Best Linear Unbiased Prediction (BLUP)
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Best Linear Unbiased Prediction (BLUP):
Suppose
y = Xβ + ε,
where
E(ε) = 0,
and Var(ε) = Σ = σ 2 V.
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Initially, we will assume the Aitken model holds so that σ 2 > 0 is
unknown and V is a known symmetric and PD matrix.
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Suppose u is a random variable with mean 0 and finite variance.
A linear predictor d + a0 y of c0 β + u is unbiased iff
E(d + a0 y) = E(c0 β + u) = c0 β
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∀ β ∈ Rp .
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c0 β + u is predictable iff ∃ an unbiased linear predictor of c0 β + u.
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Note that c0 β + u is predictable ⇐⇒ ∃ a 3 c0 = a0 X.
This follows from the same argument used to show that c0 β
estimable ⇐⇒ ∃ a 3 c0 = a0 X.
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Also, d + a0 y is an unbiased predictor of c0 β + u iff
d=0
and c0 = a0 X,
as argued for the case of an unbiased estimator.
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Suppose ŵ is a predictor of a random variable w.
The mean squared error (MSE) of ŵ as a predictor of w is
E(ŵ − w)2 .
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Suppose
Cov(ε, u) = σ 2 k
for some known vector k.
Suppose c0 β + u is predictable.
Prove that c0 β̂ GLS + û has lowest MSE among all unbiased linear
predictors of c0 β + u, where
û ≡ [Cov(ε, u)]0 [Var(ε)]−1 (y − Xβ̂ GLS ) = k0 V −1 (y − Xβ̂ GLS ).
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Proof:
First note that
c0 β̂ GLS + û = c0 β̂ GLS + k0 V −1 (y − Xβ̂ GLS )
has expectation
c0 β + k0 V −1 (Xβ − Xβ) = c0 β.
Thus, c0 β̂ GLS + û is an unbiased predictor of c0 β + u.
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Now note that
c0 β̂ GLS + û = c0 β̂ GLS + k0 V −1 (y − Xβ̂ GLS )
= (c0 − k0 V −1 X)β̂ GLS + k0 V −1 y
= (c0 − k0 V −1 X)(X0 V −1 X)− X0 V −1 y + k0 V −1 y
= [(c0 − k0 V −1 X)(X0 V −1 X)− X0 V −1 + k0 V −1 ]y
≡ b0 y (a linear estimator.)
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Because
c0 β̂ GLS + û = b0 y
is an unbiased predictor, we know
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b0 X = c0 .
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Let a0 y be any other linear unbiased predictor of c0 β + u. Then
a0 X = c0 .
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The MSE of a0 y is
E[a0 y − (c0 β + u)]2 = Var[a0 y − (c0 β + u)]
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= Var(a0 y − u)
= Var(a0 y − b0 y − u + b0 y)
= Var[(a − b)0 y] + Var(b0 y − u)
+2Cov[(a − b)0 y, b0 y − u].
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Now
Cov[(a − b)0 y, b0 y − u] = (a − b)0 Var(y)b − Cov[(a − b)0 y, u]
= (a − b)0 σ 2 Vb − (a − b)0 Cov(y, u)
= σ 2 (a − b)0 Vb − (a − b)0 Cov(ε, u)
= σ 2 (a − b)0 Vb − σ 2 (a − b)0 k
= σ 2 (a − b)0 (Vb − k).
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Now
Vb − k = V[V −1 X[(X0 V −1 X)− ]0 (c − X0 V −1 k) + V −1 k] − k
= X[(X0 V −1 X)− ]0 (c − X0 V −1 k).
Thus, the covariance term is
σ 2 (a − b)0 X[(X0 V −1 X)− ]0 (c − X0 V −1 k),
which is 0 because
a0 X = b0 X = c0 .
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Thus, we have
MSE(a0 y) = Var[(a − b)0 y] + Var(b0 y − u)
= Var[(a − b)0 y] + Var[b0 y − (c0 β + u)]
= Var[(a − b)0 y] + E[b0 y − (c0 β + u)]2
= Var[(a − b)0 y] + MSE(b0 y).
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Thus,
MSE(a0 y) ≥ MSE(b0 y)
with equality iff
Var[(a − b)0 y] = 0 ⇐⇒ a = b.
Thus c0 β̂ GLS + û is the unique best linear unbiased predictor
(BLUP) of c0 β + u.
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In practice,
Σ = σ2V
involves unknown variance components in addition to the
unknown σ 2 .
We replace unknown variance components in c0 β̂ GLS + û by the
estimates to obtain “empirical” BLUPs (eBLUPs).
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This typically results in a nonlinear predictor whose properties
are not so well characterized.
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Example:
Suppose yij is the average monthly milk production of the jth
daughter of sire i.
Suppose
yij = µ + si + eij
for i = 1, . . . , t; j = 1, . . . , ni ,
where s1 , . . . , st are iid with mean 0 and variance σs2 ,
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independent of the eij terms, which are iid with mean 0 and
variance σe2 .
Suppose σs2 /σe2 is a known constant d.
Find an expression for the BLUP of µ + s1 .
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Let y = [y11 , . . . , y1n1 , y21 , . . . , y2n2 , . . . , ytnt ]0 .
Let X =n×1
1 where n =
Pt
i=1
ni .
Let β = [µ].


s1 + e11


..


.




s1 + e1n1 


 s2 + e21 


Let ε = 
.
..


.




s2 + e2n2 


..


.


st + etnt
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Then Var(ε) is block diagonal with blocks of sizes
n1 × n1 , n2 × n2 , . . . , nt × nt .
Each block has σs2 + σe2 on the diagonal and σs2 on the
off-diagonal.
The ith block is
σe2 I + σs2 110 = σe2 [I + d110 ].
ni ×ni
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ni ×ni
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We wish to predict c0 β + u, where c = [1], β = [µ] and u = s1 .


σs2 1
n1 ×1 
Cov(ε, u) = Cov(ε, s1 ) = 
0
(n−n1 )×1


d 1
n ×1
= σe2  1  ≡ σe2 k.
0
(n−n1 )×1
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We need β̂ GLS , a solution to the Aitken equations
X0 V −1 Xβ = X0 V −1 y,
where
V = diag( I + d 110 , . . . , I + d 110 ).
n1 ×n1
0
I + d 11
ni ×ni
ni ×ni
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n1 ×n1
−1
= I −
ni ×ni
nt ×nt
nt ×nt
d
110 .
1 + dni ni ×ni
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Thus,
X0 V −1 X = 10 V −1 1
t
X
0
=
1 I−
d
0
11 1
1
+
dn
i
i=1
t X
dn2i
=
ni −
1 + dni
i=1
=
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t
X
i=1
ni
.
1 + dni
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0
−1
XV y =
t
X
i=1
d
0
11 yi
1 I−
1 + dni
0
(where yi = [yi1 , . . . , yini ]0 )
t X
dni
yi·
=
yi· −
1
+
dn
i
i=1
t X
dni
=
1−
yi·
1
+
dn
i
i=1
=
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t
X
i=1
ni
ȳi· .
1 + dni
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Thus,
β̂ GLS = µ̂GLS = (X0 V −1 X)− X0 V −1 y
Pt
ni
i=1 1+dni ȳi·
= Pt
ni
i=1 1+dn
i
Pt
2
2 −1
(σ
/n
i + σs ) ȳi·
e
= Pi=1
.
t
2
2 −1
i=1 (σe /ni + σs )
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Note that each ȳi· is an unbiased estimator of µ with variance
σe2 /ni + σs2 .
Thus, µ̂GLS is a weighted average of independent unbiased
estimators of µ, with inverse variances of estimators as weights.
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Now,
ŝ1 = û = k0 V −1 (y − Xβ̂ GLS )
0
−1
kV y =
=
=
=
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d
0
d1 I −
11 y1
1 + dn1
d2 n1
dy1· −
y1·
1 + dn1
d + d 2 n1 − d 2 n1
y1·
1 + dn1
dn1
ȳ1· .
1 + dn1
0
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0
−1
k V Xβ̂ GLS
d
0
= d1 I −
11 1µ̂GLS
1 + dn1
d2 n21
µ̂GLS
=
dn1 −
1 + dn1
dn1
µ̂GLS .
=
1 + dn1
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0
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Thus, the BLUP of µ + s1 is
dn1
dn1
ȳ1· −
µ̂GLS
1 + dn1
1 + dn1
dn1
dn1
=
ȳ1· + 1 −
µ̂GLS
1 + dn1
1 + dn1
σe2 /n1
σs2
ȳ1· + 2
µ̂GLS .
= 2
σe /n1 + σs2
σe /n1 + σs2
µ̂GLS +
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The mean squared error (MSE) of a predictor ŵ of a random
vector w is
MSE(ŵ) = E[(ŵ − w)(ŵ − w)0 ].
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Now supposeq×1
u is a random vector with mean 0 and
Cov(ε, u) = σ 2n×q
K.
The BLUP of predictable Cβ + u is Cβ̂ GLS + û, where
û ≡ [Cov(ε, u)]0 [Var(ε)]−1 (y − Xβ̂ GLS ) = K 0 V −1 (y − Xβ̂ GLS ).
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Cβ̂ GLS + û is the “best” linear unbiased predictor in the sense that
MSE(ŵ) − MSE(Cβ̂ GLS + û)
is nonnegative definite ∀ ŵ, a linear unbiased predictor of Cβ + u.
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As for the case of univariate prediction, we replace unknown
variance components in Cβ̂ GLS + û by their estimates to obtain
an eBLUP whenever necessary.
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For example, suppose
y = Xβ + Zu + e,
E(u) = 0
Var(u) = G
where
E(e) = 0
Var(e) = R
Cov(u, e) = 0.
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Then, with ε = Zu + e, we have
Var(ε) = ZGZ0 + R
and
Cov(ε, u) = Cov(Zu + e, u)
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= Cov(Zu, u)
= ZCov(u, u)
= ZVar(u)
= ZG.
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It follows that the BLUP of u is
GZ0 (ZGZ0 + R)−1 (y − Xβ̂ GLS ).
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Similarly, if D is a known m × q matrix, the BLUP of Du is
DGZ0 (ZGZ0 + R)−1 (y − Xβ̂ GLS ).
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When G and R are unknown, we use the eBLUP
DĜZ0 (ZĜZ0 + R̂)−1 (y − Xβ̂ GLS ).
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Note that computing β̂ GLS or computing the BLUP of Du requires
inversion of the n × n matrix ZGZ0 + R.
This can be computationally expensive.
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We often assume that
R = σe2 I,
but even then
ZGZ0 + σe2 I
is a n × n matrix that may be difficult to invert.
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Until further notice, assume
R = σe2 I
for some unknown σe2 > 0.
Define H =
1
G.
σe2
Then
Var(y) = ZGZ0 + σe2 I
= σe2 (ZHZ0 + I) = σe2 V
where V = ZHZ0 + I.
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C.R. Henderson introduced the Mixed Model Equations(MME)
"
X0 X
X0 Z
Z0 X H−1 + Z0 Z
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#" #
β̃
ũ
"
=
#
X0 y
Z0 y
.
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Recall that the Aitken equations are
X0 V −1 Xβ̃ = X0 V −1 y,
which in our special case are equivalent to
X0 (ZHZ0 + I)−1 Xβ̃ = X0 (ZHZ0 + I)−1 y.
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Henderson showed that
(1) The MME are consistent.
(2) If
" #
β̃
solve the MME, then β̃ is a solution to the AE
ũ
(equivalently, Xβ̃ = Xβ̂ GLS ) and ũ is the BLUP of u.
(3) Conversely, if β̃ solves the AE, then the MME have a
solution whose leading subvector is β̃.
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A nice thing about this result is that we can find β̂ GLS and the
BLUP of u (or Du) without inverting the n × n matrix (ZGZ0 + σe2 I).
"
X0 X
X0 Z
#
Z0 X H−1 + Z0 Z
is (p + q) × (p + q). In some problems, p + q n.
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By (2), the BLUP of a predictable Cβ + Du is given by
Cβ̃ + Dũ
where
" #
β̃
ũ
solves the MME.
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To prove Henderson’s results, we begin with the following
lemma.
Lemma H1: Suppose A is a symmetric and positive definite
matrix. Suppose W is any matrix with number of columns equal
to number of rows of A. Then
AW 0 (I + WAW 0 )−1 = (A−1 + W 0 W)−1 W 0 .
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Proof:
First show that
I + WAW 0
and A−1 + W 0 W
are each nonsingular.
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∀ x 6= 0, x0 (I + WAW 0 )x = x0 x + x0 WAW 0 x
= ||x||2 + ||A1/2 W 0 x||2
≥ ||x||2 > 0.
Thus, I + WAW 0 is PD and ∴ nonsingular.
∀ x 6= 0, x0 (A−1 + W 0 W)x = x0 A−1 x + x0 W 0 Wx
= x0 A−1 x + ||Wx||2
≥ x0 A−1 x > 0
∵ A PD =⇒ A−1 PD. ∴ A−1 + W 0 W is nonsingular.
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Now show that
(A−1 + W 0 W)AW 0 = W 0 (I + WAW 0 )
and then complete the proof.
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(A−1 + W 0 W)AW 0 = A−1 AW 0 + W 0 WAW 0
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= W 0 + W 0 WAW 0
= W 0 (I + WAW 0 ).
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Multiplying on the left by
(A−1 + W 0 W)−1
gives
AW 0 = (A−1 + W 0 W)−1 W 0 (I + WAW 0 ).
Multiplying on the right by (I + WAW 0 )−1 gives the result.
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Now use Lemma H1 to prove Lemma H2:
V −1 = (I + ZHZ0 )−1
= I − Z(H−1 + Z0 Z)−1 Z0 .
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Proof:
Applying the Lemma H1 with H = A and Z = W yields
(H−1 + Z0 Z)−1 Z0 = HZ0 (I + ZHZ0 )−1 .
(∗)
Thus,
[I − Z(H−1 + Z0 Z)−1 Z0 ](I + ZHZ0 )
= [I − ZHZ0 (I + ZHZ0 )−1 ](I + ZHZ0 ) by (∗)
= I + ZHZ0 − ZHZ0 = I.
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Now let’s prove Henderson’s results beginning with (2).
Suppose
" #
β̃
ũ
solves the MME.
Then
X0 Xβ̃ + X0 Zũ = X0 y
and
Z0 Xβ̃ + (H−1 + Z0 Z)ũ = Z0 y.
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The 2nd equation
=⇒ (H−1 + Z0 Z)ũ = Z0 y − Z0 Xβ̃
=⇒ ũ = (H−1 + Z0 Z)−1 Z0 (y − Xβ̃)
=⇒ ũ = HZ0 (I + ZHZ0 )−1 (y − Xβ̃) by (∗)
=⇒ ũ = HZ0 V −1 (y − Xβ̃) = GZ0 Σ−1 (y − Xβ̃).
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Now this last expression for ũ indicates that ũ is the BLUP for u
as long as β̃ solves the Aitken equations.
To show this, we examine the 1st MME.
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The 1st MME is
X0 Xβ̃ + X0 Zũ = X0 y.
We have shown
ũ = (H−1 + Z0 Z)−1 Z0 (y − Xβ̃).
Combining these equations gives
X0 Xβ̃ + X0 Z(H−1 + Z0 Z)−1 Z0 (y − Xβ̃) = X0 y.
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By Lemma H2,
Z(H−1 + Z0 Z)−1 Z0 = I − V −1 .
Thus,
X0 Xβ̃ + X0 (I − V −1 )(y − Xβ̃) = X0 y
⇐⇒ X0 Xβ̃ + X0 y − X0 Xβ̃ − X0 V −1 y + X0 V −1 Xβ̃ = X0 y
⇐⇒ X0 V −1 Xβ̃ = X0 V −1 y.
∴ β̃ solves the AE and Henderson’s result (2) follows.
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Now let’s prove (3).
Suppose β̃ solves the AE.
Let
ũ = HZ0 V −1 (y − Xβ̃)
= HZ0 (I + ZHZ0 )−1 (y − Xβ̃)
= (H−1 + Z0 Z)−1 Z0 (y − Xβ̃) by (∗)
=⇒ (H−1 + Z0 Z)ũ = Z0 (y − Xβ̃) = Z0 y − Z0 Xβ̃
=⇒ Z0 Xβ̃ + (H−1 + Z0 Z)ũ = Z0 y.
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∴ 2nd MME is satisfied.
Now
X0 V −1 Xβ̃ = X0 V −1 y
=⇒ X0 [I − Z(H−1 + Z0 Z)Z0 ]Xβ̃ = X0 [I − Z(H−1 + Z0 Z)Z0 ]y
(By Lemma H2)
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It follows that
X0 Xβ̃ + X0 Z(H−1 + Z0 Z)−1 Z0 (y − Xβ̃) = X0 y.
Now
ũ = (H−1 + Z0 Z)−1 Z0 (y − Xβ̃)
=⇒ X0 Xβ̃ + X0 Zũ = X0 y.
∴ The 1st MME is satisfied and result (3) follows.
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Result (1) is now easy to prove as follows.
The AE are consistent.
Thus, by Henderson’s result (3), the MME are consistent.
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Now suppose y = Xβ + Zu + e, where all is as before except
that, instead of Var(e) = σe2 I, Var(e) = R, a symmetric, positive
definite variance matrix.
Let S =
1
R,
σe2
where σe2 is a positive variance parameter.
Consider the transformation
S−1/2 y = S−1/2 Xβ + S−1/2 Zu + S−1/2 e.
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Then S−1/2 e ∼ N(0, σe2 I), and our previous results with
S−1/2 y, S−1/2 X, and S−1/2 Z in place of y, X, and Z
yield the mixed model equations
"
X0 S−1 X
X0 S−1 Z
#" #
β̃
Z0 S−1 X H−1 + Z0 S−1 Z
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ũ
"
=
#
X0 S−1 y
Z0 S−1 y
.
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Multiplying both sides by 1/σe2 gives
"
X0 R−1 X
X0 R−1 Z
#" #
β̃
Z0 R−1 X G−1 + Z0 R−1 Z
ũ
"
=
#
X0 R−1 y
Z0 R−1 y
.
This is the more general form of Henderson’s Mixed Model
Equations.
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Finally, let Ĝ and R̂ denote G and R with unknown parameters
replaced by their REML estimates.
In practice, we solve
"
X0 R̂−1 X
X0 R̂−1 Z
#" #
β̃
Z0 R̂−1 X Ĝ−1 + Z0 R̂−1 Z
ũ
"
=
#
X0 R̂−1 y
Z0 R̂−1 y
,
to get solutions needed for computation of BLUEs and BLUPs.
c
Copyright 2012
Dan Nettleton (Iowa State University)
Statistics 611
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