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So I have this homework problem that I am struggling a little bit with coming to a solid answer on. The problem goes like this:

Suppose X~Beta($\theta,\theta), (\theta>0)$, and let $\{X_1, X_2 , \ldots , X_n \}$ be a sample. Show that T=$\Pi_i(X_i*(1-X_i)$ is a sufficient statistic for $\theta$.

I started out with my Beta distribution as:

$f(x_i,\theta)=\frac{\Gamma(\alpha + \beta)}{\Gamma(alpha)\Gamma(\beta)}x^{(\alpha-1)}(1-x)^{(\beta-1)}$

$=\frac{\Gamma(\theta + \theta)}{\Gamma(theta)\Gamma(\theta)}x_1^{(\theta-1)}(1-x_1)^{(\theta-1)} ***\frac{\Gamma(\theta + \theta)}{\Gamma(\theta)\Gamma(\theta)}x_n^{(\theta-1)}(1-x_n)^{(\theta-1)} $

$=\frac{\Gamma(2\theta)}{\Gamma(\theta)^2}x_1^{(\theta-1)}(1-x_1)^{(\theta-1)} ***\frac{\Gamma(2\theta)}{\Gamma(\theta)^2}x_n^{(\theta-1)}(1-x_n)^{(\theta-1)}$

$={(\frac{\Gamma(2\theta)}{\Gamma(\theta)^2})}^n \Pi_i (x_i)(1-x_i)^{(\theta-1)}$

I know that in order for my statistic to be sufficient by factorization, I need to have a g(T,$\theta$) and a h($x_1,x_2,...,x_n)$.

What I have above is my g(T,$\theta$), but I am not so sure about my h($x_1,x_2,...,x_n)$. I have seen other places where the suggestion is to use 1 for my h($x_1,x_2,...,x_n)$. Could I do this here with this problem? It just seems a little too easy to do that, but I will be happy if it is that easy.

If anyone could let me know, that would be greatly appreciated.

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what if $h(x_1, x_2, \dots, x_n) = 1$? I think that suggestion should work. –  Enzo Mar 6 '13 at 3:27
    
So I can do that? I didn't know if I could or not. –  Perdue Mar 6 '13 at 4:21
    
yes you can! that's justified –  Enzo Mar 6 '13 at 14:11
    
What if I took $\Pi_i (x_i)(1-x_i)^{(\theta-1)}$ and broke that up to $\Pi_i (x_i)(1-x_i)^{(\theta)} \frac{1}{\Pi_i (x_i)(1-x_i)}$? Then I would have my statistic right there instead of having to multiply everything by 1. –  Perdue Mar 6 '13 at 18:06

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