The specific problem I need to answer is
"Let X and Y be independent Exp(1) random variables. Determine the pdf of U = X/(X + Y )."
I know that my goal is to find the CDF of U (then differentiate) by describing {U<=u}.
I have seen examples where Y = X^2 or even Z = X + Y. However, the calculus required is apparently too difficult for the stat course I'm taking, so we're using transforms. In this case the laplace transform because our variables are continuous.
We know that if I have L(X) and L(Y) (easily found) that X + Y = L(X)*L(Y), but I have no idea how the transform translates to dividing random variables? I'm finding it hard to google this question, and I'm on school break right now so I'm asking it on here :)
So that's my question, or, an explanation about why the question I'm asking is stupid would also be helpful, really any googleable methods would be helpful, thanks
[no full solutions please!]
Hint: for any $t\in(0,1)$, $$\mathbb{P}\left[\frac{X}{X+Y}\leq t\right]=\mathbb{P}\left[\frac{1}{1+\frac{Y}{X}}\leq t\right]=\mathbb{P}\left[\frac{Y}{X}\geq\frac{1}{t}-1\right]$$ and for any $\rho>0$ $$\mathbb{P}[Y\geq \rho X] =\int_{0}^{+\infty}e^{-x}\int_{\rho x}^{+\infty}e^{-y}\,dy\,dx=\int_{0}^{+\infty}e^{-(\rho+1)x}\,dx=\frac{1}{\rho+1}$$ hence for any $t\in(0,1)$ we have $$ \mathbb{P}\left[\frac{X}{X+Y}\leq t\right]= t $$ and $U=\frac{X}{X+Y}$ has a uniform distribution over $(0,1)$.