= Solution
Differentiate the <cumulative distribution function> for $y>0$ using the <chain rule> and the <standard normal density>:
$$
\begin{aligned}
f_Y(y)&=2\Phi'(\sqrt{2y})\frac1{\sqrt{2y}}\\
&=\frac{2}{\sqrt{2\pi}\sqrt{2y}}e^{-y}
=\frac{e^{-y}}{\sqrt{\pi y}}.
\end{aligned}
$$
Thus a <probability density function> is
$$
\boxed{f_Y(y)=\begin{cases}
\dfrac{e^{-y}}{\sqrt{\pi y}},&y>0,\\
0,&y\le0.
\end{cases}}
$$
The value assigned at zero is immaterial. The <probability density function> is unbounded near zero, but $F_Y(0)=0$ and continuity of the <cumulative distribution function> there show that there is no point mass. This is a <gamma distribution> with shape $1/2$ and rate $1$.
Back to article page
