 Solution

For a continuous random variable with probability density function $f_X$, its cumulative distribution function isConsequently $F_X^{\prime }=f_X$ wherever the density is continuous, and almost everywhere for an absolutely continuous distribution.

Here $Y=X^2\ge 0$. Thus $F_Y(y)=0$ for $y<0$. For $y\ge 0$, the event $X^2\le y$ is $-\sqrt{y}\le X\le \sqrt{y}$. Symmetry of the Laplace distribution givesTherefore the square of a standard Laplace variable has cumulative distribution functionIt is continuous at zero, so there is no point mass there. As a check, its probability density function for $y>0$ is $e^{-\sqrt{y}}/(2\sqrt{y})$, whose integral becomes ${\int }_0^{\infty }{e}^{-u}du=1$ after setting $u=\sqrt{y}$.