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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

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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.

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  • $\begingroup$ Wow. I am sorry but could you maybe fill in some gaps about how you arrived at this. It must have been a struggle and I don't want you to write all steps but more than 1 would be nice $\endgroup$
    – rostader
    Commented Jul 25, 2021 at 5:13
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    $\begingroup$ @rostader, since MathOverflow is focused on research-level math, I’ll leave the answer as it is. $\endgroup$
    – user44143
    Commented Jul 25, 2021 at 8:40

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