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I would like to know whether $C \perp\kern-5pt\perp D | A $ in the following two graphical models and would like to know if my reasoning is correct:

enter image description here

For the left model (Belief Network), here's my work: $$ \begin{align} P(C,D|A) &\propto \sum_{B,E} P(B)P(A|B)P(C|A)P(D|B,C)P(E|D)\\ &=\sum_B P(B)P(A|B)P(C|A)P(D|B,C). \end{align} $$

We can't factorize it anymore. So this means that we can't factorize the probability into a product of functions of $f(C)$ and $f(D)$, so $C$ and $D$ are not independent given $A$.

For the right graphical model, which is a Markov Network, I use the following method:

  1. Remove all edges from $A$
  2. Check if there is a path leading from $C$ to $D$

We can see that if we remove the edges from $A$, we still have the path $C-D$ left, so this means that they are not independent given $A$.

Is this correct? To me it seems like for the left network, it's not correct, since $A$ is not a collider for $C$ and $D$, so they should be independent given $A$.

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    $\begingroup$ You just asked the same question 20 mins ago — maybe you could try to edit it instead of asking a new question. $\endgroup$
    – Uduru
    Jul 27, 2022 at 10:29
  • $\begingroup$ @Uduru hey, it's a different question. I thought that it would be appropriate to ask a new question rather than have 2 questions in one $\endgroup$
    – user
    Jul 27, 2022 at 10:30
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    $\begingroup$ Ah, my fault then. Sorry. $\endgroup$
    – Uduru
    Jul 27, 2022 at 10:31
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    $\begingroup$ If they are not the same question, could you give them different titles? Also, there is a big overlap between them, so if you don't want them to be closed as duplicates, can you fix this? Maybe you can make a generic question of them? "Is my solution correct" in general is not a good question for Q&A site, you should not expect people to check every of your solutions. $\endgroup$
    – Tim
    Jul 27, 2022 at 10:47
  • $\begingroup$ For a related question about the same graphical model see Is B ⊥ C | A for the two graphical models? $\endgroup$
    – dipetkov
    Aug 9, 2022 at 11:01

1 Answer 1

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$\newcommand{\npperp}{\not\perp\kern-8pt\perp}$ In your reasoning for the BN, you state "We can't factorize it anymore." But this would have to be proven.
Nevertheless, it is indeed true that $C\npperp D | A$, and the graph-based, d-separation, explanation would go like this: the path $C\to D$ is (trivially) d-separation open. There is no node between $C$ and $D$ that could block this path. The conditioning on $A$ would only have a chance of blocking this path if it was on this path, which it is not.

Your procedure and conclusion for the MN are correct.

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