This is a follow-up question on that one:
Could Bessel's correction make sample variance estimation even more biased?

I understand that you need Bessel's correction to get an unbiased estimate of variance in case you sample with replacement. Now in practice most sampling is done without replacement (e.g. asking voters about their preferences or studying the effects of new drugs).

My questions are:
- What is the unbiased estimator for variance in cases of samples without replacement?
- Choosing only between the classical estimators for population variance and sample variance wouldn't it be better in most cases to take the population variance for samples because it is less biased than sample variance in cases of samples without replacement?

  • 1
    $\begingroup$ It is erroneous to assume the population variance would be less biased than the sample variance, because the bias in the former (for samples with replacement) is of order $1/n$ for sample sizes of $n$ whereas the difference between sampling with and without replacement introduces errors only of size $1/(N-n)$ where $N$ is the population size which ordinarily is hugely greater than $n$. Moreover, in the case of studying drug effects there is no definite value of $N$ because we wish to extend inference even to future populations, so $N$ is effectively infinite. $\endgroup$ – whuber Sep 16 '13 at 14:46

First, if you're trying to estimate the population variance, σ², then the population variance is a perfect estimator of the population variance. I suspect that is not what you meant. Let's call Σ(x - x̄)²/n the maximum-likelihood estimator of σ².

Yes, when sampling without replacement from a finite population, s² tends to overestimate the true variance, σ², and using the maximum-likelihood estimator gives you a lower estimate than s². But doing so might over-correct or under-correct. There would still be a bias.

If you multiply s² by the factor (N-1)/N, where N is the size of the finite population, you will get an unbiased estimator of σ².


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.