The maximum likelihood estimate of a Bernoulli process is simply given by $\hat{\theta}=\frac{\sum X_i}{N}$, where N is the total number of bernoulli trial and $X_i$ is the outcome of each trial.

This is an unbiased estimator and the variance of this estimator can be easily computed to be $Var(\hat{\theta}) = \frac{\theta(1-\theta)}{N}$. However, the actual $\theta$ is unknown.

So how do we estimate the variance of the estimator then ? Also, I would like an unbiased estimate of this variance. Its possible that this question has already been asked. If someone can give a pointer that would be great.


  • $\begingroup$ For $N\gt 1$, use $\frac{\hat{\theta}(1-\hat{\theta})}{N-1}$. $\endgroup$ – André Nicolas Aug 17 '15 at 4:00
  • $\begingroup$ $\hat{\theta}$ here apears to be sample mean. Maybe it is possible to use an unbiased sample variance $s^2 = \frac{1}{N-1}\sum_{i=1}^{N}(X_i-\hat{\theta})^2$ ? It is also consistent. $\endgroup$ – Slowpoke Aug 17 '15 at 4:12

Community wiki answer based on the comments to allow the answer to be accepted:

In this case the parameter happens to be the mean. The unbiased estimator $\hat\theta$ of the parameter is the usual unbiased estimator of the mean, whose variance is $1/N$ times the population variance, so estimating its variance is equivalent to estimating the population variance, which can be done without bias using Bessel's correction of multiplying by $N/(N-1)$ for $N\gt1$.


Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

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