Uniform distribution on $[0,1]$ and random variable $Y=\frac{U}{e^{1-U}}$

Question

$U$~$Unif[0,1]$ and we have the random variable $Y=\frac{U}{e^{1-U}}$. Find the density function of $Y$.

So far I have that $0\le Y\le1$ and that... $$F_Y(t)=P(\frac{U}{e^{1-U}}\le t)=P(\ln (\frac{U}{e^{1-U}})\le \ln (t))=P(U+ \ln(U)-1\le \ln(t))=P(U+ \ln(U) \le ln(t) + 1) $$ But now I have no idea what to do with two $U's$ in the right hand side of the inequality. I feel like I am ignoring the interval, so missing something in that sense.

Answer 1

The function $f:x\mapsto x e^{x-1}$ is increasing over $(0,1)$ and its inverse function is given by $W(ey)$, with $W$ being the Lambert $W$ function. That gives: $$ F_Y(t) = \mathbb{P}[ U e^{U-1}\leq t ] = W(et) \tag{1}$$ and by differentiating the previous line: $$ f_Y(t) = \frac{\mathbb{1}_{(0,1)}(t)}{t+\frac{t}{W(et)}}.\tag{2} $$