Expectation Inequality for Positive Random Variables

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SUMMARY

The discussion centers on the proof of the inequality E(X) > a.P(X > a) for positive random variables. Participants clarify that E(X) represents the expectation of a non-negative random variable X, and a is a positive constant. The conversation highlights attempts to prove the statement using specific distributions, such as the exponential and normal distributions, while addressing the need for a general proof. The consensus is that the inequality should be interpreted as E(X) ≥ a.P(X > a) due to the properties of non-negative variables.

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Homework Statement


Prove that
E(X) > a.P(X>a)


Homework Equations


E(X) is expectation, a is a positive constant and X is the random variable.
(Note, > should be 'greater than or equal to' but I'm not too sure how to do it)


The Attempt at a Solution


Well I can show it easy enough for X~Exp(h), but of course this is not a general proof.
And I played around a bit with P(X>a)=1-F(a) where F is the cdf, but yeah. That's all I could really do. I just don't really get how the a factor comes in.
 
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Take P(x) to be a normal distribution centered at 0. The E(x)=0. Yet a.P(x>a) is clearly positive for a>0. There's something wrong with the problem statement. '>=' is fine for greater than or equal to.
 
Dick said:
Take P(x) to be a normal distribution centered at 0. The E(x)=0. Yet a.P(x>a) is clearly positive for a>0. There's something wrong with the problem statement. '>=' is fine for greater than or equal to.

Sorry, X is non-negative.
 
Ok then. Define E(X) as an integral and split the integral into the ranges 0-a and a-infinity. Drop the first integral and think about approximating the second by something smaller.
 

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