# Is $B \perp\kern-5pt\perp C | A$ for the two graphical models?

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:

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?

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.
• 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
• Sorry, it is the path $B\to D \leftarrow C$ , I fixed it. Jul 27, 2022 at 11:03