Econ 741 2010 Spring
Suggested answers for Exercise 1
(1)
(a) The formal proof is simple but tedious, so I skip it here.
(b) Assume AB = I and since AA1 = I we know B = A1 .
!
!
A11 A12
B1 B3
1
=
A21 A22
B2 B4
0
0
1
!
.
(1)
From equation (1), we can write
Econ 741 2010 Spring
Suggested answers for Exercise 4
1.)
y1 = x1 + u1
(1)
y2 = 2 y1 + x2 + u2
(2)
From (1), we get the least square estimate of 1 , 1 = (x0 x)1 x0 y1 .
plimn (1 1 ) = plimn (x0 x)1 x0 u1 = plimn
x0 x
n
1
x0 u1
n
= 0.
Therefore, 1 is a co
Econ 741 2010 Spring
Suggested answers for Exercise 2
(1)
(a) From the restriction R1 1 + R2 2 = q, we get 1 = R11 (q R2 2 ). Then brings it back
into the regression model, we get
y = X1 1 + X2 2 + u
= X1 R11 (q R2 2 ) + X2 2 + u
y X1 R11 q = X1 R11 R2 2
\ , .,
l
x 1
,~-.,W V. \
A x
1 i M3
2 J \ i as bl
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Spring 2010 C "544 2
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#9
(1) Let A and B be two partitioned matrices as
_ A1 A2 _ Bl Ba
A<A3 A4) AB; (Bz B4)-
6"
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a
is
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I; K 51
Show that
AB = <