Let $\hat{\theta}_m$ be an unbiased estimate the average value $\theta$ of Bernoulli distribution $X_i$ $\hat{\theta}_m = \frac{1}{m}\sum_{i=1}^{m}X_i$ $\theta = \mathbb{E}(X_i)$ Therefore the Variance of $\hat{\theta}_m$ is as follows: $Var(\hat{\theta}_m) = Var(\frac{1}{m}\sum_{i=1}^{m}X_i)$ $= \frac{1}{m^2}Var(\sum_{i=1}^{m}X_i)$ $= \frac{1}{m^2}\sum_{i=1}^{m}Var(X_i)$ $= \frac{1}{m}Var(X_i)$ $= \frac{1}{m}\theta(1 - \theta)$ Note that for this example as $m$ increases the variance of the estimator approaches zero in the limit, so long as the value for $\theta$ is defined.