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A book I'm reading claims that the statistic:

$\frac{(RSS_0 - RSS_1) / (p_1 - p_0)}{RSS_1 / (N - p_1 - 1)}$ has an F distribution. Why is this? I know that an F distribution is something like $\frac{\chi^2_p / p}{\chi^2_q / q}$, where the two chi-square distributions are independent, but I fail to see why $RSS_0 - RSS_1$ is chi-squared and also why $RSS_0 - RSS_1$ and $RSS_1$ are independent.

For some context, $RSS_1$ is the $RSS$ of a least squares model with $p_1 + 1$ parameters, and $RSS_0$ is a smaller model with $p_1 - p_0$ of the parameters in the first model set to 0.

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    $\begingroup$ One of the main assumptions of the OLS model is that the error term is normally distributed. As you might know the square of a normal random variable is distributed as a Chi square. The sum of Chi squares is once again Chi square $\endgroup$ – RScrlli Mar 27 at 21:36
  • $\begingroup$ @RScrlli the sum of chi squares is chi square, but I don't think the difference is. $\endgroup$ – serendipity Mar 27 at 21:59
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    $\begingroup$ In this case, the difference can be expressed as a sum of squares. $\endgroup$ – whuber Mar 27 at 22:26
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    $\begingroup$ @serendipity Difference of (independent) chi-square variables is definitely not another chi-square. But this is not the setup here. $\endgroup$ – StubbornAtom Mar 28 at 10:30
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    $\begingroup$ And I think your result follows from a theorem on quadratic forms related to Fisher-Cochran theorem. $\endgroup$ – StubbornAtom Mar 28 at 11:16

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