Is there a way to divide random variables? Let $X$ and $Y$ be two independent, continuous random variables. I want to find the distribution of $\frac{X}{X+Y}$. 
I read a bit myself. I know for $S=X+Y$, I can just use a convolution operation between the density functions.  I also found the term ratio distribution
https://en.wikipedia.org/wiki/Ratio_distribution
However, I'm unclear if this requires independence. So for instance, if $S=X+Y$ and I want to find $\frac{X}{S}$, $S$ would not be independent of $S$, so would this ratio distribution concept still apply? 
 A: Convolution is really special - it doesn't really apply to any situations other than the addition of independent random variables. However, that doesn't mean you can't convert other operations into addition - in particular, if you just had
$$Z=\frac{X}Y$$
you would have
$$\log(Z)=\log(X)-\log(Y)$$
whose distribution can be given by a convolution of the distributions of $\log(X)$ and of $-\log(Y)$, if that is convenient.
Your example is just a bit of algebra away from such a situation, since while the denominator is a bit bad, being a sum, the numerator is fine - so we just need to exchange them:
$$\frac{X}{X+Y}=\frac{1}{\frac{X+Y}X}=\frac{1}{1+\frac{Y}X}.$$
This gives your random variable in terms of $\frac{Y}X$ which is perhaps more palatable, being a ratio of independent variables.
Whether rearranging this way is helpful is perhaps questionable, but it is possible. Of course, you could also do all this by noting that, if $X$ and $Y$ are positive independent continuous random variables with probability distribution functions $x$ and $y$, the quotient $X/Y$ has a distribution described by
$$P(X/Y \leq k ) = \int_0^{\infty}\int_{0}^{kt_2}y(t_1)x(t_2)\,dt_1\,dt_2$$
which, differentiating, would give, so long as the distributions are not too unreasonable that the value of PDF of $X/Y$ at $k$ is
$$\int_{0}^{\infty}ty(t)x(kt)\,dt.$$
If $x$ and $y$ are nice functions (e.g. exponential or rational functions), you might even be able to evaluate this explicitly.
A: In the special case that $X,Y\stackrel{\mathrm{i.i.d.}}\sim\mathrm{Expo}(1)$, we have for $t\in(0,1)$:
\begin{align}
\mathbb P\left(\frac X{X+Y}\leqslant t\right) &= \mathbb P\left(Y\geqslant X\left(\frac1t-1\right)\right)\\
&= \int_0^\infty f_X(x)\ \mathbb P\left(Y\geqslant x\left(\frac1t-1\right)\right)\ \mathsf dx\\
&= \int_0^\infty  e^{-\ x} e^{- x\left(\frac 1t-1\right)}\ \mathsf dx\\
&=  \int_0^\infty e^{-\frac x t}\ \mathsf dx\\
&= t.
\end{align}
Hence $\frac X{X+Y}$ has uniform distribution on $(0,1)$.
