Q:

CDF of Random Variable (RV) with two exponentially distributed RV

In the research article, I am reading $\Omega = h_1/h_2$ where $h_1 \sim e^{-u}$ and $h_2 \sim e^{-u}$, i.e. both $h_1$ and $h_2$ are exponentially distributed random variable with mean $u$. Now the author says that the CDF of $\Omega$ can be calculated to be
$$ F_\Omega(\omega) = \int_0^\infty yf_\Omega(y \omega,y) = \frac{\omega}{1 + \omega}$$ 
I really don't understand how is this thing done. Any help would be much appreciated.

A:

For two non-negative random variables $Y$ and $Z$ we have $$ P(Y/Z\le w) = P(Y\le wZ) = \int_0^\infty\int_0^{wz}f_{Y,Z}(y,z)dy\;dz.$$ Plugging in the the PDF for independent exponentials, $f_{Y,Z}(y,z) = e^{-y}e^{-z}$ and doing the integrals gives $$ \int_0^\infty e^{-z}\int_0^{wz}e^{-y}dy\;dz = \frac{w}{1+w}.$$
I'm not sure what they were getting at in the paper, though research articles often have errors regarding minor points like this. It is true that the formula they wrote down resembles that for the PDF, as you can see by differentitation: $$ \frac{d}{dw}\int_0^\infty \int_0^{wz}f_{Y,Z}(y,z)dy\;dz  = \int_0^\infty \frac{d}{dw}\int_0^{wz}f_{Y,Z}(y,z)dy\;dz = \int_0^\infty zf_{Y,Z}(wz,z)dz.$$ 

