Let $X$ be a random variable with uniform distribution in $[0,2]$. Find the distribution of $Y=\frac{X}{|X-1|}$.
I wrote:
$F_Y(y)=\mathbb{P}(Y\leq y)=\mathbb{P}(\frac{X}{|X-1|}\leq y)=\mathbb{P}(\frac{|X-1|}{X}\geq \frac{1}{y})=\mathbb{P}(|X-1|\geq \frac{x}{y})=\mathbb{P}[(X\geq \frac{y}{y-1},X>1)\cup (X\leq \frac{y}{y+1},X<1)]$
where, for example, $\mathbb{P}(X\leq \frac{y}{y+1},X<1)=\int_{0}^{\operatorname{min}(1,\frac{y}{y+1})}\frac{1}{2}dx=\frac{1}{2}[\frac{2y+1}{y+1}]$. Unfortunately it seems that I'm out of the way. Where am I wrong?
Thanks in advance for any help.
You're on the right path! I think it's best to plot $Y$ in order to see which is the support of the distribution.
As you probably have already noticed, the support of $Y$ is $\mathbb{R}_{++}$.
Let $x\sim\mathcal{U}[0,2]$ and $Y=\frac{X}{|X-1|}$
$$ F_{Y}(y)=P\left(Y\leq y\right)=P\left(\frac{X}{|X-1|}\leq y\right)=P\left(\frac{X}{X-1}\leq y,X>1\right)+P\left(\frac{X}{1-X}\leq y,X\leq 1\right) $$
Let's compute each term separetly: $$ P\left(\frac{X}{X-1}\leq y,X>1\right)=P\left(X\leq \frac{y}{1-y},X>1\right) $$ $$ P\left(\frac{X}{1-X}\leq y,X\leq 1\right)=P\left(X\leq \frac{y}{1+y},X\leq 1\right)=P\left(X\leq \frac{y}{1+y}\right) $$ Now you have to compute those integrals. But bear in mind that the limits of integration may vary according to different values of $y$.
Please let me know if there was something I did not explain properly. Hope this helps