Let $r$ be the observed number of successes in $n$ Bernoulli trials with probability $\pi$ of success. Then M.V.U.E (Minimum Variance Unbiased Estimator) of $\pi (1-\pi)$ is ?
$n$ Bernoulli trials can be considered to follow Binomial Distribution with parameters ($n$,$\pi$).
I understand that an efficient estimator of a parameter is always an Uniform minimum Variance Unbiased Estimator. And an unbiased estimator is called an efficient estimator if it satisfies Cramer-Rao lower bound.
So is finding the Cramer-Rao lower bound the way to solve this question? And if so how is it done because I am struggling in conjuring the Likelihood Function
EDIT : As mentioned by @wuber , I went through a few prevously answered questions(Maximum Likelihood Estimation for Bernoulli distribution) pertaining to finding the Maximum Likelihood Estimator of $n$ Bernoulli Trials and this is what I got:
Let $x_{i}$ be the $i^{th}$ success such that $r = \sum_{i=1}^n x_i$
$$ \begin{align*} L(\pi) &= \prod_{i=1}^n \pi^{x_i}(1-\pi)^{(1-x_i)}\\ \ell(p) &= \log{\pi}\sum_{i=1}^n x_i + \log{(1-\pi)}\sum_{i=1}^n (1-x_i)\\ \dfrac{\partial\ell(\pi)}{\partial \pi} &= \dfrac{\sum_{i=1}^n x_i}{\pi} - \dfrac{\sum_{i=1}^n (1-x_i)}{1-\pi} \overset{\text{set}}{=}0\\ \sum_{i=1}^n x_i - \pi\sum_{i=1}^n x_i &= \pi\sum_{i=1}^n (1-x_i)\\ \pi& = \dfrac{1}{n}\sum_{i=1}^n x_i\\ \pi& = r/n\\ \dfrac{\partial^2 \ell(\pi)}{\partial \pi^2} &= \dfrac{-\sum_{i=1}^n x_i}{\pi^2} - \dfrac{\sum_{i=1}^n (1-x_i)}{(1-\pi)^2} \end{align*} $$
The Maximum Likelihood estimator of $\pi = \frac{r}{n}$
Therefore by Invariance Property $$ Estimator \ of \ \pi (1-\pi) = \frac{r}{n}(1-\frac{r}{n})$$
But back to the question at hand, can we deduce this as M.V.U.E ?
