Up: Stat 5102
Stat 5102 (Geyer) Midterm 1
The basic fact this problem uses is
(Corollary 7.25 in the notes), which, since , specializes to
|
(1) |
in this problem. This is the only exact result we have
involving both
and , so nothing else is of any use.
To use (1), we must put the event of interest
in a form related to the left hand side of (1). Clearly, this
is equivalent to
So the probability we need to find where is
a random variable ( degrees of freedom).
From Table IIIa in Lindgren
,
so
.
To use the method of moments, we first need to find some moments.
Since this is not a ``brand name'' distribution,
we must integrate to find the moments.
The obvious moment to try first is the first moment (the mean)
Solving for as a function of , we get
(the numerator and denominator are both positive because
.)
Either way, we get a method of moments estimator by plugging in
for
The asymptotic distribution of
is, as usual, by the CLT,
Plugging in
gives
The asymptotic distribution of is, by
Corollary 7.17 in the notes,
where
and
are given
in the problem statement. Plugging these in gives
The ARE is the ratio of the asymptotic variances, either
or the the reciprocal
, depending on which way you write it.
The better estimator is the one with the smaller asymptotic variance,
in this case
.
This is a problem for the delta method. We know from the properties
of the exponential distribution
and
Hence the CLT says in this case
For any differentiable function , the delta method says
The such that
is
|
(2) |
which has derivative
|
(3) |
so
and
and
There are several different ways to proceed here.
The mean is
. Thus we could just get a confidence interval
for and take reciprocals of the endpoints to get a confidence interval
for .
A confidence interval for can be found using Theorem 9.8 in the notes.
From Section B.1.8 of the notes
so by the LLN and the continuous mapping theorem
is a consistent estimator of the population standard deviation
needed for the theorem. The theorem gives
as an asymptotic 95% confidence interval for . So
is an asymptotic 95% confidence interval for
. Plugging in
the numbers gives
The obvious point estimator for is
The CLT says
Applying the delta method with the transformation
with derivative
gives
and
and
which gives an asymptotic 95% confidence interval
Plugging in the numbers gives
or
which is pretty close to the other interval.
The really hard way to do this problem is to start with
and standardize giving the asymptotically standard normal quantity
from which we conclude that the set of such that
is an asymptotic 95% confidence interval for . It turns out this is
solvable, equivalent to
Or, writing , we see the confidence interval has endpoints satisfying
the quadratic equation
which has roots
or
Up: Stat 5102
Charles Geyer
2001-03-05