# Combining different Likelihoods - Particle Filter

I am new to particle filters and have a problem when it comes to the weightening (using SIR to be precise). The problem is that my different state-variables are subject to different noise-realizations (e.g. one variable experiences additive gaussian noise while another one is log-normally distributed), how would one normally deal with these situations?

What I tried out first was to calculate the probability of each state-variable of a particle separatly according to the respective likelihood, then multiply all resulting occurence proabilities for the respective particle and afterwards noramlize the resulting final proabilities as weights.

Sadly, this does not work very well (I suspect this might be the case because it ignores possible correlations between the state-variables) so I am very thankful for any kind of advise in this matter.

EDIT: From a more mathematical perspective, what I do is $$f(\mathbf{z}\mid \mathbf{x})\approx f(z_{1},z_{2}\mid \mathbf{x})=f(z_{1}\mid \mathbf{x})f(z_{2}\mid z_{1},\mathbf{x})\approx f(z_{1}\mid \mathbf{x})f(z_{2}\mid \mathbf{x})$$

So first I assume that the multivariate distribution can be expressed by a distribution of its components, then in the second approximation step I ignore the correlation between the respective components. So I would like to know under which restrictions the first step is allowed (I for example know that it its correct for a multivariate normal distribution) and also how I could enhance the second approximation $f(z_{1}\mid \mathbf{x})f(z_{2}\mid \mathbf{x})$ to balance the lack of the correlation.

• What is the observation model? Is $z$ the observation and $z = x+$noise? Also, what's the state transition model? – Atul Ingle Sep 12 '17 at 17:59
• $z$ is the observation and the obsv. model is given by $z=x$+noise. The transition function is given by a non-linear function $f(x)$ but I dont see how it would be important for the observation likelihood if we use a sequential approach? – J. Doe Sep 13 '17 at 14:52
• Thanks. You are right, the transition model is not relevant (for now). Now if the noise components are independent, the $\approx$ symbols in your equation can all be turned into $=$. So, correlations between state variable is probably not the issue here. You may need to elaborate more on the "this does not work very well" part of your question. – Atul Ingle Sep 13 '17 at 15:42

Let $\mathbf z = [z_1, z_2]^T$ be the observations and $\mathbf x = [x_1, x_2]^T$ be the hidden states, and the observation model is $z_i = x_i + n_i$ for $i=1,2$. Here $n_i$'s are independent (but not identical, since the noise has different distributions).
By definition, the density function of a random vector is given by the joint density of its components. So $f(\mathbf z | \mathbf x) = f(z_1, z_2 | \mathbf x)$ is always true.
The second approximation is actually an equality under the observation model with independent components. Since $z_2 = x_2 + n_2$, conditioned on $x_2$, we have $z_2 \perp (z_1, x_1)$. This implies that $f(z_2 | \mathbf x, z_1) = f(z_2 | x_1, x_2, z_1) = f(z_2|x_2)$. And similarly, $f(z_1|\mathbf x) = f(z_1 | x_1)$.