2
$\begingroup$

I have a statistical model given by

$$ y_t\sim p(y_t|x_t, \theta)\\ x_t\sim p(x_t|x_{t-1},\theta)\\ \theta\sim p(\theta) $$ where $y$ is the only observed component. Using a sequential Monte Carlo algorithm, I can obtain $$ \{x_{1:T}^i, \, W_T^i\}_{i=1}^N, $$ where the target density of the SMC algorithm is $p(x_{1:T}|y_{1:T})$ (i.e. without $\theta$).

My question is: if I want draws from the predictive distribution $p(y_{T+1:T+h}|y_{1:T})$, would the following procedure be correct:

  1. Resample $x_{1:T}^i$ using multinomial sampling with probabilities $\propto W_T^i$ to obtain $\tilde{x}_{1:T}^i$
  2. For each $i=1, \dots, N$, generate $\theta^i \sim p(\theta|\tilde{x}_{1:T}^i, y_{1:T})$
  3. Generate the predictions $y_{T+1:T+h}^i\sim p(y_{T+1:T+h}|y_{1:T},\tilde{x}_{1:T}^i, \theta^i)$

The resulting array $\{y_{T+1:T+h}^i\}_{i=1}^N$ would then represent a draw from $$ p(y_{T+1:T+h}|x_{1:T}, y_{1:T}, \theta)p(\theta|x_{1:T}, y_{1:T})p(x_{1:T}|y_{1:T}) $$ so I think it checks out. My main concern is the initial resampling, but I feel that would be necessary in order to get rid of the unequal weights. Is this the appropriate way to do it? Any other ways that might be better (in some respect)?

$\endgroup$
5
  • 1
    $\begingroup$ SMC targets a distribution that’s conditional on knowing the parameter $\endgroup$
    – Taylor
    Sep 11, 2019 at 19:04
  • $\begingroup$ @Taylor Yes, that is of course the canonical case. For my situation, however, there exists an algorithm that marginalizes over the parameters. See here: dropbox.com/s/9gbrk835gkfq07f/… $\endgroup$
    – jacknick
    Sep 12, 2019 at 4:03
  • $\begingroup$ yes you're correct, my apologies. Looks like they're using the resample-move approach $\endgroup$
    – Taylor
    Sep 12, 2019 at 12:07
  • $\begingroup$ @Taylor Do you havs any thoughts on the suggested approach? An alternative, more general question would be if people ever do a final resampling step. Without it, plotting, for example, of the distribution of the latent variables doesn’t make much sense I guess. $\endgroup$
    – jacknick
    Sep 13, 2019 at 4:37
  • $\begingroup$ it's hard to know what you mean by appropriate, but it doesn't look like a terrible idea at first glance. Also, full disclosure, I have not read the paper carefully. There are many ways to resample, but multinomial-style is the most popular. And you're correct about the other thing: you can't really turn weighted samples into unweighted ones without (re-)sampling. $\endgroup$
    – Taylor
    Sep 13, 2019 at 22:17

1 Answer 1

0
$\begingroup$

Yes, this would be correct. One justification would be the following. Let $\hat{Z}_T$ be the standard estimator of $p(y_{1:T})$ which is generated by the particle filter. Then, the set $\left( \{x_{1:T}^i, \, W_T^i\}_{i=1}^N, \hat{Z}_T \right)$ is "properly-weighted" for $p(x_{1:T}|y_{1:T})$ in the following sense: Let $\textbf{PF}(\{x_{1:T}^i, \, W_T^i\}_{i=1}^N)$ be the joint law of your particle system at time $T$. Proper weighting is the statement that for all sufficiently nice functions $g$, it holds that

\begin{align} \mathbf{E} \left[ \hat{Z}_T \cdot \sum_{i = 1}^N W_T^i \, g (x_{1:T}^i) \right] = p(y_{1:T}) \cdot \int p(x_{1:T} | y_{1:T} ) \, g(x_{1:T}) dx_{1:T}. \end{align}

Proving the correctness of your approach could be established by writing down the joint distribution of your predictive scheme as

\begin{align} \Pi \left( \{x_{1:T}^i, \, W_T^i\}_{i=1}^N, k, \theta, y_{T+1:T+h} \right) &= \textbf{PF}(\{x_{1:T}^i, \, W_T^i\}_{i=1}^N) \\ &\cdot W_T^k \cdot p(\theta|x_{1:T}^k, y_{1:T}) \\ &\cdot p(y_{T+1:T+h}|x_{1:T}^k, y_{1:T}, \theta) \end{align}

where $k$ is the index of the "selected particle". You can then use the proper weighting formulation to check that, under the joint law of $\Pi$, it holds for any $G$ that

\begin{align} &\mathbf{E}_\Pi \left[ \hat{Z}_T \cdot G (x_{1:T}^k, \theta, y_{T+1:T+h}) \right] \\ = \, &p(y_{1:T}) \cdot \int p(x_{1:T}, \theta, y_{T+1:T+h} | y_{1:T} ) \, G (x_{1:T}, \theta, y_{T+1:T+h}) \, dx_{1:T} d\theta dy_{T+1:T+h}, \end{align}

i.e. that your proposed method is properly-weighted and hence appropriate for prediction purposes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.