1
14.30
Introduction
to
Statistical
Methods
in
Economics
Lecture
Notes
20
Konrad
Menzel
April
30,
2009
Confidence Intervals (continued)
The
following example illustrates
one way of
constructing
a
confidence interval
when the distribution of
the
estimator is
not normal.
Example
1
Suppose
X
1
, . . . , X
n
are
i.i.d.
with
X
i
∼
U
[0
, θ
]
, and
we
want to
construct a
90% confidence
interval for
θ
0
.
Let
θ
ˆ
=
max
{
X
1
, . . . , X
n
}
=
X
(
n
)
the
n
th
order
statistic
(as we
showed
last time, this is also
the
maximum-likelihood
estimator).
Even
though,
as
we
saw,
θ
ˆ
is not unbiased
for
θ
,
we can
use
it to
construct a
confidence
interval
for
θ
.
From
results
for
order statistics,
we
saw that the
c.d.f. of
θ
ˆ
is given
by the
c.d.f. of
θ
ˆ
is given by
⎧
0
θ
≤
0
⎪
⎨ �
θ
�
n
F
θ
ˆ
(
θ
) =
θ
0
if 0
< θ
≤
θ
0
⎪
⎩
1
if
θ > θ
0
where
we
plugged in the
c.d.f.
of
a
U
[0
, θ
0
]
random variable,
F
(
x
) =
θ
x
0
.
In order
to
obtain the
functions for
A
and
B
,
let us first find
constants
a
and
b
such
that
P
θ
0
(
a
≤
θ
ˆ
≤
b
) =
F
θ
ˆ
(
b
)
−
F
θ
ˆ
(
b
) = 0
.
95
−
0
.
05
= 0
.
9
We
can find
a
and
b
by solving
F
θ
ˆ
(
a
) = 0
.
05
and
F
θ
ˆ
(
b
) = 0
.
95
n
n
so
that we
obtain
a
=
√
0
.
05
θ
0
and
b
=
√
0
.
95
θ
0
. This doesn’t give
us a
confidence
interval
yet, since
looking
at the
definition of
a
CI,
we
want the
true
parameter
θ
0
in
the
middle
of the
inequalities, and
the
functions on either side
depend
only on the
data
and
other known
quantities.
However,
we
can
rewrite
�
n
�
θ
ˆ
θ
ˆ
�
0
.
9 =
P
θ
0
(
a
≤
θ
ˆ
≤
b
) =
√
0
.
05
θ
0
≤
ˆ
√
0
.
95
θ
0
�
=
P
θ
0
n
P
θ
0
n
θ
≤
n
√
0
.
95
≤
θ
0
≤ √
0
.
05
Therefore
�
max
{
X
1
, . . . , X
n
}
max
{
X
1
, . . . , X
n
}
�
[
A, B
] = [
A
(
X
1
, . . . , X
n
)
, B
(
X
1
, . . . , X
n
)] =
n
,
n
√
0
.
95
√
0
.
05
is
a
90%
confidence
interval for
θ
0
.
Notice
that in
this case, the
bounds of the
confidence
intervals depend
on
the
data
only
through the
estimator
θ
ˆ
(
X
1
, . . . , X
n
)
. This need
not be
true
in
general.
1