I am facing a posterior distribution in a MCMC application that aims to sample an unobservable variable $x=\{x_t\}_{t=0}^{T}$ given an observed series $y=\{y_t\}^T_{t=0}$.

However, the conditional posteriors reads as $$p(x_t | y_{t+1}, y_t, y_{t-1} ,x_{t-1}, x_{t+1}, \Theta),$$ with $\Theta$ being a vector of additional structural parameters. According to my understanding, this would be a smoothing problem, since knowledge of $y_{t+1}$ is required to infer the value of $x_t$.

However, the articles dealing with the same problem refer to the series $x$ as filtered series.

Am I missing something here?

  • $\begingroup$ Based on comments this is about usage of this terminology in a certain article - please edit to add the reference to the article into the question $\endgroup$ Nov 17, 2016 at 7:12

2 Answers 2


I guess definitions could be different, but the standard definitions I use are

  • filtering: $p(x_t|y_1,\ldots,y_t,\Theta)$
  • smoothing: $p(x_t|y_1,\ldots,y_T,\Theta)$ for $0\le t<T$

That is, filtering is the distribution of the current state given all observations up to and including the current time while smoothing is the distribution of a past state (or states) given the data up to the current time.

For me, neither filtering nor smoothing refers to $p(x_t|y_{t+1},y_t,y_{t-1},\Theta)$. I'm also guessing that you really wanted $$p(x_t|x_{t+1},y_t,x_{t-1},\Theta) = p(x_t|y_1,\ldots,y_T,x_1,\ldots,x_{t-1},x_{t+1},\ldots,x_T,\theta)$$ which is the full conditional distribution for $x_t$ which some refer to as the conditional posterior.

  • 1
    $\begingroup$ Great answer, but perhaps change the capital T in the filtering bullet to a small t? $\endgroup$
    – Tingiskhan
    Nov 2, 2016 at 19:15
  • $\begingroup$ Broadly speaking, I am looking for the term when I need among others tomorrow's observation (t+1) to infer sth about today's state (t). $\endgroup$
    – mscnvrsy
    Nov 2, 2016 at 20:08
  • $\begingroup$ @mscnvrsy maybe $y_t$ is instead something like $Y^t$ which sometimes stands for $y_1, \ldots, y_t$. Can you link us to the resource you're looking at $\endgroup$
    – Taylor
    Nov 2, 2016 at 22:48
  • $\begingroup$ Sure. I am refering to the conditional posterior of $v_{t+1}$ in Li, Wells and Yu (2006) : lib.dr.iastate.edu/cgi/… (p. 33) $\endgroup$
    – mscnvrsy
    Nov 3, 2016 at 9:17
  • $\begingroup$ Does anyone have further ideas on that? For me it's just about the naming "Smoothed vs Filtered Series", essentially to infer the state at $t$ I need information about $t+1$. $\endgroup$
    – mscnvrsy
    Nov 14, 2016 at 9:11

Original model you mentioned in the comments to the other answer: $$ Y_{t+1} = Y_t + \mu \Delta + \sqrt{v_t \Delta} \epsilon_{t+1}^y + \xi_{t+1}^y N_{t+1}^y \\ v_{t+1} = v_t + \kappa(\theta - v_t)\Delta + \sigma_v \sqrt{v_t\Delta}\epsilon_{t+1}^v $$ with $\text{corr}(\epsilon_{t+1}^y,\epsilon_{t+1}^v) = \rho$. The reference you linked to is also linked to at the end of this.


Let's call $\epsilon_{t+1}^y = e^1_{t+1}$ and $\epsilon_{t+1}^v = \rho e^1_{t+1} + \sqrt{(1-\rho^2)}e^2_{t+1}$ with $e^1_{t+1}$ and $e^2_{t+1}$ independent standard normals. Making the substitutions we get

$$ Y_{t+1} - Y_t = \mu \Delta + \sqrt{v_t \Delta} e^1_{t+1} + \xi_{t+1}^y N_{t+1}^y \\ v_{t+1} = v_t + \kappa(\theta - v_t)\Delta + \sigma_v \sqrt{v_t\Delta}\left[\rho e^1_{t+1} + \sqrt{(1-\rho^2)}e^2_{t+1} \right] $$

Then let $\phi = \sigma_v \rho$ and $w_v = \sigma^2_v(1-\rho^2)$. They say to make this transformation on page 33.

$$ Y_{t+1} - Y_t = \mu \Delta + \sqrt{v_t \Delta} e^1_{t+1} + \xi_{t+1}^y N_{t+1}^y \\ v_{t+1} = v_t + \kappa(\theta - v_t)\Delta + \phi\sqrt{v_t\Delta} e^1_{t+1} + \sqrt{v_t\Delta}\sqrt{w_v}e^2_{t+1} $$


They menton that $\Theta = \{\mu, \kappa, \theta, \sigma_v, \rho, \lambda_y, \mu_y, \sigma_y\}$. After the transformation it's actually $\Theta = \{\mu, \kappa, \theta, \phi, w_v, \lambda_y, \mu_y, \sigma_y\}$ for us now. They also describe posteriors for the following (and these must be part of the state vector at some time): $\xi^y_{t+1}$ $N_{t+1}^y$, $v_{t+1}$.

So we could define the state vector $$ x_t = [v_{t+1}, v_t, \xi^y_{t+1} N_{t+1}^y]', $$ and this would represent a state space model closer to what the other answer was talking about. But there are probably a lot of ways to do this. At the moment I can't tell if this paper does it that way.


Anyway, back to your question...I'm not sure why you relabelled everything becuase that makes it way harder to follow along, but you said in the comment that you're trying to get at the 'conditional posterior of $v_{t+1}$.' If you mean $p(v_{t+1}|y_{1:T}, \Theta)$, then that's a marginal of the smoothing distributon $p(x_{t+1}|y_{1:T}, \Theta)$ that the other answer was talking about.

On the other hand, if you were trying to sample from $p(x_{t}|y_{1:T},x_{1:t-1},x_{t+1:T})$ then \begin{align*} p(x_{t}|y_{1:T},x_{1:t-1},x_{t+1:T}) &\propto \prod_{t=2}^T p(y_t|x_t)p(x_t|x_{t-1})p(y_1|x_1)p(x_1) \\ &\propto p(x_t|x_{t-1})p(y_t|x_t)p(x_{t+1}|x_t) \\ &\propto p(x_t|x_{t-1},x_{t+1},y_t) \end{align*} which the other answer also mentioned. I think this is called a "single-site sampler," perhaps useful if you wanted to get at $p(x_{1:T}|y_{1:T},\Theta)$ Gibbs-style. I am guessing that this is what you want, actually. You would get this if you used the state vector in part 2, and used the log returns $Y_{t+1} -Y_t$ as the observations.

So I'm kind of echoing the other answer here: it's probably one of those two things. Hope this helps.

Reference: http://lib.dr.iastate.edu/cgi/viewcontent.cgi?article=1121&context=stat_las_preprints

  • 2
    $\begingroup$ @Xi'an he mentioned it in the comments. I'll make it more clear in my answer $\endgroup$
    – Taylor
    Nov 17, 2016 at 17:46

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