I have given the following Bayesian network and need to evaluate whether q is independent of s given r.

I know that the path over the node r is blocked because r is observed. But what about the other direct path? Probabilistically it states P(s|q), therefore q is directly dependent on s. But does this also apply to conditional independence?

I do not have any probabilities given and need to solve the task using D-Separation. The problem is when we want to check if two variables are d-separated, we need to check whether the path is blocked. In the case of the lower path that includes r it is, because r is observed and we have a head-to-tail connection. But what case do we have for the other path? Here we do not have any nodes in between.

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  • $\begingroup$ you probably have to check whether P(q|c)*P(s|c) = P(q and s|c) $\endgroup$
    – Alberto Sinigaglia
    Dec 17, 2021 at 11:08
  • $\begingroup$ @AlbertoSinigaglia Only the model was given. I do not have the exact probabilities and need to solve it with D-separation. (I will make the question more clear) $\endgroup$
    – Marcello Zago
    Dec 17, 2021 at 11:21

1 Answer 1


$$ \eqalign{ \textrm{Conditional independence} &\Leftrightarrow p(q,s|r) = p(q|r)p(s|r)\cr p(q,s|r) &= p(s|q,r)p(q|r)\cr p(s|q,r) &\ne p(s|r)\cr }$$ (because observing $q$ and $r$ would tell us more about the distribution of $s$ than observing $r$ alone) $$ \eqalign{ \Rightarrow p(s|q,r)p(q|r) &\ne p(q|r)p(s|r) \cr \Rightarrow p(q,s|r) &\ne p(q|r)p(s|r) }$$ $\Rightarrow$ $q$ and $s$ are not conditionally independent given $r$, I believe.

I have used only basic probability instead of D-separation in this answer, but I think it's important that you understand the basic probability before you start trying to understand D-separation.


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