What is the distribution followed by this function of random variables

Question

$X$ and $Y$ are independently distributed $U(0,1)$ random variables (Uniform distribution)

Find the Variance of the random variable

$U = \dfrac{\ln(X)}{( \ln(X)+\ln(1-Y) )}$

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I know $-\ln(X)$ follows $\exp(1)$ and $(1-Y)$ is also $U(0,1)$ Therefore it becomes $\exp(1)$ in the numerator and $\text{gamma}(1,2)$ in the denominator. I dont know what to do next.

Answer 1

Write as $$ U=\frac{-\ln X}{-\ln X-\ln (1-Y)} $$ As you noted $-\ln X\sim \exp(1)$ and because $1-Y\sim \text{Unif}(0,1)$ it folows that $-\ln(1-Y)\sim \exp(1)$. Since $-\ln X$ and $-\ln (1-Y)$ are independent it follows that $U\sim \text{Beta}(1,1)$ which is the same as a uniform random variable on $(0,1)$.

Answer 2

$G(\lambda)=e^{-x}x^{\lambda-1};x>0,\lambda>0 $ and $B_1(u,v)={x^{u-1}}(1-x)^{v-1} ; 0<x<1 ,u>0,v>0 $

$X \sim G(\lambda) $ and $Y\sim G(m)$

$\dfrac{X}{X+Y} \sim B_1(\lambda,m)$

And good luck for JAM $2019$