Given $x$ that follows a distribution $P(x)=e^{\frac{-x^2}{2\sigma^2}}$ i.e. a random Gaussian variable, can I say anything about the distribution of $x^n$ for fixed $n$?
Specifically, is there ever a case where $x^n$ also follows a Gaussian distribution or can be approximated as Gaussian? If yes, what are the assumptions being made? I see a physics paper where an argument similar to this is presented (without justification) so I'm wondering in what regime it is true.
The distribution of $Y=X^n$ if the distribution of $X$ is $f_X(x)$ for $n$ odd is $$f_Y(y)=f_X(\sqrt[n]{y})\frac{d}{dy}\sqrt[n]{y}=f_X(\sqrt[n]{y})\frac{1}{n}{y}^{1/n-1}.$$
For $n$ even, you have to consider that $Y=X^n$ has no inverse that is a function, so it's a little more tedious. Also, problems may arise at $Y=0$ depending on the value of $n$ and if $0$ is in the support of $f_X(x)$.
I don't think any of these yield a Gaussian for any $n$.
I doubt they can be approximated very well by Gaussians. However, maybe a sum of Gaussians would work ok. Also, the sample mean of $Y$ can be approximated by a Gaussian, of course, by the central limit theorem.