Student Notes To Accompany MS4214: STATISTICAL INFERENCE

Sufficient conditions for the proof of CRLB are that all the integrands are finite,
within the range of x. We also require that the limits of the integrals do not depend
on θ. That is, the range of x, here f(x|θ), cannot depend on θ. This second condition
is violated for many density functions, i.e. the CRLB is not valid for the uniform
distribution. We can have absolute assessment for unbiased estimators by comparing
their variances to the CRLB. We can also assess unbiased estimators. If its variance is
lower than CRLB then it is indeed a very good estimate, although it is bias.
Example 2.1. Consider IID random variables Xi, i = 1, . . . , n, with
fXi (xi|µ) = 1
µ exp
�
− 1
µ xi
�
.
Denote the joint distribution of X1, . . . , Xn by
n�
f = fXi (xi|µ)
� �n 1
= exp
µ
so that
i=1
ln f = −n ln(µ) − 1
µ
�
− 1
µ
n�
xi.
The score function is the partial derivative of ln f wrt the unknown parameter µ,
S(µ) = ∂
ln f = −n
∂µ µ + 1
µ 2
n�
xi
and
E {S(µ)} = E
�
− n
µ + 1
µ 2
n�
i=1
Xi
�
i=1
= − n
µ + 1
i=1
n�
i=1
xi
�
µ 2 E
�
n�
�
Xi
i=1
For X ∼ Exp(1/µ), we have E(X) = µ implying E(X1 + · · · + Xn) = E(X1) + · · · +
E(Xn) = nµ and E {S(µ)} = 0 as required.
I(θ) =
� �
∂
−E −
∂µ
n
µ + 1
µ 2
=
n�
i=1
�
n
−E
µ 2 − 2
µ 3
=
n�
�
Xi
i=1
− n
µ 2 + 2
µ 3 E
�
n�
�
Xi
Hence
i=1
= − n
µ 2 + 2nµ
µ 3 = n
µ 2
CRLB = µ2
n .
17
Xi
��
,