Seminar 2
Seemingly Unrelated Regression (SUR) II
Petr Gapko
([email protected])
2
seminar2.nb
Problem #5 from Baltagi, Chapter 10
`
(a) Show that varHb12,OLS L =
s11
mx1 x1
`
and varHb22,OLS L =
s22
mx2 x2
, where mxi x j = ⁄Tt=1 IXit - Xi M IX jt - X j M for i, j = 1, 2.
Solution:
`
-1
i iT
b12,OLS = IX1T A X1 M X1T A Y1 , A = JIT - T N, X1T A X1 = ⁄Tt=1 IX1 t - X1 M IX1 t - X1 M = mx1 x1
`
`
-1
s
varHb12,OLS L = s11 IX1T A X1 M = m 11 . Similarly we proof that varHb22,OLS L =
x1 x1
s22
mx2 x2
seminar2.nb
`
b12,GLS
s22 mx1 x1 -s12 mx1 x2
2
var `
= Is11 s22 - s12
M
-s12 mx1 x2 s11 mx2 x2
b22,GLS
(b)
-1
.
Deduce
that
`
2
2
varHb12,GLS L = Is11 s22 - s12
M s11 mx2 x2 ë Is11 s22 mx2 x2 mx1 x1 - s12
m2x1 x2 M
`
2
2
varHb22,GLS L = Is11 s22 - s12
M s22 mx1 x1 ë Is11 s22 mx1 x1 mx2 x2 - s12
m2x1 x2 M
Solution:
`
b12,GLS
`
b22,GLS
=
=
X1T A
0
0
X2T A
X1T A
0
0
X2T A
s 11 Ä⊗I s 12 Ä⊗I
s 21 Ä⊗I s 22 Ä⊗I
s 11 s 12
s 21 s 22
A X1 0
K
O
0 A X2
s 11 X1T A X1 s 12 X1T A X2
=
-1
s 21 X2T A X1 s 22 X2T A X2
`
b12,GLS
s11 X1T A X1
var `
=
s21 X2T A X1
b22,GLS
s11 mx1 x1 s12 mx1 x2
21
22
s mx2 x1 s mx2 x2
`
varHb12,GLS L =
s22 X2T A X2
m2x1 x2
-1
-1
X1T A
0
0
X2T A
X1T A
0
0
X2T A
s 11 s 12
s 21 s 22
s 11 Ä⊗I s 12 Ä⊗I
s 21 Ä⊗I s 22 Ä⊗I
K
K
Y1
O
Y2
Y1
O
Y2
s 21 X2T A Y1 s 22 X2T A Y2
2
= Is11 s22 - s21
M
s11 s22 mx1 x1 mx2 x2 -s221
A X1 0
O
0 A X2
s 11 X1T A Y1 s 12 X1T A Y2
s12 X1T A X2
Is11 s22 -s221 M s11 mx2 x2
K
and
-1
=
s22 mx1 x1 -s12 mx1 x2
-s21 mx2 x1 s11 mx2 x2
`
and similarly varHb22,GLS L =
-1
=
s11 s22 -s221
s11 s22 mx1 x1 -s221 m2x1 x2
Is11 s22 -s212 M s22 mx1 x1
Is11 s22 mx1 x1 mx2 x2 -s212 m2x1 x2 M
s11 mx2 x2 s12 mx1 x2
s21 mx2 x1 s22 mx1 x1
3
4
seminar2.nb
(c) Using r = s12 ë Hs11 s22 L1ê2 and r = mx1 x2 ë Hmx1 x1 mx2 x2 L1ê2 and the results in part (a) and (b), show that
`
`
varHb12,GLS L í varHb12,OLS L = I1 - r2 M ë I1 - r2 r2 M.
Solution:
`
`
varHb12,GLS L í varHb12,OLS L =
`
varHb12,GLS L =
Is11 s22 -s221 M s11 mx2 x2
s11 s22 mx1 x1 mx2 x2 -s221 m2x1 x2
ì
s22
mx1 x1
=
Is11 s22 -s221 M mx1 x1 mx2 x2
s11 s22 mx1 x1 mx2 x2 -s221 m2x1 x2
s212 K
=
s212
1
r2
-1O
2
1 mx1 x2
r2
r2
m2x x
1 2
1
r2
r2 r2
-s212 m2x1 x2
=
-
1
r2 r2
1
r2
-1
=
1-r2
1-r2 r2
seminar2.nb
5
`
`
(d) Differentiate varHb12,GLS L í varHb12,OLS L = I1 - r2 M ë I1 - r2 r2 M with respect to q = r2 and show that this expression is a
non-increasing function of q. Similarly, differenetiate the expression with respect to l = r2 and show that it is a non-decreasing function of l. Finally, compute this efficiency measure for various values of r2 and r2 between 0 and 1 at 0,1 intervals.
Solution:
¶∂ J
1- r 2
1- r 2 r2
N í ¶∂ r 2 =
-I1- r 2 r2 M+I1- r 2 M Ir2 M
2
I1- r 2 r2 M
=
r2 -1
2
I1-r2 r2 M
. Both r2 and r2 are from interval < 0, 1 > and thus the expression is non-
positive for all values r2 and r2 . Moreover the sign at r2 is negative and this with the power of two causes that the expression is a non-increasing function in r2 .
¶∂ J
1- r 2
1- r 2 r2
N í ¶∂ r2 =
r2 I1-r2 M
. As both r2 and r2 are from interval < 0, 1 > this function will be non-negative
2
I1-r2 r2 M
for all values of r2 and r2 and thus the bigger r2 the bigger value of the function.
Problem #8 from Baltagi, Chapter 10
a) Derive the GLS estimator for SUR with unequal number of observations given by:
`
bGLS =
-1
s12 X1 ' X2*
s11 X1 ' X1
s11 X1 ' y1 + s12 X1 ' y*2
s22 X2* ' X2* + IX20 ' X20 ë s22 M
s12 X2* ' X1
, where * denotes T
s12 X2* ' y1 + s22 X2* ' y*2 + IX20 ' y02 ë s22 M
observations common for both datasets and 0 denotes extra N observations for the second dataset.
Solution:
First we need to define the omega matrix. From Baltagi this is block diagonal:
0
X1 0
y1
s11 IT s12 IT
*
*
0
-1
W = s12 IT s22 IT
. Moreover we have: X = 0 X2
and y = y2 .
1
IN
0 X20
y02
0
0
s22
We can write:
T
h
X1 '
0
0
0
X2* '
X20 '
e
r
`
-1
bGLS = IX ' W-1 XM X ' W-1 y =
0
0
s11 IT s12 IT
12
22
s IT s IT
0
0
e
f
11
s X1 ' X1
s12 X2* ' X1
1
s22
o
X1
0
0
IN
0
s11 X1 '
X2* =
s12 X2* '
X20
r
s X1 ' X2*
s22 X2* ' X2* + IX20 '
s12 X1 '
0
1
s22
s22 X2* '
X20 '
.
e
-1
12
s11 X1 ' y1 + s12 X1 ' y*2
X20 ë s22 M
s12 X2* ' y1 + s22 X2* ' y*2 + IX20 ' y02 ë s22 M
(b) Show that if s12 = 0, SUR with unequal number of observations reduces to OLS on each equation separately.
Solution:
s11 IT
0
0
-1
0
s22 IT
0
If s12 = 0 we have: W =
0
0
s22 IN
sii =
1
sii
1
s11
and thus
W-1 =
IT
0
1
s22
0
0
0
0
IT
1
s22
0
so that s12 = 0 and
IN
for i = 1,2.
From previous example we have
`
-1
bGLS = IX ' W-1 XM X ' W-1 y =
`
-1
bGLS = IX ' W-1 XM X ' W-1 y =
X1 ' X1
s11
0
X2* '
X2*
s22
`
HX1 ' X1 L X1 ' y1
b1,GLS
= `
-1
IX2* ' X2* + X20 ' X20 M IX2* ' y*2 + X20 ' y02 M
b2,GLS
-1
-1
0
+
X1 ' y1
s11
0
X2 '
0
X2
s22
0 0
X2* ' y*2 +X2 ' y2
s22
and
.