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→实例4 - 辛普森悖论
<math>
<math>
\begin{align}
\begin{array}{lcl}
P(Y=1|do(X=0)) = \sum_Z P(Y=1|X=0, Z=z)P(Z) \\
P(Y=1|do(X=0)) = \sum_Z P(Y=1|X=0, Z=z)P(Z) \\
= P(Y=1|X=0, Z=1)P(Z=1) + P(Y=1|X=0, Z=0)P(Z=0) \\
= P(Y=1|X=0, Z=1)P(Z=1) + P(Y=1|X=0, Z=0)P(Z=0) \\