Consider the last line of this example.

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The relevant DAG is shown here.

enter image description here

Clearly, $G$ is d-separated from $\hat{S}$ given S, J because the path $G-J-S-\hat{S}$ is blocked by S since arrows meet at S in head-to-tail fashion. So $P(G|do(S),J)=P(G|S,J)$ because of the rule below

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But the last line claims otherwise. Why?

  • $\begingroup$ Based on the language in the text, it appears that the author is assuming a direct connection between $I$ and $S:$ $I\to S.$ You do not have that edge in your DAG. Is there another copy of the DAG, perhaps, that has that edge? $\endgroup$ Sep 1, 2020 at 13:53
  • $\begingroup$ Yes the original DAG has I-S edge on Pg 1015 but the author herself tells us that to reason about P(G|do(S),J) you have to construct a new graph. The book is freely available here: djsaunde.github.io/read/books/pdfs/… $\endgroup$
    – user_1_1_1
    Sep 1, 2020 at 14:20
  • 1
    $\begingroup$ Yes, no doubt. But the graph corresponding to $P(G|\operatorname{do}(S),J),$ which does indeed have (actually all) inputs to $S$ deleted, will not be equivalent to the graph corresponding to $P(G|S,J),$ which leaves the inputs intact. Removing the do operator is only valid if there is no backdoor path from $G$ to $S.$ If you condition on $I,$ then the backdoor path is blocked. If you don't condition on $I,$ the backdoor path is still there in the un-mutilated graph. $\endgroup$ Sep 1, 2020 at 15:15


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