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.
Prove that P(A|B,C)=P(A|C,B) after applying the Bayesian updating process
1
$\begingroup$
probability
probability-theory
2 Answers
1
We know that $$P(A|B,C) = \frac{P(A \cap B \cap C)}{P(B \cap C)}$$
$$ \ \ = \frac{P(A \cap C \cap B)}{P(C \cap B)}$$
$$= P(A|C,B)$$
-
0Thanks. What is the difference between P(A|B,C) and P(A|B&C)? That was what confused me. The comma verses the &. – 2011-04-08
-
0They are the same thing. – 2011-04-08
2
$$ P(A|B,C)=P(A|B |C)=\frac{P(A,B,C)}{P(B).P(C)}= \frac{P(A,C,B)}{P(C).P(B)}= P(A|C |B)= P(A|C,B)$$