 Solution

The support condition $-y<x<y$ forces $y>0$. As a check, the joint probability density integrates toFor any real $x$, the marginal density of $X$ isDividing the joint probability density by this marginal density gives the conditional density:Thus, conditionally on $X=x$, $Y-|x|$ has the rate-two exponential distribution.

For $U=X^2$, use the two inverse branches $x=\sqrt{u}$ and $x=-\sqrt{u}$. Each branch has absolute Jacobian determinant $1/(2\sqrt{u})$. Both lie in the original support exactly when $u>0$ and $y>\sqrt{u}$. The change of variables for probability densities therefore yieldsNo atom of a measure is created at $u=0$, since $\mathbb{P}(X=0)=0$. The value of the probability density function at this boundary may be assigned arbitrarily. Checking the integral over $0<u<y^2$ gives $4y{e}^{-2y}$, which integrates to one.