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In section 4.2 of this paper:

http://cseweb.ucsd.edu/~ravir/274/15/papers/a209-xu.pdf

Xu et al approximate the product of 2 anisotropic Gaussian distributions. I want to know if someone has found not an approximation but THE solution to what the resulting ASG distribution is after taking the product.

EDIT:

The goal is to either show (or prove it's not possible) that given 2 arbitrary ASG's their product is a third ASG with specified lobe $\bar z$

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  • $\begingroup$ I only skimmed through the paper. Is the problem to express $S(v;z_1,z_2)$ as a "smooth function" $S(v,\tilde z)$ for some $\tilde z$? This is in general not possible as the product of two such smooth functions contains quadratic terms in $v$. If this is not the issue, can you expand on your question by stating the mathematical problem explicitly? Why do you think that the product of two ASGs is another ASG? $\endgroup$ – Chris Jun 23 at 21:01
  • $\begingroup$ "ASGs is another ASG" Because the product of 2 Gaussian distributions in the euclidean case is another Gaussian distribution. I know that this does not guarantee that it will be the case for the spherical case too but that's why I had the hunch. $\endgroup$ – Makogan Jun 23 at 21:42
  • $\begingroup$ As a mathematician I find the terminology used in the paper quite confusing. But as far as I can see the authors do not claim that the product of two ASGs is an ASG. In fact, $\max(v,0)\cdot\max(v,0) = v^2$ for all $v\ge 0$ and there is no way you could express this as "$\max(v,\tilde z)$ times some exponential term". $\endgroup$ – Chris Jun 23 at 22:43
  • $\begingroup$ $v\cdot v = ||v||^2 = 1$ btw, remember these are vectors not scalars. More importantly, numerical simulations and a visualization of the result strongly suggest that the product of 2 ASG's is either an ASG or something very very close to an ASG. Consider the figures in this question: math.stackexchange.com/questions/3704580/… $\endgroup$ – Makogan Jun 23 at 23:04
  • $\begingroup$ Without additional assumptions on $z_1,z_2$ and $v$ it's easy to construct a counterexample for the three dimensional case as well. Consider, for example, $z_1 = <1,0,0>$ and $z_2=<0,1,0>$. $\endgroup$ – Chris Jun 24 at 11:34
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Consider $f(v) := S(v;z_1)\cdot S(v;z_2)$ with $z_1=z_2=<1,0,0>$. Then $$f(v) = \max(v_1,0)\cdot\max(v_1,0) = 0, \qquad \text{ if } v_1<0, $$ $$f(v) = \max(v_1,0)\cdot\max(v_1,0) = v_1^2 \qquad \text{ if } v_1\ge 0. $$

We want to find a $\tilde z$ such that $f(v) \stackrel{!}{=} S(v;\tilde z) = \max(v\cdot\tilde z,0)$ for all unit vectors $v$. But $v=<0,\pm 1,0>$ implies $\tilde z_2 = 0$ and similarly $v=<0,0,\pm 1>$ implies $\tilde z_3=0$. Consequently, $S(v;\tilde z) = \max(v_1 \tilde z_1, 0)$ and it is not possible to choose $\tilde z_1$ such that $$\max(v_1\tilde z_1,0) = f(v)$$ for all unit vectors $v$.

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