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So I know the Beta distribution is $$f(x) = \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\cdot x^{a-1}\cdot(1-x)^{b-1}$$

I know the $E[X^r] = \dfrac{\Gamma(a+b)\Gamma(a+r)}{\Gamma(a)\Gamma(a+r+b)}$

And I know plugging in $1$ for $r$ to get $E[X]$ (or $\mu$ mean) gives you $a/(a+b)$ because I know $\Gamma(n+1) = n\Gamma(n)$.

However, what do you do if you have $r=2$ or anything higher than $1$? How would I find $E[X^2]$?

I am a actuary major so I don't know how to correctly format the question to make it look all nice and neat. Feel free to help edit and adjust the way the question looks to make it look next (i.e. actually put the greek letter gamma or Beta or integral signs in). Again, sorry for the slopping format. You can hate all you want.

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If $r$ is a nonnegative integer, $$E(X^r) = \frac{\dfrac{\Gamma(a+r)}{\Gamma(a)}}{\dfrac{\Gamma(a+b+r)}{\Gamma(a+b)}}=\frac{(a+r-1)\cdots(a+1)a}{(a+b+r-1)\cdots(a+b+1)(a+b)}.$$ For example, $$E(X^2)=\frac{(a+1)a}{(a+b+1)(a+b)}.$$

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  • $\begingroup$ "How I got there?" I read what YOU wrote. "How to expand?" Sorry but what is there to expand? $\endgroup$
    – Did
    Oct 25, 2014 at 20:15

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