Cumulative Distribution of X/Y Let X, Y be independent exponential variables with rates $\alpha$, and $\beta$.  Find the c.d.f. of X/Y.
So far, I let Z = X/Y.
I can then show $f_Z(z) = \int_{-\infty}^{+\infty} |x|f_{X,Y}(x,xz)  \,dx$ (unless my logic is incorrect).
Then because X,Y are independent,
= $\int_{-\infty}^{+\infty} |x|f_{X}(x)f_Y(xz)  \,dx$.
Would I just then insert the c.d.f for the exponential distribution of X, Y?  
Thanks
 A: 
Would I just then insert the c.d.f for the exponential distribution of X, Y?

Assuming you mistyped c.d.f. for PDF, the answer is "yes". 
Note that the computations can be made easier if one notes from the start that $X\gt0$ and $Y\gt0$ almost surely, hence $Z\gt0$ almost surely, and that $Z=X/Y$ yields $X=ZY$. Thus, the approach you explain, while quite sound in principle, rather leads to the fact that, for every $z\geqslant0$,
$$
f_Z(z)=\int_0^\infty yf_X(yz)f_Y(y)\mathrm dy.
$$
The next step, as you said, is to plug in $f_X(x)=\alpha\mathrm e^{-\alpha x}$ and  $f_Y(y)=\beta\mathrm e^{-\beta y}$ into this, getting
$$
f_Z(z)=\int_0^\infty \alpha\beta y\exp(-(\alpha z+\beta)y)\mathrm dy.
$$
The change of variable $t=(\alpha z+\beta)y$ yields
$$
f_Z(z)=\frac{\alpha\beta}{(\alpha z+\beta)^2}\int_0^\infty t\mathrm e^{-t}\mathrm dt=\frac{\alpha\beta}{(\alpha z+\beta)^2}.
$$
If one is interested in the CDF $F_Z$ rather than in the PDF $f_Z$, one simply writes, for every $z\geqslant0$,
$$
F_Z(z)=\int_0^zf_Z(t)\mathrm dt=\int_0^z\frac{\alpha\beta}{(\alpha t+\beta)^2}\mathrm dt,
$$
that is,
$$
F_Z(z)=\left.\frac{-\beta}{\alpha t+\beta}\right|_{t=0}^{t=z}=1-\frac{\beta}{\alpha z+\beta}=\frac{\alpha z}{\alpha z+\beta}.
$$
A: Here is the solution derived using the mathStatica package for Mathematica:   
By independence, the joint pdf of $X$ and $Y$ is:

You seek the cdf of Z, namely $P(Z \le z)$   =   $P(\frac XY \le z)$:

All done. 
