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.
Best Answer
If $ X\sim Beta(\alpha,\beta)$, then $f(x;\alpha,\beta) = {\Gamma(\alpha+\beta) \over \Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1} $ and $ f(\underline x;\alpha,\beta) = ({\Gamma(\alpha+\beta) \over \Gamma(\alpha)\Gamma(\beta)})^{n}(\prod x_{i})^{\alpha-1}(\prod (1-x_{i}))^{\beta-1}$
Thus, the sufficient statisitcs is $(\prod x_{i},\prod (1-x_{i}))$ by Fisher–Neyman factorization theorem.
Here $ h(\underline x) = 1$.
If $ \alpha = \beta $, then sufficient statistics is $(\prod x_i(1-x_i))$. Again $h(\underline x) = 1$