If one has a probability distribution $p(x,y)$, then $$p(y)=\int dx\ p(x,y)$$ In the conceptually very simple case that $x$ is drawn from a unit Gaussian and then acts as the variance of another Gaussian from which $y$ is drawn one has an integral of the form ($C$ is some constant): $$p(y)=\int dx\ p(x,y)=\int dx\ p(y|x)p(x) =C\int_{\Bbb R}dx\ \frac{e^{-y^2/x^2}}{|x|}e^{-x^2}$$ Which is completely beyond what I am capable of.
Does a closed form for this integral exist?
(Remark: $p(y\mid 0)$ should correspond to $\delta_{y,0}$, but I have ignored this in all my considerations with the belief that it is irrelevant.)
Note that
$$ p(y) = \int_{-\infty}^{+\infty}{\rm d}x~\frac{e^{-y^2/x^2 + x^2}}{|x|} = 2 \int_{0}^{+\infty}{\rm d}x~\frac{e^{-y^2/x^2 + x^2}}{x} $$
Make the change $ u = x^2/y$, such that ${\rm d}u = 2x{\rm d}x/y$ and
$$ p(y) = \frac{1}{2}\int_0^{+\infty}{\rm d}x~x\frac{e^{-y(y/x^2 + x^2/y)}}{x^2} = \int_0^{+\infty}{\rm d}u~\frac{e^{-y(1/u + u)}}{u} $$
Now, I'm going to make the change $u = e^t$, with this
$$ p(y) = \int_{-\infty}^{+\infty}{\rm d}t~ \exp[-y(e^{t} + e^{-t})] = \int_{-\infty}^{+\infty}{\rm d}t~ \exp[-2y\cosh(t)] = 2\int_{0}^{+\infty}{\rm d}t~ \exp[-2y\cosh(t)] $$
On a little side note, the modified Bessel function of the second kind $K_{\alpha}(x)$ can defined as
$$ K_\alpha(x) = \int_{0}^{+\infty}{\rm d}t~ \exp[-x\cosh(t)]\cosh(\alpha t) $$
Therefore
$$ p(y) = 2K_0(2y) ~~~\mbox{for}~~~ y > 0 $$