The mean squared error of the OLS estimator is the variance plus the squared bias,
M(b
β
)
=
(
σ
2
/n)
Q
XX
1
+
Q
XX
1
γγ′
Q
XX
1
the mean squared error of the 2SLS estimator equals its variance.
For OLS to be more precise then 2SLS,
we would have to have
(
σ
2
/n)
Q
XX
1
+
Q
XX
1
γγ′
Q
XX
1
<< (
σ
2
/n)
Q
ZX
1
Q
ZZ
Q
XZ
1
.
For convenience, let
δ
=
Q
XX
1
γ
so M(b
β
)
=
(
σ
2
/n)
Q
XX
1
+
δδ′
.
If the mean squared error matrix of the
OLS estimator is smaller than that of the 2SLS estimator, then its inverse is larger.
Use (A66) to do the
inversion.
The result would be
[(
σ
2
/n)
Q
XX
1
+
δδ′
]
1
>> [(
σ
2
/n)
Q
ZX
1
Q
ZZ
Q
XZ
1
]
1
Now, use A66
[(
σ
2
/n)
Q
XX
1
+
δδ′
]
1
= (n/
σ
2
)
Q
XX

2
1
1
(
/
)
n
′
+
σ
XX
Q
δ
δ
(n/
σ
2
)
Q
XX
δδ′
(n/
σ
2
)
Q
XX
Reinsert
δ
=
Q
XX
1
γ
and the right hand side above reduces to
(n/
σ
2
)
Q
XX

2
1
1
(
/
)
n
′
+
σ
1
XX
Q
γ
γ
(n/
σ
2
)
2
γγ′
Therefore, if the mean squared error matrix of OLS is smaller, then
(n/
σ
2
)
Q
XX

2
1
1
(
/
)
n
′
+
σ
1
XX
Q
γ
γ
(n/
σ
2
)
2
γγ′
>>
(n/
σ
2
)
Q
XZ
Q
ZZ
1
Q
ZX
Collect the terms, and this implies
(n/
σ
2
)[
Q
XX

Q
XZ
Q
ZZ
1
Q
ZX
] >>
2
1
1
(
/
)
n
′
+
σ
1
XX
Q
γ
γ
(n/
σ
2
)
2
γγ′
divide both sides by (n/
σ
2
),
Q
XX

Q
XZ
Q
ZZ
1
Q
ZX
>>
2
2
(
/
)
1
(
/
)
n
n
σ
′
+
σ
1
XX
Q
γ
γ
γγ′
and divide numerator and denominator of the fraction by n/
σ
2
Q
XX

Q
XZ
Q
ZZ
1
Q
ZX
>>
2
1
(
/
)
n
′
σ
+
1
XX
Q
γ
γ
γγ′
which is the desired result.
Is it possible?
It is possible, since
Q
XX

Q
XZ
Q
ZZ
1
Q
ZX
=
plim (1/n)[
X
′
X

X
′
Z
(
Z
′
Z
)
1
Z
′
X
]
=
plim (1/n) X
′
M
Z
X
which is a positive definite matrix.
SInce
γ
varies independently of
Z
and
X
, certainly there is some
configuration of the data and parameters for which this is the case.
The result is that it is, indeed, possible
for OLS to be more precise, in the mean squared error sense, than 2SLS.
5.
The matrices are X = [i,x] and Z = [i,z].
For the OLS estimators, we know from chapter 2 that
a
=
y
bx
−
and b = Cov[x,y]/var[x].
For the IV estimator, (
Z
′
X
)
1
Z
′
y
, we obtain the result in detail.
Given the forms,
1
1
1
1
1
1
1
1
1
1
1
1
1
(
)
, (
)
,
(
)
i
z
i
n
x
n
nx
n x
nx
ny
n
x
n
n x
n
n
n y
nn
x
x
−
=
Σ
−
⎡
⎤
⎡
⎤
⎡
⎤
⎡
′
′
=
=
=
=
⎢
⎥
⎢
⎥
⎢
⎥
⎢
Σ
−
−
⎣
⎦
⎣
⎦
⎣
⎣
⎦
Z X
Z X
Z y
⎤
′
⎥
⎦
where subscript 1 indicates the mean of the observations for which z equals 1, and n
1
is the number of
observations.
Multiplying the matrix times the vector and cancelling terms produces the solutions
a
IV
=
1
1
1
1
1
and
IV
IV
x y
x y
y
y
a
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 Dr.Fang
 Regression Analysis, Standard Deviation, Standard Error, Standard Error b/St.Er.P

Click to edit the document details