In a generic particle filter, I understand the importance weights for each particle are calculated as $w_t^s \propto w_{t-1}^s \frac{p(y_t \mid z_t^s) p(z_t^s \mid z_{t-1}^s)}{q(z_t^s \mid z_{t-1}^s, y_t)}$.
I am confused about the weight of the conditioned particle in the conditional SMC (used in Particle Gibbs). Since the conditioned particle $z_t^N$ is set deterministically, does that mean that $q(z_t^N \mid z_{t-1}^N, y_t) = 1$?


1 Answer 1


Edit: I was mistaken. The particles in the path being conditioned on DO have weights. When you sample the $N-1$ ancestors at the beginning of each step, you may sample the conditioned-on particle as an ancestor. The weight formula here is roughly the same as the traditional formula you have written first.

To be more clear, the unnormalized weight for the $k$th particle, if $k\neq B_n$, at time $n>1$, assuming you just sampled the ancestor index $A_{n-1}^k$, is $$ \frac{f_{\theta}(x_n^k \mid x_{n-1}^{A^k_{n-1}})g(y_n \mid x_n^{k})}{q_{\theta}(x_n^k \mid y_n, x_{n-1}^{A_{n-1}^k}) }. $$ For the special particle, where $k=B_n$, the ancestor is not random: it is $B_{n-1}$. So the above formula would change slightly to $$ \frac{f_{\theta}(x_n^k \mid x_{n-1}^{B_{n-1}})g(y_n \mid x_n^{k})}{q_{\theta}(x_n^k \mid y_n, x_{n-1}^{B_{n-1}}) }. $$

I am using the notation from around page 10 of this paper, which describes the conditional SMC update. Sorry about the confusion earlier; I find it strange how they re-use the letter $k$ for the two different types of indices.

  • $\begingroup$ Thank you for your help. It is not yet clear to me that the conditioned particle has no weight. In the pseudocode you refer, I see that steps (a) and (b) (sampling ancestors and sampling new particles, respectively) are not carried out for the conditioned particle. However, in step (c) (calculating weights) there is no indication that the conditioned particle is left out. $\endgroup$ Commented Jun 23, 2018 at 23:23
  • $\begingroup$ @GonzaloBenegas there is an indication. The superscript is $k$, and $k$ is any index that does not equal $B_n$ at time $n$ $\endgroup$
    – Taylor
    Commented Jun 23, 2018 at 23:31
  • $\begingroup$ I'm not sure $k$ does not include $B_n$ in step (c). If the conditional particle has no weight, how does it interact with the rest? What's the point of its inclusion? $\endgroup$ Commented Jun 24, 2018 at 20:34
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    $\begingroup$ (-1) The weight of the conditioned particle is definitely used to sample indices: i) the conditioned particle may obtain descendants (in addition to the conditioned one) in each resampling step ii) in the end, the conditioned path may be selected as the "final result" of the cSMC step. $\endgroup$ Commented Jun 25, 2018 at 6:26
  • $\begingroup$ @JuhoKokkala Thanks for the clarification. Regarding my original question, do you have any insights on how the weight of the conditioned particle is calculated? $\endgroup$ Commented Jun 25, 2018 at 17:04

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