Discussiones Mathematicae
Probability and Statistics 36 (2016) 93–113
doi:10.7151/dmps.1182
BEST UNBIASED ESTIMATES FOR PARAMETERS OF
THREE-LEVEL MULTIVARIATE DATA WITH DOUBLY
EXCHANGEABLE COVARIANCE STRUCTURE AND
STRUCTURED MEAN VECTOR
Arkadiusz Koziol
Faculty of Mathematics, Computer Science and Econometrics
University of Zielona Góra
Szafrana 4a, 65–516 Zielona Góra, Poland
e-mail: [email protected]
Abstract
In this article author obtain the best unbiased estimators of doubly
exchangeable covariance structure. For this purpose the coordinate freecoordinate approach is used. Considered covariance structure consist of
three unstructured covariance matrices for three-level m−variate observations with equal mean vector over v points in time and u sites under the
assumption of multivariate normality. To prove, that the estimators are
best unbiased, complete statistics are used. Additionally, strong consistency
is proven. Under the proposed model the variances of the estimators of
covariance components are compared with the ones in the model in [11].
Keywords: best unbiased estimator, doubly exchangeable covariance structure, three-level multivariate data, coordinate free approach, structured
mean vector.
2010 Mathematics Subject Classification: 62H12, 62J10, 62F10.
1.
Introduction
In this article doubly exchangeable covariance structure for three-level multivariate observations is considered. In this type of multivariate observations m
dimensional observation vector is repeatedly measured over v time points and u
locations. Additionally, author assumes that the mean vector remains constant
over time points and over sites (locations) which means that vectors µ = 1vu ⊗ µ
with µ = (µ1 , µ2 , . . . , µm ) ∈ Rm .
94
A. Koziol
In this paper, optimal unbiased estimators for fixed and covariance parameters will be constructed from sufficient and complete statistics. These statistics
will be derived using the free coordinate approach (see [2, 6, 14, 16, 4] and [17]).
2.
Doubly exchangeable covariance structure
The (vum × vum)−dimensional doubly exchangeable covariance structure is defined as


Σ0 Σ1 . . . Σ1
 ..
.. 
..
 .
.
. 


Γ =  .
. 
..
 ..
. .. 
Σ1 Σ1 . . . Σ0
(2.1)
= I v ⊗ (Σ0 − Σ1 ) + J v ⊗ Σ1 .
The above doubly exchangeable covariance structure Γ can equivalently be written as follows
(2.2)
Γ = I v ⊗ Σ0 + (J v − I v ) ⊗ Σ1 ,
where I v is the v × v identity matrix, 1v is a v × 1 vector of ones, J v = 1v 10v is
matrix of ones, ⊗ represents the Kronecker product and


Γ0 Γ1 . . . Γ1
 .. . .
.. 
 .
.
. 


Σ0 =  .
. 
..
 ..
. .. 
Γ1 Γ1 . . . Γ0
(2.3)
= I u ⊗ (Γ0 − Γ1 ) + J u ⊗ Γ1 .



Σ1 = 


(2.4)

Γ2 . . . . . . Γ2
.. . .
.. 
.
.
. 

..
.. 
..
. . 
.
Γ2 . . . . . . Γ2
= J u ⊗ Γ2 .
Thus writing the doubly exchangeable covariance structure in terms of gammas
(2.5)
Γ = I v ⊗ I u ⊗ (Γ0 − Γ1 ) + I v ⊗ J u ⊗ (Γ1 − Γ2 ) + J v ⊗ J u ⊗ Γ2 ,
Best unbiased estimates for parameters of three-level ...
95
which can equivalently be written as
(2.6)
Γ = I v ⊗ I u ⊗ Γ0 + I v ⊗ (J u − I u ) ⊗ Γ1 + (J v − I v ) ⊗ J u ⊗ Γ2 .
The last form will be used to build orthogonal with respect to trace of inner
product base for components of matrix Γ.
Γ0 is assumed to be a positive definite symmetric m × m matrix, Γ1 and Γ2
are assumed to be a symmetric m×m matrices, and Γ0 −Γ1 , Γ0 +(u−1)Γ1 −uΓ2 ,
Γ0 + (u − 1)Γ1 + (v − 1)uΓ2 are positive definite matrices, so that the vum × vum
matrix Γ is positive definite for a proof, see [8] and [9] . The m × m block
diagonals Γ0 in Γ represent the variance-covariance matrix of the m response
variables at any given time point and at any given site, whereas the m × m block
off diagonals Γ1 in Γ represent the covariance matrix of the m response variables
at any given time point and between any two sites. The m×m block off diagonals
Γ2 in Γ represent the covariance matrix of the m response variables between any
two time points. In view of form of matix Γ presented in (2.6) is clear that Γ0
is constant for all time points and sites, Γ1 is same between any two sites and
for all time points and Γ2 is assumed to be the same for any pair of time points,
irrespective of the same site or between any two sites.
3.
Best unbiased estimators of µ and Γ
Let y r,ts be a m-variate vector of measurements on the rth individual at the
tth time point and at the sth site; r = 1, . . . , n, t = 1, . . . , v, s = 1, . . . , u.
The n individuals are all independent. Let y r = (y 0r,11 , . . . , y 0r,vu )0 be the vumvariate vector of all measurements corresponding to the rth individual. Finally,
let y 1 , y 2 , . . . , y n be a random sample of size n drawn from the population
Nvum (1vu ⊗ µ, Γ), where E[y r ] = 1vu ⊗ µ ∈ Rvum and Γ is assumed to be a
vum × vum positive definite matrix.
In this section, optimal properties of unbiased estimators for parameters of
the probability distribution of the following column vector
y
= vec( Y 0 ) ∼ N (1nvu ⊗ I m )µ, I n ⊗ Γvum
vum×n
nvum×1
are presented. This means that n independent random column vectors are identically distributed (vum × vum)−dimensional variance covariance matrix
Γ = I v ⊗ I u ⊗ Γ0 + I v ⊗ (J u − I u ) ⊗ Γ1 + (J v − I v ) ⊗ J u ⊗ Γ2 .
Define the projection matrix P as follows
(3.7)
P =
1
1
1
J n ⊗ J v ⊗ J u ⊗ I m.
n
v
u
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A. Koziol
It is clear that P is an orthogonal projector on the subspace of the mean vector
of y. If I n ⊗ I vum ∈ ϑ, from [3] it follows that P y is the best linear unbiased
estimator (BLUE) if and only if P commutes with all covariance matrices V .
Therefore, we have the following results.
Result 1. The projection matrix P commutes with the covariance matrix V ,
i.e., P V = V P , where V = I n ⊗ Γ, the covariance matrix of y.
Proof. Now,
P = (1n ⊗ 1v ⊗ 1u ⊗ I m )(1n ⊗ 1v ⊗ 1u ⊗ I m )+
1 0
1 0
1 0
= (1n ⊗ 1v ⊗ 1u ⊗ I m )
1 ⊗ 1 ⊗ 1 ⊗ Im
n n v v u u
1
1
1
= J n ⊗ J v ⊗ J u ⊗ I m.
n
v
u
Note that
1
1
1
PV =
J n ⊗ J v ⊗ J u ⊗ I m (I n ⊗ Γ)
n
v
u
1
1
1
Jv ⊗ Ju ⊗ Im
= Jn ⊗
n
v
u
(I v ⊗ I u ⊗ Γ0 + I v ⊗ (J u − I u ) ⊗ Γ1 + (J v − I v ) ⊗ J u ⊗ Γ2 )
1
1
1
1
u−1
= J n ⊗ J v ⊗ J u ⊗ Γ0 + J v ⊗
J u ⊗ Γ1
n
v
u
v
u
v−1
+
J v ⊗ J u ⊗ Γ2
v
is symmetric. It implies that matrix P commutes with the covariance matrix
of y.
Lemma 1. Let ϑ denote the subspace spanned by V , i.e., ϑ = sp{V }. Then,
ϑ is a quadratic subspace, meaning that ϑ is a linear space and if V ∈ ϑ then
V 2 ∈ ϑ (see [12] for the definition).
Proof. See Lemma 4.1 in [13].
Because orthogonal projector on the space generated by the mean vector
commutes with all covariances matrices, there exists BLUE for each estimable
function of mean. Moreover BLUE are least squares estimators (LSE), in view
e is the unique solution of the following normal equation:
of Result 1. Thus, µ
Best unbiased estimates for parameters of three-level ...
97
(1n ⊗ 1v ⊗ 1u ⊗ I m )0 (1n ⊗ 1v ⊗ 1u ⊗ I m )µ = (1n ⊗ 1v ⊗ 1u ⊗ I m )0 y or
nvuI m µ = [I m , I m , . . . , I m ]y,
which means that
n
e=
µ
v
u
1 XXX
y r,ts .
nvu
r=1 t=1 s=1
Let M = I n ⊗ I v ⊗ I u ⊗ I m − P . So, M is idempotent. Now, since P V = V P ,
and ϑ is a quadratic space, M ϑM = M ϑ is also a quadratic space. We now
construct a base for the quadratic subspace ϑ. Define
Aii = E ii
and Aij = E ij + E ji ,
for i < j; and j = 1, . . . , m,
as a base for symmetric matrices Γ. The (m × m)−dimensional matrices E ij has
1 only at the ijth element, and 0 at all other elements. Then it is clear that
the base for diagonal matrices of the form I n ⊗ I v ⊗ I u ⊗ Γ0 is constituted by
matrices
(3.8)
(0)
K ij = I n ⊗ I v ⊗ I u ⊗ Aij , for i ≤ j, j = 1, . . . , m,
the base for matrices of the form I n ⊗ I v ⊗ (J u − I u ) ⊗ Γ1 is constituted by
matrices
(3.9)
(1)
K ij = I n ⊗ I v ⊗ (J u − I u ) ⊗ Aij , for i ≤ j, j = 1, . . . , m
and the base for matrices of the form I n ⊗ (J v − I v ) ⊗ J u ⊗ Γ2 is constituted by
matrices
(3.10)
(2)
K ij = I n ⊗ (J v − I v ) ⊗ J u ⊗ Aij , for i ≤ j, j = 1, . . . , m.
It is clear from (2.2) that above base is orthogonal with respect to trace of inner
product. See also [11].
Result 2. The complete and minimal sufficient statistics for the mean vector
and the variance-covariance matrix are
(3.11)
(10nvu ⊗ I m )y
and
(3.12)
(l)
y 0 M K ij M y, l = 0, 1, 2,
where M = I nvum − P and P is given in (3.7), see [1, 14] and [17].
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A. Koziol
e vum is the best quadratic unbiased estiNow it is necessary to prove that Γ
mator (BQUE) for Γ. Since P commutes with the covariance matrix of y, for
each parameter of covariance there exists BQUE if and only if
sp{M V M },
where
M = I nvum − P ,
is a quadratic subspace (see [15, 17] and [3]) or Jordan algebra (see [5]), where
V stands for covariance matrix of y. It is clear that if sp{V } is a quadratic
subspace and if for each Σ ∈ sp{V } commutativity P Σ = ΣP holds, then
sp{M V M } = sp{M V } is also a quadratic subspace. According to the coordinate free approach, the expectation of M yy 0 M can be written as a linear
(0)
(1)
(2)
combination of matrices M K ij , M K ij and M K ij with unknown coefficients
(0)
(1)
(2)
σij , σij and σij , respectively. Note also that identity covariance operator of
yy 0 belongs to sp{cov(yy 0 )}. It implies that the ordinary best quadratic estima(0)
(1)
(2)
tors are least square estimators for corresponding parameters σij , σij and σij .
(0)
(1)
They cannot be calculated independently (as in [11]) because M K ij , M K ij
(2)
(l)
and M K ij are not orthogonal. Defining m(m+1)
column vectors σ (l) = [σij ]
2
for i ≤ j = 1, . . . , m; l = 0, 1, 2, the normal equations have the following block
diagonal structure
(3.13)


  (0)   (0) 
r
σ
a b c
 b d e  ⊗ I m(m+1)  σ (1)  = r (1) ,
2
c e f
r (2)
σ (2)
(0) (0) (1) where for i ≤ j = 1, . . . , m; a = tr M (K ij )2 , b = tr M K ij K ij , c =
(0) (2) (1) (1) (2) (2) tr M K ij K ij , d = tr M (K ij )2 , e = tr M K ij K ij , f = tr M (K ij )2
0 (l) 1
while r (l) = 2−δ
r K ij r for l = 0, 1, 2 is m(m+1)
×1 vector, δ ij is the Kronecker
2
ij
delta and r stands for the residual vector, i.e., r = M y = (I nvum − P )y. Now
to prove (3.13), consider the following six cases
Case 1. for l = 0, i = j,
h
i
1
1
1
(0)
tr M (K ii )2 = tr I nvum − J n ⊗ J v ⊗ J u ⊗ I m (I nvu ⊗ Aii )2
n
v
u
h
i
1
1
1
= tr I nvum − J n ⊗ J v ⊗ J u ⊗ I m (I nvu ⊗ A2ii )
n
v
u
1
1
1 = tr(A2ii )tr I nvu − J n ⊗ J v ⊗ J u
n
v
u
= nvu − 1.
Best unbiased estimates for parameters of three-level ...
Case 2. for l = 0, i < j,
h
i
1
1
1
(0)
tr M (K ij )2 = tr I nvum − J n ⊗ J v ⊗ J u ⊗ I m (I nvu ⊗ Aij )2
n
v
u
h
i
1
1
1
= tr I nvum − J n ⊗ J v ⊗ J u ⊗ I m (I nvu ⊗ A2ij )
n
v
u
1
1
1 = tr(A2ij )tr I nvu − J n ⊗ J v ⊗ J u = 2(nvu − 1).
n
v
u
Case 3. for l = 1, i = j,
h
1
1
(1)
tr M (K ii )2 = tr I nvum − J n ⊗ J v ⊗
n
v
i
(I nv ⊗ (J u − I u ) ⊗ Aii )2
h
1
1
= tr I nvum − J n ⊗ J v ⊗
n
v
1
Ju ⊗ Im
u
1
Ju ⊗ Im
u
i
(I nv ⊗ ((u − 2)J u + I u ) ⊗ A2ii )
= tr(A2ii )tr I nv ⊗ ((u − 2)J u + I u )
1
1
(u − 1)2 − Jn ⊗ Jv ⊗
Ju
n
v
u
= nvu(u − 1) − (u − 1)2
= [(nv − 1)u + 1](u − 1).
Case 4. for l = 1, i < j,
h
1
1
(1)
tr M (K ij )2 = tr I nvum − J n ⊗ J v ⊗
n
v
i
(I nv ⊗ (J u − I u ) ⊗ Aij )2
h
1
1
= tr I nvum − J n ⊗ J v ⊗
n
v
1
Ju ⊗ Im
u
1
Ju ⊗ Im
u
i
(I nv ⊗ ((u − 2)J u + I u ) ⊗ A2ij )
= tr(A2ij )tr I nv ⊗ ((u − 2)J u + I u )
1
1
(u − 1)2 − Jn ⊗ Jv ⊗
Ju
n
v
u
= 2(nvu(u − 1) − (u − 1)2 )
= 2[(nv − 1)u + 1](u − 1).
99
100
A. Koziol
Case 5. for l = 2, i = j,
h
1
1
1
(2)
tr M (K ii )2 = tr I nvum − J n ⊗ J v ⊗ J u ⊗ I m
n
v
ui
(I n ⊗ (J v − I v ) ⊗ J u ⊗ Aii )2
h
1
1
1
= tr I nvum − J n ⊗ J v ⊗ J u ⊗ I m
n
v
u
i
(I n ⊗ ((v − 2)J v + I v ) ⊗ uJ u ⊗ A2ii )
= tr(A2ii )tr I n ⊗ ((v − 2)J v + I v ) ⊗ uJ u
1
(v − 1)2
− Jn ⊗
J v ⊗ uJ u
n
v
= nv(v − 1)u2 − (v − 1)2 u2
= [(n − 1)v + 1](v − 1)u2 .
Case 6. for l = 2, i < j,
h
1
1
1
(2)
tr M (K ij )2 = tr I nvum − J n ⊗ J v ⊗ J u ⊗ I m
n
v
ui
(I n ⊗ (J v − I v ) ⊗ J u ⊗ Aij )2
h
1
1
1
= tr I nvum − J n ⊗ J v ⊗ J u ⊗ I m
n
v
u
i
(I n ⊗ ((v − 2)J v + I v ) ⊗ uJ u ⊗ A2ij )
= tr(A2ij )tr I n ⊗ ((v − 2)J v + I v ) ⊗ uJ u
1
(v − 1)2
J v ⊗ uJ u
− Jn ⊗
n
v
= 2(nv(v − 1)u2 − (v − 1)2 u2 )
= 2[(n − 1)v + 1](v − 1)u2 .
Case 7. for l1 = 0, l2 = 1, i = j,
h
1
1
1
(0) (1)
tr M K ii K ii
= tr I nvum − J n ⊗ J v ⊗ J u ⊗ I m
n
v
u
i
(I nvu ⊗ Aii )(I nv ⊗ (J u − I u ) ⊗ Aii )
h
1
1
1
= tr I nvum − J n ⊗ J v ⊗ J u ⊗ I m
n
v i u
(I nv ⊗ (J u − I u ) ⊗ A2ii )
1
1
u−1 = tr(A2ii )tr I nv ⊗ (J u − I u ) − J n ⊗ J v ⊗
Ju
n
v
u
= −(u − 1).
Best unbiased estimates for parameters of three-level ...
101
Case 8. for l1 = 0, l2 = 1, i < j,
h
1
1
1
(0) (1)
tr M K ij K ij = tr I nvum − J n ⊗ J v ⊗ J u ⊗ I m
n
v
u
i
(I nvu ⊗ Aij )(I nv ⊗ (J u − I u ) ⊗ Aij )
h
1
1
1
= tr I nvum − J n ⊗ J v ⊗ J u ⊗ I m
n
v i u
(I nv ⊗ (J u − I u ) ⊗ A2ij )
1
1
u−1 = tr(A2ij )tr I nv ⊗ (J u − I u ) − J n ⊗ J v ⊗
Ju
n
v
u
= −2(u − 1).
Case 9. for l1 = 0, l2 = 2, i = j,
h
1
1
1
(0) (2)
= tr I nvum − J n ⊗ J v ⊗ J u ⊗ I m
tr M K ii K ii
n
v
u
i
(I nvu ⊗ Aii )(I n ⊗ (J v − I v ) ⊗ J u ⊗ Aii )
h
1
1
1
= tr I nvum − J n ⊗ J v ⊗ J u ⊗ I m
n
v
ui
2
I( n ⊗ (J v − I v ) ⊗ J u ⊗ Aii )
v−1
1
Jv ⊗ Ju
= tr(A2ii )tr I n ⊗ (J v − I v ) ⊗ J u − J n ⊗
n
v
= −(v − 1)u.
Case 10. for l1 = 0, l2 = 2, i < j,
h
1
1
1
(0) (2)
tr M K ij K ij = tr I nvum − J n ⊗ J v ⊗ J u ⊗ I m
n
v
u
i
(I nvu ⊗ Aij )(I n ⊗ (J v − I v ) ⊗ J u ⊗ Aij )
h
1
1
1
= tr I nvum − J n ⊗ J v ⊗ J u ⊗ I m
n
v
ui
2
(I n ⊗ (J v − I v ) ⊗ J u ⊗ Aij )
1
v−1
= tr(A2ij )tr I n ⊗ (J v − I v ) ⊗ J u − J n ⊗
Jv ⊗ Ju
n
v
= −2(v − 1)u.
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A. Koziol
Case 11. for l1 = 1, l2 = 2, i = j,
h
1
1
1
(1) (2)
tr M K ii K ii
= tr I nvum − J n ⊗ J v ⊗ J u ⊗ I m
n
v
u
i
(I nv ⊗ (J u − I u ) ⊗ Aii )(I n ⊗ (J v − I v ) ⊗ J u ⊗ Aii )
h
1
1
1
= tr I nvum − J n ⊗ J v ⊗ J u ⊗ I m
n
v
u
i
(I n ⊗ (J v − I v ) ⊗ (u − 1)J u ⊗ A2ii )
= tr(A2ii )tr I n ⊗ (J v − I v ) ⊗ (u − 1)J u
1
v−1
− Jn ⊗
J v ⊗ (u − 1)J u
n
v
= −(v − 1)u(u − 1).
Case 12. for l1 = 1, l2 = 2, i < j,
h
1
1
1
(1) (2)
tr M K ij K ij = tr I nvum − J n ⊗ J v ⊗ J u ⊗ I m
n
v
u
i
(I nv ⊗ (J u − I u ) ⊗ Aij )(I n ⊗ (J v − I v ) ⊗ J u ⊗ Aij )
h
1
1
1
= tr I nvum − J n ⊗ J v ⊗ J u ⊗ I m
n
v
u
i
(I n ⊗ (J v − I v ) ⊗ (u − 1)J u ⊗ A2ij )
= tr(A2ij )tr I n ⊗ (J v − I v ) ⊗ (u − 1)J u
1
v−1
J v ⊗ (u − 1)J u
− Jn ⊗
n
v
= −2(v − 1)u(u − 1).
e for Γ, the following normal
Thus, to find the best quadratic unbiased estimator Γ
equation has to be solved

nvu − 1
 −(u − 1)
−(v − 1)u
−(u − 1)
[(nv − 1)u + 1](u − 1)
−(v − 1)u(u − 1)
 (0)   (n−1)vu+1
σ
n(n−1)v 2 u2

1
(3.14) σ (1)  =  n(n−1)v
2 u2
(2)
1
σ
2 2
n(n−1)v u

  (0)   (0) 
σ
r
−(v − 1)u
−(v − 1)u(u − 1)  ⊗ I m(m+1)  σ (1)  = r (1) ,
2
[(n − 1)v + 1](v − 1)u2
σ (2)
r (2)
1
n(n−1)v 2 u2
(n−1)vu+u−1
n(n−1)v 2 u2 (u−1)
1
n(n−1)v 2 u2

1
n(n−1)v 2 u2

1
⊗
n(n−1)v 2 u2
nv−1
n(n−1)v 2 (v−1)u2


r (0)

I m(m+1)r (1) .
2
r (2)
Best unbiased estimates for parameters of three-level ...
103
The right hand side of this equation can be expressed by C 0 , C 1 and C 2 defined
in the following way
(3.15)
C0 =
v X
u X
n
X
e
y r,ts − µ
0
e ,
y r,ts − µ
t=1 s=1 r=1
(3.16)
C1 =
v X
u X
u X
n
X
t=1 s=1 s∗ =1 r=1
s6=s∗
(3.17)
C2 =
0
e y r,ts − µ
e ,
y r,ts∗ − µ
v X
v X
u X
u X
n
X
t=1 t∗ =1 s=1 s∗ =1 r=1
t6=t∗
e=
where µ
1
nvu
Pn
r=1
Pv
t=1
Pu
s=1 y r,ts
e
y r,t∗ s∗ − µ
0
e ,
y r,ts − µ
and then we have

   
Γ0
C0
nvu − 1
−(u − 1)
−(v − 1)u
 −(u − 1) [(nv − 1)u + 1](u − 1)
−(v − 1)u(u − 1)  ⊗ I m  Γ1  = C 1  .
C2
−(v − 1)u
−(v − 1)u(u − 1)
[(n − 1)v + 1](v − 1)u2
Γ2

Solving this equation we get
  

−1  
Γ0
nvu − 1
−(u − 1)
−(v − 1)u
C0
Γ1  =  −(u − 1) [(nv − 1)u + 1](u − 1)
−(v − 1)u(u − 1)  ⊗ I m C 1 ,
Γ2
−(v − 1)u
−(v − 1)u(u − 1)
[(n − 1)v + 1](v − 1)u2
C2
   (n−1)vu+1
Γ0
n(n−1)v 2 u2

1
Γ1  = 
 n(n−1)v
2 u2
1
Γ2
2 2
n(n−1)v u
1
n(n−1)v 2 u2
(n−1)vu+u−1
n(n−1)v 2 u2 (u−1)
1
n(n−1)v 2 u2
1
n(n−1)v 2 u2

1

n(n−1)v 2 u2
nv−1
n(n−1)v 2 (v−1)u2

 
C0

⊗ I m  C 1  .
C2
It is worth noting that all elements off diagonal of matrix on the right hand side
in the above equation are equal. Estimators for Γ0 , Γ1 and Γ2 are
1
1
e 0 = (n − 1)vu + 1 C 0 +
Γ
C1 +
C 2,
n(n − 1)v 2 u2
n(n − 1)v 2 u2
n(n − 1)v 2 u2
1
(n − 1)vu + u − 1
1
e1 =
Γ
C0 +
C1 +
C 2,
2
2
2
2
n(n − 1)v u
n(n − 1)v u (u − 1)
n(n − 1)v 2 u2
1
nv − 1
1
e2 =
Γ
C0 +
C1 +
C 2.
2
2
2
2
n(n − 1)v u
n(n − 1)v u
n(n − 1)v 2 (v − 1)u2
104
A. Koziol
Now using Result 2 we are ready to formulate the following theorem.
Theorem 2. Assume that y nvum×1 ∼ N (1nvu ⊗ I m )µ, I n ⊗ Γ with doubly
exchangeable covariance structure on Γ, i.e.,
Γ = I v ⊗ I u ⊗ (Γ0 − Γ1 ) + I v ⊗ J u ⊗ (Γ1 − Γ2 ) + J v ⊗ J u ⊗ Γ2 ,
where Γ0 , Γ1 and Γ2 are m × m unknown symmetric matrices such that Γ is
positive definite. Then
n
(3.18)
e=
µ
v
u
1 XXX
y r,ts ,
nvu
r=1 t=1 s=1
y
= (y 01 , y 02 , . . . , y 0n )0
nvum×1
y r,ts = (y 0r,ts1 , . . . , y 0r,tsm )0 and
where
(3.19)
with y r = (y 0r,11 , . . . , y 0r,vu )0 with
e = I v ⊗ I u ⊗ (Γ
e0 − Γ
e 1 ) + I v ⊗ J u ⊗ (Γ
e1 − Γ
e 2) + J v ⊗ J u ⊗ Γ
e 2,
Γ
e0 =
where Γ
1
(C 0
n(n−1)v 2 u2
(n−1)vu+1
C
n(n−1)v 2 u2 0
1
e 1 = (n−1)vu+u−1
(C 1 + C 2 ), Γ
C
n(n−1)v 2 u2
n(n−1)v 2 u2 (u−1) 1
nv−1
1
e2 =
Γ
C + n(n−1)v
2 u2 (C 0 + C 1 )
n(n−1)v 2 (v−1)u2 2
+
+
+ C 2 ) and
are the best unbiased estimators (BUE) for µ and Γ, respectively. Here C 0 , C 1
and C 2 are defined in (3.15), (3.16) and (3.17), respectively.
Proof. These estimators for µ and Γ are BLUE and BQUE, respectively. Now,
because they are function of complete statistics from Result 2 it follows that they
are BUE.
Now we are able to make a statement that estimators presented in Theorem 2
are consistent and obviously the family of distribution of above estimators is
complete.
Theorem 3. Estimators given in (3.18) and (3.19) are consistent. Moreover, the
family of distributions of these estimators is complete.
Proof. Note that the variance of the quadratic forms y 0 Ay, where y ∼ N (µ, V ),
is given by the following formula
(3.20)
var(y 0 Ay) = 2tr (AV AV ) + (AV Aµµ0 + µµ0 AV A) .
Best unbiased estimates for parameters of three-level ...
105
In a special case, if A = M AM , and if M V = V M then Aµµ0 = 0, and (3.20)
reduces to the following form
var(y 0 Ay) = 2tr(M AV AV ).
(3.21)
Now making an use of (3.21) of doubly exchangeable covariance structure of the
covariance matrix of y and from (3.8) it follows that for any fixed Γ if n → ∞
then
2((n − 1)uv + 1)
4(u − 1)
tr(Aij Γ0 Aij Γ0 ) +
tr(Aij Γ0 Aij Γ1 )
2
2
(n − 1)nu v
(n − 1)nu2 v 2
(0)
var(e
σij ) =
+
4(v − 1)
tr(Aij Γ0 Aij Γ2 )
(n − 1)nuv 2
+
2(u − 1)((n − 1)uv + u − 1)
tr(Aij Γ1 Aij Γ1 )
(n − 1)nu2 v 2
+
4(u − 1)(v − 1)
tr(Aij Γ1 Aij Γ2 )
(n − 1)nuv 2
+
2(v − 1)(nv − 1)
tr(Aij Γ2 Aij Γ2 ) → 0.
(n − 1)nv 2
Similarly it follows that we get for each fixed Γ if n tends to ∞ then
(1)
var(e
σij ) =
2((n − 1)uv + u − 1)
tr(Aij Γ0 Aij Γ0 )
(n − 1)n(u − 1)u2 v 2
+
4((u − 2)u((n − 1)v + 1) + 1)
tr(Aij Γ0 Aij Γ1 )
(n − 1)n(u − 1)u2 v 2
+
4(v − 1)
tr(Aij Γ0 Aij Γ2 )
(n − 1)nuv 2
+
2(u((u − 3)u + 3)((n − 1)v + 1) − 1)
tr(Aij Γ1 Aij Γ1 )
(n − 1)n(u − 1)u2 v 2
+
4(u − 1)(v − 1)
tr(Aij Γ1 Aij Γ2 )
(n − 1)nuv 2
+
2(v − 1)(nv − 1)
tr(Aij Γ2 Aij Γ2 ) → 0
(n − 1)nv 2
106
A. Koziol
and also we get that for each fixed Γ if n → ∞ then
(2)
var(e
σij ) =
2(nv − 1)
tr(Aij Γ0 Aij Γ0 )
(n − 1)nu2 (v − 1)v 2
+
4(u − 1)(nv − 1)
tr(Aij Γ0 Aij Γ1 )
(n − 1)nu2 (v − 1)v 2
+
4(n(v − 2)v + 1)
tr(Aij Γ0 Aij Γ2 )
(n − 1)nu(v − 1)v 2
+
2(u − 1)2 (nv − 1)
tr(Aij Γ1 Aij Γ1 )
(n − 1)nu2 (v − 1)v 2
+
4(u − 1)(n(v − 2)v + 1)
tr(Aij Γ1 Aij Γ2 )
(n − 1)nu(v − 1)v 2
+
2(nv((v − 3)v + 3) − 1)
tr(Aij Γ2 Aij Γ2 ) → 0.
(n − 1)n(v − 1)v 2
To finish the proof, note that estimators for µ and estimators for elements of
covariance matrix are one-to-one functions of minimal sufficient statistic given
by (3.11) and (3.12).
4.
Comparison of BUE in two models
In this paragraph author compares variances of covariance parameters in two
models Mo1 and Mo2. Both with a doubly exchangeable covariance structure, in
which the first one has the unstructured mean vector 1n ⊗ µ where µ has vum
components and the second one has the structured mean vector 1nvu ⊗ µ where
µ has m components. We compare variances of estimators under model Mo1,
(0)
(1)
(2)
(0)
(1)
(2)
σ
e[1]ij , σ
e[1]ij and σ
e[1]ij , with model Mo2, σ
e[2]ij , σ
e[2]ij and σ
e[2]ij . It is clear that the
(0)
(1)
(2)
expectation of σ
e[1]ij , σ
e[1]ij and σ
e[1]ij calculated for Mo1 are unbiased under Mo2.
From the Lehmann-Scheffé theorem (see [7]) it follows that the variance of all
estimators both covariance and expectation parameters have smaller variances
in Mo2. The inverse conclusion is not true because estimators for covariance
parameters have different expectation under Mo1.
In model Mo1, estimators of mean vector µ[1] and components of variancecovariance matrix Γ[1] , i.e., Γ[1]0 , Γ[1]1 , Γ[1]2 are, respectively (see Theorem 2 in
[11]).
Best unbiased estimates for parameters of three-level ...
107
n
e [1]
µ
1X
=
yr ,
n
r=1
1
e [1]0 =
Γ
C ,
(n − 1)vu [1]0
1
e [1]1 =
C ,
Γ
(n − 1)vu(u − 1) [1]1
1
e [1]2 =
Γ
C ,
(n − 1)v(v − 1)u2 [1]2
where C [1]0 , C [1]1 and C [1]2 are the following matrices
C [1]0 =
C [1]1 =
C [1]2 =
v X
u X
n
X
y r,ts − y •,ts
t=1 s=1 r=1
v
u
u
n
XX
X
X
t=1 s=1 s6=s∗ =1 r=1
v
u
v
u X
n
XX
X
X
0
y r,ts − y •,ts ,
y r,ts∗ − y •,ts∗
0
y r,ts − y •,ts ,
y r,t∗ s∗ − y •,t∗ s∗
0
y r,ts − y •,ts ,
t=1 s=1 t6=t∗ =1 s∗ =1 r=1
where y •,ts =
1
n
Pn
r=1 y r,ts ,
for t = 1, . . . , v and s = 1, . . . , u.
Alternatively, it can also calculate and present graphically the difference of
variances for both models.
(0)
(0)
var(e
σ[2]ij ) − var(e
σ[1]ij ) = −
2
((vu − 1)tr(Aij Γ0 Aij Γ0 )
(n − 1)nu2 v 2
− 2(u − 1)tr(Aij Γ0 Aij Γ1 )
− 2(v − 1)utr(Aij Γ0 Aij Γ2 )
+ (u − 1)((v − 1)u + 1)tr(Aij Γ1 Aij Γ1 )
− 2(v − 1)u(u − 1)tr(Aij Γ1 Aij Γ2 )
+ (v − 1)u2 tr(Aij Γ2 Aij Γ2 )).
108
A. Koziol
After simple calculations we get
(0)
(0)
var(e
σ[2]ij ) − var(e
σ[1]ij ) = −
2
((u − 1)tr(Aij Γ0 Aij Γ0 )
(n − 1)nu2 v 2
−2(u − 1)tr(Aij Γ0 Aij Γ1 )
+ (u − 1)tr(Aij Γ1 Aij Γ1 )
+ (v − 1)utr(Aij Γ0 Aij Γ0 )
− 2(v − 1)utr(Aij Γ0 Aij Γ2 )
+ (v − 1)utr(Aij Γ2 Aij Γ2 )
+ (v − 1)u(u − 1)tr(Aij Γ1 Aij Γ1 )
− 2(v − 1)u(u − 1)tr(Aij Γ1 Aij Γ2 )
+ (v − 1)u(u − 1)tr(Aij Γ2 Aij Γ2 )),
(0)
(0)
thus var(e
σ[2]ij ) − var(e
σ[1]ij ) < 0 if
tr(Aij Γ0 Aij Γ0 ) + tr(Aij Γ1 Aij Γ1 ) > 2tr(Aij Γ0 Aij Γ1 ),
tr(Aij Γ0 Aij Γ0 ) + tr(Aij Γ2 Aij Γ2 ) > 2tr(Aij Γ0 Aij Γ2 ),
tr(Aij Γ1 Aij Γ1 ) + tr(Aij Γ2 Aij Γ2 ) > 2tr(Aij Γ1 Aij Γ2 )
which holds for any fixed Γ0 , Γ1 , Γ2 .
Similarly, it is easy to see that
(1)
(1)
var(e
σ[2]ij ) − var(e
σ[1]ij ) = −
2
(n − 1)n(u − 1)u2 v 2
((u(v − 1) + 1)tr(Aij Γ0 Aij Γ0 )
− 2(1 − u(u − 2)(v − 1))tr(Aij Γ0 Aij Γ1 )
− 2(v − 1)u(u − 1)tr(Aij Γ0 Aij Γ2 )
+ (u((u − 3)u + 3)(v − 1) + 1)tr(Aij Γ1 Aij Γ1 )
− 2(v − 1)u(u − 1)2 tr(Aij Γ1 Aij Γ2 )
+ (v − 1)u2 (u − 1)tr(Aij Γ2 Aij Γ2 )).
Best unbiased estimates for parameters of three-level ...
109
After simple calculations we get
(1)
(1)
var(e
σ[2]ij ) − var(e
σ[1]ij ) = −
2
(n − 1)n(u − 1)u2 v 2
((1 − (u − 2)u(v − 1))tr(Aij Γ0 Aij Γ0 )
− 2(1 − (u − 2)u(v − 1))tr(Aij Γ0 Aij Γ1 )
+ (1 − (u − 2)u(v − 1))tr(Aij Γ1 Aij Γ1 )
+ (v − 1)u(u − 1)tr(Aij Γ0 Aij Γ0 )
− 2(v − 1)u(u − 1)tr(Aij Γ0 Aij Γ2 )
+ (v − 1)u(u − 1)tr(Aij Γ2 Aij Γ2 )
+ (v − 1)u(u − 1)2 tr(Aij Γ1 Aij Γ1 )
− 2(v − 1)u(u − 1)2 tr(Aij Γ1 Aij Γ2 )
+ (v − 1)u(u − 1)2 tr(Aij Γ2 Aij Γ2 )),
(1)
(1)
thus var(e
σ[2]ij ) − var(e
σ[1]ij ) < 0 if
tr(Aij Γ0 Aij Γ0 ) + tr(Aij Γ1 Aij Γ1 ) > 2tr(Aij Γ0 Aij Γ1 ),
tr(Aij Γ0 Aij Γ0 ) + tr(Aij Γ2 Aij Γ2 ) > 2tr(Aij Γ0 Aij Γ2 ),
tr(Aij Γ1 Aij Γ1 ) + tr(Aij Γ2 Aij Γ2 ) > 2tr(Aij Γ1 Aij Γ2 )
which holds for any fixed Γ0 , Γ1 , Γ2 .
(2)
(2)
var(e
σ[2]ij ) − var(e
σ[1]ij ) = −
2
(n − 1)nu2 (v − 1)v 2
(tr(Aij Γ0 Aij Γ0 ) − 2(1 − u)tr(Aij Γ0 Aij Γ1 )
− 2utr(Aij Γ0 Aij Γ2 ) + (u − 1)2 tr(Aij Γ1 Aij Γ1 )
− 2u(u − 1)tr(Aij Γ1 Aij Γ2 ) + u2 tr(Aij Γ2 Aij Γ2 )).
110
A. Koziol
After simple calculations we get
(2)
(2)
var(e
σ[2]ij ) − var(e
σ[1]ij ) = −
2
((1 − u)tr(Aij Γ0 Aij Γ0 )
(n − 1)nu2 (v − 1)v 2
− 2(1 − u)tr(Aij Γ0 Aij Γ1 )
+ (1 − u)tr(Aij Γ1 Aij Γ1 ) + utr(Aij Γ0 Aij Γ0 )
− 2utr(Aij Γ0 Aij Γ2 ) + utr(Aij Γ2 Aij Γ2 )
+ u(u − 1)tr(Aij Γ1 Aij Γ1 )
− 2u(u − 1)tr(Aij Γ1 Aij Γ2 )
+ u(u − 1)tr(Aij Γ2 Aij Γ2 )),
(2)
(2)
thus var(e
σ[2]ij ) − var(e
σ[1]ij ) < 0 if
tr(Aij Γ0 Aij Γ0 ) + tr(Aij Γ1 Aij Γ1 ) > 2tr(Aij Γ0 Aij Γ1 ),
tr(Aij Γ0 Aij Γ0 ) + tr(Aij Γ2 Aij Γ2 ) > 2tr(Aij Γ0 Aij Γ2 ),
tr(Aij Γ1 Aij Γ1 ) + tr(Aij Γ2 Aij Γ2 ) > 2tr(Aij Γ1 Aij Γ2 )
which holds for any fixed Γ0 , Γ1 , Γ2 .
For graphical illustration of these differences, author fixed Γ0 = I and Γ1 =
Γ2 = 0. For each figure, values for n are chosen from 3 to 25 and for u and v from
2 to 10. For the plot of n and u, v is treated as constant and v = 2. Similarly,
For the plot of n and v, u is treated as constant and u = 2.
(0)
(0)
Figure 1. var(e
σ[2]ij ) − var(e
σ[1]ij ) for n and u, plotting separate figure for parameters n
and v is redundant because difference is symmetric with respect to u and v.
(1)
(1)
(1)
(1)
(2)
(2)
(2)
(2)
Figure 2. (A) var(e
σ[2]ij ) − var(e
σ[1]ij ) for n and u, and (B) var(e
σ[2]ij ) − var(e
σ[1]ij ) for n
and v.
Figure 3. (A) var(e
σ[2]ij ) − var(e
σ[1]ij ) for n and u, and (B) var(e
σ[2]ij ) − var(e
σ[1]ij ) for n
and v.
Best unbiased estimates for parameters of three-level ...
111
All three figures reveal the fact that differences between variances of estimators for
σ (0) , σ (1) and σ (2) in models Mo2 and Mo1 are negative and if n → ∞ then tend to
0, thus variances of estimators for sigmas in Mo2 are smaller than corresponding
variances of estimators for sigmas in Mo1.
5.
Conclusions
Under multivariate normality, the free-coordinate approach was used to obtain
linear and quadratic estimates for parameters that are sufficient, complete, unbiased and consistent in the model with doubly exchangeable covariance structure
with structured mean vector. Comparison with the model with the same covariance structure but unstructured mean vector shows that estimators of covariance
parameters in model with structured mean vector have smaller variances.
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Received 15 September 2016

On the estimation of the second order parameter in extreme

Numerical Abstract Interpretation