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Statistics 358b : Comments on Fisher’s
Information
February 9, 2011
We showed under the regularity conditions,
f
being pdf (or pmf and sum
mation accordingly)
E
θ
0
[
∂
log (
f
(
X
;
θ
0
))
∂θ
]
=
∫
R
[
∂
log (
f
(
x
;
θ
0
))
∂θ
]
f
(
x
;
θ
0
)
dx
=
∫
R
[
1
f
(
x, θ
0
)
∂f
(
x
;
θ
0
)
∂θ
]
f
(
x
;
θ
0
)
dx
=
∫
R
∂f
(
x
;
θ
0
)
∂θ
dx
=
∂
±
R
f
(
x
;
θ
)
dx
∂θ
²
²
²
²
θ
=
θ
0
=
∂
(1)
∂θ
=
0
Notice the integration is done with respect to
θ
0
, and this ﬁnal statement is
therefore true for all
θ
0
. Thus in our statement we can also write
∫
R
[
∂
log (
f
(
x
;
θ
))
∂θ
]
f
(
x
;
θ
)
dx
= 0
(1)
for all
θ
.
Deﬁne I(
θ
) in both the 1D and higher dimensional parameter case, that is
as a matrix.
Re : Lemma 8.5.2 A proof.
Remark : Equation (1) is true for all
θ
but it is not true that
∂
Λ(
θ
)
∂θ
= E
θ
0
[
∂
log (
f
(
X
;
θ
))
∂θ
]
=
∫
R
[
∂
log (
f
(
x
;
θ
))
∂θ
]
f
(
x
;
θ
0
)
dx
is equal to 0 for all
θ
. In each integral the integrand is a product of two func
tions, one with
parameters
θ
and
θ
0
(not necessarily matching) and in the other
integral with
parameters
θ
and
θ
(necessarily matching). If we calculate the
1
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2
derivative is this second integral, which is valid, it is not the case that the
derivative is equal to 0. However if we diﬀerentiate (1) with respect to
θ
, it is
the case that this derivative is equal to 0.
The correct method is therefore to diﬀerentiate (1).
For real
θ
∂
∂θ
E
θ
[
∂
log (
f
(
X
;
θ
))
∂θ
]
=
∂
∂θ
∫
R
[
∂
log (
f
(
x
;
θ
))
∂θ
]
f
(
x
;
θ
)
dx
=
∂
∂θ
0 = 0
This last line is not true if we calculate
E
θ
0
[
∂
log (
f
(
X
;
θ
))
∂θ
]
=
∫
R
[
∂
log (
f
(
x
;
θ
))
∂θ
]
f
(
x
;
θ
0
)
dx
even though the diﬀerentiation is valid.
Remark : A variance matrix (also called a variancecovariance or covariance
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 Spring '11
 qqqq

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