# Is $C \perp\kern-5pt\perp D | A$ for the two graphical models? [duplicate]

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:

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 $$A$$.

• You just asked the same question 20 mins ago — maybe you could try to edit it instead of asking a new question. Jul 27, 2022 at 10:29
• @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
– user
Jul 27, 2022 at 10:30
• Ah, my fault then. Sorry. Jul 27, 2022 at 10:31
• 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.
– Tim
Jul 27, 2022 at 10:47
• For a related question about the same graphical model see Is B ⊥ C | A for the two graphical models? Aug 9, 2022 at 11:01

$$\newcommand{\npperp}{\not\perp\kern-8pt\perp}$$ 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.