# Variance estimator of Bernoulli RV (in CI)

Assume n samples from Bernoulli distribution with unknown parameter p,

i. e. $x_{1},....,x_{n}\underset{iid}{\sim}Ber\left(p\right)$

It is known that the confidence interval is given by:

$CI=\hat{p}\pm z_{1-\frac{\alpha}{2}}\cdot\sqrt{\hat{p}\left(1-\hat{p}\right)/n}$

where $\hat{p}=\frac{1}n\sum_{i=1}^{n}x_{i}$

This means that the estimator of the variance being used in this formula is:

$s^2 = \hat{p}\left(1-\hat{p}\right)$

The latter is not trivial for me. I've managed to prove that this estimator is biased (with a finite sample size) but failed to prove consistency.

Since it's not clear (for me) if this estimator is consistent, and it's clearly biased, why should we use it and not the usual unbiased estimator?

Is there a full mathematical reason for this? (A reference could be great also). Maybe I'm failing to understand something else?

• Surely $\hat{p}$ needs to be divided by $n$? May 2, 2017 at 19:11
• You are right, forgot it. May 2, 2017 at 19:20

$\hat{p}$ converges almost surely to $p$ by SLLN. Since the product of two almost-sure convergent random variable converges almost-surely to the product of their limits, it follows that $s^2$ converges almost surely to $p(1-p)$, which also implies it converges in probability, hence consistency.

• Thank you! Last thing - why should you prefer this estimator (which is biased) over the usual unbiased $\frac{1}{n-1}\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}$ estimator? May 2, 2017 at 20:35
• @MaorSH you can show that for Bernoulli $X_i$ $\frac 1n \sum_i(x_i-\bar x)^2 = \bar x (1 - \bar x)$, so this reduces to the usual $\frac 1n$ vs $\frac 1{n-1}$ debate, as discussed here for instance: stats.stackexchange.com/questions/3931/…
– jld
May 2, 2017 at 21:46