# How can I find a complete, minimal sufficient statistic from a $Beta(\sigma,\sigma)$ distribution?

Let $$X_1,\cdots,X_n$$ be a random sample from $$Beta(\sigma,\sigma)$$, where $$\sigma > 0$$ is unknown. Is the minimal sufficient statistic for $$\sigma$$ complete?

My work

I found the minimal sufficient statistic to be $$T(\mathbf{X})=\prod^n_{i=1}(x_i-x_i^2)$$ with the following work:

$$\frac{f(\mathbf{x}|\sigma)}{f(\mathbf{y}|\sigma)} = \frac{\Gamma(2\sigma)^n/\Gamma(\alpha)^{2n} \cdot\prod^n_{i=1}[x_i^{\sigma-1}(1-x_i)^{\sigma-1}]}{\Gamma(2\sigma)^n/\Gamma(\alpha)^{2n} \cdot\prod^n_{i=1}[y_i^{\sigma-1}(1-y_i)^{\sigma-1}]} = \frac{(\prod^n_{i=1}(x_i-x_i^2))^{\sigma-1}}{(\prod^n_{i=1}(y_i-y_i^2))^{\sigma-1}}$$,

where $$\prod^n_{i=1}(x_i-x_i^2)=\prod^n_{i=1}(y_i-y_i^2)$$ if we want the likelihood ratio to be constant as a function of $$\sigma$$. So, we get $$T(\mathbf{X})=\prod^n_{i=1}(x_i-x_i^2)$$.

Then, I obtain the complete statistic by recognizing this is an exponential family:

$$f(x|\sigma)=\frac{\Gamma(2\sigma)}{\Gamma(\sigma)^2}x_i^{\sigma-1}(1-x_i)^{\sigma-1} = \frac{\Gamma(2\sigma)}{\Gamma(\sigma)^2} \cdot exp[(\sigma-1)log(x_i-x_i^2)]$$, so a complete statistic is $$\sum^n_{i=1}log(x_i-x_i^2)=log(\prod^n_{i=1}(x_i-x_i^2))$$

Is it enough to say that there exists a 1-1 function from $$log(\prod^n_{i=1}(x_i-x_i^2))$$ to $$T(\mathbf{X})=\prod^n_{i=1}(x_i-x_i^2)$$, so $$T(\mathbf{X})$$ is a complete, minimal sufficient statistic?

Yes, it is enough. If two statistics are in 1-1 correspondence $$T_1(\mathbf X)=h(T(\mathbf X))$$ with 1-1 function $$h$$, then $$\mathbb E_\sigma[g(T_1)]\equiv 0 \iff \mathbb E_\sigma[g(h(T))]\equiv 0 \iff g(h(T)) = 0 \text{ a.s. } \iff g(T_1)=0 \text{ a.s.}$$
You can say that $$T_1=\log(\prod^n_{i=1}(x_i-x_i^2))$$ is complete. It means that for any Borel function $$g$$, if for each $$\sigma>0$$ $$\mathbb E_\sigma[g(T_1)]=0$$ then $$\mathbb P_\sigma(g(T_1)=0)=1$$ for all $$\sigma>0$$.
We need to say the same for $$T=e^{T_1}$$. Let $$g(x)$$ is a Borel function and suppose that for each $$\sigma>0$$ $$\mathbb E_\sigma[g(T)]=0.$$ It means that $$\mathbb E_\sigma[g(T)]=\mathbb E_\sigma[g(e^{T_1})]=\mathbb E_\sigma[g_1(T_1)]=0$$ with new Borel function $$g_1(x)=g(e^x)$$. Since $$T_1$$ is complete, for each $$\sigma>0$$, $$1=\mathbb P_\sigma(g_1(T_1)=0)=\mathbb P_\sigma\bigl(g(e^{T_1})=0\bigr) = \mathbb P_\sigma(g(T)=0).$$ Then $$T$$ is complete too.