Sampling from Likelihood: Likelihood Particle Filter

Question

I am particularly baffled with the idea of sampling from the likelihood. To give some context, I am studying particle filters and am investigating the "Likelihood Particle Filter". I am reading this Tutorial(Appendix Section) at the appendix section. My questions are written in bold.

In the context discussed, it is assumed that $g(x_{k})=x_{k}^2$. However, to keep things simple I want to make the derivation of the algorithm using $g(x_{k})= x_k$. I feel that for other examples that the addition of the auxiliary variable $s_k=x_k^2$ is done for the purpose of simplifying the selection of a non-linear $g$. More specifically, $g$ (chosen by the author) is a many to one function, so we run into the problem that $x_t$ may be generated by either $x_{t-1}$ which is negative or positive, hence the inclusion of the expression of $p(x_k|s^i_k)=\frac{\delta(x_k+\sqrt(s^i_k))+\delta(x_k-\sqrt(s^i_k))}{2}$. My guess is that for an invertible $g$,e.g. $g(x_{k})= x_k$, the problem becomes significantly simpler plus we do not need to include the auxiliary variable $s_k$.Here is my derivation.

$$x_k=f(x_{k-1})+v_k=x_{k-1}+v_k$$

$$y_k=g(x _{k-1})+w_k=x_{k-1}+w_k$$

where, $v_k$ and $w_k$ are both gaussian with fixed variance. The idea is to select the proposal $q(x_k|x_{k-1},y_k)=p(x_k|y_k)$. By bayes theorem we can re-formulate $p(x_k|y_k)$

$$p(x_k|y_k)= \frac{p(y_k|x_k)p(x_k)}{p(y_k)}\propto p(y_k|x_k)p(x_k)$$

This equation becomes easier because we select a uniform prior, whose density is constant. Take note of this point because I made an arbitrary choice of prior.

$$p(x_k|y_k)\propto p(y_k|x_k)$$

Therefore, I can sample from $p(x_k|y_k)$ from literally just the measurement model. This is the reason invertiblity of $g$ was so important. We can sample by

$$X^i_k|y_k \sim y_k-w_k$$

Is this inversion I made by the measurement model valid? I am apprehensive about it and I am not sure whether there is some missing assumptions. Most of all, I am curious whether the authors in the paper meant anything when including the auxiliary variable $s_k$.

So moving on. We are aiming at deriving a weighted recursion so that we can incorporate the proposals into the filter.

$$w^i_k=w^i_{k-1}\frac{p(x_k^i|x_{k-1}^i)p(y_k|x^i_k)}{q(x^i_k|x^i_{k-1},y_k)}=w^i_{k-1}\frac{p(x_k^i|x_{k-1}^i)p(y_k|x^i_k)}{p(x^i_k|y_k)}=w^i_{k-1}\frac{p(x_k^i|x_{k-1}^i)p(x^i_k|y_k)p(y_k)}{p(x^i_k|y_k)p(x^i_k)}$$

I know that I can cut $p(x^i_k|y_k)$. But, Can I simplify any further? Can I just remove $p(y_k)$ because it is a constant and can I just remove $p(x^i_k)$ because I chose the density to be uniform. In which case, I will obtain

$$w^i_k \propto w^i_{k-1}p(x_k^i|x^i_{k-1})$$

I am slightly wary because the solution in the paper gave me $w^i_k \propto w^i_{k-1}p(x_k^i|x^i_{k-1})x^i_k$ but then again I used a different measurement model. Are these the same solution? However, my solution to the weighted recursion at least follows the notion that I am weighting by the prior.

Answer 1

It makes it easier for others to follow your writing if you define notation you're using. Personally I would prefer not to read a long paper to gather this information.

1.) When you say:

This equation becomes easier because we select a uniform prior, whose density is constant. Take note of this point because I made an arbitrary choice of prior.

this does not necessarily work with your model. The distribution $p(x_k)$ might not be the same for all time points. It would be if a) your hidden Markov Chain was stationary, and b) the stationary distribution it admitted was your uniform distribution.

2.) Then you say, "$X^i_k|y_k \sim y_k-w_k$." This would be true if the previous was true, and if $f$ is the identity function, which I don't think you specified.

3.) When you write $w^i_{k-1}\frac{p(x_k^i|x_{k-1}^i)p(x^i_k|y_k)p(y_k)}{p(x^i_k|y_k)p(x^i_k)}$, you have made it less simple, because you probably do not have the ability to evaluate $p(y_k)$ or $p(x_k^i|y_k)$.

4.) Lastly, if you choose $g$ and $f$ to be linear, then you can test your derivations empirically. This is the situation that a Kalman filter would work in, and a Kalman filter is able to directly evaluate the conditional likelihoods at each time step.