Simulating ratio of two independent normal variables I have two non centered independent normally distributed variables $X$ and $Y$.  Specifically
$$ X \sim N(\mu_1, \sigma_1)$$ 
$$ Y \sim N(\mu_2, \sigma_2)$$ 
I am interested in sampling from the ratio $X/Y$. I know that the ratio is Cauchy distributed if the denominator distribution is centered at 0 which mine is not. My question is is there an approximate distribution that I can sample the ratio $X/Y$ from using the means and standard deviations from the normal distributions?

An exact expression for the ratio of two correlated normal random variables was given by Hinkley in 1969. The notation is taken from the stackexchange post
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 probability density function (pdf) 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 of the CDF based on an asymptotic behaviour is: (for $\theta_Y/\sigma_Y \to \infty$): $$
 F(z) \to \Phi\left( \frac{z - \mu_X/\mu_Y}{\sigma_X \sigma_Y a(z)/\mu_Y} \right)
 $$
You end up with the Delta method result when you insert the approximation $a(z) = a(\mu_X/\mu_Y)$ $$a(z) \sigma_X \sigma_Y /\mu_Y \approx a(\mu_X/\mu_Y)  \sigma_X \sigma_Y /\mu_Y = \left( \frac{\mu_X^2\sigma_Y^2}{\mu_Y^4} - \frac{2 \mu_X \sigma_X \sigma_Y}{\mu_Y^3} + \frac{\sigma_X^2}{\mu_Y^2} \right) ^{\frac{1}{2}}$$
