# Bias corrrection for MLE when dealing with normally distributed small samples

When estimating the standard-deviation for samples of normally distributed data, it is sometimes necessary to account for bias in whatever estimator one chooses -- which is usually related to the number of observations in the sample.

For example if we take the corrected-sample standard deviation, $$s = \sqrt{\frac{1}{n-1} \sum_{i}^{N} \left(x_{i} - \bar{x} \right)^{2}}$$ we account for sample-size bias with $$C_{4}(N) = \sqrt{\frac{2}{N-1}}\frac{\Gamma(N/2)}{\Gamma((N-1)/2)}$$ with an unbiased estimator for the standard deviation being $$\sigma_{\rm{est}} = s / c_{4}(n).$$

For other estimators such as $$\rm{MAD}$$, $$S_{n}$$, and $$Q_{n}$$ there are also bias correction factors which also scale with the sample size.

Is there a similar correction factor one can apply when using maximum-likelihood methods? In the plot below I have simulated normally distributed, $$\mathcal{N}(0, 1)$$, for different sample sizes and estimated the standard deviation with $$s$$, $$s/c_{4}$$, and with a maximum-likelihood evaluation. We can see the bias behaviour:

The difference between the MLE for the variance and $$s^2$$ is dividing by $$N$$ instead of by $$N-1$$. Therefore, the MLE is $$\sqrt{\frac{N-1}{N}}s$$. Hence, $$\sqrt{\frac{N}{N-1}}\times MLE/c_4(N)=s/c_4(N)$$ is unbiased.

• Allow me to play around and I'll accept.
– N.B.
Dec 31, 2020 at 21:35
• Thanks for this, I'll award the bounty when the timer allows -- can you give me a source or name for this? I haven't been able to find this on Wikipedia.
– N.B.
Jan 1, 2021 at 15:32