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Given the above Bayesian Network ($A,B,C,D$ are events), how can I prove the following equality?

$$ \begin{align} P(D|A) &= P(D|B \cap C)P(B|A)P(C|A)+ \\ &\ \ \ \ \ \ P(D|B \cap C^c)P(B|A)P(C^c|A)+ \\ &\ \ \ \ \ \ P(D|B^c \cap C)P(B^c|A)P(C|A)+ \\ &\ \ \ \ \ \ P(D|B^c \cap C^c)P(B^c|A)P(C^c|A) \end{align} $$


(Add) I have no idea about using $$\def\P{\mathop{\sf P}}\P(A,B,C,D)=\P(D\mid B,C)\P(B\mid A)\P(C\mid A)\P(A)$$

I can't progress over this:

$$P(D|A) = \frac{P(D \cap A)}{P(A)}$$ $$ \begin{align} P(D \cap A) &= P(B \cap C)P(D \cap A|B,C) +\\ & P(B \cap C^c)P(D \cap A|B,C^c) +\\ & P(B^c \cap C)P(D \cap A|B^c,C) +\\ & P(B^c \cap C^c)P(D \cap A|B^c,C^c) \\ \end{align} $$

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how can I prove following equality?

The Factorisation encoded by that DAG is: $$\def\P{\mathop{\sf P}}\P(A,B,C,D)=\P(D\mid B,C)\P(B\mid A)\P(C\mid A)\P(A)$$

And likewise for the terms involving complements.

Begin with this.

Also use the Law of Total Probability and the Definition for Conditional Probability.

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  • $\begingroup$ I have add more question @Graham Kemp $\endgroup$ – firia2000 Sep 16 at 7:13
  • $\begingroup$ Try the Law of Total Probability: $\mathsf P(A,D)={\mathsf P(A,B,C,D)+\mathsf P(A,B,C^{\small\complement},D)+\mathsf P(A,B^{\small\complement},C,D)+\mathsf P(A,B^{\small\complement},C^{\small\complement},D)}$, @firia2000 . $\endgroup$ – Graham Kemp Sep 16 at 8:43

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