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I am reading McElreath's Statistical Rethinking book and I'm wondering if anyone can help clarify my doubts on confronting confounding with DAGs. I'll specify 2 examples:

  1. Taken from link. The DAG below is an example on p.186 of the book. $U$ is an unobserved variable. To find the causal influence of $X$ on $Y$, we find all paths from $X$ to $Y$, classify them as open/closed, and do conditioning if necessary. The path with the node $B$ has a collider so we want to exclude it. The paths that we care about are then given by $X \leftarrow U \leftarrow A \rightarrow C \rightarrow Y$ and $X \rightarrow Y$. Am I therefore correct in saying that to estimate the total causal influence of $X$ on $Y$, I regress $Y$ on the variables $X,A,C$?

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  1. Taken from link. The DAG below is from Problem 6M3 on p.189. Here, the two non-direct paths from $X$ to $Y$ contain $Z$ which is a collider. In this case, how do we then estimate the total causal influence of $X$ on $Y$? Including $Z$ as predictor in the regression would confuse the model since it's a collider, i.e. all coefficients of the model except that for $Z$ would be conditioned on $Z$ (and some other variables). Do we just regress $Y$ on $X$?

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Thanks for reading and I hope to get some clarifications!

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For example 1, one of $(A, C)$ would suffice to close the open path, but controlling for both is also fine. In example 2 there are no open paths, so there is nothing to control for (and indeed controlling for $Z$ would be wrong), just as you say.

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  • $\begingroup$ Thanks for the response but I'm confused by "control". For 1), I know that it suffices to condition on $C$ or $A$ only. So does that mean you only regress $Y$ on $C,X$? For 2) When you say "there is nothing to control", does that mean we just regress $Y$ on $X$? $\endgroup$ Aug 7 at 14:35
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    $\begingroup$ In a linear regression setting, "controlling for" or "conditioning on" is completely equivalent to "including on the RHS of the model". I would also add that in Example 2 it is necessary NOT to condition on Z - I would state it stronger. $\endgroup$ Aug 7 at 15:07
  • $\begingroup$ @AdrianKeister I included the point on how controlling for $Z$ in example 2 would be wrong. Thanks for pointing this out! $\endgroup$
    – Scriddie
    Sep 14 at 18:19

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