Importance weight of conditioned particle in conditional SMC

Question

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 $q(z_t^N \mid z_{t-1}^N, y_t) = 1$?

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.