MLE vs MME (or MOM)

[USER]

Hello all,

I have been doing some self-study in point estimates and looking at the
method of moments vs. maximum likelihood.

I have two questions I hope to get some help with.

Question 1:
I would like to know if anyone can give an example where the MLE (max.
likelihood estimate) has variation that is greater than the MME (method
of moments estimate). I have read in a book that this is possible,
although no book seems to give an example of this.

Question 2:
I noticed that when using the MME for the well-known example of the
continuous uniform distribution, U(0,theta) [with theta as the unknown
parameter] that the first moment for the parameter \theta gives

\theta-hat = 2*x-bar,

but we can use the second moment to get

\theta-hat = sqrt( 3*M_2 ), with M_2 as the second sample moment.

I generalized this to

\theta_r -hat = ( (r+1) * M_r )^(1/r),

with M_r as the r^{th} sample moment.
( sample moment: M_r = (1/n) * sum_{i=1}^n x_i^r )

This seems(?) to yield unbiased estimates (for any positive integer r )
for \theta, but with variance that decreases with increasing r, although
I have not found a formula for the variance of \theta_r -hat.

Is this the case? This seems counterintuitive in that we get "better"
estimates by using larger sample moments, i.e. somewhat more work
(calculations) yields much better estimates (unbiased, very small variance).

I hope the notation is understandable and that someone can give answers
to these questions.

Thank you for any help,
A stats student/teacher

[Jack Tomsky]

> Hello all,
>
> I have been doing some self-study in point estimates
> and looking at the
> method of moments vs. maximum likelihood.
>
> I have two questions I hope to get some help with.
>
> Question 1:
> I would like to know if anyone can give an example
> where the MLE (max.
> likelihood estimate) has variation that is greater
> than the MME (method
> of moments estimate). I have read in a book that this
> is possible,
> although no book seems to give an example of this.
>
>

Here's a simple example. Suppose you have a single observation x from

x ~ N(mu, 1).

Let theta = mu^2.

Then the MLE of theta is

MLE = x^2.

For MOM, set

E(x^2) = mu^2 + 1 = theta + 1.

Then

MOM = x^2 - 1.

While MLE and MOM have the same variance, MOM has a smaller mean-squared error because it's an unbiased estimate of theta.

MSE(MLE) = Var(MLE) + 1
MSE(MOM) = Var(MLE).

Jack
www.tomskystatistics.com