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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: $P(A\,|\,B)={P(A \cap B)\over P(B)}$? $\endgroup$ – David Mitra Jul 11 '13 at 12:45
  • $\begingroup$ @DavidMitra Shouldn't that be $P(A \cap B)$? $\endgroup$ – Andrew D Jul 11 '13 at 12:48
  • $\begingroup$ @AndrewD Of course, thanks. $\endgroup$ – David Mitra Jul 11 '13 at 12:48
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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 '17 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 '17 at 15:52

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