# [SOLVED] MM of Beta

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• April 2nd 2010, 08:14 PM
Statistik
[SOLVED] MM of Beta
Hi,

I'm trying to find the MM of a Beta(a,b) using the first and second raw moments.

I use

$
E[X] = \frac {a} {a+b}
$

and
$
E[X^2] = \frac {a * (a+1)} {(a+b) * (a+b+1)}
$

I then set those equal to $\bar X_n$ and $\overline {X^2_n}$, respectively, and (try to) solve for a and b.

Does that sound right? (I'm getting stuck.)

Thanks!
• April 2nd 2010, 09:33 PM
matheagle
Quote:

Originally Posted by Statistik
Hi,

I'm trying to find the MM of a Beta(a,b) using the first and second raw moments.

I use

$
E[X] = \frac {a} {a+b}
$

and
$
E[X^2] = \frac {a * (a+1)} {(a+b) * (a+b+1)}
$

I then set those equal to $\bar X_n$ and $\overline {X^2_n}$, respectively, and (try to) solve for a and b.

Does that sound right? (I'm getting stuck.)

Thanks!

what is ... $\overline {X^2_n}$?

you should set the sample moments equal to the pop moments

you set

$\hat E[X] = \bar X$

and

$\Hat E[X^2] = {\sum_{i=1}^n X^2_i\over n}$
• April 2nd 2010, 10:34 PM
Statistik
Matheagle,

Thanks. That's what I am trying - sorry for not clarifying. I am using

$
\bar X_n = \frac {\sum_{i=1}^n X_i} {n}
$

and
$
\overline {X^2_n} = \frac {\sum_{i=1}^n X^2_i} {n}
$

and then am setting

$
E[\bar X_n] = \frac {a} {a+b}
$

and
$
E[\overline {X^2_n}] = \frac {a * (a+1)} {(a+b) * (a+b+1)}
$

Does that mean I'm just stuck with the math? ;-(

Thanks!
• April 2nd 2010, 11:14 PM
matheagle
you can find the answer at
Beta distribution - Wikipedia, the free encyclopedia
Note that they use the centered second moment, but you can expand v and use your two sample moments.