# How to prove if P(A|B)>P(A) then P(B|A)>P(B) [closed]

How to prove that If P(A|B)>P(A) then P(B|A)>P(B)

• Do you know the definition of conditional probability: $P(A\,|\,B)={P(A \cap B)\over P(B)}$? – David Mitra Jul 11 '13 at 12:45
• @DavidMitra Shouldn't that be $P(A \cap B)$? – Andrew D Jul 11 '13 at 12:48
• @AndrewD Of course, thanks. – David Mitra Jul 11 '13 at 12:48

$$P(A|B)>P(A)$$
$$\frac{P(A \wedge B)}{P(B)}>P(A)$$ $$\frac{P(B|A)P(A)}{P(B)}>P(A)$$ Now, since $P(A)$ and $P(B)$ are positive. $$\frac{P(B|A)P(A)}{P(A)}>P(B)$$ $$P(B|A)>P(B)$$ $$\square$$ It should be noted this works with all other comparison operators as well.
• It used the definition of conditional probability. $P(B|A)=P(A \land B)÷P(A)$ – PyRulez Oct 3 '17 at 15:52