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The picture above is from Frigola et al (2013) - Bayesian Inference and Learning in Gaussian Process State-Space Models with Particle MCMC.

In this paper, the authors later define $\mathbf{f}_t=f(\mathbf{x}_{t-1})$, and here lies my misunderstanding...

Let me explain better my doubt. When $f(\mathbf{x}_t)$ follows a unidimensional GP, we have for any $n\in \mathbb{N}$

$(f(\mathbf{x}_1),...,f(\mathbf{x}_n))\sim \mathcal{N}(\mu,k(X,X'))$, where $f(\cdot) \in \mathbb{R}$, and $\mu = (m(\mathbf{x}_1),...,m(\mathbf{x}_1))'$.

However, when we write $\mathbf{f}_t=f(\mathbf{x}_{t-1})$, it seems we are using a univariate function to define vector $\mathbf{f}_t$. How can this be? Any help would be appreciated.


Although I did not find this explicitly stated in the paper, it seems that $f$ is actually intended to be a multiple output Gaussian process (possibly with independent components) - so that a realization is a function from $\mathbb{R}^{n_x}$ to $\mathbb{R}^{n_x}$ and thus $\mathbb{f}_t = f(\mathbf{x}_t) \in \mathbb{R}^{n_x}$. Note that:

  1. In the first author's PhD thesis [1], Section 3.1, similar notation is used to define a Gaussian process state-space model, and on p. 27 the following remark appears:

Note that the covariance function $\kappa(\cdot, \cdot)$ returns a matrix of the same size as the state in an analogous manner to multi-output Gaussian processes (Rasmussen and Williams, 2006).

  1. In an earlier paper considering Gaussian process state-space models, Turner et al. [2] (reference [2] of the case paper), "independent GPs are used for each target dimension of $f$" (p. 869, last paragraph of the left-hand-side column). Which is a special case of a multiple-output GP.


[1] Roger Frigola-Alcalde (2015) Bayesian Time Series Learning with Gaussian Processes. PhD thesis, University of Camridge. http://www.rogerfrigola.com/doc/thesis.pdf

[2] Ryan Turner, Marc Deisenroth, and Carl Rasmussen (2010). "State-space inference and learning with Gaussian processes." AISTATS 2010, http://proceedings.mlr.press/v9/turner10a/turner10a.pdf.


You need to understand the notation here. When the author say \begin{align} x_{t+1}| f_t \sim \mathcal{N}(x_{t+1}|f_t, Q), \end{align} this implies that $x_{t+1}$ is distributed with mean $f_t$ and covariance $Q$, where $f_t$ is obtained from the state evolution equation given in (1a), i.e., \begin{align} x_{t+1} = f(x_t,u_t) + v_t. \end{align} As $f$ depends on $t$, they denote the mean of $x_{t+1}$ as $f_t$ which is nothing but $\mathbb{E}(x_{t+1}) = \mathbb{E}(f(x_t,u_t)) = f_t$ if $v_t \sim \mathcal{N}(0,Q)$. Therefore $x_{t+1}| f_t \sim \mathcal{N}(x_{t+1}|f_t, Q)$.

An easy way to understand this by assuming $f(x_t,u_t)$ takes the form \begin{align} f(x_t, u_t) = A x_t + B u_t \end{align}

  • 1
    $\begingroup$ Maxtron, thank you for your answer. I hope I'm not being too blunt, but do you know what's a gaussian process? I have no problem understanding what you've just written. My doubt is that there seems to exist an inconsistency in the dimensions of what you've just written and the definition of $f(x_t)$ being a Gaussian Process with a scalar covariance function $k(x_t, x_t')$ $\endgroup$ – An old man in the sea. Sep 12 '18 at 0:07
  • $\begingroup$ The Gaussian distributions here are not univariate. They are multivariate Gaussian distributions where $x_t$ is an $n$-dimensional vector. Does this make things clear? $\endgroup$ – Maxtron Sep 12 '18 at 1:12
  • $\begingroup$ Here's a tutorial on Gaussian processes. If you are looking for more depth in understanding Gaussian processes, read GPML textbook written by Carl Edward Rasmussen and Christopher K. I. Williams. I'd definitely recommend Chap. 2 and 3. $\endgroup$ – Maxtron Sep 12 '18 at 1:16
  • $\begingroup$ Maxtron, I've edited my question. See if it's now easier to understand my doubt. $\endgroup$ – An old man in the sea. Sep 12 '18 at 7:19
  • 1
    $\begingroup$ I understand what's a GP (I think), and what's a Multivariate Gaussian Distribution. $\endgroup$ – An old man in the sea. Sep 12 '18 at 7:26

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