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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?

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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 $$ \mathbb E_\sigma[g(T_1)]\equiv 0 \iff \mathbb E_\sigma[g(h(T))]\equiv 0 \iff g(h(T)) = 0 \text{ a.s. } \iff g(T_1)=0 \text{ a.s.} $$

Step-by-step:

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.

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  • $\begingroup$ Perfect - thank you for your great answer! $\endgroup$ – Ron Snow Feb 13 '20 at 6:06

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