# Thread: [SOLVED] MM of Beta

1. ## [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

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

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

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

Thanks!

2. 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

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

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

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

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

you should set the sample moments equal to the pop moments

you set

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

and

$\displaystyle \Hat E[X^2] = {\sum_{i=1}^n X^2_i\over n}$

3. Matheagle,

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

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

and then am setting

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

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

Thanks!

4. 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.