Distribution of an RBF-transformed normal variable My question might be related to this or this one, but I have reasons to hope my problem is more benign.
Assume I have a normally distributed variable $X \sim  N(0, 1)$. What can be said of $Y = \exp(-\frac{X^2}{2 \sigma^2})$? It is obviously bounded to $(0, 1]$ and it certainly looks like beta-distribution:


However, can this be shown mathematically? How do $\alpha$ and $\beta$ depend on $\sigma$? And, assuming that $Y$ is really beta distributed: What if $X$ is multidimensional? I reckon that in that case we'd be dealing with $Y = \exp(-\frac{Z}{2 \sigma^2})$, with $Z$ having a $\chi^2$ distribution. Would $Y$ still have beta distribution and how would degrees of freedom come into play?

To get to the essence of this question, let's generalize it a little.  Suppose $X$ is a random variable supported on a set (of real numbers) $\mathcal A$ where it has a density proportional to
$$f_X(x) = C \exp(-h(x))$$
for some (almost everywhere) differentiable function $h$ defined on $\mathcal A;$ and let $Y = \exp(-h(X)).$
Because $Y$ is a (differentiable) transformation of $X,$ it also is a continuous random variable.  Let's find its density.  To do so, we must analyze the entire probability element by changing the variable in $f_X(x)\mathrm{d}x$ to $y = \exp(-h(x)).$  One key calculation is the new differential
$$\mathrm{d}y = \mathrm{d}\left(\exp(-h(x))\right) = -\exp(-h(x))h^\prime(x)\mathrm{d}x = -y h^\prime(x) \mathrm{d}x,$$
from which we may equate a differential form in $y$ with a differential form in $x$ via
$$\frac{\mathrm{d} y}{y} = -h^\prime(x)\mathrm{d}x.$$
Exploit this to compute
$$f_X(x)\mathrm{d}x = C\exp(-h(x)) |\mathrm{d}x| = Cy \bigg|\frac{-1}{h^\prime(x)} \frac{\mathrm d y}{y} \bigg| = \frac{C}{|h^\prime(x)|}\mathrm{d}y.$$
We have to deal with the $h^\prime(x)$ term on the right: it needs to be expressed in terms of $y.$  Consider any possible value $y$ of $Y.$  Corresponding to it via the transformation $h$ is the set of $x,$ written $h^{-1}(y),$ that $h$ maps to this particular $y.$  Because $h$ is differentiable, $h^{-1}(y)$ is at most countable and from the previous equality we obtain

$$f_Y(y) = C\sum_{x\ \in\ h^{-1}(y)} \frac{1}{|h^\prime(x)|}.\tag{*}$$

That's almost as far as we can take the analysis at this level of generality, so let's specialize to the case of the question where $h(x) = x^2 / (2\sigma^2)$ and $\mathcal A$ is the set of real numbers.  Taking the derivative is easy; $$h^\prime(x) = x/\sigma^2.$$  The image of $h$ is the set of non-negative numbers, almost all of which are positive.  When $y$ is a positive number, there are two corresponding $x$ values obtained by solving
$$y = h(x) = \exp(-x^2/(2\sigma^2));\quad x = \pm \sigma\sqrt{-2\log(y)}.$$
The general formula $(*)$ and the fact $C = 1/(|\sigma|\sqrt{2\pi})$ gives
$$\begin{aligned}
f_Y(y) &= \frac{C}{|h^\prime( \sigma\sqrt{-2\log(y)} )|} +  \frac{C}{|h^\prime(-\sigma\sqrt{-2\log(y)} )|} \\
&= \frac{C\sigma^2}{|\sigma\sqrt{-2\log(y)} |} +  \frac{C\sigma^2}{|-\sigma\sqrt{-2\log(y)}|} \\
&= \frac{1}{\sqrt{-\pi\log(y)}}.
\end{aligned}$$
(It's easy to show in general that any scale factor $\sigma$ in $f_X$ will disappear in the final formula.)
Here is a histogram of one million draws of $Y,$ on which the plot of $f_Y$ is superimposed in red to show their close agreement.

Notice that this does not give rise to any Beta distribution: near $0$ it does not behave like any distribution in that family.
