Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 0 down vote accepted

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

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

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

share|improve this answer
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

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

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.