I was wondering if it is possible to get 3 intersection points if use Bayes's theorem
$$P(B|x) = \frac{P(x\mid B) \times P(B)}{P(x)}$$ Where $P(x\mid B)$ is a gaussian function.$$ P(x\mid B) = \frac{1}{\sigma_b\sqrt{2\pi}} \exp\left[-\frac{1}{2} \frac{(x - \mu_b)^2}{\sigma_b^2}\right]$$
I already found it was possible to get 1 intersection if the variance is equal and the prior probability is equal. But i was wondering if it is possible to get 3 intersection points?
(to be exact only 2 functions will be compared)
Update:
(I don't know how to explain it better but maybe this image helps?)
No, your posterior distributions can only cross each other at two points. This is because they're both Gaussian, so the equation for their intersection will look like this $$\frac1{\sigma_1\sqrt{2\pi}}\exp\left[-\frac{(x-\mu_1)^2}{2\sigma_1^2}\right]=\frac1{\sigma_2\sqrt{2\pi}}\exp\left[-\frac{(x-\mu_2)^2}{2\sigma_2^2}\right]$$ We can rearrange this as follows $$\exp\left[-\frac{(x-\mu_1)^2}{2\sigma_1^2}\right]\div\exp\left[-\frac{(x-\mu_2)^2}{2\sigma_2^2}\right]=\frac{\sigma_1}{\sigma_2}$$ $$\exp\left[-\frac{(x-\mu_1)^2}{2\sigma_1^2}+\frac{(x-\mu_2)^2}{2\sigma_2^2}\right]=\frac{\sigma_1}{\sigma_2}$$ $$-\frac{(x-\mu_1)^2}{2\sigma_1^2}+\frac{(x-\mu_2)^2}{2\sigma_2^2}=\log\left(\frac{\sigma_1}{\sigma_2}\right)$$ $$\left(\frac{1}{2\sigma_2^2}-\frac{1}{2\sigma_1^2}\right)x^2+\left(-\frac{\mu_2}{\sigma_2^2}+\frac{\mu_1}{\sigma_1^2}\right)x+\left(\frac{\mu_2^2}{2\sigma_2^2}-\frac{\mu_1^2}{2\sigma_1^2}\right)=\log\left(\frac{\sigma_1}{\sigma_2}\right)$$
This is a quadratic for $x$ and so it has either 0, 1, or 2 roots.