# Thread: How to prove P(A|B)=P(A) implies P(B|A)=P(B) for independent events

1. ## How to prove P(A|B)=P(A) implies P(B|A)=P(B) for independent events

I know that the answer is yes (my professor stated so in class). Yet I'm struggling to prove how it is so. I can prove how P(A|B)=P(A) for independent events A and B, yet I don't know how to use that knowledge to prove P(B|A)=P(B).

Thanks for any help!

2. ## Re: How to prove P(A|B)=P(A) implies P(B|A)=P(B) for independent events

Originally Posted by nyy321
You know that \begin{align*}\mathscr{P}(B|A)&=\dfrac{\mathscr{P }(A\cap B)}{\mathscr {P}(A)}\\&=\dfrac{ \mathscr{P}(A|B)\mathscr{P}(B)}{\mathscr{P}(A)}\\& =\dfrac{ \mathscr{P}(A)\mathscr{P}(B)}{\mathscr{P}(A)}\\&= \mathscr {P} (B) \end{align*}
$P(A|B) = \dfrac{P(A\cap B)}{P(B)} = P(A)$ implies $P(A\cap B) = P(A)P(B)$
Now $P(B|A) = \dfrac{P(B \cap A)}{P(A)} = \dfrac{P(A \cap B)}{P(A)} = \dfrac{\cancel{P(A)}P(B)}{\cancel{P(A)}} = P(B)$.