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The proof of theorem 11.1.1 in greg rempala's stat 9220 lecture 11, which covers the lse (least squares estimator) and umvue (uniform minimum variance unbiased estimator) in the context of a linear regression model. The theorem states that the lse is the umvue for any estimable parameter, and that both attain the crlb (cramér-rao lower bound).
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Theorem 11.1.1. Consider model
X = Zβ + (11.1)
with assumption (A1) ( is distributed as Nn(0, σ
2
In) with an unknown σ
2
0 ).
(i) The LSE l
β is the UMVUE of l
β for any estimable l
β.
(ii) The UMVUE of σ
2 is σˆ
2 = (n − r)
− 1 ‖X − Z
β‖
2 , where r is the rank of Z
(iii) The UMVUE’s in (i) and (ii) both attain CRLB.
Proof. Let
β be an LSE of β. By Z
Z
β = Z
X,
β)
Z(
β − β) = (X
Z − X
Z)(
β − β) = 0
and, hence,
‖X − Zβ‖
2
= ‖X − Z
β + Z
β − Zβ‖
2
β‖
2
β − Zβ‖
2
β‖
2
− 2 β
Z
Z
β + ‖Zβ‖
2
β‖
2
β‖
2
− 2 β
Z
X + ‖Zβ‖
2
β‖
2
.
Therefore
β‖
2
= E(X − Z
β)
(X − Z
β) = EX
(X − Zβ + Z(β −
β))
(X − Zβ) − EX
Z(
β − β) = E(X − Zβ)
(X − Zβ) − E(Z
Z
β)
(
β − β)
= E‖X − Zβ‖
2
− E(
β − β)
Z
Z(
β − β)
= tr (V ar(X − Zβ)) − tr
V ar(Z(
β − β)
= tr(V ar(X)) − tr(V ar(Z
β))
= σ
2
[n − tr
V ar(ZZ
X(Z
Z)
−
= σ
2
[n − tr
Z)
−
Z
Z(Z
Z)
−
Z
= σ
2
[n − tr
Z)
−
Z
Z
since tr(AB) = tr(BA). We evaluate tr
Z)
− Z
Z
using any choice of (Z
Z)
− .
From the theory of linear algebra, there exists a p×p matrix C such that CC
= I p
and
(Z
Z)C =
where Λ is an r × r diagonal matrix whose diagonal elements are positive. Then,
a particular choice of (Z
Z)
− is
Z)
−
= C
− 1 0
Z)(Z
Z)
−
(Z
Z) = (Z
Z)).
It follows that
tr((Z
Z)
−
Z
Z) = tr
r
= tr
Ip
r
= r
Hence,
ˆσ
2
=
β‖
2
n − r
= σ
2
.
and ˆσ
2 is the UMVUE of σ
2 as a function of the complete sufficient statistic.
(iii) It follows from the result on unbiased estimators in exponential families.
Theorem 11.1.2. Consider model (11.1) with assumption (A1). For any es-
timable parameter l
β, the UMVUE’s l
β and σˆ
2 are independent. Moreover, the
distribution of l
ˆ β is N (l
β, σ
2
l
(Z
Z)
−
l); and (n − r)ˆσ
2
/σ
2
∼ χ
2
n−r
Linear algebra refreshment.
(1) tr(A) =
n
i=
aii, tr(AB) = tr(BA),
(2) P is called projection matrix if P
2 = P ,
(3) Let P be symmertic projection matrix, then the only possible eigenvalues of
P are 0, 1.
(4) Let A be symmetric, then there exist orthogonal matrices C, C
such that
= I and CAC
= Λ, where Λ is diagonal matrix,
(5) If EX = 0 then EX
X = tr(V ar(X)).
Note: Let Y = (Y 1
n−r
). Under (A1) Y is normal, V arY = σ
2 I n−r
since
V arY = EY Y
= E(GPnX)(GPnX)
= GPn(EXX
)P
n
n
σ
2
I n
n
= σ
2
GP n
n
= σ
2
GP n
= σ
2
I n−r
We need to show that EY = 0. But EY j
j
X = 0, since for j ≤ n − r,
j
j
j
n
Zβ.
It suffices to show that G j
n
Z = 0 for every j ≤ n. But
(GPnZ)
(GPnZ) = Z
P
n
GPnZ = Z
PnZ
(I n
Z)
−
Z
)Z = Z
Z − Z
Z(Z
Z)
−
Z
Z = 0.
Example 11.1.1. In on-way ANOVA:
m ∑
i=
n i ∑
j=
ij
i·
2
;
And UMVUE’s of estimable l
β in poly-regression and ANOVA are the LSE’s.
Theorem 11.1.3 (Gauss-Markov). Consider model (11.1) with assumption (A2).
(i) A necessary and sufficient condition for the existence of a linear unbiased es-
timator of l
β (i.e., an unbiased estimator that is linear in X) is l ∈ lin(Z). (ii)
If l ∈ lin(Z), then the LSE l
β is the best linear unbiased estimator (BLUE) of
l
β in the sense that it has the minimum variance in the class of linear unbiased
estimators of l
β of the form T = c
X.
Proof. (i) The sufficiency has been established already. To show necessity,
suppose a linear function of X, c
X with c ∈ R
n , is unbiased for l
β. Then
l
β = E(c
X) = c
EX = c
Zβ.
hence l ∈ lin(Z).
(ii) Let l ∈ lin(Z) = lin(Z
Z). Then l = (Z
Z)α for some α and l
β =
α
(Z
Z)
β = α
Z
X by Z
Zb = Z
X.
Let c
X be any linear unbiased estimator of l
β. From the proof of (i), Z
c = l.