15
Solutions
•
∂ log p
−2(x − θ)
=
∂θ
1 + (x − θ)2
1
I(θ) =
π
Z −2(x − θ)
1 + (x − θ)2
2
1
4
dx =
2
1 + (x − θ)
π
•
f (θ, x1 , . . . , xn ) = θn exp[−θ
X
Z
1
x2
dx =
2
3
(1 + x )
2
xi ]
n X
∂ log f (θ, x1 , . . . , xn )
= −
xi
∂θ
θ
i
n
θ̂ = P
xi
Z
E[θ̂] =
It is biased and t̂ =
2 2
V ar(t̂) = (n − 1) θ
∞
0
n−1
θ̂
n
n θn
nθ
exp[−θ t]tn−1 dt =
t Γ(n)
n−1
is unbiased.
1
1
θ2
1
θ2
−
≥
=
=
2
(n − 1)(n − 2) (n − 1)
n−2
n
nI(θ)
It is asymptotically efficient.
• MLE is the Median. By symmetry it is unbiased.
∂ log f
= −sign(x − θ)
∂θ
I(θ) = 1
Cramer-Rao lower bound is n1 . Asymptotic variance of the MLE is
1
4
=
2
n[f (θ)]
n
36
• A statistic is T (0), . . . , T (n). Unbiased means
X n 3 j 1 n−j
j
and
j
4
4
X n 1 j 3 n−j
j
j
4
4
T (j) =
3
4
T (j) =
1
4
Two equations. n + 1 unknowns. Lots of solutions. Easy to construct
unbiased estimators with variance that is very small. For example if n
is odd T (x) = a if x < n2 and T (x) = b if x > n2 can be made unbiased
by proper choice of a and b. If we denote by
X n 3 j 1 n−j X n 1 j 3 n−j
pn =
=
j 4
j 4
4
4
n
n
j< 2
j> 2
we need
apn + b(1 − pn ) =
and
1
4
3
4
n −1
and b = 4p
. The variance is given by
8pn −4
a(1 − pn ) + bpn =
giving us a =
4pn −3
8pn −4
1
1
σ 2 = pn (a − )2 + (1 − pn )(b − )2
4
4
and is seen to be very very small for large n.
• The log-likelihood is
n
1 X
1 X 2 X
n
n
− log θ −
(xi − θ)2 = − log θ −
xi +
xi − θ
2
2θ
2
2θ
2
i
Clearly Un =
1
n
P
x2i is sufficient and the likelihood equation is
1 Un
− + 2 −1=0
θ
θ
37
or
θ2 + θ = Un
This gives
r
1
1
+ Un
θn = − +
2
4
which is consistent because
r
1
1
− +
+ θ2 + θ = θ
2
4
Has an asymptotic variance
1
var (x2 )[f 0 (θ2 + θ)]2
n
with
1
f (y) = − +
2
and
f 0 (θ2 + θ) =
r
1
+y
4
1
2θ + 1
2
2θ
On simplification this reduces to n1 (1+2θ)
The Cramer-Rao lower
2.
bound is eaxctly the same. The eficiency of the mean with variance nθ
2θ
is given by (1+2θ)
2.
• The log-likelihood function is
−n log Γ(p) −
X
xi + (p − 1)
X
Tha likelihood equation is
1X
Γ0 (p)
= G(p) =
log xi
Γ(p)
n
and the MLE is
θ̂n = G−1 (
It is consistent because
1X
log xi )
n
θ̂n → G−1 (m)
38
log xi
Z
where
m=
and
1 −x p−1
e x log xdx = G(p)
Γ(p)
G−1 (G(p)) = p
θ̂n is asymptotically normal with variance
is calculate easily as
var (log xi )
n[G0 (p)]2
. The quantity I(p)
I(p) = E[(log x − G(p))2 ] = var (log x) = G0 (p)
• To test f0 (x) = 2xR against f1 (x) = 2(1−x) the
critical region is 1−x
>c
x
√
c
2
or x < c. Size is 0 2xdx = c = α or c = α. Power is calculated to
be
Z √α
√
2(1 − x)dx = 2 α − α
0
• The critical region can be any subset of [0, 12 ] if α < 12 or the entire
[0, 12 ] along with a subset of [ 12 , 1] if α > 12 . The power is 2α if α < 12
and 1 if α > 12 .
√
• The critical region is of the form √
n|x̄n | > c and c = 1.96 from the
tables. Power at µ = 1 is√P [|z − n| > 1.96] is essentially P [z <
√
n−1.96] and this is .95 if n > 1.96+1.64 = 3.61 or n > (3.61)2 = 14
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