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May 19, 2015 at 2:03 comment added DavidR Can you follow the same pattern as the first situation? Have a new node related to the distribution of D, which connects to both D and E?
May 18, 2015 at 21:44 comment added Tom This works for the simple situation I described. Any thoughts on how it could be extended to handle a case such as: $A \to B \to C \to D \to E$. I would want, $\Pr(E | D=d)$ to also depend on the distribution $\Pr(D)$, not just on the realization $d$. In this situation, the only input that relates to your suggestion goes into $A$ and then this induces a distribution $\Pr(D)$.
May 17, 2015 at 15:08 history edited DavidR CC BY-SA 3.0
added 34 characters in body
May 17, 2015 at 14:47 history answered DavidR CC BY-SA 3.0