I have some trouble with score functions in likelihood calculation. I'm not good at statistics or probability, so I'm still confused on formalism and mathematical-probabilistic language.

Some background: I'm on particle filter, so inferring a pdf that is analytically intractable by a Monte Carlo sampling: put some random particles, evaluate them with a likelihood function, and the normalised-weighted set of particles can be an approximation of the pdf.

I have to give a weight to each particle and compute the likelihood with a likelihood function or score function.

If my hidden state described by this difficult pdf that I have to infer is denoted as $X$, and I have some observation $z$, the likelihood can be formalised by:

$$ p( X | z ) = \text{likelihood} $$

What I have now: a nice distance function that give me the distance between the particle and the target. It works. It is the Bhattacharyya distance (I'm working with images, and colour histogram is a relevant feature). So that:

$$ d = \text{Bhattacharyya(particle, target)} $$

This $d$ is a distance, and for a likelihood function I need a probability, so I'm putting it into a Gaussian with zero means and variance = 0.4:

$$ \text{likelihood} = \frac{e^{-\frac{(d-\mu)^2}{2 \sigma^2}}}{\sqrt[2]{ \pi \sigma}} $$

What I want: Now I'd like to improve my experiments, adding similarly another score function so that the total likelihood is calculated with two score functions (let's say a shape-distance or something else) $s_1$ and $s_2$.

I remember from school (but don't remember exactly why) that if you want to combine 2 probabilities, you have only to multiply, not sum. So I know that:

$$ \text{likelihood} \propto s_1 s_2 $$

where $s_1$ is the Gaussian with the Bhattacharyya with the colour histogram, and $s_2$ is the shape distance.

The two score functions are Gaussian, and they are not independent (they depend on the same hidden state).

Question 1: Is it true I have to multiply them? Why? So that my final formula will be:

$$ \text{likelihood} = \frac{e^{-\frac{(d_1-\mu)^2}{2 \sigma^2}}}{\sqrt[2]{ \pi \sigma}} \frac{e^{-\frac{(d_2-\mu)^2}{2 \sigma^2}}}{\sqrt[2]{ \pi \sigma}} $$

Question 2: What if I want to empirically add some more importance to one of the two scores because I think it is more descriptive of my data? Like:

$$ \text{likelihood} = \alpha s_1 (1-\alpha)s_2 $$

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    $\begingroup$ $\sigma$ should not be under the square root in the denominator of the likelihood. $\endgroup$ – Glen_b Mar 10 '16 at 12:26

Question 1: There are plenty of instances when you would sum probabilities, and also plenty when you multiply them. It depends what you're doing with the probabilities. If you're this unsure about the basics operations you're using, I'd recommend taking a look at Wikipedia and Googling some online tutorials about manipulating probabilities. A couple of hours spent on that might save you many more further down the road (especially if you want to use Monte Carlo methods).

Without getting too in depth, multiplication is like an AND operation – I want the probability of this AND this happening together. Summation is more like an OR operation. In this case you want something which matches the colour histogram and which fits your shape distribution, so you multiply. There's plenty of caveats I'm leaving out here though.

Question 2: A simple way to achieve this is to raise each to a different power. $$ \text{likelihood} = s_1^\alpha s_2^{1-\alpha} $$ This is equivalent to a weighted sum in log likelihood space.

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