I am trying to find the distribution of a RV that should seem relatively straight forward but is eluding me. Let $X \sim N(0,\sigma_1)$ and $Y \sim N(0,\sigma_2)$ while $X$ and $Y$ are independent. I am looking for the distribution of $Q = \frac{Y}{X-Y}$.

Now I know that $\frac{X}{Y} \sim \text{Cauchy}\left(0,\frac{\sigma_1}{\sigma_2}\right)$ so I tried to start from there.

I used the following transformation to try to deal with the weird denominator: $U=X-Y$.

This provides the following transformed RV: $\frac{X-U}{U} = \frac{X}{U}-1$

Now this should be Cauchy distributed too. When I run numerical calculation in Excel (0ver 20,000 data points) and test the hypothesis whether the distribution is Cauchy it says each time it is.

The problem I am facing however is the parameters. The distribution does not have a zero median and the variance I expect to get does not match the numerical calculations from the data.

I'll illustrate with an example. Let $X \sim N(0,1)$ and $X \sim N(0,4)$. I expect the distribution to be $Q\sim \text{Cauchy}\left(0,0.8\right)$. However, from the data I obtained $Q\sim \text{Cauchy}\left(-0.94215,0.23406\right)$. After dealing with data sets of 20,000 data points it seems as if

$\frac{X-U}{U}\sim \text{Cauchy} \left(\frac{\sigma_1^2}{\sigma_1^2+\sigma_2^2}-1,\frac{\sigma_1\sigma_2}{\sigma_1^2 + \sigma_2^2}\right)$

This, however, I am not able to show analytically yet.

I suspect I am missing something about covariance however I am not quite sure. Any help to determine the issue and provide the correct formulas for the parameter estimations would be greatly appreciated.

  • 1
    $\begingroup$ $\mathrm{Cov}(X-U, U) = \mathrm{Cov}(Y, X-Y) = \mathrm{Cov}(Y, X) - \mathrm{Cov}(Y, Y) = - \sigma_2^2$, so the identity you tried to use doesn't apply – that's only for the ratio of independent normals. I don't know whether the ratio is truly Cauchy or not. $\endgroup$
    – Danica
    Oct 4, 2015 at 19:06
  • 3
    $\begingroup$ This could get you started. $\endgroup$
    – Gumeo
    Oct 4, 2015 at 19:20
  • 1
    $\begingroup$ You don't say that X and Y are independent but you seem to be assuming they are. Note that $\frac{X-U}{U} = \frac{X}{U} - 1$ ... so it's not clear why a negative median would surprise you. $\endgroup$
    – Glen_b
    Oct 5, 2015 at 10:22
  • 2
    $\begingroup$ See Marsaglia, G (2006) "Ratios of Normal Variables" - Journal of Statistical Software, vol 16. jstatsoft.org/issue/view/v016 $\endgroup$
    – Glen_b
    Oct 5, 2015 at 11:47
  • $\begingroup$ Thanks for the help so far! I have read Hinkley's paper "On the Ratio of Two Correlated Normal Random Variables" and I have looked at Marsaglia's paper "Ratios of Normal Variables". Unless I am missing something, neither of these two papers provide a way to solve my question. I am still scratching my head what the parameters of the supposed Cauchy distribution is which I present empirical justification for. Alas the mathematical derivation eludes me still. I have thought about expanding the ratio as a series and going from there but that seems like a larger problem than the original... $\endgroup$
    – dsmalenb
    Oct 10, 2015 at 2:20

1 Answer 1


First we have the following result:

Lemma If $A\sim Cauchy(\mu,\sigma)$ then $1/A \sim Cauchy(\mu/(\mu^2+\sigma^2),\sigma/(\mu^2+\sigma^2))$.


\begin{equation} \begin{split} f_{\frac{1}{A}}(a) & = \frac{1}{\pi} \frac{1}{1+\left(\frac{\frac{1}{a}-\mu}{\sigma}\right)^2}\frac{1}{a^2}\\ & = \frac{1}{\pi\sigma} \frac{\sigma^2}{\sigma^2 a^2+1-2\mu a +\mu^2 a^2}\\ & = \frac{1}{\pi\sigma} \frac{\sigma^2}{1-2\mu a + \left(\mu^2+\sigma^2\right)a^2}\\ & = \frac{1}{\pi\sigma} \frac{\sigma^2}{1-\frac{\mu^2}{\mu^2+\sigma^2}+\left(a\sqrt{\mu^2+\sigma^2}-\frac{\mu^2}{\sqrt{\mu^2+\sigma^2}}\right)^2}\\ & = \frac{1}{\pi\sigma} \frac{\sigma^2}{\frac{\mu^2+\sigma^2}{\mu^2+\sigma^2}-\frac{\mu^2}{\mu^2+\sigma^2}+\left(a\sqrt{\mu^2+\sigma^2}-\frac{\mu}{\sqrt{\mu^2+\sigma^2}}\right)^2}\\ & = \frac{1}{\pi\sigma} \frac{\sigma^2}{\frac{\sigma^2}{\mu^2+\sigma^2}+\left(a\sqrt{\mu^2+\sigma^2}-\frac{\mu}{\sqrt{\mu^2+\sigma^2}}\right)^2}\\ & = \frac{1}{\pi\sigma} \frac{1}{\frac{1}{\mu^2+\sigma^2}+\frac{1}{\sigma^2}\left(a\sqrt{\mu^2+\sigma^2}-\frac{\mu}{\sqrt{\mu^2+\sigma^2}}\right)^2}\\ & = \frac{1}{\pi\frac{\sigma}{\mu^2+\sigma^2}} \frac{1}{1+\frac{\mu^2+\sigma^2}{\sigma^2}\left(a\sqrt{\mu^2+\sigma^2}-\frac{\mu}{\sqrt{\mu^2+\sigma^2}}\right)^2}\\ & = \frac{1}{\pi\frac{\sigma}{\mu^2+\sigma^2}} \frac{1}{1+\left(\frac{a\sqrt{\mu^2+\sigma^2}-\frac{\mu}{\sqrt{\mu^2+\sigma^2}}}{\frac{\sigma}{\sqrt{\mu^2+\sigma^2}}}\right)^2}\\ & = \frac{1}{\pi\frac{\sigma}{\mu^2+\sigma^2}} \frac{1}{1+\left(\frac{a-\frac{\mu}{\mu^2+\sigma^2}}{\frac{\sigma}{\mu^2+\sigma^2}}\right)^2}\\ & = Cauchy\left(\frac{\mu}{\mu^2+\sigma^2},\frac{\sigma}{\mu^2+\sigma^2}\right) \end{split} \end{equation}

We wish to determine the distribution of $A = Y/(X-Y)$. It is easier to first determine the distribution of $1/A$ and then that of $A$.

\begin{equation} \frac{1}{A} = \frac{X}{Y} - 1 \end{equation}

Since $X,Y \sim N$ and $X$ and $Y$ are independent, we know that $1/A$ is distributed as $Cauchy(-1, \sigma_1/\sigma_2)$. For the case of Cauchy random variables, if $X\sim Cauchy$ then $1/X$ is also Cauchy distributed. So we end up with:

\begin{equation} \frac{Y}{X-Y} \sim Cauchy\left(-\frac{\sigma_2^2}{\sigma_1^2+\sigma_2^2},\frac{\sigma_1\sigma_2}{\sigma_1^2+\sigma_2^2}\right) \end{equation}


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