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I'm reading a documentation on causal inference on graph and I'm currently on the chapter about identification. In this section, the authors give examples of invalid adjustment sets, one of which is as in Figure 4a below, followed by a description of why conditioning on a collider biases the estimate of P(B|do(A)): enter image description here

However, I'm confused about the first sentence of this paragraph. Why is A and B statistically independent of each other/why are they d-separated in this graph? A is shown to be adjacent to B/they are connected by a directed edge. Am I understanding this incorrectly? Is it the case that the authors put this statement in a slightly wrong way?

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Taking the DAGs from left to right:

  1. Left DAG: $A$ and $B$ are d-separated because there are no backdoor paths open between them. Unconditioned, the path $A\leftarrow X_1 \rightarrow Z \leftarrow X_2 \rightarrow B$ is blocked by the collider $Z$. The point they are making in their comment is that conditioning on $Z$ opens the flow of association through $X_1 \rightarrow Z \leftarrow X_2$.

  2. Middle DAG: $A$ and $B$ are d-separated because there are no backdoor paths, and 'front door paths' (i.e. no conditioning on a common descendant of both $A$ and $B$). The point they are making in their comment is that conditioning on $Z$ closes the flow of association from $A \rightarrow Z \rightarrow B$ (conditioning on a non-collider path closes it to the flow of association).

  3. Right DAG: $A$ and $B$ are d-separated because there are no backdoor or 'front door' paths open between them. The point they are making in their comment is that conditioning on $Z$ opens the 'front door' path from $A \rightarrow Z \leftarrow X$ (and thence to $B$ via $B$'s backdoor path).

I think the statement about d-separation is that any contrast measure relating $A$ and $B$ produces an unbiased estimate of the causal effect of $A$ on $B$. Maybe that's a bit sloppy: clearly if $A \rightarrow B$, then $A$ and $B$ are not d-separated. So I think your question "Is it the case that the authors put this statement in a slightly wrong way?" must be answered "Yes."

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  • $\begingroup$ Thanks so much! $\endgroup$ Mar 20, 2023 at 1:26

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