# Determining degrees of freedom on Chi Sq distr

• Jan 15th 2010, 02:09 PM
garymarkhov
Determining degrees of freedom on Chi Sq distr
I have in my notes that $\frac{\hat u^T \hat u}{T-k} = \sigma ^2$. I also have that $\frac{\hat u^T \hat u}{ \sigma ^2} \sim \chi^2_{T-k}$. I know how to derive the first part, but how the heck do we know the second thing is true? Does it come from the derivation of the first equation or... what? Help!
• Jan 15th 2010, 09:38 PM
matheagle
$\frac{\hat u^T \hat u}{T-k} = \sigma ^2$

is incomplete, I need to know what the distribution is, probably normal
then ${\hat u^T \hat u}$ is a chi-square...
AND I'm sure you're missing an expectation, because you have a random variable equal to a constant.
-----------------------------------------------------------------------

If $\frac{\hat u^T \hat u}{ \sigma ^2} \sim \chi^2_{T-k}$

Then $E\left(\frac{\hat u^T \hat u}{ \sigma ^2}\right)=T-k$.

Flipping constants gives.... $E\left(\frac{\hat u^T \hat u}{T-k}\right)=\sigma ^2$.
• Jan 16th 2010, 07:35 AM
garymarkhov
Quote:

Originally Posted by matheagle
$\frac{\hat u^T \hat u}{T-k} = \sigma ^2$
is incomplete, I need to know what the distribution is, probably normal

Sorry, yeah, the error terms are assumed to be normally distributed.

Quote:

AND I'm sure you're missing an expectation, because you have a random variable equal to a constant.
Right, I agree with that. But must the expectation encompass the $T-k$ part too?