Square of a standard Laplace variable ($F_{X^2}(y)=1-e^{-\sqrt{y}}(y\ge 0)$)

For a random variable $X$ with probability density function $e^{-|x|}/2$, the event $X^2\le y$ is $-\sqrt{y}\le X\le \sqrt{y}$ for $y\ge 0$. Integrating the symmetric density gives the displayed cumulative distribution function; it is zero for negative $y$. The density on $y>0$ is $e^{-\sqrt{y}}/(2\sqrt{y})$. Its integrable singularity at zero does not create an atom, because the cumulative distribution function is continuous there.