ORIE 3500/5500, Fall ’10
HW 12 Solution
Assignment 12 Solution
Problem 1
The joint pdf of
X
1
, . . . , X
n
is given by
f
X
1
,...,X
n
(
x
1
, . . . , x
n
;
θ
) =
n
Y
i
=1
f
X
i
(
x
i
;
θ
) =
n
Y
i
=1
1
2
e

x
i

θ

=
1
2
n
e

∑
n
i
=1

x
i

θ

.
Taking the logarithm of this expression, we obtain
log
f
X
1
,...,X
n
(
x
1
, . . . , x
n
;
θ
) =

n
log 2

n
X
i
=1

x
i

θ

.
Let us call this function
h
(
θ
) for notational convenience.
Note that
h
is not
differentiable at
θ
=
x
1
, . . . , x
n
, so we cannot simply set the derivative of
h
equal to zero to find out where
h
is maximized.
Instead, we note that the
derivative of
h
, when it exists, is equal to
h
0
(
θ
) =
n
X
i
=1
sgn(
x
i

θ
)
,
where sgn denotes the sign function
sgn(
x
) =

1
if
x <
0
,
0
if
x
= 0
,
1
if
x >
0
.
Let
x
(1)
< x
(2)
< . . . < x
(
n
)
denote the order statistics of the sample. Observe
that
•
for
θ < x
(1)
, we have sgn(
x
i

θ
) = 1 for all
i
, so
h
0
(
θ
) =
n
,
•
for
x
(1)
< θ < x
(2)
, we have
sgn(
x
i

θ
) =
(

1
if
x
i
=
x
(1)
,
1
otherwise
,
so
h
0
(
θ
) =

1 + (
n

1)(1) =
n

2,
•
for
x
(2)
< θ < x
(3)
,
h
0
(
θ
) =
n

4,
•
...
•
for
θ > x
(
n
)
,
h
0
(
θ
) =

n
.
Thus we see that the graph of
h
is a continuous polygonal curve with vertices
at
θ
=
x
1
, . . . , x
n
. If
n
is odd,
h
is strictly increasing for
θ
≤
x
(
n
+1
2
)
and strictly
decreasing for
θ
≥
x
(
n
+1
2
)
. So
h
is maximized at
θ
=
x
(
n
+1
2
)
. If
n
is even,
h
is
1
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ORIE 3500/5500, Fall ’10
HW 12 Solution
strictly increasing for
θ
≤
x
(
n
2
)
, constant for
x
(
n
2
)
≤
θ
≤
x
(
n
2
+1
)
, and strictly
decreasing for
θ
≥
x
(
n
2
+1
)
.
So
h
is maximized at any
θ
∈
h
x
(
n
2
)
, x
(
n
2
+1
)
i
.
In either case, setting
θ
equal to ˜
x
, the sample median, maximizes
h
.
So we
conclude that the MLE of
θ
is
ˆ
θ
=
e
X
= median
{
X
1
, . . . , X
n
}
.
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 '10
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 Derivative, Conditional Probability, Trigraph, Estimation theory, Yi

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