Density function of angle between two positive random N-dimensional vectors

Question

For the more general case of angles between random N-dimensional vectors, which are not restricted in any direction, the probability distribution for the angles is $f_\phi \propto \sin(\phi)^{N-2}$ in $[0,\pi]$.

But since I'm looking at a high dimensional space where negative contributions do not make sense, I want to take the absolute value of all vector elements. Consequently all random vectors are within the first quadrant (or its higher dimensional equivalent) and the angle between two vectors is bound to $[0,\frac{\pi}{2}]$.

So far I wasn't able to adapt the $\sin(\phi)^{N-2}$ dependency to the case of positive vectors or derive the angle distribution by construction from the underlying distributions of the generation of random positive vectors.

Does anybody have an idea on how to solve this or maybe how to come up at least with a decent approximation?

Edit: One example how the angle distribution changes when the absolute value of the vector components are taken (for dimension 3):

$\sin(\phi)^{3-2}$ distribution for ${\rm I\!R^3}$

Distribution for ${\rm I\!R^3_+}$

Answer 1

I think you are looking for some kind of truncation of the given distribution. You want to consider only angles in the range $[0,\frac{\pi}{2}]$, therefore you should use the density $$f_{\phi}^{\ast}=\begin{cases} \frac{f_{\phi}}{\int_0^{\frac{\pi}{2}}{f_{\alpha}\,\text{d}\alpha}}, &0 \le \phi \le \frac{\pi}{2},\\ \;\quad 0, &\text{else}.\end{cases}$$ Now you only consider angles in the "upper half" of your sphere. Due to symmetry arguments in shouldn't matter that you reduce your examination to the first quadrant, because all quadrants have the same probability.