TITLE: How calculate the probability density function of $Z = X_1/X_2$
QUESTION [14 upvotes]: Let $X_1$ and $X_2$ be two continuous r.v., my question is: what is the p.d.f of $Z=X_1/X_2$?

REPLY [2 votes]: @Sasha
Could someone elaborate on why:
\begin{equation}
F_Z(z)=\mathbb{P}(X_1/X_2≤z)=\mathbb{P}(X_1≤X_2z;X_2>0)+\mathbb{P}(X_1≥X_2z;X_2<0)\\
\mathbb{E}(F_{X_1}(zX_2);X_2>0)+\mathbb{E}(1−F_{X_1}(zX_2);X_2<0)
\end{equation}
In the first line the R.H.S. has both $(X_1\leq X_2z)$ and $(X_1\geq X_2z)$. I get that we consider the cases when $X_2 \gt 0$ and $X_2 \lt 0$ but the $X_1$ inequality shouldn't remain $\leq$?
And in the second line why $\mathbb{P}(X_1≤X_2z;X_2>0) = \mathbb{E}(F_{X_1}(zX_2);X_2>0)$
Thanks in advance.