> What is the distribution of the ratio of two normals?

A related question is https://stats.stackexchange.com/questions/398436 The following is from a part of an answer to that question.

(You state that both variables are positive. And thus you do not have real normal distributions, which can take the value zero. You will have something that resembles the Cauchy distribution, but it does not have the undefined mean.)

For two (asymptomatically) Gaussian distributed variables (that are potentially correlated) you can use the Delta method or use an exact expression to express the ratio distribution.

> The use of the Delta method for the estimation of ratio's is described [here][1]. The result of this application of the Delta method actually coincides with an approximation of [Hinkley's result][2], an exact expression for the ratio of two correlated normal distributed variables (Hinkley D.V., 1969,  On the Ratio of Two Correlated Normal Random Variables, *Biometrica* vol. 56 no. 3).  
>
> For $Z = \frac{X}{Y}$ with $$
\begin{bmatrix}X\\Y\end{bmatrix} \sim N\left(\begin{bmatrix} \mu_x \\ \mu_y \end{bmatrix} , \begin{bmatrix} \sigma_x^2 & \rho \sigma_x \sigma_y \\ \rho \sigma_x \sigma_y  & \sigma_y^2   \end{bmatrix} \right)
$$  The exact result is:  $$
f(z) = \frac{b(z)d(z)}{a(z)^3} \frac{1}{\sqrt{2\pi} \sigma_X\sigma_Y} \left[  \Phi \left( \frac{b(z)}{\sqrt{1-\rho^2}a(z)} \right) - \Phi \left( - \frac{b(z)}{\sqrt{1-\rho^2}a(z)} \right) \right] + \frac{\sqrt{1-\rho^2}}{\pi \sigma_X \sigma_Y a(z)^2} \exp \left( -\frac{c}{2(1-\rho^2)}\right)
$$ with $$ \begin{array}{}
a(z) &=& \left( \frac{z^2}{\sigma_X^2} - \frac{2 \rho z}{\sigma_X \sigma_Y} + \frac{1}{\sigma_Y^2} \right) ^{\frac{1}{2}} \\ 
b(z) &=& \frac{\mu_X z}{ \sigma_X^2} - \frac{\rho (\mu_X+ \mu_Y z)}{ \sigma_X \sigma_Y} + \frac{\mu_Y}{\sigma_Y^2} \\
c &=& \frac{\mu_X^2}{\sigma_Y^2} - \frac{2 \rho \mu_X \mu_Y + }{\sigma_X \sigma_Y} + \frac{\mu_Y^2}{\sigma_Y^2}\\
d(z) &=& \text{exp} \left(  \frac {b(z)^2 - c a(z) ^2}{2(1-\rho^2)a(z)^2}\right) 
 \end{array}$$ 
 And an approximation based on an assymptotic behaviour is: (for $\mu_Y/\sigma_Y \to \infty$): $$
 F(z) \to \Phi\left( \frac{z - \mu_X/\mu_Y}{\sigma_X \sigma_Y a(z)/\mu_Y} \right)
 $$
 You end up with the Delta method result when you insert the approximation $a(z) = a(\mu_X/\mu_Y)$ $$a(z) \sigma_X \sigma_Y /\mu_Y \approx a(\mu_X/\mu_Y)  \sigma_X \sigma_Y /\mu_Y = \left( \frac{\mu_X^2\sigma_Y^2}{\mu_Y^4} - \frac{2 \mu_X \rho \sigma_X \sigma_Y}{\mu_Y^3} + \frac{\sigma_X^2}{\mu_Y^2} \right) ^{\frac{1}{2}}$$
>

(Sidenote: The above quote has been updated with some references to earlier descriptions and simpler descriptions. An earlier description of the exact expression was given by George Marsaglia 1965 in the [JASA Vol. 60, No. 309](https://doi.org/10.2307/2283145). A simple modern description is given in 2006 in [Jstatsoft Volume 16 Issue 4](http://dx.doi.org/10.18637/jss.v016.i04))

  [1]: https://stats.stackexchange.com/a/291652
  [2]: https://www.jstor.org/stable/2334671