The following example is taken from Hogg Introduction to Mathematical Statistics 7e and the exercise is to show that the MLE is a minimal sufficient statistic. I am not 100% sure about my argument.
Given: $X_1, \dots, X_n \overset{iid}{\sim} Beta(1, \theta)$
- Deriving the ML-Estimator:
$$ f_X(x) = \frac{ \Gamma(1+\theta) }{\Gamma(1)\Gamma(\theta)} x^0(1 - x)^{\theta -1} = \frac{\theta!}{(\theta-1)!}(1 - x)^{\theta -1} = \theta(1 - x)^{\theta -1} \quad x \in (0,1] $$
Likelihood function (let $I := {1,\dots, n})$: \begin{align*} L(\theta|x_{i \in I}) &= \prod_{i \in I} \theta\cdot(1 - x_i)^{\theta-1} = \theta^n\prod_{i \in I}(1 - x_i)^{\theta-1} \\ \Rightarrow logL(\theta|x_{i \in I})&= n log(\theta) + (\theta - 1)\sum_{i \in I}log(1-x_i) \rightarrow max! \\ &\Rightarrow\frac{n}{\theta} + \sum_{i \in I}log(1-x_i) \overset{!}{=} 0 \\ &\Rightarrow \hat{\theta}_{ML} = -\frac{1}{n \sum_{i \in I}log(1-x_i)} \end{align*}
which will be positive since $log(x) \leq 0 \quad \forall x \in (0, 1)$.
Factorizing:
\begin{align*} L(\theta|X_{i \in I}) &= \prod_{i \in I} \theta\cdot(1 - x_i)^{\theta-1} = \theta^n\prod_{i \in I}(1 - x_i)^{\theta-1} = \theta^n\prod_{i \in I}(1 - x_i)^{\theta} \prod_{i \in I}(1 - x_i)^{-1} \\ &\Rightarrow log(L(\theta|X_{i \in I})) =nlog(\theta)+ \theta\sum_{i \in I}log(1 - x_i) - \sum_{i \in I}log(1 - x_i) \end{align*}
My argument:
Thus by the factorization theorem $(\sum_{i \in I}log(1-x_i))$ is a sufficient statistic. Since $\hat{\theta}$ is a function of the sufficient statistic, it is in fact a minimal sufficient statistic (without proof).