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

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

    Does P(A|B)=P(A) imply P(B|A)=P(B)? Please prove your answer.

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

    Quote Originally Posted by nyy321 View Post
    Does P(A|B)=P(A) imply P(B|A)=P(B)? Please prove your answer.
    You know that $\begin{align*}\mathscr{P}(B|A)&=\dfrac{\mathscr{P }(A\cap B)}{\mathscr {P}(A)}\\&=\dfrac{ \mathscr{P}(A|B)\mathscr{P}(B)}{\mathscr{P}(A)}\\& =\dfrac{ \mathscr{P}(A)\mathscr{P}(B)}{\mathscr{P}(A)}\\&= \mathscr {P} (B) \end{align*}$
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    Re: How to prove P(A|B)=P(A) implies P(B|A)=P(B) for independent events

    $P(A|B) = \dfrac{P(A\cap B)}{P(B)} = P(A)$ implies $P(A\cap B) = P(A)P(B)$

    Now $P(B|A) = \dfrac{P(B \cap A)}{P(A)} = \dfrac{P(A \cap B)}{P(A)} = \dfrac{\cancel{P(A)}P(B)}{\cancel{P(A)}} = P(B)$.
    Q.E.D.
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