Ratio of Two Zero-Mean Normal R.Vs

Question

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.

Answer 1

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}

Answer 2

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))}
optim(1,L)$par
}

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 $\gamma = \frac{s_1}{s_2}$.