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Relating the CDF to the probability density

  1. Aug 30, 2014 #1
    1. The problem statement, all variables and given/known data
    If ##X## is any random variable defined on ##[0,\infty]## with continuous CDF ##F_X(t)##. Prove that ##E(X)=\int_1^\infty (1-F_X(t)) dt##.
    .
    2. Relevant equations



    3. The attempt at a solution
    I am not sure how to go about this. I think double integration can be used to prove it, but I don't want to go down that path. Is there another which I can prove this. Thanks.
     
    Last edited by a moderator: Sep 1, 2014
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  3. Aug 30, 2014 #2

    haruspex

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    Are you sure about those limits? I don't see how E(X) can be independent of the behaviour of F between 0 and 1.
     
  4. Aug 30, 2014 #3
    Well it says that ##X## is a random variable with range ##[0, \infty]##.
     
  5. Aug 30, 2014 #4

    Ray Vickson

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    Somebody has asked you to prove a false result. The correct version is
    [tex] EX = \int_0^{\infty} [1 - F_X(x)] \,dx .[/tex]

    Exactly how you should prove it depends on what you know about Lebesgue-Stieltjes integrals, or whether you even need to use them. The proof is easiest if you assume that ##F_X(\cdot)## is absolutely continuous, in which case there is a density ##f_X(\cdot)## on ##(0,\infty)## giving ##F_X(x) = \int_0^x f_X(t) \, dt## for all ##x > 0##. If ##F_X## is not absolutely continuous, it can be continuous but not have a density function; then the proof is harder.

    I suggest you try it first for the case where a density function exists. Think integration by parts.
     
    Last edited: Aug 30, 2014
  6. Aug 30, 2014 #5

    pasmith

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    I agree; it should be [itex]E(X) = \int_0^\infty 1 - F_X(t)\,dt[/itex].

    This is an application of integration by parts. You want to calculate [tex]
    E(X) = \int_0^\infty xF_X'(x) \,dx.[/tex] Note that [tex]
    F_X'(x) = -\frac{d}{dx}(1 - F_X(x)).
    [/tex]
     
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