Is the Point Estimate for variance of Bernoulli Unbiased?

Question

Let $X_1,...,X_n$ be a random sample from the Bernoulli family with parameter $p$. Let $\sigma^2$ be the variance of $X_1$. Consider the point estimate for $\sigma^2$ given by $\overline{X}(1-\overline{X})$. Is it unbiased?

Since $\overline{X}(1-\overline{X}) = \overline{X}-(\overline{X})^2$, we have $\mathbb{E}(\overline{X}(1-\overline{X})) = \mathbb{E}(\overline{X}-(\overline{X})^2)= \mathbb{E}\overline{X}-\mathbb{E}(\overline{X})^2$.

We know that $\mathbb{E}\overline{X}=\mu$ and $\mathbb{E}(\overline{X})^2=\frac{\sigma^2}{n}+\mu^2$.

Putting them in $\mathbb{E}\overline{X}-\mathbb{E}(\overline{X})^2$ we find that it equals to $\mu-\frac{\sigma^2}{n}-\mu^2$ which is not equal to $\sigma^2$ and thus it is not unbiased. Is this correct?

Answer 1

$\newcommand{\e}{\operatorname{E}}$We have $\mu = p$ and $\sigma^2 = p(1-p) = \mu(1-\mu) = \mu-\mu^2.$

So \begin{align} \e\big( \overline X(1-\overline X) \big) = \e\left(\overline X\right) - \e\left( \overline X ^2 \right) & = \mu - \left( \mu^2 + \frac{\sigma^2} n \right) \\[10pt] & = (\mu-\mu^2) - \frac{\sigma^2} n \\[10pt] & = \sigma^2 - \frac{\sigma^2} n = \frac{n-1} n \sigma^2. \end{align} So this estimator is biased. If I'm not mistaken, it has a slightly smaller mean squared error than does the best unbiased estimator.

Answer 2

$$\mathbb{E}(\overline{X}(1-\overline{X})) = \mathbb{E}(\overline{X}-(\overline{X})^2)= \mathbb{E}[\overline{X}]-\mathbb{E}[(\overline{X})^2] =p-\mathbb{E}[(\overline{X})^2] < p-(\mathbb{E}\overline{X})^2 = p-p^2$$ thanks to Jensen's Inequality. Thus it is biased.