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

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For the left graphical model (Belief Network):

Checking $B \perp\kern-5pt\perp C | D,A $: We have that $$ \begin{align} P(B,C|D,A) &\propto \sum_E P(B)P(A|B)P(C|A)P(D|B,C)P(E|D)\\ &= P(B)P(A|B)P(C|A)P(D|B,C). \end{align} $$

We can see that there is a term, $P(D|B,C)$ which contains both $B$ and $C$, so this means that the probability does not factorize into a product of a function of $B$ and a function of $C$, hence $B$ and $C$ are not independent given $D$ and $A$.

Question 1 Is this correct? And also, would it be correct to just say that since $D$, which is in the conditioning set, is a collider for $B$ and $C$, hence no independence?

Checking $D \perp\kern-5pt\perp A | B,C $: We have that $$ \begin{align} P(D,A|B,C) &\propto \sum_E P(B)P(A|B)P(C|A)P(D|B,C)P(E|D)\\ &= \underbrace{P(B)P(A|B)P(C|A)}_{f(A)}\underbrace{P(D|B,C)}_{f(D)}. \end{align} $$

We can see that the probability factorizes into a product of functions of $A$ and $D$, hence $A$ and $D$ are independent given $B$ and $C$.

Question 2 Is this correct? Would it be correct to just say that we have no colliders in the conditioning set, hence independence?

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

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

Checking $B \perp\kern-5pt\perp C | D,A $: we can see that if we remove all edges connected to $A$ and $D$, then there are no paths from $B$ to $C$. Similar reasoning to deduce that $D \perp\kern-5pt\perp A | B,C $.

Question 3: Is this a correct way to approach this problem?

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Your Question 1:
I don't think your first reasoning is correct, because you would still have to show that $p(D|B, C)$ cannot be factored into $f(B)g(C)$ for some functions $f,g$.
Your second reasoning is correct: Since $D$ is a collider on the path $B\to D \leftarrow C$, conditioning on $D$ results in a d-separation open path from $B$ to $C$, so they can be dependent.

Your Question 2:
This is correct. The fact that you don't have to sum over $B$ and $C$, because they are conditioned on, creates the factorization we require. And the graph-based explanation is also correct: both paths $A\leftarrow B\to D$ and $A\to C\to D$ are (d-separation) blocked by conditioning on $B, C$ (because they are no colliders).

Your Question 3:
This is correct.

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