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If we have a latent state space $\mathbf{X}$ and the observations $\mathbf{Y}$ and the transition function between two states $\mathbf{x}_{t-1}$ and $\mathbf{x}_{t}$ is given by $\mathbf{f}$ which is a gaussian process with mean function $m_f$ and covariance $k_f$, where its graphical model is as follows:

enter image description here

The generative model is given by $$f(x)\sim\mathcal{GP}(m_f(\mathbf{x}),k_f(\mathbf{x},\mathbf{x}'))\\ \mathbf{x}_0\sim p(\mathbf{x}_0)\\ \mathbf{f}_t=f(\mathbf{x}_{t-1})\\ \mathbf{x}_{t}|\mathbf{f}_{t}\sim\mathcal{N}(\mathbf{f}_{t},\mathbf{Q})\\ \mathbf{y}_t|\mathbf{x}_t\sim p(\mathbf{y}_t|\mathbf{x}_t,\boldsymbol{\theta}_y)$$

I would like to know how the following probability can be rederived $$\mathbf{f}_2|\mathbf{f}_1,\mathbf{x}_{0:1}\sim\mathcal{N}(\mathbf{f}_2|m_f(\mathbf{x}_1)+k_f(\mathbf{x}_1,\mathbf{x}_0)k_f(\mathbf{x}_0,\mathbf{x}_0)^{-1}(\mathbf{f}_1-m_f(\mathbf{x}_0)),k_f(\mathbf{x}_1,\mathbf{x}_1)-k_f(\mathbf{x}_1,\mathbf{x}_0)k_f(\mathbf{x}_0,\mathbf{x}_0)^{-1}k_f(\mathbf{x}_0,\mathbf{x}_1))$$ Thanks!

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  • $\begingroup$ According to the specification of your generative model, $\mathbf{f}_2$ is a deterministic function of $\mathbf{x}_1$, so the "distribution" of $\mathbf{f}_2 | \mathbf{f}_1, \mathbf{x}_{0:1}$ would just be a point-mass on $\mathbf{f}_2 = f(\mathbf{x}_1)$. I take it this is not what you intended for the question? $\endgroup$ – Ben Apr 4 at 6:54
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    $\begingroup$ @Ben note that $f$ itself is a random function, so $\mathbf{f}_2$ is random even when $\mathbf{x}_1$ is fixed. $\endgroup$ – Martin Modrák Apr 4 at 11:15
  • $\begingroup$ Ah yes, sorry, I missed that bit. $\endgroup$ – Ben Apr 4 at 11:33
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If we assume a Gaussian Process prior on the function values, $$P(f_1|\mathbf{x}_1)\sim\mathcal{GP}(m_f(\mathbf{x}_1),k_f(\mathbf{x}_1,\mathbf{x}_1))$$ then for the new state, we have $$\begin{bmatrix} f_1\\ f_2\\ \end{bmatrix} =\mathcal{N}\Bigg( \begin{bmatrix} m_f(\mathbf{x}_1) \\ m_f(\mathbf{x}_2)\\ \end{bmatrix}, \begin{bmatrix} k_f(\mathbf{x}_0,\mathbf{x}_0) & k_f(\mathbf{x}_0,\mathbf{x}_1) \\ k_f(\mathbf{x}_1,\mathbf{x}_0) & k_f(\mathbf{x}_1,\mathbf{x}_1) \\ \end{bmatrix}\Bigg) $$ the conditional distribution of $f_2$ given $f_1$ is itself Gaussian-distributed, we can use the Schur complement and write this as $$f_2|f_1\sim\mathcal{N}(m_f(\mathbf{x}_1)+k_f(\mathbf{x}_1,\mathbf{x}_0)k_f(\mathbf{x}_0,\mathbf{x}_0)^{-1}(f_1-m_f(\mathbf{x}_0)),k_f(\mathbf{x}_1,\mathbf{x}_1)-k_f(\mathbf{x}_1,\mathbf{x}_0)k_f(\mathbf{x}_0,\mathbf{x}_0)^{-1}k_f(\mathbf{x}_0,\mathbf{x}_1))$$

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