Where is my flaw with the calculation of this cdf? I want to calculate the ratio distribution $X/Y$ of two continuous random variables $X$ and $Y$ with each having support $(0, \infty)$
I was starting like that:
$$\mathbb P\left(\frac{X}{Y}\leq z\right)=\int_{0}^\infty f_Y(y) F_{X\mid Y}(zy \mid y)dy$$
Now:
$$F_{X\mid Y}(zy \mid y)=\int_{0}^{zy} \frac{f_{X,Y}(x,y)}{f_Y(y)}dx   =\frac{\frac{dF_{X,Y}(zy,y)}{dy}}{f_Y(y)}$$
and hence:
$$\mathbb P\left(\frac{X}{Y}\leq z\right)=\int_{0}^\infty \frac{dF_{X,Y}(zy,y)}{dy}dy=\left[F(zy,y)\right]_0^\infty=1,$$
which of course is not correct. Anyone can see my mistake? Thank you very much in advance
 A: Putting together your first equation and the first equality in the second, you are esentially getting
$$ p=P( X/Y \le z)= \int_0^\infty \int_0^u f_{X,Y}(x,y) \, dx \, dy, \hspace{1cm} u=u(y)=zy \tag{1}$$
which is right, of course (there was no need to use a conditional for that, though).
Your problem in what follows is that you are mixing total derivatives with partial derivatives.
When we write (in general) the formula
$$ \frac{d F_{X,Y}(x,y)}{dy } = \int_{-\infty}^x f_{X,Y}(x',y) dx' \tag{2}$$
we are implicity assuming that the derivative is done by varying $y$ and keeping $x$ constant, i.e., it's actually a partial derivative. When the variables have some arbitrary functional dependence, we should be more careful and write that partial derivative explicitly:
$$ \frac{\partial F_{X,Y}(u,v)}{\partial v } = \int_{-\infty}^u f_{X,Y}(u',v) \, du' \tag{3}$$
In our case, plugging $(3)$ into $(1)$ we get
$$p= \int_0^\infty \frac{\partial F_{X,Y}(u,y)}{\partial y }  dy \tag{4} $$
with $u=u(y)=zy $. Now, the integrand is not a total derivative, hence you cannot apply the fundamental theorem of calculus directly. The relation is
$$ \frac{d F_{X,Y}(u,y)}{d y } = \frac{\partial F_{X,Y}(u,y)}{\partial u } \frac{du}{dy}+
\frac{\partial F_{X,Y}(u,y)}{\partial y } = \frac{\partial F_{X,Y}(u,y)}{\partial u } z +
\frac{\partial F_{X,Y}(u,y)}{\partial y } \tag{5}$$
You could isolate from this equation the integral in $(4)$ and replace it there... but you will get something similar (probably not simpler) than $(1)$.
