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It is well known that the sum of a series $m$ of squared standard independent normal random variables follows a $\chi^2$ dstribution with $m$ degrees of freedom. It is also true that the ratio of two such independent distributions follows a F distribution, after the $\chi^2$ variables have been divided by $m$.

But how is this ratio distributed when the two chi-squared random variables are not independent?

For example, say there is a multivariate normal data-generating process $P = \mathcal{N}(\mathbf{0,\Sigma})$, and the data are $N$ draws from this distribution. The mahalanobis distance $D$ of any given observation follows a $\chi^2_m$ distrbution, where $m$ is the number of dimensions.

Say we take the same $N$ draws and calculate the "inverted" squared mahalanobis distance $(\mathbf{x-\mu})'\mathbf{S}^{-1^{-1}}(\mathbf{x-\mu})$. Call these distances $D_i$. They'll be correlated with $D$, because the inverted covariance matrix has the same eigenvectors, only with inverted eigenvalues.

Given that $D$ and $Di$ have a non-zero correlation $\rho$, what is the CDF of $\frac{D}{Di},m,\Sigma$?? (And do I need all of the information in $\Sigma$, or do I simply need $\rho$?)

edit: I've found the following paper on correlated bivariate $\chi^2$ distributions, but it gives a result for a bivariate pdf, rather than their univariate ratio. I'm working on seeing whether I can take that result and get a Cdf of the ratio, but it may be beyond me: [Moments of the product and ratio of two correlated chi-square variables][1]

That PDF is given as $$ \frac{2^{-(m+1)}(xy)^{(m-2)/2}e^{\frac{-x-y}{2*(1-\rho^2)}}}{\sqrt{\pi}\Gamma(\frac{m}{2})(1-\rho^2)^{m/2}}\displaystyle\sum_{i=0}^{\infty}(1+(-1)^i)\left(\frac{\rho \sqrt{xy}}{1-\rho^2}\right)^i\left(\frac{\Gamma((k+1)/2)}{k!\Gamma((k+m)/2)}\right) $$ (yikes)

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    $\begingroup$ Please double-check your first paragraph. The normal random variables need to be standard normal. And they need to be squared before taking the sum. And to get the F from two independent $\chi^2$ random variables, they need to be divided by their degrees of freedom before taking their ratio. $\endgroup$ – Wolfgang Sep 25 '15 at 9:01

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