Download Final Exam Suggested Solutions - Mathematical Statistics | STAT 710 and more Exams Mathematical Statistics in PDF only on Docsity!
STAT 710 2007 Final Exam Suggested Solutions
Department of Statistics, University of Wisconsin-Madison
May 13, 2008
Problem 1.
(a) The likelihood function is
`(θ) =
(2π)
−n/ 2
exp{−
1
2
n
i=
X
2
i
}, θ = 0,
−n exp{−
n
i=
|X
i
|}, θ = 1.
Then the MLE of θ is
θ =
0 , (0) ≥(1),
1 , (0) <(1).
(b) If θ = 0, X 1
,... , X
n
∼ N (0, 1), then EX
2
1
= 1 and E|X 1
2 /π, so
n
log `(0) = −
log(2π) −
2 n
n ∑
i=
X
2
i
→p −
log(2π) −
n
log `(1) = − log(2) −
n
n ∑
i=
|X
i
p
− log(2) −
2 /π.
Then P (
θ = 0) = P (
1
n
log `(0) ≥
1
n
log `(1)) → 1. Similarly one can prove that when
θ = 1, P (
θ = 1) → 1. So
θ is consistent.
(c) p(θ) = π
1 −θ
0
(1 − π 0
θ , and the likelihood function can be written as `(0)
1 −θ `(1)
θ , then
the posterior density of θ given X 1
,... , X
n
is
`(0)π 0
(0)π 0 +(1)(1 − π 0 )
1 −θ
`(1)(1 − π 0
(0)π 0 +(1)(1 − π 0 )
θ
(d) From the definition of the Bayes action, it is seen that
θ
B
=
0 , `(0)π 0
≥ `(1)(1 − π 0
1 , `(0)π 0
< `(1)(1 − π 0
(e) From Neyman-Pearson Lemma, a UMP test is given by
T =
1 , (1) > c(0),
0 , (1) ≤ c(0).
where c > 0 is determined by P θ=
((1) > c(0)) = α.
Problem 2.
(a) The likelihood function is `(θ) = θ
n
Π
n
i=
(1 − Xi)
θ− 1
Π
n
i=
I(0,1)(Xi), then
log `(θ) = n log θ + (θ − 1)
n ∑
i=
log(1 − X i
∂θ
log `(θ) =
n
θ
n ∑
i=
log(1 − X i
2
∂θ
2
log `(θ) = −
n
θ
2
So the MLE
θ = −n/(
n
i=
log(1 − X i
)). The acceptance region of the LR test is
`(θ 0
`(
θ)
= (θ 0
θ)
n
exp(−n(θ 0
θ − 1)) ≥ exp(−χ
2
1 ,α
Then the 1 − α asymptotically correct confidence interval is
{θ : (θ/
θ)
n
exp(−n(θ/
θ − 1)) ≥ exp(−χ
2
1 ,α
(b) In Wald’s test, R(θ, p) = θ − θ 0
, the Fisher information matrix is I n
(θ) = n/θ
2
. So
the Wald’s statistic is
W
n
= n(
θ − θ 0
2
/
θ
2
= n(1 − θ 0
θ)
2
.
Then the 1 − α asymptotically correct confidence interval is
{θ : W n
≤ χ
2
1 ,α
} = [(1 −
χ
2
1 ,α
/n)
θ, (1 +
χ
2
1 ,α
/n)
θ].
(c) In Rao’s test, s n
(θ) = n/θ − n/
θ, then the Rao’s statistic is
R
n
= (n/θ 0
− n/
θ)
2
θ
2
0
/n = n(1 − θ 0
θ)
2
.
Then the 1 − α asymptotically correct confidence interval is
{θ : R n
≤ χ
2
1 ,α
} = [(1 −
χ
2
1 ,α
/n)
θ, (1 +
χ
2
1 ,α
/n)
θ].
(c) It is seen that X/θ has the density
h(x) = 2e
−x
/(1 + e
−x
)
2
I (0,∞)
(x),
so it is pivotal. We get the same class of confidence intervals as in part (b).
(d) We show that x
2 h(x) is unimodal. We see that
d
dx
(x
2
h(x)) = (1 + e
x
)
− 3
(x + 2 + 2e
x
− xe
x
) = (1 + e
x
)
− 3
g(x).
Since
d
2
dx
2
g(x) = −xe
x
< 0 ,
g(x) is a concave function, and by the fact g(0) = 4, there exist a x 0
0 such that
g(x) > 0 when x < x 0
and g(x) < 0 when x > x 0
. Then x
2 h(x) is unimodal.
By Theorem 7.3, we choose b > a > 0 such that a
2
h(a) = b
2
h(b) and
b
a
h(x) = 1 − α,
resulting in the shortest confidence interval [X/b, X/a].
