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)?


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.