# Prove that P(A|B,C)=P(A|C,B) after applying the Bayesian updating process

Prove that P(A|B,C)=P(A|C,B) after applying the Bayesian updating process. That is, prove that the order in which the information is presented does not matter.

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We know that $$P(A|B,C) = \frac{P(A \cap B \cap C)}{P(B \cap C)}$$

$$\ \ = \frac{P(A \cap C \cap B)}{P(C \cap B)}$$

$$= P(A|C,B)$$

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Thanks. What is the difference between P(A|B,C) and P(A|B&C)? That was what confused me. The comma verses the &. –  user7435 Apr 8 '11 at 0:35
They are the same thing. –  PEV Apr 8 '11 at 0:55

$$P(A|B,C)=P(A|B |C)=\frac{P(A,B,C)}{P(B).P(C)}= \frac{P(A,C,B)}{P(C).P(B)}= P(A|C |B)= P(A|C,B)$$

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