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Math Help - Determining degrees of freedom on Chi Sq distr

  1. #1
    Member garymarkhov's Avatar
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    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!
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  2. #2
    MHF Contributor matheagle's Avatar
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    \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.
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    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.
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  3. #3
    Member garymarkhov's Avatar
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    Quote Originally Posted by matheagle View Post
    \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.

    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?
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