Second Edition
813
The Type II error probability is
1

P
(1
< X <
3

θ
= 1) = 1

Z
3
1
1
π
1
1+(
x

1)
2
dx
= 1

1
π
arctan(
x

1)
±
±
±
±
3
1
≈
.
6476
.
c. We will not have
f
(1

θ
)
/f
(1

0) =
f
(3

θ
)
/f
(3

0) for any other value of
θ
6
= 1. Try
θ
= 2, for
example. So the rejection region 1
< x <
3 will not be most powerful at any other value of
θ
. The test is not UMP for testing
H
0
:
θ
≤
0 versus
H
1
:
θ >
0.
8.30 a. For
θ
2
> θ
1
>
0, the likelihood ratio and its derivative are
f
(
x

θ
2
)
f
(
x

θ
1
)
=
θ
2
θ
1
θ
2
1
+
x
2
θ
2
2
+
x
2
and
d
dx
f
(
x

θ
2
)
f
(
x

θ
1
)
=
θ
2
θ
1
θ
2
2

θ
2
1
(
θ
2
2
+
x
2
)
2
x.
The sign of the derivative is the same as the sign of
x
(recall,
θ
2
2

θ
2
1
>
0), which changes
sign. Hence the ratio is not monotone.
b. Because
f
(
x

θ
) = (
θ/π
)(
θ
2
+

x

2
)

1
,
Y
=

X

is suﬃcient. Its pdf is
f
(
y

θ
) =
2
θ
π
1
θ
2
+
y
2
,
y >
0
.
Diﬀerentiating as above, the sign of the derivative is the same as the sign of
y
, which is
positive. Hence the family has MLR.
8.31 a. By the KarlinRubin Theorem, the UMP test is to reject
H
0
if
∑
i
X
i
> k
, because
∑
i
X
i
is suﬃcient and
∑
i
X
i
∼
Poisson(
nλ
) which has MLR. Choose the constant
k
to satisfy
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 Spring '12
 Dr.Hackney
 Statistics, Probability, Statistical hypothesis testing, y1, Type I and type II errors

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