Bayes' rule gives us the following equation:
$$P(X|Y) = \frac{P(X,Y)}{P(Y)} = \frac{1}{Z}P(X)P(Y|X)$$
where Z is a normalizer.
Now I saw that the following two equations holds according to Bayes' rule:
$$P(A|B,C,D) = \frac{1}{Z} P(B,C|A,D)P(A,D)$$
$$P(A|B,C) = \frac{P(B|A,C)P(A|C)}{P(B|C)}$$
Why is this true according to Bayes' rule? In opposite to Bayes' rule we condition on 2 (B,C,D) and 3 (B,C) variables and it is not clear to me how in this case Bayes' rule works.
Best Answer
The two equations can obtained by applying several times Bayes' rule and grouping the event : \begin{align*} \mathbb P[A|B,C,D] &= \frac{\mathbb P[A,(B,C,D)]}{\mathbb P[B,C,D]}\\ &=\frac{\mathbb P[A,B,C,D]}{\mathbb P[B,C,D]}\\ &=\frac{\mathbb P[(B,C),(A,D)]}{\mathbb P[B,C,D]}\\ &=\frac{\mathbb P[B,C|A,D]\mathbb P[A,D]}{\mathbb P[B,C,D]} \end{align*} and \begin{align*} \mathbb P[A|B,C] &= \frac{\mathbb P[A,(B,C)]}{\mathbb P[B,C]}\\ &=\frac{\mathbb P[B,(A,C)]}{\mathbb P[B,C]}\\ &=\frac{\mathbb P[B|A,C]\mathbb P[A,C]}{\mathbb P[B|C]\mathbb P[C]}\\ &=\frac{\mathbb P[B|A,C]\mathbb P[A|C]\mathbb P[C]}{\mathbb P[B|C]\mathbb P[C]}\\ &=\frac{\mathbb P[B|A,C]\mathbb P[A|C]}{\mathbb P[B|C]}\\ \end{align*}