Find 1 dimensional sufficient statistic for $Beta(\alpha, 2\alpha)$ I am currently getting $\prod x_i$ and $\prod (1-x_i)$ but a 1 dimensional sufficient statistic is asked for and considering that there is only a single parameter ($\alpha$), I think we should get a 1-dimensional sufficient statistic. I am not sure how to go about doing so though.
 A: Already answered in comments...
PDF of $X\sim\mathcal{Be}(\alpha,2\alpha)$ is
$$f(x;\alpha)=\frac{x^{\alpha-1}(1-x)^{2\alpha-1}}{B(\alpha,2\alpha)}\mathbf 1_{0<x<1},\quad\alpha>0$$
Suppose $(X_1,X_2,\cdots,X_n)$ is a random sample drawn from the above distribution.
Joint PDF of $(X_1,X_2,\cdots,X_n)$ is
\begin{align}f_{\alpha}(x_1,x_2,\cdots,x_n)&=\frac{1}{(B(\alpha,2\alpha))^n}\left(\prod_{i=1}^nx_i\right)^{\alpha-1}\left(\prod_{i=1}^n(1-x_i)\right)^{2\alpha-1}\mathbf1_{0<x_1,\cdots,x_n<1}
\\\implies\ln f_{\alpha}(x_1,x_2,\cdots,x_n)&=-n\ln B(\alpha,2\alpha)+(\alpha-1)\sum_{i=1}^n\ln x_i+(2\alpha-1)\sum_{i=1}^n\ln(1-x_i)
\\\implies f_{\alpha}(x_1,x_2,\cdots,x_n)&=\exp\left[(\alpha-1)\sum_{i=1}^n\ln x_i+(2\alpha-1)\sum_{i=1}^n \ln(1-x_i)+c(\alpha)\right]
\\&=\exp\left[\alpha\sum_{i=1}^n\left(\ln x_i+2\ln (1-x_i)\right)+c(\alpha)+d(x_1,x_2,\cdots,x_n)\right]
\end{align}
for some $c$ and $d$.
Clearly, $\mathcal{Be}(\alpha,2\alpha)$ belongs to the one-parameter exponential family. 
Hence our sufficient statistic for $\alpha$ is
\begin{align}
T(X_1,X_2,\cdots,X_n)&=\sum_{i=1}^n\left(\ln X_i+2\ln (1-X_i)\right)
\\&=\ln \left[\left(\prod_{i=1}^n X_i\right)\left(\prod_{i=1}^n(1-X_i)\right)^2\right]
\end{align}

The claim that $\displaystyle T^*(X_1,\cdots,X_n)=\left(\prod_{i=1}^n X_i,\prod_{i=1}^n (1-X_i)\right)$ is sufficient for $\alpha$ is not exactly correct.  If one was working with the $\mathcal Be(\alpha,\beta)$ density, then  $T^*$ would have been the sufficient statistic for $(\alpha,\beta)$ where $\alpha\ne \beta$.
