Student Notes To Accompany MS4214: STATISTICAL INFERENCE

As this example demonstrates, maximum likelihood estimation does not automati-
cally produce unbiased estimates. If it is thought that this property is (in some sense)
desirable, then some adjustments to the MLEs, usually in the form of scaling, may be
required. We conclude this example with the following tedious (but straightforward)
calculations.
�
1
E
Z2 �
We have already calculated that
However,
�∞
1
= θ
Γ(n)
0
n z n−3 exp −θzdz
= θ2
�∞
u
Γ(n)
n−3 exp −θudu
0
= θ2 Γ(n − 2)
Γ(n)
=
θ 2
(n − 1)(n − 2)
⇒ Var[ ˜ θ] = E[ ˜ θ 2 �
] − E[ ˜ � �
2
2 (n − 1)
θ] = E
Z2 �
− θ 2
=
(n − 1) 2θ2 (n − 1)(n − 2) − θ2 = θ2
n − 2 .
I(θ) = n
θ2 ⇒ E [I(θ)] = n
θ
eff( ˜ θ) =
(E [I(θ)])−1
Var[ ˜ θ]
2 �=
θ2 θ2
= ÷
n n − 2
�
Var[ ˜ �−1 θ] .
= n − 2
n
which although not equal to 1, converges to 1 as n → ∞, and ˜ θ is asymptotically
efficient. �
Example 2.8 (Lifetime of a component). The time to failure T of components has an
exponential distributed with mean µ. Suppose n components are tested for 100 hours
and that m components failed at times t1, . . . , tm, with n − m components surviving
the 100 hour test. The likelihood function can be written
m� 1
L(θ) =
µ
i=1
e−ti/µ
n�
P (Tj > 100) .
j=m+1
� �� � � �� �
components failed components survived
Clearly P (T ≤ t) = 1 − e −t/µ implies P (T > 100) = e −100/µ is the probability of a
component surviving the 100 hour test. Then
L(µ) =
�
m�
1
µ e−ti/µ
�
�e−100/µ �n−m
,
i=1
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