Show that random variable $U=\frac{X}{X+Y}$ has uniform distribution on [0,1] when X & Y are independent random variables with same exp. distribution.

Question

Show that the random variable $U=\frac{X}{X+Y}$ has a uniform distribution on the range [0,1] when X and Y are independent random variables with the same exponential distribution.

I'm stuck at

$F_U\left(u\right)=P\left(U\le u\right)=P\left(\frac{X}{X+Y}\le u\right)$

Even if I assume that X=1 I get a divergent integral in the end

And If I assume that

$P\left(0\le \frac{X}{X+Y}\le 1\right)=P\left(0\le \:X\le \:X+Y\right)=P\left(-X\le Y\right)=1-P\left(Y\le X\right)$

Then I still get an divergent integral

So what should I do?

Answer 1

You want to prove $P(U\le u)=u$ for $u\in[0,\,1]$. Indeed,$$\begin{align}P(X\le uX+uY)&=P(X\le\tfrac{uY}{1-u})\\&=\int_0^\infty\lambda e^{-\lambda y}(1-e^{-\lambda uy/(1-u)})dy\\&=1-\int_0^\infty\lambda e^{-\lambda y/(1-u)}dy\\&=1-(1-u)\\&=u.\end{align}$$You can also do it by noting the problem is equivalent to $Z:=Y/X$ having PDF $\frac{d}{dz}[1-(1+z)^{-1}]=(1+z)^{-2}$ on $\Bbb R^+$, because $U=1/(1+Z)$. Indeed, $Z$'s PDF is$$\int_0^\infty x\lambda^2e^{-\lambda(1+z)x}dx=(1+z)^{-2}.$$