Wooldridge J., Introductory Econometrics - A Modern Approach (Instructors Manual)
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regressions are the same, which means the residuals must be the same for all t. (The dependent
variable is the same in both regressions.) Therefore,
2
σ
%
=
2
ˆ
σ
. Further, as we showed in part (i),
(Z′Z)
-1
= A
-1
(X′X)
-1
(A′)
-1
, and so
2
σ
%
(Z′Z)
-1
=
2
ˆ
σ
A
-1
(X′X)
-1
(A
-1
)′, which is what we wanted to
show.
(iv) The
j
β
%
are obtained from a regression of y on XA, where A is the k × k diagonal matrix
with 1,
a
2
, K , a
k
down the diagonal. From part (i),
β
= A
%
-1
ˆ
β
. But A
-1
is easily seen to be the
k × k diagonal matrix with 1, , K ,
1
2
a
− 1
k
a
−
down its diagonal. Straightforward multiplication
shows that the first element of
A
-1
ˆ
β
is
1
ˆ
β
and the j
th
element is
ˆ
j
β
/a
j
, j = 2, K , k.
(v) From part (iii), the estimated variance matrix of
β
is
%
2
ˆ
σ
A
-1
(X′X)
-1
(A
-1
)′. But A
-1
is a
symmetric, diagonal matrix, as described above. The estimated variance of
j
β
%
is the j
th
diagonal element of
2
ˆ
σ
A
-1
(X′X)
-1
A
-1
, which is easily seen to be =
2
ˆ
σ
c
jj
/ , where c
2
j
a
−
jj
is the j
th
diagonal element of (
X′X)
-1
. The square root of this,
ˆ
jj
c
σ
/|a
j
|, is se(
j
β
%
), which is simply
se(
j
β
%
)/|a
j
|.
(vi) The t statistic for
j
β
%
is, as usual,
j
β
%
/se(
j
β
%
) = (
ˆ
j
β
/a
j
)/[se(
ˆ
j
β
)/|a
j
|],
and so the absolute value is (|
ˆ
j
β
|/|a
j
|)/[se(
ˆ
j
β
)/|a
j
|] = |
ˆ
j
β
|/se(
ˆ
j
β
), which is just the absolute value
of the t statistic for
ˆ
j
β
. If a
j
> 0, the t statistics themselves are identical; if a
j
< 0, the t statistics
are simply opposite in sign.
E.4 (i)
垐
E( | ) E( | ) E( | ) .
===δ XGβ XGβ XGβδ=
(ii)
21 2 1
垐
Var( | ) Var( | ) [Var( | )] [ ( ) ] [( ) ] .
σσ
−−
′
′′ ′′
== = =δ XGβ XG β XGG XXG GXXG
(iii) The vector of regression coefficients from the regression
y on XG
-1
is
1111 1 111
11 1
1
[( ) ] ( ) [( ) ] ( )
( ) [( ) ] ( )
ˆ
( ) ( ) ( ) .
−−−− − −−−
−− −
−
′ ′ ′′ ′′
=
′′ ′ ′ ′
=
′′′′ ′ ′′′
=
==
XG XG XG y G X XG G X y
GXX G G Xy
GXX G G Xy GXX Xy δ
Further, as shown in Problem E.3, the residuals are the same as from the regression
y on X, and
so the error variance estimate,
2
ˆ
,
σ
is the same. Therefore, the estimated variance matrix is
203
21112 1
垐
[( ) ] ( ) ,
σσ
−−− −
′
′′
=XG XG G X X G
which is the proper estimate of the expression in part (ii).
(iv) It is easily seen by matrix multiplication that choosing
123
1 0 0 ... 0
0 1 0 ... 0
...
0 ... 0 1 0
...
k
ccc c
⋅⋅ ⋅⋅
⋅⋅
⋅
⎛⎞
⎜⎟
⎜⎟
⎜⎟
=
⋅
⋅
⋅⋅
⎜⎟
⎜⎟
⎜⎟
⎜⎟
⎝⎠
G
⋅
k
does the trick: if δ = Gβ then
δ
j
=
β
j
, j = 1,…,k−1, and
11 2 2
... .
kk
cc c
δ
ββ β
=
+++
(v) Straightforward matrix multiplication shows that, for the suggested choice of
G
-1
,
Also by multiplication, it is easy to see that, for each t,
1
.
n
−
=GG I
1
11 2 2 ,1 1
[ ( / ) , ( / ) ,..., ( / ) , / ].
t t ktkt ktk tk k ktktk k
x
ccxx ccx x c cxx c
−
−−
=− − −xG
E.5 (i) By plugging in for y, we can write
11
() ()( ) ()
−−
′′′′ ′′
== +=+β ZX Zy ZX Z Xβ u β ZX Zu
%
1
.
−
1−
Now we use the fact that
Z is a function of X:
11
E(|) E[( ) |] ( ) E(|) .
−−
′′ ′′
=+ =+ =β X β ZX Zu X β ZX Z u X β
%
(ii) We start from the same representation in part (i):
11
12 12 1
Var( | ) ( ) [Var( | )] [( ) ]
( ) ( ) ( ) ( ) ( ) .
n
σσ
−−
−−−
′′ ′′
=
′′ ′ ′′′
==
β XZXZ uXZZX
ZX Z I Z XZ ZX ZZ XZ
%
(iii) The estimator
β
is linear in y and, as shown in part (i), it is unbiased (conditional on X).
Since the Gauss-Markov assumptions hold, the OLS estimator, , is best linear unbiased. In
particular, its variance-covariance matrix is “smaller” (in the matrix sense) than
Var
Therefore, we prefer the OLS estimator.
%
ˆ
β
( | ).β X
%
204