# Do-calculus example from PGM Book by Daphne Koller

Consider the last line of this example.

The relevant DAG is shown 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

But the last line claims otherwise. Why?

• 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? Sep 1, 2020 at 13:53
• 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/… Sep 1, 2020 at 14:20
• 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. Sep 1, 2020 at 15:15