STAT 510
Homework 6 Solutions
Spring 2016
1. [55pts]
(a) [10pts] The corresponding design matrix

1
1

1

1
X=
1

1

1
1
is
1
1
1
1
0
0
0
0
0
0
0
0
1
1
1
1
1
1
0
0
1
1
0
0

0
0

1

1
.
0

0

1
1
(b) [10pts] Recall that a scalar c0 β is estimable if there exists a vector a0 such that
c0 β = a0 E(y),
since E(y) = Xβ. Hence, τ1 − τ2 is estimable if it can be written as a linear combination
of the expected value of y. Put
a0 = (1, 1, −1, −1, 1, 1, −1, −1)/4.
Then
2
2
1 XX
E(yi1k ) − E(yi2k )
a E(y) =
4 i=1 k=1
0
2
2
2
2
1 XX
=
E(µ + λi + τ1 + εi1k ) − E(µ + λi + τ2 + εi2k )
4 i=1 k=1
1 XX
=
(µ + λi + τ1 + 0) − (µ + λi + τ2 + 0)
4 i=1 k=1
1
= (2 · 2) τ1 − τ2
4
= τ1 − τ2 .
Alternatively, and more simply, E(y111 ) − E(y121 ) = τ1 − τ2 , so τ1 − τ2 is estimable.
(c) [10pts] Notice that rank(X) = 3, so that X ∗ needs to be 8 × 3 to have full column rank
and the same column space as X. One possible choice is


1
1
1
 1
1
1 


 1

1
−1


 1

1
−1
∗
.
X =
 1 −1
1 


 1 −1

1


 1 −1 −1 
1 −1 −1
1
(d) [10pts] Let β ∗ = (β1∗ , β2∗ , β3∗ )0 . Using X ∗ as given in part (c), equating expected values
gives
β1∗ + β2∗ + β3∗
β1∗ + β2∗ − β3∗
β1∗ − β2∗ + β3∗
β1∗ − β2∗ − β3∗
= µ + λ1 + τ1 ,
= µ + λ1 + τ2 ,
= µ + λ2 + τ1 ,
= µ + λ2 + τ2 .
Then
2β1∗ = 2µ + λ1 + λ2 + τ1 + τ2 ,
2β2∗ = λ1 − λ2 ,
2β3∗ = τ1 − τ2 ,
which implies
β1∗ = µ + (λ1 + λ2 )/2 + (τ1 + τ2 )/2,
β2∗ = (λ1 − λ2 )/2,
β3∗ = (τ1 − τ2 )/2.
(e) [15pts] Since τ1 − τ2 = 2β3∗ = 0, 0, 2 β ∗ ,
∗
\
τ1 − τ2
= 0, 0, 2 β̂OLS
OLS
= 0, 0, 2 (X ∗ 0 X ∗ )−1 X ∗ 0 y

−1
8
0
0
= 0, 0, 2 0 8 0 X ∗ 0 y
0 0 8


1/8
0
0
= 0, 0, 2  0 1/8 0  X ∗ 0 y
0
0 1/8


1
1
1
1
1
1 1 1
1 −1 −1 −1 −1 y
= 0, 0, 1/4 1 1 1
1 1 −1 −1 1
1 −1 −1


y111
y112 


y121 


y122 
1

1 1 −1 −1 1 1 −1 −1 
=
y211 
4


y212 


y221 
y222
2
2
1 XX
=
[yi1k − yi2k ]
4 i=1 k=1
= ȳ1 − ȳ2
2
2. [15pts] By slide 51 of set 12, the BLUE of µ is a weighted average of independent linear
unbiased estimators, where the weights are proportional to the inverse variances of the linear
unbiased estimators.
We can divide y into two subsets that each have homogeneous variances by considering y5
separately
from y1 , . . . , y4 . By the hint given, the BLUE of µ based only on y1 , . . . , y4 is
P4
1
i=1 yi . Clearly, the BLUE of µ based on only y5 is y5 itself. These two estimators are
4
independent, with variances
!
4
1 0
1X
0
yi = Var
1 (y1 , y2 , y3 , y4 )
Var
4 i=1
4 4×1
1 0
1 Var (y1 , y2 , y3 , y4 )0 14×1
16 4×1 

5 1 1 1
1 5 1 1
1

= 104×1 
1 1 5 1 14×1
16
1 1 1 5
32
=
16
= 2,
=
Var(y5 ) = 4.
Then,
µ̂BLUE =
Var(
1
4
1
P4
1
4
i=1 yi )
Var( 14
=
11
24
P4
1
P4
i=1
P4
i=1
yi )
1
i=1 yi + 4 y5
1
+ 14
2
4
1X
1
=
yi + y5 .
6 i=1
3
3
yi +
+
1
y
Var(y5 ) 5
1
Var(y5 )
3. [30pts]
(a) [15pts]
EM Sou(xu,trt) =
=
1
dfou(xu,trt)
E SSou(xu,trt)
1
E
tnm − tn
t X
n X
m
X
!
(yijk − ȳij )2
i=1 j=1 k=1
t X
n X
m
X
1
E
([µ + τi + uij + eijk ] − [µ + τi + uij + ēij ])2
=
tnm − tn
i=1 j=1 k=1
!
t X
n
m
X
X
1
=
E
(eijk − ēij )2
tnm − tn i=1 j=1
k=1
t
n
XX
1
(m − 1) σe2
=
tnm − tn i=1 j=1
iid
since eijk ∼ N 0, σe2
1
tn (m − 1) σe2
tnm − tn
= σe2 .
=
(b) [15pts]
From slide 6, the sum of squares can be written as y 0 (I − P3 )y, where
−1
0
[Itn×tn ⊗ 1m×1 ]0
P3 = [Itn×tn ⊗ 1m×1 ] [Itn×tn ⊗ 1m×1 ] [Itn×tn ⊗ 1m×1 ]
−1
= [Itn×tn ⊗ 1m×1 ] mItn×tn
[Itn×tn ⊗ 1m×1 ]0
=
1
Itn×tn ⊗ 110m×m .
m
Let
A=
Itnm×tnm − m1 Itn×tn ⊗ 110m×m
I − P3
=
.
tnm − tn
tn(m − 1)
4
!
Then, by linearity of trace,
Itnm×tnm − m1 Itn×tn ⊗ 110m×m
2
0
2
tr(AΣ) = tr
σu Itn×tn ⊗ 11m×m + σe Itnm×tnm
tn(m − 1)
1
1
0
2
0
2
tr Itnm×tnm − Itn×tn ⊗ 11m×m σu Itn×tn ⊗ 11m×m + σe Itnm×tnm
=
tn(m − 1)
m
1
tr Itnm×tnm σu2 (Itn×tn ⊗ 110m×m ) + Itnm×tnm σe2 Itnm×tnm
=
tn(m − 1)
1
1
0
2
2
0
0
− (Itn×tn ⊗ 11m×m )(σu Itn×tn ⊗ 11m×m ) − (Itn×tn ⊗ 11m×m )σe Itnm×tnm
m
m
1
=
tr σu2 Itn×tn ⊗ 110m×m + σe2 Itnm×tnm
tn(m − 1)
σu2
σe2
0
0
0
− (Itn×tn Itn×tn ) ⊗ (11m×m 11m×m ) − Itn×tn ⊗ 11m×m
m
m
1
=
tr σu2 Itn×tn ⊗ 110m×m + σe2 Itnm×tnm
tn(m − 1)
σe2
σu2
0
0
− Itn×tn ⊗ m11m×m − Itn×tn ⊗ 11m×m
m
m
1
=
σ 2 tr(Itn×tn ⊗ 110m×m ) + σe2 tr(Itnm×tnm )
tn(m − 1) u
σe2
0
2
0
tr(Itn×tn ⊗ 11m×m )
− σu tr(Itn×tn ⊗ 11m×m ) −
m
1
σe2
2
0
=
σ tr(Itnm×tnm ) −
tr(Itn×tn ⊗ 11m×m )
tn(m − 1) e
m
1
σe2
2
=
σe (tnm) − (tnm)
since Itnm×tnm and Itn×tn ⊗ 110m×m have 1’s on the diagonal
tn(m − 1)
m
2
= σe ,
and
Itnm×tnm − m1 Itn×tn ⊗ 110m×m
E(y)
E(y) A E(y) = E(y)
tn(m − 1)
1
1
0
0
=
E(y) Itnm×tnm − Itn×tn ⊗ 11m×m E(y)
tn(m − 1)
m
1
1
0
0
0
E(y) Itnm×tnm − E(y) (Itn×tn ⊗ 11m×m ) E(y)
=
tn(m − 1)
m
1
1
0
0
E(y) − m E(y) E(y)
=
tn(m − 1)
m
1
00 E(y)
=
tn(m − 1)
= 0.
0
0
5
Now,
I − P3
y
=E y
tnm − tn
= E (y 0 Ay)
= tr(AΣ) + E(y)0 A E(y)
= σe2 + 0
= σe2 ,
EM Sou(xu,trt)
0
which is the same result as in part (a).
6
by slide 19 of set 12