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Rule 2: Conditional Equivalence of do(x)do(x) and see(x)see(x)

If a set Z of variables blocks all back-door paths from XX to YY, then conditional on ZZ, do(X)do(X) is equivalent to see(X)see(X). We can, therefore, write P(Ydo(X),Z)=P(YX,Z)P(Y | do(X), Z) = P(Y | X, Z) if ZZ satisfies the back-door criterion.

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Updated 2020-06-20

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