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For all discrete probabilities $P$ and all events $X$ and $Y$, prove or disprove:

If $0 < P(X) \leq P(Y)$, then $P(X|Y) \leq P(Y|X)$.

My attempt goes something like this (which I am sure is wrong):

Consider $X$ and $Y$ are independent of each other.
Since $P(Y) \geq P(X)$, $p(Y|X) = 0$ and $P(X|Y) \leq 0$.

Ottavio
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  • Hint: $\Pr(X\mid Y)=\dfrac{\Pr(X\cap Y)}{\Pr(Y)}$ – JMoravitz Dec 07 '20 at 14:25
  • As for your attempt... it is way off. First, you may never assume independence. You will have only proven things about independent events and will have missed all of the infinitely many other scenarios where the events are not independent. Next... you say something about things equaling zero? Rubbish. $X$ and $Y$ independent would have meant that $\Pr(Y\mid X)=\Pr(Y)$ which is explicitly stated in the problem statement as being nonzero. – JMoravitz Dec 07 '20 at 14:26
  • My only guess as to what you were grasping at which relates to zero is in the event that $X$ and $Y$ are mutually exclusive of one another in which case yes, you would have $\Pr(Y\mid X)=0$... Mutual exclusivity is not independence. Do not confuse these terms, they mean completely different things. The same problem however occurs if you were to follow this route, you will have only proven things for mutually exclusive $X,Y$ and missed the infinitely many other examples of events $X,Y$ which are not mutually exclusive. – JMoravitz Dec 07 '20 at 14:31

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