Homework 5 Solutions
(1) By change of variables,
E
φ
(
X
2
) =
Z
1
0
φ
(
x
2
)
dx
=
Z
1
0
φ
(
y
)
2
√
y
dy
E
φ
(1

X
2
) =
Z
1
0
φ
(1

x
2
)
dx
=
Z
1
0
φ
(
y
)
2
√
1

y
dy.
By the generalized NeymanPearson Lemma, the maximizer takes the form
φ
*
(
x
) =
1 1
>
k
1
2
√
x
+
k
2
2
√
1

x
0 otherwise
,
where
k
1
and
k
2
are determined to satisfy the constraints. Because
g
(
x
) =
k
1
2
√
x
+
k
2
2
√
1

x
is convex,
k
1
and
k
2
have to be chosen such that
{
1
> g
(
x
)
}
=
{
b
1
< x < b
2
}
for some
b
1
and
b
2
satisfying the constraints. In order to satisfy the constraints,
E
φ
*
(
X
2
) =
Z
b
2
b
1
1
2
√
y
dy
=
p
b
2

p
b
1
=
1
2
E
φ
*
(1

X
2
) =
Z
b
2
b
1
1
2
√
1

y
dy
=
p
1

b
1

p
1

b
2
=
1
2
,
from which we obtain
b
1
=
1
8
(4

√
7) and
b
2
=
1
8
(4 +
√
7). Hence, we have
φ
*
(
x
) =
1
1
8
(4

√
7)
< x <
1
8
(4 +
√
7)
0 otherwise
.
(2)
(i) The joint density of (
X
1
,X
2
,...,X
n
) is
f
(
x
1
,x
2
,...,x
n
) =
λ
n
e

nλ
x
.
For
λ
1
< λ
0
,
f
1
((
X
1
,X
2
,...,X
n
))
f
0
((
X
1
,X
2
,...,X
n
))
=
±
λ
1
λ
2
²
n
e
n
X
(
λ
0

λ
1
)
,
which is increasing in
X
.
1
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View Full Document(ii) Reparametrize as
θ
=

λ
. Then, the equivalent form of the test is H
0
:
θ
=
θ
0
against
θ
=
θ
1
> θ
0
. By (i), we know that when
θ
1
=

λ
1
>

λ
0
=
θ
0
,
f
1
/f
0
is increasing in
X
. Thus, by the NeymanPearson lemma, the UMP test rejects when
X > c
for some
c
.
Note that 2
λn
X
=
∑
n
i
=1
2
λX
i
∼
χ
2
2
n
. Thus,
α
= P
λ
0
(
X > c
)
= P
λ
0
(2
λ
0
n
X >
2
λ
0
nc
)
= 1

q

1
2
n
(2
λ
0
nc
)
.
(Note that
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 Fall '09
 moulib
 Probability theory, Exchangeability

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