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Solution to Homework 6
(The problem numbers refer to numbers in 7
th
edition)
8.1
:
Let
)
ˆ
(
θ
=
B
B
.
Then,
[
]
[
]
(
29
[
]
)
ˆ
(
ˆ
2
)
(
)
ˆ
(
ˆ
)
)
ˆ
(
ˆ
(
)
ˆ
(
)
ˆ
(
2
2
2
2
θ

θ
×
+
+
θ

θ
=
+
θ

θ
=
θ

θ
=
θ
E
E
B
B
E
E
E
B
E
E
E
MSE
2
)
ˆ
(
B
V
+
θ
=
.
8.6
:
a.
θ
=
θ

+
θ
=
θ

+
θ
=
θ
)
1
(
)
ˆ
(
)
1
(
)
ˆ
(
)
ˆ
(
2
1
3
a
a
E
a
aE
E
.
b.
2
2
2
1
2
2
2
1
2
3
)
1
(
)
ˆ
(
)
1
(
)
ˆ
(
)
ˆ
(
σ

+
σ
=
θ

+
θ
=
θ
a
a
V
a
V
a
V
, since it was assumed that
1
ˆ
θ
and
2
ˆ
θ
are independent.
To minimize
)
ˆ
(
3
θ
V
, we can take the first derivative (with
respect to
a
), set it equal to zero, to find
2
2
2
1
2
2
σ
+
σ
σ
=
a
.
(One should verify that the second derivative test shows that this is indeed a minimum.)
8.7
:
Following Ex. 8.6 but with the condition that
1
ˆ
θ
and
2
ˆ
θ
are not independent, we find
c
a
a
a
a
V
)
1
(
2
)
1
(
)
ˆ
(
2
2
2
1
2
3

+
σ

+
σ
=
θ
.
Using the same method w/ derivatives, the minimum is found to be
8.8
:
a.
Note that
1
ˆ
θ
,
2
ˆ
θ
,
3
ˆ
θ
and
5
ˆ
θ
are simple linear combinations of
Y
1
,
Y
2
, and
Y
3
.
So, it is
easily shown that all four of these estimators are
unbiased
.
b.
It is easily shown that
V
(
1
ˆ
θ
) = θ
2
,
V
(
2
ˆ
θ
) = θ
2
/2,
V
(
3
ˆ
θ
) = 5θ
2
/9, and
V
(
5
ˆ
θ
) = θ
2
/9, so the
estimator
5
ˆ
θ
is unbiased and has the smallest variance.
8.12
:
a.
For the uniform distribution given here,
E
(
Y
i
) = θ + .5.
Hence,
E
(
Y
) = θ + .5 so
that
B
(
Y
) = .5.
b.
Based on
,
Y
the unbiased estimator is
Y
– .5.
1
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 Fall '08
 Staff
 Economics

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