I am working on some statistic project and trying to sort out the properties and n-root of Normal pdf. Here is my question and thought process.
Suppose the pdf $p$ of $\beta$ is
$ p_\beta(\beta) = \frac{1}{(2pi)^\frac{k}{2}\sigma^k} e^{-\frac{1}{2\sigma^2} \beta^T\beta}, \forall \beta \in R^k$
(or equivalently, $\beta$ follows a multivariate normal with mean $0$ and covariance matrix $ \sigma^2 I$ )
Then what will be the distribution of random variable $ X$ with pdf $ p_X(x) =\sqrt[n]{p_\beta(x)}$?
My hypothesis is $X$ will be multivariable normal with mean $0$ and covariance matrix $n\sigma^2I$ using the following argument:
Since
$ p_X(x) =\sqrt[n]{p_\beta(x)} \propto e^{-\frac{1}{2n\sigma^2}x^Tx} $
As $X$ only depends on that quadratic form, $X $ ~ $ N(0, n\sigma^2I)$ (something like the characterization of normal distribution here)
Will you guys got another view or rigorous proof for distribution of $X$?
Thanks in advance.