= Square of a standard Laplace variable
{title2=$F_{X^2}(y)=1-e^{-\sqrt y}\quad(y\geq0)$}
For a <random variable> $X$ with <probability density function> $e^{-|x|}/2$, the event $X^2\leq y$ is $-\sqrt y\leq X\leq\sqrt y$ for $y\geq0$. 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.
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