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Math Help - [SOLVED] MM of Beta

  1. #1
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    [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

     <br />
E[X] = \frac {a} {a+b} <br />
    and
     <br />
E[X^2] = \frac {a * (a+1)} {(a+b) * (a+b+1)} <br />

    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!
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  2. #2
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Statistik View Post
    Hi,

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

    I use

     <br />
E[X] = \frac {a} {a+b} <br />
    and
     <br />
E[X^2] = \frac {a * (a+1)} {(a+b) * (a+b+1)} <br />

    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}
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  3. #3
    Junior Member
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    Matheagle,

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

     <br />
\bar X_n = \frac {\sum_{i=1}^n X_i} {n}<br />
    and
    <br />
\overline {X^2_n} = \frac {\sum_{i=1}^n X^2_i} {n}<br />

    and then am setting

    <br />
E[\bar X_n] = \frac {a} {a+b}<br />
    and
    <br />
E[\overline {X^2_n}]  = \frac {a * (a+1)} {(a+b) * (a+b+1)}<br />

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

    Thanks!
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  4. #4
    MHF Contributor matheagle's Avatar
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    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.
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