Estimable Functions and Their Least
Squares Estimators in Reparameterized
Models
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Once again consider the linear models
y = Wα + ε
and
y = Xβ + ε,
where E(ε) = 0 and C(W) = C(X).
Suppose W = XT and X = WS.
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We have
E(y) = Xβ = Wα
= WSβ = XTα.
Note the correspondence between
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β
and Tα
α
and
Sβ.
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Result 3.4:
Suppose c0 β is estimable and W 0 W α̂ = W 0 y. Then c0 T α̂ is the least
squares estimator of c0 β.
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Dan Nettleton (Iowa State University)
Statistics 611
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Proof of Result 3.4:
W 0 W α̂ = W 0 y ⇒ X0 XT α̂ = X0 y by Result 2.9.
∵ T α̂ solves the NE X0 Xb = X0 y,
c0 T α̂ is LS estimator of c0 β by definition.
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Statistics 611
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Example:
Consider again the case where
"
X=
"
W=
1n1
1n1
0n1
1n2
0n2
1n2
1n1
0n1
1n2
1n2
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Dan Nettleton (Iowa State University)
#
#

1


T = 0
0
"
1
S=
0
0


0

1
1 0
−1 1
#
.
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Recall that the unique solution to the NE
W 0 W α̂ = W 0 y
is
"
α̂ =
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ȳ1·
ȳ2· − ȳ1·
#
.
Statistics 611
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Suppose we denote the components of β by µ, τ1 , τ2 so that
 
µ
 

β=
τ1 
τ2
and
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E(y) =
#
"
(µ + τ1 )1n1
(µ + τ2 )1n2
.
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τ1 − τ2 = (µ + τ1 ) − (µ + τ2 ) is estimable ∵ it is a LC of elements of E(y).
τ1 − τ2 = c0 β where
c0 = [0, 1, −1].
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Statistics 611
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Result 3.4 implies that


#
1 0 "


ȳ
1·

c0 T α̂ = [0, 1, −1] 
0 0 ȳ − ȳ
2·
1·
0 1
"
#
ȳ1·
= [0, −1]
ȳ2· − ȳ1·
= ȳ1· − ȳ2·
is LSE of c0 β = τ1 − τ2 .
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Statistics 611
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Result 3.5:
If d0 α is estimable in the model y = Wα + ε, then d0 Sβ is estimable in
the model y = Xβ + ε, and its LSE is d0 α̂ = d0 Sβ̂, where α̂ and β̂ are
solutions to
W 0 Wa = W 0 y and X0 Xb = X0 y,
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respectively.
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Proof of Result 3.5:
By Result 3.1, d0 α is estimable ⇐⇒ ∃ a 3 d0 = a0 W.
Multiplying on the right by S leads to ∃ a 3 d0 S = a0 WS = a0 X.
∴ By Result 3.1, d0 Sβ is estimable in the model y = Xβ + ε.
By definition, we know d0 α̂ is LS estimate of d0 α and d0 Sβ̂ is LS
estimate of d0 Sβ.
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Statistics 611
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To see that d0 α̂ = d0 Sβ̂, note that Result 3.4 implies
d0 Sβ̂ = d0 ST α̂
= a0 WST α̂
= a0 XT α̂
= a0 W α̂
(d0 = a0 W)
(X = WS)
(W = XT)
= d0 α̂.
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Returning to our example, rank(W) = 2 ⇒ d0 α is estimable ∀ d ∈ R2 .
For example, d0 α is estimable for d0 = [1, 0], and the LSE is
"
d0 α̂ = [1, 0]
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ȳ1·
ȳ2· − ȳ1·
#
= ȳ1·
.
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According to Result 3.5,
 
µ


1
1
0
τ1 
d0 Sβ = [1, 0]
0 −1 1  
τ2
 
µ
 

= [1, 1, 0] τ1 

τ2
"
= µ + τ1
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#
is also estimable.
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The LSE is

"
0


1 0 
ȳ1· 


0 −1 1
ȳ2·
 
0
 

= [1, 1, 0] 
ȳ1· 
ȳ2·
d0 Sβ̂ = [1, 0]
= ȳ1·
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1
#
.
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