Let $X$ be a multivariate normal $\mathcal{N}(\mu, \Sigma^2)$ and let $X$ be anisotropic, that is I am considering $\Sigma$ to be a diagonal matrix but the elements on the diagonal might be different.


I am interested in finding the distribution of $X/\|X\|\_2$.


As a start let $X$ be isotropic. Then $\|X\|\_2^2$ will be Gamma distributed. But then $\|X\|\_2$ will perhaps have to follow what I found to be a Nakagami distribution (<https://en.wikipedia.org/wiki/Nakagami_distribution>). So I need to find the ratio of a normal and this Nakagami distribution. However for the anisotropic case $\|X\|\_2^2$ will not be Gamma but a mixture of Gamma distributions (<https://stats.stackexchange.com/questions/72479/generic-sum-of-gamma-random-variables>) and this seems more complicated to be honest.


On the other hand it feels intuitively somewhat similar to a truncated normal distribution, but with the truncation happening over a unit ball and this also seems complicated.


Any ideas/ references/ hints/ thoughts


Thanks



Norm contrained Gaussian distribution
In two dimensions, with
$$\mu=\begin{pmatrix} m \\ n \end{pmatrix},\ \ 
\Sigma=\begin{pmatrix} v & 0 \\ 0 & w \end{pmatrix},$$
an integration over all possible radii gives the distribution of $X/\|X\|\_2$ as
$$f(\cos t,\sin t)=\frac{
1+\sqrt{\pi}u
\exp(u^2)
(1+\text{erf}(u))
}{
\exp(a)c\pi \sqrt{v w}
}
$$
where
$$a=\frac{m^2}{2v}+\frac{n^2}{2w},\ \
b=\frac{m\cos t}{v}+\frac{n\sin t}{w},\ \ 
c = \frac{2\cos^2t}{v}+\frac{2\sin^2 t}{w},\ \
u = \frac{b}{\sqrt{c}}$$
and the expression is indeed complicated enough to deter more exact computations.


