Seminar 3
Seemingly Unrelated Regression (SUR) II
Petr Gapko
([email protected])
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2
seminar3_SUR2.nb
Problem #5 from Baltagi, Chapter 10

(a) Show that var1,OLS  
11
mx1 x1

and var2,OLS  
22
mx2 x2
, where mxi x j  Tt1 Xit  Xi  X jt  X j  for i, j  1, 2.
Solution:

1
i iT
1,OLS  X1T A X1  X1T A Y1 , A  IT  T , X1T A X1  Tt1 X1 t  X1 X1 t  X1   mx1 x1




1
var1,OLS   11 X1T A X1   m 11 . Similarly we proof that var2,OLS   m 22
x1 x1
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x2 x2
seminar3_SUR2.nb

1,GLS
22 mx1 x1 12 mx1 x2
var 
 11 22  212 

11 mx2 x2
12 mx1 x2
2,GLS
(b)
1
.
Deduce
that

var1,GLS   11 22  212  11 mx2 x2  11 22 mx2 x2 mx1 x1  212 m2x1 x2 

var2,GLS   11 22  212  22 mx1 x1  11 22 mx1 x1 mx2 x2  212 m2x1 x2 
Solution:

1,GLS


2,GLS
X1T A
0
X1T A
0
X2T
11 I 12 I
21
A
11 12
0
X2T
0
A
21
I
11 X1T A X1 12 X1T A X2
1
21 X2T A X1 22 X2T A X2
11 22  221 

var1,GLS  
I
A X1
0
0 A X2
22

1,GLS
11 X1T A X1
var 

21 X2T A X1
2,GLS
22
0
1
22 X2T A X2
m2x1 x2
X1T A
0
X2T
0
11 12
0
X2T
21
A
22
11 I 12 I
21
A
I
22
I
Y1

Y2
21 X2T A Y1  22 X2T A Y2
22 mx1 x1 12 mx1 x2
21 mx2 x1 11 mx2 x2
11 22 mx1 x1 mx2 x2 221
1
X1T A
11 X1T A Y1  12 X1T A Y2
12 X1T A X2
11 22 221  11 mx2 x2
1
A X1
0
0
A X2
and

11 mx1 x1 12 mx1 x2
1
21 mx2 x1 22 mx2 x2
1

11 22 221
11 22 mx1 x1 mx2 x2 221

and similarly var2,GLS  
m2x1 x2

11 mx2 x2 12 mx1 x2
21 mx2 x1 22 mx1 x1
11 22 212  22 mx1 x1
11 22 mx1 x1 mx2 x2 212 m2x1 x2 
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Y1

Y2
3
4
seminar3_SUR2.nb
(c) Using   12  11 22 12 and r  mx1 x2  mx1 x1 mx2 x2 12 and the results in part (a) and (b), show that


var1,GLS  var1,OLS   1  2  1  2 r2 .
Solution:


var1,GLS  var1,OLS  
11 22 221  11 mx2 x2
11 22 mx1 x1 mx2 x2 221
m2x1 x2

11
mx1 x1

11 22 221  mx1 x1 mx2 x2
11 22 mx1 x1 mx2 x2 221
m2x1 x2

212 
212
1
2
1
1
m2x x
1 2
2
r2
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m2x
1
1 x2
r2
212 m2x1 x2


2 r 2
1
2 r 2
1
r2
1

12
12 r2
seminar3_SUR2.nb
5


(d) Differentiate var12,GLS   var12,OLS   1  2  1  2 r2  with respect to   2 and show that this expression is a non-increasing function of . Similarly, differenetiate the expression with respect to   r2 and show that
it is a non-decreasing function of . Finally, compute this efficiency measure for various values of 2 and r2
between 0 and 1 at 0,1 intervals.
Solution:

1  2
1  2 r
   2 
2
 1  2 r2 1  2  r2 
1  r 
2 2 2

r2 1
12 r2 
2
. Both 2 and r2 are from interval  0, 1  and thus the expression
is non-positive for all values 2 and r2 . Moreover the sign at 2 is negative and this with the power of two causes
that the expression is a non-increasing function in 2 .

1  2
2 2
1  r
   r2 
2 12 
12 r2 
2
. As both 2 and r2 are from interval  0, 1  this function will be non-
negative for all values of 2 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:

GLS 
1
12 X1 ' X2
11 X1 ' X1
22 X2 ' X2  X20 ' X20  22 
12 X2 ' X1
11 X1 ' y1  12 X1 ' y2
12 X2 ' y1  22 X2 ' y2  X20 ' y02  22 
, where *
denotes T 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
11 IT 12 IT


0
1
12
22
   IT  IT
. Moreover we have: X  0 X2
and y  y2 .
1
0 X20
y02
IN
0
0
22
We can write:
X1 '
0
0
0
X2 '
X20 '
12 IT 22 IT
0
0

1
Therefore GLS  X ' 1 X  X ' 1 y 
11 X1 ' X1
12 X2 ' X1
0
0
11 IT 12 IT
1
22
12 X1 ' X2

IN
11 X1 '
12
12 X1 '
X2 '
22
0
1
22
X2 '
.
X20 '
.
22 X2 ' X2  X20 ' X20  22 
1
11 X1 ' y1  12 X1 ' y2
12 X2 ' y1  22 X2 ' y2  X20 ' y02  22 
(b) Show that if 12  0, SUR with unequal number of observations reduces to OLS on each equation separately.
Solution:
11 IT
0
0
0
22 IT
0
If 12  0 we have: 1 
0
0
22 IN
and ii 
1
ii
1
11
and thus
1

IT
0
1
22
0
0
IT
0
0
0
1
22
so that 12  0
IN
for i = 1,2.

1
From previous example we have GLS  X ' 1 X  X ' 1 y 

1
and GLS  X ' 1 X  X ' 1 y 
X1 ' X1
11
0

1
0
X2 ' X2
22

X20 ' X20
22

1,OLS
X1 ' X1 1 X1 ' y1
 
1
X2 ' X2  X20 ' X20  X2 ' y2  X20 ' y02 
2,OLS
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
X1 ' y1
11
X2 ' y2 X20 ' y02
22