conditional independence on Bayes graph

Question

I am confusing on conditional independence on Bayes graph.

a graph:

          P6
          ↓
P1 → P3 → P4 → P5
      ↓
     P7

Please kindly let me know if below understanding is correct or not?

(1) P1 and P6 given P5 is NOT conditional independence because:

knowing P5 gave info on P4, so P1 and P6 are related. Does P3 matter?

(2) P7 and P6 given P5 is NOT conditional independence because:

knowing P5 gave info on P4, so P1 and P7 are related. Does P3 matter?

Answer 1

D-separation is short for "directionally dependent separation." The direction of the edges leon the path through the node are important.

We say nodes X and Y are d-separated by conditioning set S, if and only if, on every path between the two nodes, there is some node V where one of the following occurs:

• V is in S, and one edge on path leads into V and one edge in path leads out of V.
• V is in S, and both edges on the path lead out of V.
• Neither V nor any descendant of V is in S, and both edges on path leads into V.

D-separation implies conditional independence.

(1) P1 and P6 given P5 is NOT conditional independence because:

knowing P5 gave info on P4, so P1 and P6 are related. Does P3 matter?

There is only one path between the nodes. On it are P3 and P4.

P3 is not in {P5}, and one edge leads into P3 and one edge leads out. It does not "matter".

P4 is not in {P5}, both edges lead into P4, and its decendent P5 is in {P5}. It d-seperates the path.

(2) P7 and P6 given P5 is NOT conditional independence because:

knowing P5 gave info on P4, so P1 and P7 are related. Does P3 matter?

There is only one path between the nodes. On it is P3.

P3 is not in {P5}, and one edge leads into P3 and one edge leads out. It does not "matter".