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

enter image description here

For the left graphical model, which is a Belief Network, here's how I deduce it:

$P(B,C|A) \propto \sum_{D,E} P(B)P(A|B)P(C|A)P(D|B,C)P(E|D) = \underbrace{P(B)P(A|B)}_{f(B)}\underbrace{P(C|A)}_{f(C)}$

Since the probability factors into a product of functions of $B$ and $C$, we can say that they are 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 $B$ to $C$

So we remove all edges from $A$ and we can see that we have a path, $B-D-C$ and hence $B$ and $C$ are not independent given $A$.

Is my reasoning correct? Also, for the Belief network deduction, is there a faster way to see this? My reasoning, without writing anything, would be to just say that since $A$ is not a collider for $B$ and $C$, then they are independent. Would this be correct reasoning?


1 Answer 1


Your arguments both for the BN and the MN are correct, provided those graphs are perfect maps for your two distributions (which are then necessarily different).

The formula-free explanation for the BN is correct, too, although I would add that the path $B\to D\leftarrow C$ is blocked because of $D$ being a collider.

  • $\begingroup$ Thank you. I just don't understand how is $A\rightarrow D \rightarrow C$ a path? It seems like in the BN it doesn't exist $\endgroup$
    – user
    Jul 27, 2022 at 11:01
  • 1
    $\begingroup$ Sorry, it is the path $B\to D \leftarrow C$ , I fixed it. $\endgroup$
    – frank
    Jul 27, 2022 at 11:03
  • $\begingroup$ Hi Frank, do you mind taking a look at this and this Q's. $\endgroup$
    – user
    Aug 3, 2022 at 10:11

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