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recently I am studying PF. And I am stuck in APF for a few days, though I derived many times. Here is my question:

I followed the framework of this paper. The APF is defined in Algorithm 1: enter image description here

The authors define the weight function $w_n(x_{n-1:n}) = \frac{p(y_n|x_n)p(x_n|x_{n-1})}{q(x_n, y_n|x_{n-1})} = \frac{p(y_n|x_n)p(x_n|x_{n-1})}{q(x_n|y_n, x_{n-1})q(y_n|x_{n-1})}$

The $q_\theta(y_{n+1}|X_n^i)$ makes me very confused. I see some papers define it as predictive likelihood $q_\theta(y_{n+1}|X_n^i) \equiv p(y_{n+1}|X_n^i)$.

  1. In $n$-th step, the $y_{n+1}$ should not be observed, how do I calculate this term?
  2. In some contexts, $p(y_{n}|X_{n-1}^i) \equiv p(y_{n}|\mu(X_{n-1}^i)) $, where $\mu(\cdot)$ is the characterization of $\mu(X_{n-1}^i)$ like mode or mean, but $X_{n-1}^i$ is known in $n$ step right? Why do I need another function?
  3. In sequential monte carlo, the weights are designed for $\frac{p(x_{1:n}|y_{1:n})}{q(x_{1:n}|y_{1:n})}$, but in filtering problem, the target distribution is the marginal distribution $p(x_n|y_{1:n})$, how is the equivalence of two terms?

Many thanks for your help!

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  • $\begingroup$ The auxiliary particle filter is the same as a regular SISR but with a specific proposal distribution that is meant to resemble the optimal proposal distribution using only the two distributions of the model—the state transition and the observation density. The downside is that you have to iterate over your particles twice per iteration instead of once. What’s confusing you is probably that writing down the APF proposal looks weird because it’s a mixture distribution whose wrights rely on the observation density. $\endgroup$
    – Taylor
    Jul 12, 2023 at 15:52
  • $\begingroup$ @Taylor Hi thanks for your comment, I did some some contexts did two resampling per iteration. But the framework I provided said only one resampling is enough. Thus I am confused. Could you write an answer? $\endgroup$ Jul 12, 2023 at 16:13
  • $\begingroup$ They're presenting it differently from the original paper. The original paper emphasizes the auxiliary variable more. In this review paper, I think of step 2 in their algorithm 1 as getting the "stage 1 weights" of the proposal distribution. $\endgroup$
    – Taylor
    Jul 12, 2023 at 16:30
  • $\begingroup$ @Taylor I have added the definition of the weight in this review paper. It seems that the step 2 is a mix of stage 1 and 2? $\endgroup$ Jul 12, 2023 at 17:01

1 Answer 1

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If you're trying to learn about this for the first time, I'd check out the original paper: https://www.jstor.org/stable/2670179 Speaking for myself, I don't like to mix references on this topic because explanations vary.

  • There can be equivocation on what "resampling" means,
  • people can de-emphasize the auxiliary variable (even though that's where the algorithm gets its name from),
  • alternative descriptions of the algorithm can describe the last step of an iteration as the first step of the iteration, or vice versa, and
  • there is a weaker incentive for recent papers to re-explain everything in as much detail as older papers.

It also helps to code it up yourself. Here's my implementation in c++.

To answer your questions:


  1. In $n$-th step, the $y_{n+1}$ should not be observed, how do I calculate this term?

Yes, that's right. But in the $n+1$-th step, it is, and at some point you're starting this step with "old" particles from time $n$. There's another problem, though. And that is that the predictive density is usually intractable. That's why they do the $\mu(\cdot)$ thing that you talk about in the next bullet point.

  1. In some contexts, $p(y_{n}|X_{n-1}^i) \equiv p(y_{n}|\mu(X_{n-1}^i))$, where $\mu(\cdot)$ is the characterization of $\mu(X_{n-1}^i)$ like mode or mean, but $X_{n-1}^i$ is known in $n$ step right? Why do I need another function?

$p(y_{n}|\mu(X_{n-1}^i))$ is just an approximation. It's not the same. You're just really trying to use the most up-to-date observation in any way you can. You pick $\mu$ yourself. It's a tuning choice. Then you apply it to all the particles you have, yes.

  1. In sequential monte carlo, the weights are designed for $\frac{p(x_{1:n}|y_{1:n})}{q(x_{1:n}|y_{1:n})}$, but in filtering problem, the target distribution is the marginal distribution $p(x_n|y_{1:n})$, how is the equivalence of two terms?

This is a minor issue. The filtering distribution is the marginal of the smoothing distribution. Writing the smoothing means you just don't have to write a bunch of integrals or sums.

Algorithmically you can save memory and only save the last stage of the trajectories, or you can save the entire paths.

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  • $\begingroup$ thank you for your answer. Still confused at the first point you answered. The review paper integrated $p(y_{n+1}|X^i_{n})$ in n-th step weight, at this moment, the $y_{n+1}$ is unknown. You are saying that at next step $y_{n+1}$ is known. I want to know what is $p(y_{n+1}|X^i_{n})$ whether it is intractable. Thanks $\endgroup$ Jul 13, 2023 at 6:42
  • $\begingroup$ $p(y_{n+1}|X^i_{n}) = \int p(y_{n+1} \mid x_{n+1}) p(x_{n+1} | x_n) dx_{n+1}$. It's a mixture distribution using the observation density and the state transition density. The APF does not assume it is tractable, but it does try to approximate it with $ \int p(y_{n+1} \mid \mu(x_{n})) p(x_{n+1} | x_n) dx_{n+1} = p(y_{n+1} \mid \mu(x_{n}))$. $\endgroup$
    – Taylor
    Jul 13, 2023 at 14:22
  • $\begingroup$ I see, thank you! $\endgroup$ Jul 14, 2023 at 3:02

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