞𝑝 ≤ 𝑠𝑢𝑝𝑖 𝛿𝑖2 𝑛𝑞𝑝∑

𝑛𝑞𝑝𝜎𝛿2


𝑛𝑞𝑝𝑖=1 = 𝑂(𝑛𝑞𝑝 𝑆𝑛𝑞𝑝,1), then 𝑤𝑛𝑞𝑝 = 𝑂(𝑛𝑞𝑝

2).

Now we can consider 𝛿𝑚2 > 0 , where 𝑚 ≥ 1 is a fixed positive integer and a sufficiently large 𝑛𝑞𝑝 with

𝑛𝑞𝑝𝜎𝛿2

1+𝑛𝑞𝑝𝜎𝛿2 ≥

1

2 . Then for such 𝑛𝑞𝑝 and 𝑚.

𝑤𝑛𝑞𝑝 = ∑𝑛𝑞𝑝2𝜎𝛿

2𝛿𝑖2


𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)𝑖=1 ≥

𝑛𝑞𝑝2𝜎𝛿2

1+𝑛𝑞𝑝𝜎𝛿2 𝛿𝑚

2

= 𝑛𝑞𝑝 𝑛𝑞𝑝𝜎𝛿

2

1+𝑛𝑞𝑝𝜎𝛿2 𝛿𝑚

2 ≥ 𝑛𝑞𝑝𝛿𝑚2

2.

Similarly to the previous calculation, we have

𝐸(𝑌𝑛𝑞𝑝)𝑇𝑄𝑛𝑞𝑝2 𝐸(𝑌𝑛𝑞𝑝) = (𝜇, 𝜏

𝑇 , 𝛾𝑇 , (𝜏𝛾)𝑇 , 𝛿𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)𝑇 )𝑀𝑛𝑞𝑝

𝑇1

(𝑛𝑞𝑝)3𝑀𝑛𝑞𝑝




Vol.4, No.11, 2014

154

[ 0 01×𝑞 01×𝑝 01×𝑞𝑝 01×𝑛𝑞


0𝑝×1

0𝑞𝑝×1

0𝑛𝑞×1

0𝑝×𝑞

0𝑞𝑝×𝑞

0𝑛𝑞×𝑞



2𝜎𝛿2


𝛿

2)𝜎𝜖2 )2

𝐼𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)]

𝑀𝑛𝑞𝑝𝑇 𝑀𝑛𝑞𝑝 (𝜇, 𝜏𝑇, 𝛾𝑇, (𝜏𝛾)

𝑇, 𝛿𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)𝑇

)𝑇

=1

𝑛𝑞𝑝𝛿𝑛𝑞𝑝−((1+𝑞+𝑝+𝑞𝑝)𝑇 (



2)𝜎𝜖2 )

2𝐼𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)𝛿𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)

=1

𝑛𝑞𝑝∑


(1+𝑛𝑞𝑝𝜎𝛿2)2

𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)𝑖=1 .

Since 𝐸(𝑌𝑛𝑞𝑝)𝑇𝑄𝑛𝑞𝑝2 𝐸(𝑌𝑛𝑞𝑝) ≤ ∑



𝑛𝑞𝑝𝑖=1 ≤ 𝑛𝑞𝑝 𝑠𝑢𝑝𝑖 𝛿𝑖

2∑(𝑛𝑞𝑝)2𝜎𝛿

4


𝑛𝑞𝑝𝑖=1 = 𝑂( 𝑛𝑞𝑝𝑆𝑛𝑞𝑝,2) (as in case

of 𝐸(𝑌𝑛𝑞𝑝)𝑇𝑄𝑛𝑞𝑝 𝐸(𝑌𝑛𝑞𝑝)).

Now for previous 𝑛𝑞𝑝 and 𝑚 we can find

𝐸(𝑌𝑛𝑞𝑝)𝑇𝑄𝑛𝑞𝑝2 𝐸(𝑌𝑛𝑞𝑝) = ∑



𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)𝑖=1 ≥ 𝑛𝑞𝑝 𝛿𝑚

2 (𝑛𝑞𝑝𝜎𝛿

2

1+𝑛𝑞𝑝𝜎𝛿2)2

≥ 𝛿𝑚2

4 𝑛𝑞𝑝.

By combining value of 𝑆𝑛𝑞𝑝,1 and the previous result, get

𝑣𝑎𝑟(𝑌𝑛𝑞𝑝𝑇 𝑄𝑛𝑞𝑝𝑌𝑛𝑞𝑝 ) ≤ 𝑂(𝑛𝑞𝑝 𝑆𝑛𝑞𝑝,1).

Furthermore, note that 𝐸(𝑌𝑛𝑞𝑝)𝑇𝑄𝑛𝑞𝑝 𝐸(𝑌𝑛𝑞𝑝) ≥ 𝐸(𝑌𝑛𝑞𝑝)

𝑇𝑄𝑛𝑞𝑝

2 𝐸(𝑌𝑛𝑞𝑝).

By the Chebshev inequality[13] get

𝑝𝑟 {|1

𝑤𝑛𝑞𝑝𝑌𝑛𝑞𝑝𝑇 𝑄𝑛𝑞𝑝𝑌𝑛𝑞𝑝 −

𝐸(𝑌𝑛𝑞𝑝𝑇 𝑄𝑛𝑞𝑝𝑌𝑛𝑞𝑝)

𝑤𝑛𝑞𝑝| > 휀 } ≤

𝑣𝑎𝑟(𝑌𝑛𝑞𝑝𝑇 𝑄𝑛𝑞𝑝𝑌𝑛𝑞𝑝)

𝑤𝑛𝑞𝑝2 2 = 𝑂(𝑛𝑞𝑝−2)

𝑛𝑞𝑝→∞→ 0, where 휀 > 0 .

Since lim𝑛𝑞𝑝→∞𝐸(𝑌𝑛𝑞𝑝

𝑇 𝑄𝑛𝑞𝑝𝑌𝑛𝑞𝑝)

𝑤𝑛𝑞𝑝= lim𝑛𝑞𝑝→∞ (

𝑆𝑛𝑞𝑝,1

𝑤𝑛𝑞𝑝+𝑤𝑛𝑞𝑝

𝑤𝑛𝑞𝑝) = 1

∴ lim𝑛𝑞𝑝→∞1

𝑤𝑛𝑞𝑝𝑌𝑛𝑞𝑝𝑇 𝑄𝑛𝑞𝑝𝑌𝑛𝑞𝑝 = 1 , in probability ∎

Lemma(5):

Suppose we have the mixed one- way repeated measurements model, then

lim𝑛𝑞𝑝→∞1

𝑤𝑛𝑞𝑝log

det(𝜎𝜖2𝐼𝑛𝑞𝑝+𝑋∑0𝑋



𝑇+𝑍𝑟∑1,𝑟𝑍𝑟𝑇)= 0

Proof:

Since det(𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 + 𝑍∑1𝑍𝑇) ≤ ∏ (𝜎𝜖

2 + 𝜎𝜇2𝑥𝑖2 + ∑ 𝜎𝜏

2𝑥𝑖𝑘2 + ∑ 𝜎𝛾

2𝑥𝑖𝑘2 + ∑ 𝜎(𝜏𝛾)

2𝑥𝑖𝑘2𝑞𝑝

𝑘=1

𝑝

𝑘=1

𝑞

𝑘=1 +𝑛𝑞𝑝𝑖=1

∑ 𝜎𝛿2𝑛𝑞

𝑘=1 𝑥𝑖𝑘2 ) (from the property of the determinant of the positive definite matrix)

= (𝑛𝑞𝑝)𝑛𝑞𝑝∏ (𝜎𝜖2

𝑛𝑞𝑝+(𝜎𝜇2𝑥𝑖2+∑ 𝜎𝜏

2𝑥𝑖𝑘2 +∑ 𝜎𝛾

2𝑥𝑖𝑘2 +∑ 𝜎(𝜏𝛾)

2 𝑥𝑖𝑘2𝑞𝑝

𝑘=1𝑝𝑘=1

𝑞𝑘=1 +∑ 𝜎𝛿

2𝑛𝑞𝑘=1 𝑥𝑖𝑘

2

𝑛𝑞𝑝 )

𝑛𝑞𝑝𝑖=1 .

And we have


𝑇 + 𝑍𝑟∑1,𝑟𝑍𝑟𝑇) = (𝑛𝑞𝑝)𝑛𝑞𝑝(

𝜎𝜖2


2)(𝜎𝜖2


2)𝑞(𝜎𝜖2


2)𝑝(𝜎𝜖2



𝑛𝑞𝑝+

𝜎𝛿2)𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝) (from lemma 1).

→ logdet(𝜎𝜖

2𝐼𝑛𝑞𝑝+𝑋∑0𝑋𝑇+𝑍∑1𝑍

𝑇)


𝑇+𝑍𝑟∑1,𝑟𝑍𝑟𝑇)= logdet(𝜎𝜖


𝑇) − log det(𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 +

𝑍𝑟∑1,𝑟𝑍𝑟𝑇)

≤ {(𝑛𝑞𝑝)log(𝑛𝑞𝑝) + ∑ log (𝜎𝜖2

𝑛𝑞𝑝+(𝜎𝜇2𝑥𝑖2+∑ 𝜎𝜏

2𝑥𝑖𝑘2 +∑ 𝜎𝛾

2𝑥𝑖𝑘2 +∑ 𝜎(𝜏𝛾)

2 𝑥𝑖𝑘2𝑞𝑝

𝑘=1𝑝𝑘=1

𝑞𝑘=1 +∑ 𝜎𝛿

2𝑛𝑞𝑘=1 𝑥𝑖𝑘

2

𝑛𝑞𝑝 )

𝑛𝑞𝑝𝑖=1 } − {𝑛𝑞𝑝log(𝑛𝑞𝑝) +

log (𝜎𝜖2


2) + 𝑞log (𝜎𝜖2


2) + 𝑝log (𝜎𝜖2


2) + 𝑞𝑝log (𝜎𝜖2


2 ) + 𝑛𝑞𝑝 − (1 + 𝑞 + 𝑝 +

𝑞𝑝)log (𝜎𝜖2


2)} = 𝑂(1) − 𝑂(1) = 𝑂(1).

Then

1

𝑤𝑛𝑞𝑝log





𝑛𝑞𝑝→∞→ 0

Theorem(2):




Vol.4, No.11, 2014

155

Suppose that the true model is the mixed one- way repeated measurements model (the model under 𝐻1in 5),

and let 𝑃1𝑛𝑞𝑝

denote the true distribution of the data, with probability density function

𝑝1(yijk|μτj, δi(j), γk, (τγ)jk) = 𝛷{(yijk − (μ + τj + δi(j) + γk+ (τγ)jk))/𝜎𝜖}, where 𝛷(∙) is the standard normal

density. Then the Bayes factor is consistent under the alternative hypothesis 𝐻1 i.e

lim𝑛𝑞𝑝→∞ 𝐵01 = 0 in probability 𝑃1𝑛𝑞𝑝

.

Proof:

log 𝐵01 = 1

2log




𝑇)−1


2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋𝑇) −1−(𝜎𝜖


𝑇)−1)𝑌𝑛𝑞𝑝

=

1

2log




𝑇+𝑍𝑟∑1,𝑟𝑍𝑟𝑇)+

1

2log




𝑇)−1



2𝐼𝑛𝑞𝑝 +

𝑋∑0𝑋𝑇 + 𝑍𝑟∑1,𝑟𝑍𝑟

𝑇)−1)𝑌𝑛𝑞𝑝 −

1



𝑇)−1

−(𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 + 𝑍∑1𝑍𝑇)−1)𝑌𝑛𝑞𝑝

= log�̃�01 −1




𝑇)−1)𝑌𝑛𝑞𝑝 −

1

2𝑌𝑛𝑞𝑝𝑇

((𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 + 𝑍𝑟∑1,𝑟𝑍𝑟𝑇)−1−(𝜎𝜖


𝑇)−1)𝑌𝑛𝑞𝑝 , where

log�̃�01 =

1

2log




𝑇)−1




𝑇)−1)𝑌𝑛𝑞𝑝 .

From lemma (1), we have

det(𝜎𝜖

2𝐼𝑛𝑞𝑝+𝑋∑0𝑋𝑇+𝑍𝑟∑1,𝑟𝑍𝑟

𝑇)


𝑇)=

(𝑛𝑞𝑝)𝑛𝑞𝑝(𝜎𝜖2


2)(𝜎𝜖2


2)𝑞(𝜎𝜖2


2)𝑝(𝜎𝜖2





𝑛𝑞𝑝1+𝑞+𝑝+𝑞𝑝(𝜎𝜖2


2)(𝜎𝜖2


2)𝑞(𝜎𝜖2


2)𝑝(𝜎𝜖2



=𝑛𝑞𝑝𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)(

𝜎𝜖2


2)


𝜎𝜖2𝑛𝑞𝑝−(1+𝑞+𝑝+𝑞𝑝)

= (𝜎𝜖2+𝑛𝑞𝑝𝜎𝛿

2

𝜎𝜖2 )


=(1 +𝑛𝑞𝑝𝜎𝛿

2

𝜎𝜖2 )


= 𝑂(𝑛𝑞𝑝)

⟹ log (1 +𝑛𝑞𝑝𝜎𝛿

2

𝜎𝜖2 )


= 𝑂(𝑛𝑞𝑝).

And from lemma (4) 𝑤𝑛𝑞𝑝 = 𝑂(𝑛𝑞𝑝2). Then

1

𝑤𝑛𝑞𝑝log




𝑇)= 𝑂(𝑛𝑞𝑝−1)

𝑛𝑞𝑝→∞→ 0.

Then,

1

𝑤𝑛𝑞𝑝log�̃�01 =

1

2(

1

𝑤𝑛𝑞𝑝 log

det(𝑉1,𝑛𝑞𝑝)


𝑇)−

1

𝑤𝑛𝑞𝑝𝑌𝑛𝑞𝑝𝑇 ((𝜎𝜖


2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋𝑇 +

𝑍𝑟∑1,𝑟𝑍𝑟𝑇)−1)𝑌𝑛𝑞𝑝).


𝑇) −1 − (𝜎𝜖2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋

𝑇 + 𝑍𝑟∑1,𝑟𝑍𝑟𝑇)−1= 𝑄𝑛𝑞𝑝, then

1




𝑇)−1)𝑌𝑛𝑞𝑝 =

1

𝑤𝑛𝑞𝑝𝑌𝑛𝑞𝑝𝑇 𝑄𝑛𝑞𝑝𝑌𝑛𝑞𝑝. from lemma

(4), lim𝑛𝑞𝑝→∞1

𝑤𝑛𝑞𝑝𝑌𝑛𝑞𝑝𝑇 𝑄𝑛𝑞𝑝𝑌𝑛𝑞𝑝 = 1 , then

1

𝑤𝑛𝑞𝑝log�̃�01

𝑛𝑞𝑝→∞→ =

1

2(

1

𝑤𝑛𝑞𝑝 log




𝑇)−

1



2𝐼𝑛𝑞𝑝 + 𝑋∑0𝑋𝑇 +

𝑍𝑟∑1,𝑟𝑍𝑟𝑇)−1)𝑌𝑛𝑞𝑝)

𝑛𝑞𝑝→∞→ =

1

2(0 − 1) =

−1

2

∴1

𝑤𝑛𝑞𝑝log�̃�01

𝑛𝑞𝑝→∞→

−1

2 , in probability 𝑃1

𝑛𝑞𝑝.

Now from lemma (5),

1

𝑤𝑛𝑞𝑝 1

2log




𝑇+𝑍𝑟∑1,𝑟𝑍𝑟𝑇) 𝑛𝑞𝑝→∞→ 0, and −

1



𝑇)−1−(𝜎𝜖

2𝐼𝑛𝑞𝑝 +

𝑋∑0𝑋𝑇 + 𝑍∑1𝑍

𝑇)−1)𝑌𝑛𝑞𝑝 is nonpositive random variable since (𝜎𝜖


𝑇)−1−


𝑇 + 𝑍∑1𝑍𝑇)−1

is a nonnegative definite matrix.




Vol.4, No.11, 2014

156

Therefore, by combining the above results, there exist 𝐶2 > 0 such that

lim𝑛𝑞𝑝→∞ sup1

𝑤𝑛𝑞𝑝log 𝐵01 ≤

−𝐶2

2, with probability tending to 1.

Then , 𝐵01𝑛𝑞𝑝→∞→ 0, in probability 𝑃1

𝑛𝑞𝑝. ∎

4.Conclusions

1- The posterior distribution 𝜋1(𝜉𝑛𝑞𝑝|𝑌𝑛𝑞𝑝) is the multivariate normal with

𝐸(𝜉𝑛𝑞𝑝|𝑌𝑛𝑞𝑝 , 𝜎𝜖2) = ∑2(∑2 + 𝜎𝜖

2𝐼𝑁)−1𝑌𝑛𝑞𝑝 , where N=1+q+p+qp+nq and

𝑉𝑎𝑟(𝜉𝑛𝑞𝑝|𝑌𝑛𝑞𝑝, 𝜎𝜖2) = (∑2

−1 + (𝜎𝜖2𝐼𝑁)

−1)−1 = ∑2(∑2 + 𝜎𝜖2𝐼𝑁 )

−1𝜎𝜖2𝐼𝑁, where

𝜉𝑖𝑗𝑘 = μ+ τj + δi(j) + γk+ (τγ)jk + eijk,for, 𝑖 = 1,2, … , 𝑛, 𝑗 = 1,… , 𝑞 𝑎𝑛𝑑 𝑘 = 1,… , 𝑝, and

∑2 = 𝑋∑0𝑋𝑇 + 𝑍∑1𝑍

𝑇 = 𝑋

[

𝜎𝜇2 01×𝑞 01×𝑝


0𝑝×1 0𝑝×𝑞 𝜎𝛾2𝐼𝑝×𝑝



𝑋𝑇 + 𝑍

[ 𝜎𝛿112 0 ⋯ 0

0 𝜎𝛿122 ⋯ 0

⋮0

⋮0


2]

𝑍𝑇

2-The Bayes factor for testing problem 𝐻0: yijk = μ + τj + γk + (τγ)jk + eijk versus

𝐻1: yijk= μ + τj + δi(j) + γk + (τγ)jk + eijk is given by:

𝐵01 =



2)(𝜎𝜖2


2)𝑞

(𝜎𝜖2


2)𝑝

(𝜎𝜖2


2 )𝑞𝑝

(𝜎𝜖2


2)𝑛𝑞



2)(𝜎𝜖2


2)𝑞(𝜎𝜖2


2)𝑝(𝜎𝜖2



exp{ −1




−1


𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 )𝑌𝑛𝑞𝑝}.

3-The approximation of 𝐵01 with design matrix of the random effects (𝑍) to the first 𝑛𝑞𝑝 terms as following

�̃�01 =



2)(𝜎𝜖2


2)𝑞(𝜎𝜖2


2)𝑝(𝜎𝜖2







2)(𝜎𝜖2


2)𝑞(𝜎𝜖2


2)𝑝(𝜎𝜖2



exp{−1




−1


𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 )𝑌𝑛𝑞𝑝}.

4- If we have the one- way repeated measurements model, then

1


1


𝜎𝜖2


−1

𝑀𝑛𝑞𝑝𝑇 − [𝑋 𝑍] [

𝜎𝜖2


−1

[𝑋 𝑍]𝑇)𝑛𝑞𝑝→∞→ 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 𝐻0 . This implies that, lim𝑛𝑞𝑝→∞ 𝐵01 = ∞ in probability 𝑃0𝑛𝑞𝑝

.

6- If we have the mixed one- way repeated measurements model, then

lim𝑛𝑞𝑝→∞1

𝑤𝑛𝑞𝑝log




𝑇+𝑍𝑟∑1,𝑟𝑍𝑟𝑇)= 0.

7- If the true model is the mixed one- way repeated measurements model, then the Bayes factor is consistent

under the alternative hypothesis 𝐻1.This implies that, lim𝑛𝑞𝑝→∞ 𝐵01 = 0 in probability 𝑃1𝑛𝑞𝑝

.

5.References

[1] Aerts, M. , Claeskens, G. and Hart, J.D., (2004). Bayesian –motivated 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 –Way Multivariate Repeated Measurements analysis of

Variance Model, Applied Mathematics a Journal of Chinese Universities, 19,4,pp435- 448.




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 –of-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 –of-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

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