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Math Help - [SOLVED] Moments and MME

  1. #1
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    [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?
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  2. #2
    MHF Contributor matheagle's Avatar
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    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.
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  3. #3
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    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.
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  4. #4
    MHF Contributor matheagle's Avatar
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    just apply parts to the gamma function u=x^{\alpha-1}

    I'd bet your teacher went over it.
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  5. #5
    MHF Contributor matheagle's Avatar
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    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}
    Last edited by matheagle; March 10th 2010 at 08:11 AM.
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