Function of random variables: the ratio of two bounded random variables

If x,y and z are continuous random variables.

$$f_X(x)=2ax\exp(-ax^2+aR^2)\ \ \ R where $$R$$,$$a$$ and $$b$$ are positive real numbers. $$z=\frac{x}{y}$$ and we want to find the $$CDF$$ 'Cumulative Distribution Function' of z?

This is the question. I tried to solve it but without success. Here is my attempt:

First we need to find the range of z which is $$1

Then, $$F_Z(z)=\int\limits_{0}^{R}\int\limits_{R}^{zy} f_X(x)f_Y(y)dxdy$$ And the final answer using mathematica: $$F_Z(z)=\frac{b e^{R^2 (a+b)} \left(e^{-R^2 \left(a z^2+b\right)}-1\right)}{\left(e^{b R^2}-1\right) \left(a z^2+b\right)}+1$$ Since this is a CDF then it must start at $$0$$ and end at $$1$$.

@ $$z= \infty$$, $$F_Z(z) =1$$ which is correct, but when $$z=1$$ it doesn't equal $$0$$!! Where is the mistake.

I plotted the function using some values for a,b and R (these values don't make a difference since it should be valid for all the values) and here is the plot: The limits of the inner integral defining $F_Z(z)$, from $x=R$ to $x=zy$, are incorrect. If $zy\gt R$, no problem. But, if $zy\lt R$, this integral should be zero. Using the limits $x=R$ and $x=zy$ yields a negative contribution which explains the negative values of your plot. Equivalently, the inner integral should go from $x=R$ to $x=\max\{zy,R\}$.
• Makes sense, I changed the outer integration limits to make sure that $zy>R$ always, so I set it from $R/z$ to $R$, and it worked, Thanks. May 27 '14 at 17:13