# Thread: [SOLVED] Moments and MME

1. ## [SOLVED] Moments and MME

Given pdf is $f_y(y)=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}y^{a-1}(1-y)^{b-1}$ for $0\le y\le1$ $a>0$ $b>0$

a) Compute the moments about the origin $E(Y^m)$
b) Find the method of moments estimators for a and b (Using a sample of size n).

Attempt:
a) $E(Y^m) = \displaystyle\int_0^1 y^m \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}y^{a-1}(1-y)^{b-1}$
$=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}*\beta(m+a, b)$
$=\frac{\Gamma(a+b)\Gamma(a+m)}{\Gamma(a+b+m)\Gamma (a)}$

b) Am i supposed to use part a to find the first and second central moments? I am not sure how to do this. Even so, by looking them up I try the following:

$E(Y) = \frac{a}{a+b} = m_1$
$E(Y^2) = \frac{a(a+1)}{(a+b)(a+b+1)} = m_2$

where m1 and m2 are the sample moments.

Basically when I try to solve this system of equations I get gross quadratics that I don't know how to simplify. Is there an elegant way of solving these, or if not, what is the best strategy for attacking these?

2. Use the fact that $\Gamma(a+1)=a\Gamma(a)$
which comes from integration by parts

Hence $E(X)=\frac{\Gamma(a+b)\Gamma(a+1)}{\Gamma(a+b+1)\G amma(a)}$

$=\left({\Gamma(a+b)\over \Gamma(a+b+1)}\right)\left({\Gamma(a+1)\over \Gamma(a)}\right)$

$=\left({\Gamma(a+b)\over (a+b)\Gamma(a+b)}\right)\left({a\Gamma(a)\over \Gamma(a)}\right)$

${a\over a+b}$

Now do it when m=2.

3. Ah, I did not know that fact. Thank you.
I'm still clueless on solving for the MME though.

From $\frac{a}{a+b} = m_1$

I've got that $b=\frac{a-m_1a}{m_1}$

and

From $\frac{a(a+1)}{(a+b)(a+b+1)}$

Assuming I did all my algebra right, I get

$a=\frac{-1+m_2+2bm_2\pm\sqrt{(1-m_2-2bm_2)^2-4(1-m_2)(m_2b^2-m_2b)}}{2(1-m_2)}$

How should I approach this to get it simplified? I suppose this may belong in an algebra section.

4. just apply parts to the gamma function $u=x^{\alpha-1}$

I'd bet your teacher went over it.

5. IF your moments are correct I would use a little observation here...

From $\frac{a}{a+b} = m_1$

and

$\frac{a(a+1)}{(a+b)(a+b+1)}=m_2$

insert m1 and FLIP it and divide......

$\frac{m_1(a+1)}{(a+b+1)}=m_2$

that gives ${a+b+1\over a+1}={m_1\over m_2}$