What is the variance of the average of $n$ Bernoulli distributed random variables $Bernoulli(p)$?
So $\text{Var}\left[\frac{\sum_i x_i}{n}\right]$
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$\begingroup$ Sorry, clarified that a bit $\endgroup$– user1507844Feb 13, 2014 at 4:51
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$\begingroup$ Wouldn't it be the variance of the binomial distribution? $\endgroup$– Brash EquilibriumFeb 13, 2014 at 4:54
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$\begingroup$ Don't think so because the binomial distribution is the sum(xi) without the / N term $\endgroup$– user1507844Feb 13, 2014 at 5:02
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$\begingroup$ 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. $\endgroup$– Brash EquilibriumFeb 13, 2014 at 5:05
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1 Answer
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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, $${\rm Var}[S/n] = \frac{1}{n^2}\cdot np(1-p) = \frac{p(1-p)}{n}.$$
