I am studying Bayesian Networks using the Neapolitan book (Learning Bayesian Networks). In section 1.3.2 it is stated the following:

Definition 1.9 Suppose we have a joint probability distribution $P$ of the random variables in some set $V$ and a DAG $G = (V, E)$. We say that $(G, P )$ satisfies the Markov condition if for each variable $X \in V$, ${X}$ is conditionally independent of the set of all its non-descendants given the set of all its parents. Using the notation established in Section 1.1.4, this means if we denote the sets of parents and non-descendants of $X$ by $PA_X$ and $ND_X$ respectively, then $I_P ({X}, ND_X |PA_X )$.

Then he proceeds to give some examples and he gives four simple cases:

  1. $V \gets C \to S $
  2. $V \to C \to S$
  3. $V \gets C \gets S$
  4. $V \to C \gets S$

i.e. the final item is the collider. When talking about the examples, the author says:

$(G, P)$ satisfies the Markov condition if $G$ is the DAG in Figure 1.3 (1), (2), or (3). However, $(G, P)$ does not satisfy the Markov condition if $G$ is the DAG in Figure 1.3 (4) because $I_P ({V}, {S})$ is not the case."

And here I could not understand. He is stating that in the collider, the parents are not independent. However, to my knowledge, the parents in the collider are indeed independent and are d-connected conditional to the common descendent. It is clear that $p(v,c,s) = p(c|v,s)p(v)p(s)$. This implies that $V$ and $S$ are in fact independent. So, my question is: why the graph in (4) does not obey the Markov condition as said by the author?

  • 2
    $\begingroup$ Does this answer your question? Graphical dependence in the DAG X->Z<-Y $\endgroup$ Apr 9, 2021 at 0:13
  • $\begingroup$ Partially. I am updating my question to be more specific. $\endgroup$
    – donut
    Apr 9, 2021 at 2:01
  • $\begingroup$ It sounds like a mixup about the difference between 'independent' and 'conditionally independent'. $\endgroup$ Apr 9, 2021 at 13:35


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