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This seems intuitively obvious to me, yet I can't seem to convince myself algebraically that it is.

Let $B,C>0$

Given $P(X)=P(B>C)$

Is it true that $P(X) \leq P(B \geq C)$ ? If so, what is the easiest was to demonstrate it ?

PS this is not homework.

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2 Answers 2

up vote 3 down vote accepted

Yes, it is: $P(B\geq C) = P(B>C)+P(B = C) = P(X)+P(B=C)\geq P(X)$.

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Thanks. So simple ! –  Joe King May 29 '12 at 11:23

I would add that the sum can be taken because the events (B>C} and {B=C} are mutually exclusive. Also it is implicit that B and C are random variables that are positive. If they were constants then P(B>=C) is either 0 or 1. If it is 0 then P(B>C)=0. If it is 1 it could be because B=C in which case P(B=C)=1 and P(B>C)=0 or B>C in which case P(B>C)=1 and P(B=C)=0. So in this trivial case P(B>=C)>=P(B>C)=P(X).

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