Chapter 18
The Generalized Method of Moments
1
.
For the normal distribution
µ
2
k
=
σ
2
k
(2
k
)!/(
k
!2
k
) and
2
k
+1
= 0,
k
= 0, 1, . . . . Use this result to analyze
the two estimators
3
4
12
3/2
2
22
and
m
m
bb
mm
==
.
where
1
1
(
n
ki
i
n
x
=
=Σ
−
)
k
x
m
.
The following result will be useful:
211
11
.[
,
]
j
k
j
k
j
k
jk
k j
Asy Cov
nm
nm
jk
j
k
µµµ
+−
−
−
+
=− +
−
−
.
−
+
Use the delta method to obtain the asymptotic variances and covariance of these two functions assuming
the data are drawn from a normal distribution with mean
and variance
2
. (Hint: Under the assumptions,
the sample mean is a consistent estimator of
, so for purposes of deriving asymptotic results, the
difference between
x
and
may be ignored. As such, no generality is lost by assuming the mean is zero,
and proceeding from there.
Obtain
V
, the 3
×
3 covariance matrix for the three moments, then use the delta
method to show that the covariance matrix for the two estimators is
60
02
4
′ =
JVJ
where
J
is the 2
×
3 matrix of derivatives.
The elements of
J
are
23
5/2
(3
/2
)
32
2
0
m
−−
−
∂∂
1
4
b
m
=
∂
∂
(2
)
42
0
m
b
m
−
∂
∂
2
4
2
=
Using the formula given for the moments, we obtain,
µ
2
=
σ
2
,
µ
3
= 0,
µ
4
= 3
σ
4
.
Insert these in the
derivatives above to obtain
3
00
24
σσ
−
−
=
J
.
Since the rows of J are orthogonal, we know that the off diagonal term in
JVJ
′
will be zero, which
simplifies things a bit.
Taking the parts directly, we can see that the asymptotic variance of
1
b
will be
σ
6
Asy.Var[m
3
], which will be
Asy.Var[
1
b
]
=
σ
6
(
µ
6

µ
3
2
+ 9
µ
2
3
 3
µ
2
µ
4
 3
µ
2
µ
4
).
The parts needed, using the general result given earlier, are
µ
6
= 15
σ
6
,
µ
3
= 0,
µ
2
=
σ
2
,
µ
4
= 3
σ
4
.
Inserting
these in the parentheses and multiplying it out and collecting terms produces the upper left element of JVJ
′
equal to 6, which is the desired result.
The lower right element will be
Asy.Var[b
2
] = 36
σ
4
Asy.Var[m
2
] +
σ
8
Asy.Var[m
4
]  2(6)
σ
6
Asy.Cov[m
2
,m
4
].
The needed parts are
Asy.Var[m
2
] = 2
σ
4
93