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This is the third question I am asking about these notes http://www.stat.cmu.edu/~larry/=sml/DAGs.pdf .This time it is about the proof of a small theorem (page 426), that I report:

Theorem: Let $G$ be a DAG and P a distribution that is faithful to $G.$ If $X_i$ and $X_j$ are adjacent in $G,$ than the conditional independence test $X_i \mathrel{\unicode{x2AEB}} X_j |A$ fails for $A \in V-\{i,j\}$. On the other hand, if $X_i$ and $X_j$ are not adjacent in $G,$ then either $X_i \mathrel{\unicode{x2AEB}} X_j |\pi(X_j)$ or $X_i \mathrel{\unicode{x2AEB}} X_j |\pi(X_i)$, where $\pi(X_i)$,$\pi(X_j)$ are the parent sets of $X_i$ and $X_j$ in the DAG $G$.

Proof: For the first part, [...]. For the second part we consider two cases. (i) $X_j$ is a descendant of $X_i$ and (ii) $X_j$ is not a descendant of $X_i$. By definition of $d$-separation in the first case we can show that $X_i \mathrel{\unicode{x2AEB}} X_j |\pi(X_j)$ and in the second case that $X_i \mathrel{\unicode{x2AEB}} X_j |\pi(X_i)$.

But I had a few doubts. First of all in the first case (i) I understand that the path from $X_i$ to $X_j$ making $X_j$ a descendant of $X_i$ is blocked when we observe $\pi(X_j)$ but why can't we have other paths that are open?

For (ii), In the DAG below:

enter image description here

$X_j$ is not a descendant of $X_i$ but $X_i$ and $X_j$ are not d-separated given $\pi(X_i)$, because the only collider (in the only path between them) is observed. Isn't it a counterexample to the reasoning or am I applying $d$-separation in a wrong way ?

(note I am not saying that the theorem is wrong, just that I do not understand the proof ;) )

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Having a collider's vertex observed or measured does not mean that causal information can flow through it. What makes causal information flow through a collider is when you condition on the collider. And note that this is exactly the opposite behavior from chains and forks. In summary: $$\begin{array}{cccc} \text{Label} &\text{Diagram} &B \text{ Conditioned?} &\text{Causal Information flow-through?} \\ \hline \text{Chain} &A\to B\to C &\text{No} &\text{Yes} \\ & &\text{Yes} &\text{No} \\ \hline \text{Fork} &A\leftarrow B\to C &\text{No} &\text{Yes} \\ & &\text{Yes} &\text{No} \\ \hline \text{Collider} &A\to B\leftarrow C &\text{No} &\text{No} \\ & &\text{Yes} &\text{Yes} \\ \end{array}$$ To condition on a variable, you can do all sorts of things such as backdoor adjustment, frontdoor adjustment, instrumental variable, stratified analysis, including the variable in the RHS of a linear regression model, and probably a few others.

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  • $\begingroup$ Ok so you distinguish the concepts of "conditioning on a variable" and of "observing the variable". I use them to indicate similar things since when we "observe" something is like taking the joint and "conditioning" on the observed value. Or not? $\endgroup$
    – Thomas
    Dec 9, 2021 at 20:02
  • $\begingroup$ Anyway in the dag that I described we cannot deduce, according to your rules, that $X_i \mathrel{\unicode{x2AEB}} X_j |\pi(X_i)$, as the proof claims, since the only collider is a parent of $X_i$, on which we are conditioning. Therefore we are on the "Yes" side and casual information can flow. Is there some mistake in this reasoning ? $\endgroup$
    – Thomas
    Dec 9, 2021 at 20:03
  • $\begingroup$ I would not consider that standard usage. By "to observe" or "to measure" a variable, I mean simply that we have data for it. In a data table, there would be a column of values for that variable. This is to distinguish between measured and unmeasured variables - the latter meaning that we do not (often cannot) have any values in a table for that variable. In your DAG, the collider is not a parent of $X_i,$ but a descendant of it. If you condition on the collider or any of its descendants, you open up the collider. However, conditioning on the fork would close up the fork. $\endgroup$ Dec 9, 2021 at 20:28
  • $\begingroup$ The only parent of $X_j$ in your DAG is the fork immediately to its left. If you condition on that, then $X_i$ and $X_j$ are $d$-separated, regardless of what happens at the collider. $\endgroup$ Dec 9, 2021 at 20:30
  • $\begingroup$ The article claims that in case (b) we are d-separated conditioning on $X_i$ (see text) not $X_j$. Something is wrong there and I am not sure it is just a typo (or maybe yes?)... their proof looks quite done by intimidation... $\endgroup$
    – Thomas
    Dec 9, 2021 at 20:36
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I will try to post a complete solution showing that if $X_i$ and $X_j$ are not adjacent, than $X_i$ and $X_j$ are independent given either $\pi(X_j)$ or $\pi(X_i)$. As Adrian Keister noticed, my counterexample for point (ii) is wrong since I had mistakenly taken a child for a father...

Case when $X_j$ is a descendant of $X_i$.

From the fact that $X_j$ is a descendant of $X_i$ we know that there is a directed path from $X_i$ to $X_j$. This is blocked conditioning on $\pi(X_j)$. Now take an other path $p$: we have to show that it is blocked as well. If the path $p$ reaches $X_j$ pointing outwards than, since we cannot have a directed cycle, we will have a collider along $p$. Take the first collider that one meets going from $X_j$ to $X_i$ and call it $v$. $v$ cannot be a father of $X_j$ (again we would have a directed cycle) and is therefore not conditioned when we condition on $\pi(X_j)$, therefore the path is blocked. If the path $p$ reaches $X_j$ pointing inwards, than we have a father of $X_j $ at the last previous-to-last vertex along the path, which is in a "chain" or "fork" configuration, then conditioning on it the path is blocked also in this configuration.

Case when $X_j$ is not a descendant of $X_i$.

We consider a generic path from $X_i$ to $X_j$ and want to check that it is blocked conditioning on $\pi(X_i)$. If the path starts outwards from $X_i$, since $X_j$ is not a descendant we must have a collider in the path. Take the first one. This one cannot be a father of $X_i$ (because otherwise we would have a directed cycle). Therefore the path is blocked. Instead, if the path starts with an arrow towards $X_i$, the second element of the path counting from $X_i$ is a father of $X_i$ in either a chain or fork configuration, which is therefore blocked after conditioning.

These solutions are much clearer with some drawing :). The hypothesis that $X_i$ and $X_j$ are not ajacent is used implicitly so that each path is at least of length 2. So I guess all doubts solved. Thanks again to Adrian Keister for pointing out my mistake.

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