I know that a question with the same topic has been asked a few years ago (How do you derive the conditional variance for $s^2$, the OLS estimator of $\sigma^2$?) however, My question is kind of different.

I have to prove that $var(s^2|\textbf{X})=\frac{2\sigma^4}{(n-k)}$ where $s^2=\frac{e'e}{n-k}$ and $e=y-X\beta$

So far I'm given the term $s^2$ which is unconditional. I have to prove that it's conditional variance on $\textbf{X}$ is equal to what I have stated earlier. The first step in order to solve this is this

$var(s^2|\textbf{X})= var\big(\frac{e'e}{n-k}\big)$

What I fail to understand is when it is algebraically correct to set that the conditional variance of $s^2$ is equal to its unconditional variance. I know that once this is solved, we can see that the variance of $s^2$ does not depend on $\textbf{X}$, so this assumption makes sense, but how can I know that before hand? Is there even a way?

The rest is easy to compute and the solution is:

multiplying and dividing by $\sigma^2$ I get:

$var(s^2|\textbf{X})= \frac{\sigma^4}{(n-k)^2}var\big(\frac{e'e}{\sigma^2}\big)$

Knowing that $\big(\frac{e'e}{\sigma^2}\big)$ follows a Chi-Squared distribution with $(n-k)$ degrees of freedom, I get that $var\big(\frac{e'e}{\sigma^2}\big)=2(n-k)$. Thus I can simplify my expression

$var(s^2|\textbf{X}) = \frac{2\sigma^4}{(n-k)}$

  • $\begingroup$ True, it was a typo from my side. Corrected now. $\endgroup$
    – Aurel
    Commented Dec 27, 2015 at 12:04
  • 1
    $\begingroup$ "So far I'm given the term $s^2$ which is unconditional." ... no, $s^2$ is not the unconditional variance. $\endgroup$
    – Glen_b
    Commented Dec 27, 2015 at 12:07
  • 2
    $\begingroup$ The notation $|X$ is not a conditionning. You can drop it everywhere, consider $X$ is fixed, not random. This is totally useless and puzzling. $\endgroup$ Commented Dec 27, 2015 at 13:55


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.