# Variance of the $\hat{\sigma}$ of a Maximum likelihood estimator

With a random sample having size $$n$$ from a normal distribution $$\mathcal{N}(\mu,\sigma^2)$$, and the ML estimate of $$g(\mu,\sigma^2) = \mu +\sigma$$ being $$\widehat{g(\mu,\sigma^2)}=\overline{x}+\hat{\sigma} \:\:\:\:\:\bigg(\text{where}\:\:\hat{\sigma}=\sqrt{\hat{\sigma}^2}\;\bigg)$$ I need to find an estimate of the standard deviation of the estimate of $$g(\mu,\sigma^2)$$.

In order to do this, I think I need to "simply" find Var$$(\overline{x}) +$$ Var$$(\hat{\sigma})$$

Var$$(\overline{x})$$ I am not too concerned about, but I am a little stumped on how to find Var$$(\hat{\sigma})$$. From the help here, I think to use the fact that

$$\frac{n\hat{\sigma}^2}{\sigma^2}\sim \chi_{n-1}^2$$

Following this through, I get down to $$Var(\hat{\sigma}^2)=\frac{2\sigma^4(n-1)}{n^2}$$

From here I am a little bit stuck though - can I simply just take the square root of both sides, take the square root inside the variance, and then have my result? I was kind of hoping to not see $$\sigma$$ appearing in the function for variance, but I suppose it has to?

• Is this homework? I have omitted some details on the assumption that this is an exercise for you be doing, to at least leave some of it for you. Commented Sep 5, 2021 at 2:09
• Thank you Glen, I will work through this soon. It is homework, so thank you for leaving some details out. Commented Sep 5, 2021 at 2:44
• By our rules here at CrossValidated, your post should include the self-study tag (so we aren't left in two minds whether to provide complete answers or hints and guidance) Commented Sep 5, 2021 at 2:56
• Thank you @Glen_b. Is there a tag I should put for it if it is homework? Commented Sep 5, 2021 at 3:00
• The one I mentioned covers homework as well. See stats.stackexchange.com/help/on-topic -- in particular, the paragraph beginning with "Homework...". Commented Sep 5, 2021 at 3:00

You will not get the correct variance of the standard deviation by working out the variance of the variance and taking its square root $$-$$ that's the population standard deviation of the sample variance not the population variance of the sample standard deviation. You also have to be careful about what $$\hat{\sigma}$$ is!

At the normal, the MLE of $$\sigma$$ is $$s_n$$, the $$n$$-denominator version of sample standard deviation (rather than the Bessel-corrected $$s_{n-1}$$).

$$E(s_n) = E\left(\sqrt{\frac{n-1}{n}}s_{n-1}\right) = \sqrt{\frac{n-1}{n}}E(s_{n-1})$$

and $$E(s_{n-1})$$ can be found in Macro's answer here

So now

$$\begin{eqnarray} \text{Var}(s_n) &=& E(s_n^{2}) - E(s_n)^2 \end{eqnarray}$$

and the rest is substitution (take care with the first term as well as with the second).

The variance result will have $$\sigma^2$$ multiplied by a constant (being a function of $$n$$ only).

• is there a reason that the MLE of $\sigma$ is $s_n$ and not simply $\sqrt{\hat{\sigma}^2}$ as seems to be suggested here? Commented Sep 5, 2021 at 6:03
• I don't understand the question. The $\hat{\,}$ indicates MLE, and $\widehat{\sigma^2} = \frac{1}{n}\sum_i (y_i-\bar{y})^2 = s_n^2$. By the theorem discussed at that link, therefore, $\hat{\sigma} = s_n$ Commented Sep 5, 2021 at 6:46
• Commented Sep 5, 2021 at 7:03