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Pardon me for the newbie question, I'm new in Bayesian network.

I'm reading Chapter 10, Directed Graphical Models (Bayes nets), of Kevin Murphy's textbook: Machine Learning, a Probabilistic Perspective.

Some discussion about using Bayes ball algorithm to test if d-separation holds between two nodes X, Y (or two sets of nodes X, Y) is not clear to me.

For example, the first figure below indicates that node X and node Z are not d-separated.

However, the rule corresponding to the boundary case (second figure below) suggests that the path between nodes x and z is blocked.

enter image description here

So, my question is, if X in the first figure is observed, are Y and Z d-separated then (since the case will be reduced to the second figure)?

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The answer is NO. If $Y$ is not observed, $X$ and $Z$ are not d-separated. The ball can simply pass through $Y$ and reach to $Z$.

Note that the second figure does not mean that the path between nodes $x$ and $z$ is blocked. It means that if $z$ is an unobserved leaf, the ball cannot pass through $z$, that is the ball cannot bounce back to $x$. It is defined to ensure that if none of the descendants of a v-structure is observed, the ball does not bounce back.

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  • $\begingroup$ Thanks for the explanation @hossein! I think what I don't fully understand is, if I focus on Y and Z nodes in the third figure above, I don't see any Bayes ball rules that tell me how to determine d-separation in that situation, although intuition tells me that Y and Z are not d-separated. $\endgroup$ – cwl Apr 2 '17 at 15:34
  • $\begingroup$ When you have $Y \rightarrow Z$, the nodes $Y$ and $Z$ cannot be d-separated at all. $\endgroup$ – Hossein Apr 2 '17 at 15:54

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