# About MLE of $\sigma$ with Normal Distribution

I tried to find answers the questions below, but I could not get clear answers for them.

For a random sample of size n, $$x_1, x_2, ..., x_n$$ from a Normal distribution where $$\sigma^2$$ is unknown.

It is quite easy to derive answers with the MLE of $$\sigma^2$$

However, how to show the biasedness and consistency for maximum likelihood estimator of $$\sigma$$? I mean, not $$\sigma^2$$

Moreover, how to show the asymptotic distribution of the MLE $$\sigma$$?

• Thank you for your reply. But the thing is, MLE of $\sigma$ and showing its consistency and asymptotic normality. – coalt Oct 10 '18 at 15:21

A nice thing about MLE is that its structure means that if you have an MLE $$\hat{p}$$ for $$p$$, then $$f(\hat{p})$$ is the MLE of $$f(p)$$ for any injective measurable function $$f$$. Thus, if $$T$$ is the MLE of $$\sigma^2$$ then $$\sqrt{T}$$ is the MLE of $$\sigma$$ for instance.
In terms of properties of $$f(\hat{p})$$ as an estimator, consistency is relatively mild. If $$\hat{p}$$ was consistent and $$f$$ is, say, uniformly continuous then $$f(\hat{p})$$ will be consistent. But $$f(\hat{p})$$ will usually be biased even if $$\hat{p}$$ wasn't. In the case of passing from a variance estimator to a standard deviation estimator you can see this by Jensen's inequality: the square root of an unbiased variance estimator is a biased standard deviation estimator.
The issue is that MLE estimates $$\sigma^2$$ as the sample mean of $$(X-\bar{x})^2$$, where $$\bar{x}$$ is the sample mean rather than the distribution's true mean. Because $$\bar{x}$$ is obtained by averaging the empirical $$x$$, it's slightly closer to them than the true mean; indeed, you can prove $$\sum_{i=1}^n (x_i-m)^2$$ is $$n\sigma^2$$ if $$m$$ is the true mean, but only $$(n-1)\sigma^2$$ if $$m$$ is the sample mean. Replacing the usual $$\frac{1}{n}$$ with $$\frac{1}{n-1}$$ addresses this. This is called the Bonferroni correction (that link proves the aforementioned $$n-1$$ result).