11.
The asymptotic variance of the MLE is, in fact, equal to the CramerRao Lower Bound for the variance
of a consistent, asymptotically normally distributed estimator, so this completes the argument.
In example 4.9, we proposed a regression with a gamma distributed disturbance,
y
i
=
α
+
x
i
′
β
+
ε
i
where,
f
(
i
)
=
[
λ
P
/
Γ
(
P
)]
i
P
1
exp(
λ
i
),
i
>
0,
λ
> 0,
P
> 2.
(The fact that
i
is nonnegative will shift the constant term, as shown in Example 4.9.
The need for the
restriction on
P
will emerge shortly.)
It will be convenient to assume the regressors are measured in
deviations from their means, so
Σ
i
x
i
=
0
.
The OLS estimator of
remains unbiased and consistent in this
model, with variance
V
a
r
[
b

X
] =
σ
2
(
X
′
X
)
1
where
σ
2
= Var[
i

X
] =
P
/
λ
2
.
[You can show this by using gamma integrals to verify that
E
[
i

X
] =
P
/
λ
and
E[
i
2

X
] =
P
(
P
+1)/
λ
2
.
See B39 and (E1) in Section E2.3.
A useful device for obtaining the variance is
Γ
(
P
) = (
P
1)
Γ
(
P
1).]
We will now show that in this model, there is a more efficient consistent estimator of
.
(As we saw in Example 4.9, the constant term in this regression will be biased because
E
[
i

X
] =
P
/
λ
;
a
estimates
α
+
P
/
λ
.
In what follows, we will focus on the slope estimators.
The log likelihood function is
L
n
L
=
1
ln
ln ( )
(
1)ln
n
ii
i
PP
P
=
λ
−Γ + − ε
−
λ
ε
∑
The likelihood equations are
∂
ln
L
/
∂α
=
Σ
i
[(
P
1)/
i
+
λ
] = 0,
∂
ln
L
/
∂β
=
Σ
i
[(
P
1)/
ε
i
+
λ
]
x
i
=
0
,
∂
ln
L
/
∂λ
=
Σ
i
[
P
/
λ

ε
i
] = 0,
∂
ln
L
/
∂
P
=
Σ
i
[ln
λ

ψ
(
P
) 
i
] = 0.
The function
ψ
(
P
) = dln
Γ
(
P
)/d
P
is defined in Section E2.3.)
To show that these expressions have
expectation zero, we use the gamma integral once again to show that
E
[1/
i
] =
λ
/(
P
1).
We used the result
E
[ln
i
] =
ψ
(
P
)
λ
in Example 15.5.
So show that
E
[
∂
ln
L
/
∂β
] =
0
, we only require
E
[1/
i
] =
λ
/(
P
1) because
x
i
and
i
are independent.
The second derivatives and their expectations are found as follows:
Using the
gamma integral once again, we find
E
[1/
i
2
] =
λ
2
/[(
P
1)(
P
2)].
And, recall that
Σ
i
x
i
=
0
.
Thus, conditioned
on
X
, we have

E
[
∂
2
lnL/
∂α
2
]
=
E
[
Σ
i
(
P
1)(1/
i
2
)]
=
n
λ
2
/(
P
2),

E
[
∂
2
lnL/
∂α∂
]
=
E
[
Σ
i
(
P
1)(1/
i