$X,Y \sim N(0,1)$, which is the distribution of $W = \frac X Y$? Problem: Let $X,Y \sim N(0,1)$ independent random variables.

Which is the distribution of $W = \frac X Y$?

Attempt:
Something wierd happens. In fact $$ \mathbb P\left(\frac X Y \leq c\right) = \mathbb P(X -cY \leq 0) = \mathbb P(N(0,c^2+1) \leq 0)= \frac 1 2 $$ which is quite impossible. What's wrong?
The region of the plane in which $X/Y\le c$ is that in which $X\le cY$ if $Y&gt;0$ and $X\ge cY$ if $Y&lt;0.$ In polar coordinates, half of that region is described by $\pi &gt; \theta &gt; \arctan (1/c),$ if $c&gt;0,$ and the other half has the same probability assigned to it, so let's double the probability assigned to the case $Y&gt;0.$

The probability measure is $\displaystyle \frac 1 {2\pi} e^{-(x^2+y^2)/2} \, dx \, dy.$ In polar coordinates that becomes $\displaystyle \frac 1 {2\pi} e^{-r^2/2} r\,dr\,d\theta.$ Therefore
$$
\Pr\left( \frac X Y \le c \right) = \frac 1 {2\pi} \int_{\arctan(1/c)}^\pi \int_0^\infty e^{-r^2/2} r\,dr\,d\theta = \frac 1 \pi \left(\pi - \arctan\frac 1 c \right).
$$
Differentiating with respect to $c,$ we get the density:
$$
\frac 1 {\pi(1+c^2)}, 
$$
i.e. this is a standard Cauchy distribution.
You know that X~N(0,1) so you know the density function of X (fx(x)). You know also that Y~N(0,1) so you know fy(y).
Furthemore you have that X,Y are independent so you know the density function of f_(X,Y)(x,y) (i.e. f_(X,Y)= fx(x)*fy(y) )
Now you need to calculate the distribution function of X/Y so what you re gonna do is: P( W&lt;= c) = P(X/Y &lt;= c) = P(X&lt;= Yc) and now you need to calculate the double integral where: y will be in all R and x will be bounded by (-infinite, xc).