2
$\begingroup$

What is the variance of the average of $n$ Bernoulli distributed random variables $Bernoulli(p)$?

So $\text{Var}\left[\frac{\sum_i x_i}{n}\right]$

$\endgroup$
4
  • $\begingroup$ Sorry, clarified that a bit $\endgroup$ Commented Feb 13, 2014 at 4:51
  • $\begingroup$ Wouldn't it be the variance of the binomial distribution? $\endgroup$ Commented Feb 13, 2014 at 4:54
  • $\begingroup$ Don't think so because the binomial distribution is the sum(xi) without the / N term $\endgroup$ Commented Feb 13, 2014 at 5:02
  • $\begingroup$ The average of n Bernoulli distributed random variables is a proportion. That proportion is the maximum likelihood estimate of the Bernoulli distribution's probability parameter. The Bernoulli distribution is a special case of the binomial, which has the same parameter, same interpretation. $\endgroup$ Commented Feb 13, 2014 at 5:05

1 Answer 1

8
$\begingroup$

If $X_1, X_2, \ldots, X_n \sim {\rm Bernoulli}(p)$ are IID, then $S = \sum_{i=1}^n X_i \sim {\rm Binomial}(n,p)$. Therefore, $${\rm Var}[S/n] = \frac{1}{n^2}\cdot np(1-p) = \frac{p(1-p)}{n}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.