# 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.

• Do you not understand the formula, or how to do the integral? Feb 20, 2018 at 0:31
• the formula makes zero sense to me.. It looks like the correct formula for the PDF (but the answer is indeed the CDF). Also, $f_{\Omega}(y\omega,y)$ is not coherent notation for anything. If I'm right about them incorrectly writing the formula for the PDF is true then they must mean $f_{h_1,h_2}(y\omega,y).$ Feb 20, 2018 at 0:53

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: $$\frac{d}{dw}\int_0^\infty \int_0^{wz}f_{Y,Z}(y,z)dy\;dz = \int_0^\infty \frac{d}{dw}\int_0^{wz}f_{Y,Z}(y,z)dy\;dz = \int_0^\infty zf_{Y,Z}(wz,z)dz.$$

• No, this is the CDF as they have said in the paper, and rightly so as you have derived it. The CDF would be $\frac{d}{df}\frac{w}{1+w} = \frac{1}{(1+w)^2}$. (as mentioned by the authors as well).
– SJa
Feb 20, 2018 at 1:12
• I think the statement $yf_\Omega(y,w,y)$ may be a typo then.
– SJa
Feb 20, 2018 at 1:13
• By the way, I canot understand why you have taken limits of the second integral from 0 to $infty$ and the first integral from 0 to $wz$. Really appreciate your time for the help
– SJa
Feb 20, 2018 at 1:15
• @Sjaffry The reason for the integration range is that in order to get $P(Y\le w Z),$ you integrate the joint pdf over the region $y < wz$ (i.e. the region of the first quadrant of the $z-y$ plane below the line $y=wz$). The integration bounds $\int_0^\infty dz \int_0^{wz} dy$ is one way to do this. You are letting $z$ range over everything but making sure $y$ only ranges from $0$ to $wz.$ Feb 20, 2018 at 2:20
• @Sjaffry You forgot the factor of $e^{-z}.$ It should be $\int_0^\infty e^{-z}(1-e^{-wz})dz.$ Feb 22, 2018 at 6:18