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Given factorizations of two joint densities $p(x_1,...,x_n)=\prod_{i=1}^n p(x_i\mid \textrm{cond}(x_i))$ and $q(x_1,...,x_n)=\prod_{i=1}^n q(x_i\mid \textrm{cond}(x_i))$, where $\textrm{cond}(\bullet)$ denotes the set of conditioning variables, does the KL-divergence decompose, i.e., does

$\textrm{KL}(p\Vert q)= \sum_{i=1}^n \textrm{KL}\left(p(x_i\mid \textrm{cond}(x_i))\ \Vert\ q(x_i\mid \textrm{cond}(x_i))\right)$

hold?

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    $\begingroup$ If it makes things easier, you can assume $\textrm{cond}(x_i) \in \{x_1,\ldots,x_n\}$ . More generally, the two factorisations are Bayesian networks, which means that $\textrm{cond}(x_i)$ can be any subset of $\{x_1,\ldots,x_n\}$ so that the induced graph structure is a directed acyclic graph. $\endgroup$ – ASML Jan 23 '16 at 14:23
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    $\begingroup$ How do you define $\textrm{KL}\left(p(x_i\mid \textrm{cond}(x_i))\ \Vert\ q(x_i\mid \textrm{cond}(x_i))\right)$ ? The ambiguous part is how you integrate out the $cond(x_i)$ variables in there $\endgroup$ – Guillaume Dehaene Feb 3 '16 at 13:42
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As ASML suggested you can represent the factorization of the joint probability distribution according to a Bayesian network. Then, as it is pointed in [Tong, S., & Koller, D. (2001)] (page 4) the KL-divergence decomposes with the graphical structure of the network.

[Tong, S., & Koller, D. (2001)] Tong, S., & Koller, D., Active learning for parameter estimation in Bayesian networks. In Advances in neural information processing systems (pp. 647-653).

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