I have spent a fair amount of time trying to solve this problem but I can't find the solution. More specifically, I have the following matrix:

P(D|E&F) = [ 0.5 0.3 0.5 0.2 ; 
             0.5 0.7 0.5 0.8 ]

All the variables are binary (two states) D, E and F are nodes of a Bayesian network. E and F are the parents of D and they are independent.

Now, new evidence comes in and we know that D and E are independent as well (the link between D and F remains but the arc from E to D is removed). How do I go about finding P(D|F) from P(D|E&F) ?


2 Answers 2


You should add the Bayesian network to your question since that implies a further independence assumption: E and F are independent give D.

There might still be something missing. Do you know P(E)? In that case, you could simply marginalize over E: $P(D|F) = \sum_i P(E=i)P(D|E=i \cap F)$.

  • $\begingroup$ @Neil this does not help to answer the question. $\endgroup$ Commented Apr 23, 2012 at 8:06
  • $\begingroup$ @StéphaneLaurent: Please have a look at the whole question including the probability table given. Your answer misses half the question. $\endgroup$
    – Neil G
    Commented Apr 23, 2012 at 8:43
  • $\begingroup$ $\Pr(D \mid E\cap F)$ is a number, so what is this matrix ? $\endgroup$ Commented Apr 23, 2012 at 8:49
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    $\begingroup$ Ok - strange notations. More rigorously this is $\Pr(D=i \mid E=j \cap F=k)$. Therefore your notation $\Pr(F)$ is not clear because $F$ is a r.v. and not an event. $\endgroup$ Commented Apr 23, 2012 at 9:15
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    $\begingroup$ Hm..I think @NeilG is right because E and F are also independent (they're the parents of D). Therefore, if I'm not mistaken P(E=i∣F) will become P(E=i) $\endgroup$ Commented Apr 23, 2012 at 10:12

You cannot conclude anything. For example, consider two independent r.v.\ $X$ and $Y$ taking positive integer values. Take $D=\{X=1\}$, $E=\{Y=2\}$, and $F=\{X+Y=3\}$. Then the equality $\Pr(D \mid F)=\Pr(D \cap E\cap F)$ is equivalent to $\Pr(X+Y=3)=\Pr(X=1 \cap Y=2)$, which can be true in some cases, and false in other cases.

By the way, I do not understand why $\Pr(D \mid E\cap F)$ is a matrix in your question ?

EDIT: my interpretation of the question was not the right one - see the answer by Neil G

EDIT2: Neil G has now completely deleted the above mentioned answer...

  • $\begingroup$ My question doesn't have any Pr(D∩E∩F) matrix!!! $\endgroup$ Commented Apr 23, 2012 at 8:09
  • $\begingroup$ Sorry: $\Pr(D \mid E\cap F)$, I will edit my post. And my answer does not satisfy you ? $\endgroup$ Commented Apr 23, 2012 at 8:22
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    $\begingroup$ Ok. So this question's taken from a past exam paper for Bayesian Networks. I was given that matrix because there is a child node D with two parents E and F. However, at some point the question specifies that the link between E and D must be removed since E and D are independent. And the question asks to deduce P(D|F) from P(D|E&F). Therefore, I assume that the question must have an answer but you said that we cannot conclude anything. Thanks for answering tho! $\endgroup$ Commented Apr 23, 2012 at 8:29
  • $\begingroup$ Sorry I don't know about the notion of "child node". You could remove $E$ if $D$ and $E$ are conditionally independent given $F$, for instance. $\endgroup$ Commented Apr 23, 2012 at 8:33
  • $\begingroup$ You're right George, you can remove the link between D and E. Stephane's answer assumes that F can render D and E dependent, but that's inconsistent with your graphical model. $\endgroup$
    – Neil G
    Commented Apr 23, 2012 at 9:05

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