1. Finding a sufficient statistic

Let $f(x|\alpha)=\frac{\Gamma(2\alpha)}{\Gamma(\alpha)^ 2}[x(1-x)]^{\alpha -1}$

Then the joint pdf= $\prod\limits_{i=1}^n f(x|\alpha)$
= $(\frac{\Gamma(2\alpha)}{\Gamma(\alpha)^2})^n \prod\limits_{i=1}^n[x_i(1-x_i)]^{\alpha -1}$

by factorization theorem, $g(t,\alpha)h(x)$
$t=\prod\limits_{i=1}^n[x_i(1-x_i)]$
$g(t,\alpha)=(\frac{\Gamma(2\alpha)}{\Gamma(\alpha) ^2})^n t^{\alpha -1}$
$h(x)=1$

Can I conclude that $T(X)=\prod\limits_{i=1}^n[X_i(1-X_i)]$ is a sufficient statistic?

2. I assume 0<x<1 and that this is a beta distribution with $\alpha =\beta$
I don't think I've seen this question in regards to the beta, but this seems true.

that matches up with 9 from http://www.ds.unifi.it/VL/VL_EN/point/point6.html

3. Originally Posted by matheagle
I assume 0<x<1 and that this is a beta distribution with $\alpha =\beta$
I don't think I've seen this question in regards to the beta, but this seems true.

that matches up with 9 from Sufficient, Complete and Ancillary Statistics
Thanks just to clarify a bit more, so if $\alpha$ in this case is increased, the distribution will go to normal symmetrical shape. Am I right? thoguh I guess the center is not 0.