# PC algorithm in "Causation, Prediction, and Search"

In the page 120 in "Causation, Prediction, and Search", why the edge B-D will not be removed if the edge E-D is mistakenly removed from the initial complete graph. I think B-D is D-connected given C since C is a collider in the path B-C-E-D.

• Your last sentence doesn't appear to be complete? Commented Dec 1, 2023 at 14:19
• The missing sentence is "If an edge is mistakenly left in the graph and there are no additional errors in the list of d-separations in the input, the only further errors that result are that some edges which theoretically could be oriented, will not be oriented."
– Nick
Commented Dec 2, 2023 at 12:39
• I'm probably missing something obvious, but I don't see an edge $B-D$ in the graph? Commented Dec 3, 2023 at 17:00
• The graph is complete graph, which means all the points is connected to the other. In this stage, we want to remove some faluse edges according to the conditional independence test. The question in this page is that edge B-D will not be removed if the edge E-D is mistakenly removed from the initial complete graph. I think B-D is D-connected given C since C is a collider in the path B-C-E-D, then B-D will not be removed.
– Nick
Commented Dec 5, 2023 at 0:42
• I see, so Figure 3 is the true graph. Thanks for the explanation :) Commented Dec 5, 2023 at 12:32

Once, starting from the fully connected graph, the edge $$E-D$$ is mistakenly removed, it would seem that controlling for every subset of $$\{A, C\}$$ is sufficient to see if $$B$$ and $$D$$ are conditionally independent, since there can be no open backdoor path between $$B$$ and $$D$$ that includes $$E$$ if $$E$$ and $$D$$ are independent.
I reality and just as you are saying, this is not the case because conditioning on $$C$$ closes the path $$B-C-D$$ but opens the path $$B-C-E-D$$, so the edge $$B-D$$ would not be removed although it should be.