So I know that the ratio of two normal distributions is Cauchy. My question is, what are the resulting parameters of the Cauchy R.V., assuming the normal RVs are centered at zero and independent? I'm assuming location is zero, but what about the scale parameter? Thank you in advance.

  • 2
    $\begingroup$ The scale of a ratio of zero-centered variables is the ratio of the scales, as the units calculus will tell you. $\endgroup$
    – whuber
    Dec 29, 2022 at 20:34

2 Answers 2


Suppose $X \sim N(0, \sigma_1^2)$ and $Y \sim N(0, \sigma_2^2)$ are independent, then $$\frac{X}{Y} = \frac{\sigma_1Z_1}{\sigma_2Z_2} := \gamma\frac{Z_1}{Z_2},$$ where $Z_1, Z_2 \text{ i.i.d.} \sim N(0, 1)$. Since you already knew that $Z_1/Z_2 \sim \text{Cauchy}(0, 1)$, therefore $X/Y \sim \text{Cauchy}(0, \gamma)$, as the Cauchy family is a location-scale family (you can also verify it by directly calculating the pdf of $X/Y$ using the pdf of $Z_1/Z_2$). In other words, the pdf of $X/Y$ is \begin{align} f(x) = \frac{1}{\pi\gamma[1 + (x/\gamma)^2]}, \quad x \in \mathbb{R}^1. \end{align}


Here is some basic R code which will allow you to experiment,

f = function(s1,s2){
x1 = rnorm(1e5, mean = 0, sd = s1)
x2 = rnorm(1e5, mean = 0, sd = s2)
y = x1/x2
L = function(p){-sum(dcauchy(y, location = 0, scale = p, log=T))}

What this code does it use MLE to estimate the scale parameter after you specify two sigma values $s_1,s_2$ for the two normals. It uses $10^5$ simulations. For example, $f(2,3) = 0.66$, and $f(2,1) = 2$.

Therefore, the reasonable guess is that the location parameter is $\gamma = \frac{s_1}{s_2}$.

  • 1
    $\begingroup$ Thank you! I did the first four lines but hadn;t yet thought of how to formally pinpoint the scale parameter. Much appreciated. $\endgroup$
    – Supermang
    Dec 29, 2022 at 20:37

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