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In some papers, they proposed a model in which the joint posterior probability was modeled as[1]: $$ P(Y|\textbf{X},G) = \frac{P(\textbf{X},G|Y)P(Y)}{P(X,G)} \propto P(\textbf{X}|Y) \cdot P(Y|G)$$ where $\textbf{X}$ is attribute vectors set and $Y$ is the labels of all the edges in the network.

I know the meaning of the first equal sign which can be applied with Bayesian Theorem. But I feel confused about the third term, $P(\textbf{X}|Y) \cdot P(Y|G)$. How this term comes out?

[1]: J. Tang, T. Lou, and J. Kleinberg, “Inferring social ties across heterogenous networks,” in Proceedings of the fifth ACM international conference on Web search and data mining - WSDM ’12, 2012, p. 743.

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This model seems a little confusing due to the role of $G$ and I didn't find their definition of $P(Y|G)$ very sound.

You can consider it the following way:

$$P(Y|\textbf{X},G) \propto P(\textbf{X}|Y) \cdot P(G|Y) \cdot P(Y) = P(\textbf{X}|Y) \cdot [P(G|Y) \cdot P(Y)] \propto P(\textbf{X}|Y) \cdot P(Y|G)$$

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  • $\begingroup$ $P(Y|G)$ is defined as the probability of labels given the structure of the network. According to your inference, I find that you have the assumption: given the labels of edges, the attribute vectors of all nodes $\textbf{X} $ is conditional independent to the network structure $G$. e.g. $P(X,G|Y) = P(X|Y)P(G|Y)$. @emab $\endgroup$ – luck Jul 29 '15 at 9:09
  • $\begingroup$ @luck independent or almost independent. You cannot consider the impact of network structure two times (one for the labels and one for the attributes). This is how I digest this model. However, it is totally reasonable to consider $X$ and $G$ to be dependent, but you might end up with a different formula. $\endgroup$ – orezvani Jul 30 '15 at 1:57
  • $\begingroup$ It seems that it is more reasonable to consider $\textbf{X}$ is independent to $G$ in [1], since the attribute vector $\textbf{x}_i$ belongs to a pair involved in an edge $y_i$ in a social network. $\endgroup$ – luck Jul 30 '15 at 2:27
  • $\begingroup$ Based on the paper [1], I would say yes. $\endgroup$ – orezvani Jul 30 '15 at 2:32

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