How do you get the variance of the residual variance in a simple linear regression model? In a book I see that ${\rm Var}(S_R^2)=\frac{2\sigma^4}{n-2}$, where $S_R^2=\sum\frac{e_i^2}{n-2}$ and $e_i=y_i-\hat{y}_i$. How can I get this result?

I already know that $\frac{\sum e_i^2}{\sigma^2}\sim\chi^2_{n-2}$, but I need to know how I get the variance of $S_R^2$.

  • $\begingroup$ Could you add the self-study tag? I think it applies here. $\endgroup$ – Xi'an Feb 4 '15 at 20:24
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    $\begingroup$ Hint: the variance of a $\chi^2_{n-2}$ distribution is $2(n-2)$. That's the only additional information you need. $\endgroup$ – whuber Feb 4 '15 at 20:24
  • $\begingroup$ Have you visited the Wikipedia page on the $\chi^2$ distribution? It contains all the relevant items of information. $\endgroup$ – Xi'an Feb 4 '15 at 20:25
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    $\begingroup$ @will198, if you've figured it out, why not answer your own question? $\endgroup$ – gung - Reinstate Monica Feb 4 '15 at 20:38
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    $\begingroup$ Maybe you could post this as an answer instead of a comment. $\endgroup$ – Richard Hardy Feb 4 '15 at 20:51


  • $\frac{\sum e_i^2}{\sigma ^2}\sim \chi^2_{n-2} $
  • and knowing that $var(\chi^2_{n-2}=2(n−2)$
  • I can get that $var[\frac{\sum e_i^2}{\sigma ^2}]=2(n−2)$ dividing for $(n−2)^2$ and multipliying for $\sigma^4$ I get $var[\frac{\sum e_i^2}{(n-2)}]=\frac{2\sigma^4}{n-2}=var[S^2_R]$. I applied that $a^2 \cdot Var(x)=Var[a \cdot x]$
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