IEOR 165: Engineering Statistics, Quality Control and Forecasting, Spring 2008
Homework 3 Solution
Extra Question
(a) The first sample moment is
μ
=
∑
n
i
=1
X
i
n
.
Looking at the expected value of the first moment for a single uniform random variable,
E
[
X
] =
a
2
⇒
a
= 2
·
E
[
X
]
.
Therefore
A
MOM
= 2
μ
= 2
∑
n
i
=1
X
i
n
.
(b)
E
[
A
MOM
]
=
2
n
n
X
i
=1
E
[
X
i
] =
2
n
n
·
a
2
=
a
V ar
(
A
MOM
)
=
4
n
2
n
X
i
=1
V ar
(
X
i
) =
4
n
n
2
·
a
2
12
=
a
2
3
n
.
The estimator is unbiased.
(c) Noting that the density of
X
i
is
f
X
i
(
x
) =
1
|
a
|
I
{
a < x <
0
}
:
a
MLE
=
argmax
a
f
(
x
1
, . . . , x
n
|
a
) =
argmax
a
n
Y
i
=1
f
X
i
(
x
i
|
a
) =
argmax
1
|
a
|
!
n
n
Y
i
=1
I
{
a < x
i
<
0
}
=
argmax
I
{
a <
min
{
x
1
, . . . , x
n
}}
|
a
|
n
= min
{
x
1
,
. . . .
x
n
}
Thus,
A
MLE
= min
{
X
1
,
. . . .
X
n
}
.
(d)
P
(min
{
X
1
, . . . , X
n
} ≤
u
)
=
1
-
P
(min
{
X
1
, . . . , X
n
}
> u
) = 1
-
n
Y
i
=1
P
(
X
i
> u
) = 1
-
u
a
n
f
min
{
X
1
,...,X
n
}
(
u
)
=
-
nu
n
-
1
a
n
⇒
E
[
A
MLE
] =
-
n
Z
0
a
u
u
n
-
1
a
n
du
=
-
n
n
n
+1
(
n
+ 1)
a
n
!
0
u
=
a
=
n
n
+ 1
a
E
[
A
2
MLE
]
=
-
n
Z
0
a
u
2
u
n
-
1
a
n
du
=
n
n
+ 2
a
2
V ar
(
A
MLE
)
=
n
n
+ 2
a
2
-
n
n
+ 1
a
2
=
n
(
n
+ 2)(
n
+ 1)
2
a
2
This estimator is biased. The unbiased alternative is:
n
+1
n
A
MLE
(e) We use the mean square error
MSE
(
ˆ
θ
) =
E
(
ˆ
θ
-
θ
)
2
=
V ar
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- Spring '04
- MichaelChrist
- Statistics, Forecasting, Normal Distribution, Estimation theory, Bias of an estimator
-
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