In section 4.2 of this paper:


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.


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 $\bar z$

  • $\begingroup$ 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? $\endgroup$ – user9485 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". $\endgroup$ – user9485 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 $z_2=<0,1,0>$. $\endgroup$ – user9485 Jun 24 at 11:34

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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