UC Berkeley
Department of Statistics
STAT 210A: Introduction to Mathematical Statistics
Problem Set 9 Solutions
Fall 2007
Issued:
Thursday, December 1
Due:
Thursday, December 8
Problem 9.1
We want to prove that, for all
θ
∈
Θ,
P
θ
(
θ
∈
S
(
X
))
≥
1

α
. To prove that, we first notice
that, according to the definition of
S
(
X
),
θ
∈
S
(
X
)
⇔
X
∈
A
(
θ
).
It follows that:
P
θ
(
θ
∈
S
(
X
)) =
P
θ
(
X
∈
A
(
θ
))
≥
1

α,
for all
θ
∈
Θ
where the inequality follows from the definition of a acceptance region for a level
α
test.
Problem 9.2
a) Let
T
(
X
) = max
1
≤
i
≤
n
X
i
and
φ
(
X
)
∈ {
0
,
1
}
be any function such that:
H
0
is rejected if
φ
(
X
) = 1 and accepted otherwise;
E
θ
(
φ
(
X
))
≥
α
for all
θ
≤
θ
0
;
We want to prove:
E
θ
1
(
φ
(
X
))
≤
E
θ
1
(
δ
(
X
))
,
∀
θ
1
> θ
0
To prove that, first notice that for any
θ
:
E
θ
(
φ
(
X
))
=
E
θ
(
φ
(
X
)
I
(
T
(
X
)
≤
θ
0
)) +
E
θ
(
φ
(
X
)
I
(
T
(
X
)
> θ
0
))
E
θ
(
δ
(
X
))
=
E
θ
(
δ
(
X
)
I
(
T
(
X
)
≤
θ
0
)) +
E
θ
(
δ
(
X
)
I
(
T
(
X
)
> θ
0
))
Since
φ
(
X
)
≤
1 and
T
(
X
)
> θ
0
implies
δ
(
X
) = 1, we have:
E
θ
1
(
δ
(
X
)
I
(
T
(
X
)
> θ
0
))
≥
E
θ
1
(
φ
(
X
)
I
(
T
(
X
)
> θ
0
))
It is then enough to prove that:
E
θ
1
[(
δ
(
X
)

φ
(
X
))
I
(
T
(
X
)
≤
θ
0
)]
≥
0
,
for all
θ
1
> θ
0
1
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We have:
E
θ
1
[(
δ
(
X
)

φ
(
X
))
I
(
T
(
X
)
≤
θ
0
)]
=
R
(
δ
(
X
)

φ
(
X
))
θ
n
1
I
(
x
i
≤
θ
0
)
I
(
x
i
≤
θ
1
)
dx
=
1
θ
n
1
•Z
(
δ
(
X
)

φ
(
X
))
I
(
x
i
≤
θ
0
)
dx
‚

{z
}
h
(
δ,φ,θ
0
)
=
h
(
δ, φ, θ
0
)
θ
n
1
and so, it is enough to prove that
h
(
δ, φ, θ
0
)
≥
0. To prove that, notice:
h
(
δ, φ, θ
0
)
θ
n
0
=
E
θ
0
[(
δ
(
X
)

φ
(
X
))
I
(
T
(
X
)
≤
θ
0
)]
=
E
θ
0
[(
δ
(
X
)

φ
(
X
))]
≥
0
where:
the second equality is due to the fact that
P
θ
0
(
X
i
≥
θ
0
) = 0;
the inequality follows from the fact that
E
θ
0
φ
(
X
)
≤
E
θ
0
δ
(
X
) =
α
.
b) First, we show that
E
θ
0
(
δ
(
X
)) =
α
. To do that notice that:
P
θ
(
X
(
n
)
≤
x
)
=
0
,
for
x
∈
(
∞
,
0);
(
x
θ
)
n
,
for
x
∈
[0
,
1];
1
,
for
x
∈
(1
,
∞
);
so:
E
θ
0
(
δ
(
X
))
=
P
θ
0
‡
T
(
X
)
≤
θ
0
α
1
n
·
+
P
θ
0
(
T
(
X
)
> θ
0
)

{z
}
=0
=
ˆ
θ
0
α
1
n
θ
0
!
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 Statistics, Normal Distribution, standard normal distribution, zi

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