Ratio of Cauchy: ChatGPT fails on this simple question

It is well known that the ratio of two independent standard normals \(Z_1/Z_2\) follows a Cauchy distribution. But what about the ratio of two (independent) Cauchy distributions? ChatGPT confidently tells me that it is still a Cauchy distribution. I almost agreed it with it and went on to have a productive day, but got suspicious after the following simulation:

set.seed(1)
x <- rcauchy(n=10^4); y <- rcauchy(n=10^4)
r <- x/y

#Quantiles of cauchy 
quantile(x, c(0.025, 0.05, 0.25, 0.5, 0.75, 0.95, 0.975))

        2.5%           5%          25%          50%          75%          95% 
-12.20648869  -6.24480097  -0.93010302   0.01295798   1.04038961   6.28112571 
       97.5% 
 13.14378604 

#Quantiles of ratio of cauchy
quantile(r, c(0.025, 0.05, 0.25, 0.5, 0.75, 0.95, 0.975))

         2.5%            5%           25%           50%           75% 
-37.319835916 -14.631235783  -0.962655232   0.001017854   0.985213988 
          95%         97.5% 
 14.609945616  33.499447199 

Although the quantiles match, the ratio appears to have significantly heavier tails.

The density of the ratio \(U\), \(f_U(u)\), can be computed by the standard approach for (ratio distributions):

\[\begin{align} f_U(u) &= \int_{-\infty}^{\infty} \frac{|v|}{\pi^2 (1 + (uv)^2)(1+v^2)} = \frac{1}{\pi^2} \int_0^{\infty} \frac{1}{\pi^2 (1 + u^2 s)(1+s )ds } \end{align}\]

Madhav recognized that the integrand has a partial fraction decomposition:

\[\begin{equation} \frac{1}{\pi^2 (1 + u^2 s)(1+s )} = \frac{1}{\pi^2(u^2-1)} \left( \frac{u^2}{1+u^2 s} - \frac{1}{1+s} \right) \end{equation}\]

Thus

\[\begin{align} f_U(u) &= \frac{1}{\pi^2 (u^2 - 1)} \left( \lim_{M \to \infty} \int_0^M \frac{u^2}{1+u^2 s} ds - \int_0^M \frac{1}{1+s}ds \right) \\ &= \frac{1}{\pi^2 (u^2 -1)} \lim_{M \to \infty} \left( \ln(1 + u^2 M) - \ln(1+M) \right) \\ &= \frac{\log u^2}{\pi^2 (u^2-1)} \end{align}\]

which is definitely not a Cauchy distribution. By the way, this is one of the weirdest densities I have ever seen, \(f(u) \to \infty\) as \(u \to 0\) and \(f\) is indeterminate at \(u = \pm 1\) (the limit may be evaluated using L’hospital’s rule). Here is a plot comparing the Cauchy density (red) to its ratio (black).

x <- seq(-10, 10, by=0.001)
gx <- 1/(pi^2*(1+x^2)) #Standard Cauchy
fx <- (log(x^2))/(pi^2 * (x^2-1)) #Ratio

plot(x,fx,type="l",ylab="Density")
lines(x,gx,type="l", col="red")

After deriving this, I found this old paper with the same result:

Ride, P. R. (1965). Distributions of product and quotient of Cauchy variables. The American Mathematical Monthly, 72(3), 303-305.

This means ChatGPT is at least 60 years behind cutting edge statistics!