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Proving probability inequality

  1. May 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove the following
    a>0, X is a non-negative function

    [itex]Ʃ_{n\in N} P(X>an)≥\frac{1}{a}(E[X]-a)[/itex]

    [itex]Ʃ_{n\in N} P(X>an)≤\frac{E[X]}{a}[/itex]

    3. The attempt at a solution

    I know that
    [itex]\sum_{n\in N} P(X>an)=\sum_{k \in N} kP((k+1)a≥X>ka)=\sum_{k \in N} E[k1_{[(k+1)a,ka)}(X)][/itex]
    where [itex]1_{[(k+1)a,ka)}[/itex] is the indicator function.

    I'm not sure how to preceed from here.

    All I know is that for all ε>0, [itex]E[ε1_{[ε,∞)}(X)]≤E[X][/itex]
     
    Last edited: May 8, 2013
  2. jcsd
  3. May 8, 2013 #2

    haruspex

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    The second is not true. Is there a factor 1/a missing on the right?
     
  4. May 8, 2013 #3
    Sorry, my bad... I've edited it...
     
  5. May 8, 2013 #4

    haruspex

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    You can eliminate a by defining Y = X/a. Does that help?
     
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