Student Notes To Accompany MS4214: STATISTICAL INFERENCE

Also
implies that
S2(µ, v) = ∂ℓ
∂v
ˆv = 1
n
n�
i=1
n 1
= − +
2v 2v2 (xi − ˆµ) 2 = 1
n
n�
(xi − µ) 2 = 0
i=1
n�
(xi − ¯x) 2 . (2.5)
Calculating second derivatives and multiplying by −1 gives that the information matrix
I(µ, v) equals
I(µ, v) =
⎛
⎜
⎝
1
v 2
i=1
i=1
n
1
v
v2 n�
(xi − µ)
i=1
n�
(xi − µ) − n
2v2 + 1
v3 n�
(xi − µ) 2
⎞
⎟
⎠
Hence I(ˆµ, ˆv) is given by : �
n
ˆv 0
0 n
2v2 �
Clearly both diagonal terms are positive and the determinant is positive and so (ˆµ, ˆv)
are, indeed, the MLEs of (µ, v).
Go back to equation (2.4), and ¯ X ∼ N (µ, v/n). Clearly E( ¯ X) = µ (unbiased) and
Var( ¯ X) = v/n, so ¯ X achieved the CRLB. Go back to equation (2.5). Then from lemma
2.9 we have
so that
nˆv
v ∼ χ2 n−1
i=1
� �
nˆv
E = n − 1
v
� �
n − 1
⇒ E(ˆv) = v
n
Instead, propose the (unbiased) estimator
Observe that
E(˜v) =
˜v =
n
ˆv =
n − 1
� �
n
E(ˆv) =
n − 1
1
n − 1
n�
(xi − ¯x) 2
i=1
� n
n − 1
� � n − 1
and ˜v is unbiased as suggested. We can easily show that
Hence
Var(˜v) =
2v 2
(n − 1)
n
�
v = v
(2.6)
eff(˜v) = 2v2 2v2 1
÷ = 1 −
n (n − 1) n
Clearly ˜v is not efficient, but is asymptotically efficient. �
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