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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.

Thanks.

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  • $\begingroup$ 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
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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 $N\gt1$.

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