ieor_hw3_sol - IEOR 165 Engineering Statistics Quality...

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