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I'm working with two independent normal distributions $X$ and $Y$, with means $\mu_x$ and $\mu_y$ and variances $\sigma^2_x$ and $\sigma^2_y$.

I'm interested in the distribution of their ratio $Z=X/Y$. Neither $X$ nor $Y$ has a mean of zero, so $Z$ is not distributed as a Cauchy.

I need to find the CDF of $Z$, and then take the derivative of the CDF with respect to $\mu_x$, $\mu_y$, $\sigma^2_x$ and $\sigma^2_y$.

Does anyone know a paper where these have already been calculated? Or how to do this myself?

I found the formula for the CDF in a 1969 paper, but taking these derivatives will definitely be a huge pain. Maybe someone has already done it or knows how to do it easily? I mainly need to know the signs of these derivatives.

This paper also contains an analytically simpler approximation if $Y$ is mostly positive. I can't have that restriction. However, maybe the approximation has the same sign as the true derivative even outside the parameter range?

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I have added $\TeX$ for you. You wrote "sigma" but mentioned that these were variances, so I made them sigma-squared. Make sure it still says what you want to ask. – gung Jul 25 '12 at 23:17
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en.wikipedia.org/wiki/Ratio_distribution has the probability density function. – Douglas Zare Jul 26 '12 at 0:38
That's the same PDF as in the paper above. I'm trying to take the derivative of the CDF with respect to the underlying mus and sigmas. – ABC Jul 26 '12 at 1:14
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The formula of the pdf found by David Hinkley is completely in closed-form. So you can take those derivatives, one step at a time. I am actually curious about the point of doing such derivations as there is no reason the sign should be constant uniformly over the real numbers... – Xi'an Jul 26 '12 at 7:43
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@ABC You can find the density of $X/Y$ in equation 1 of this paper. I worked on it some time ago and it agrees with Hinkley's result and Marsaglia's result. It can be deduced by brute force as well as Douglas Zare suggests (I did it, only recommended if you really need to do it). – user10525 Jul 26 '12 at 9:16
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