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Expectation prob. Proof

  1. Apr 16, 2007 #1
    1. The problem statement, all variables and given/known data
    Prove that
    E(X) > a.P(X>a)

    2. Relevant 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)

    3. 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.
  2. jcsd
  3. Apr 16, 2007 #2


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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.
  4. Apr 16, 2007 #3
    Sorry, X is non-negative.
  5. Apr 17, 2007 #4


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    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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