I am reading a CG paper that came on 2019 and I am stuck at one section.
After thinking about my issue I am starting to believe the author made a typo, but the likelihood of me being wrong is much higher than the author making a mistake.
The paper is this: https://dl.acm.org/doi/10.1145/3306346.3323026
My specific problem is in section 4. Specifically equation 5.
That equation is the formula to calculate the new mean of a product of powers of Gaussian functions. i.e it gives the mean of $f(x|\Theta_{ij})$ in the following equation:
$$f(x | \Theta_{ij}) = f(x | \Theta_i) f(x | \Theta_j)$$
That much is easy to understand and I have seen this proof a couple of times.
The moment where I get lost is right under where the authors claim:
The set [...] describes a curve_{ij} connecting the Gaussian's means, as > the weights are invariant to any scaling and can thus be expressed using > a single parameter $t = \frac{\alpha_i}{\alpha_i + \alpha_j}$
I have played around with that $t = \frac{\alpha_i}{\alpha_i+\alpha_j}$ formula over and over again and I am unable to plug it into equation 5, no matter what I do. However if I modify it slightly such that:
$$t = \frac{\alpha_i\Sigma^{-1}_i}{\alpha_i\Sigma^{-1}_i + \alpha_j\Sigma^{-1}_j}$$ Then everything falls into place.
I can write:
$$\mu_{ij} = \frac{\alpha_i\Sigma^{-1}_i\mu_i + \alpha_j\Sigma^{-1}_j\mu_j}{\alpha_i\Sigma^{-1}_i + \alpha_j\Sigma^{-1}_j}$$ $$ = \frac{\alpha_i\Sigma^{-1}_i\mu_i}{\alpha_i\Sigma^{-1}_i + \alpha_j\Sigma^{-1}_j} + \frac{\alpha_j\Sigma^{-1}_j\mu_j}{\alpha_i\Sigma^{-1}_i + \alpha_j\Sigma^{-1}_j}$$ $$= t\mu_i + (1-t)\mu_j$$
Which makes sense in the context of the paper. Is there a typo in the paper? Am I ignoring something critical? Is there a simplification that allows you to remove the sigmas in the formula I provided? I am very confused.
If $s=\alpha_i+\alpha_j$ then: $$ \alpha_i=st,\quad \alpha_j=s(1-t). $$ But a factor of $s^{-1}$ can be factored out from the first term in $(5)$, and a factor of $s$ can be factored out from the second term. Hence $s$ gets simplified away, leaving: $$ \mu_{ij}=\left(t\Sigma_i^{-1}+(1-t)\Sigma_j^{-1}\right)^{-1} \left(t\Sigma_i^{-1}\mu_i+(1-t)\Sigma_j^{-1}\mu_j\right). $$