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

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    $\begingroup$ Do you know the definition of conditional probability: $P(A\,|\,B)={P(A \cap B)\over P(B)}$? $\endgroup$ Jul 11, 2013 at 12:45
  • $\begingroup$ @DavidMitra Shouldn't that be $P(A \cap B)$? $\endgroup$
    – Andrew D
    Jul 11, 2013 at 12:48
  • $\begingroup$ @AndrewD Of course, thanks. $\endgroup$ Jul 11, 2013 at 12:48

1 Answer 1

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$$P(A|B)>P(A)$$

$$\frac{P(A \wedge B)}{P(B)}>P(A)$$ $$\frac{P(B|A)P(A)}{P(B)}>P(A)$$ 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.

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  • $\begingroup$ Does the third step use Bayes' rule? $\endgroup$
    – Paradox
    Oct 3, 2017 at 15:47
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    $\begingroup$ It used the definition of conditional probability. $P(B|A)=P(A \land B)÷P(A)$ $\endgroup$
    – PyRulez
    Oct 3, 2017 at 15:52

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