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Finding E(X) from distribution function

  1. May 17, 2010 #1
    Theorem: Let F(x) be the distribution function of X.
    If X is any r.v. (discrete, continuous, or mixed) defined on the interval [a,∞) (or some subset of it), then
    E(X)=

    ∫ [1 - F(x)]dx + a
    a

    1) Is this formula true for any real number a? In particular, is it true for a<0?

    2) When is this formula ever useful (computationally)? Why don't just get the density function then integrate to find E(X)?

    Thanks for clarifying!
     
  2. jcsd
  3. May 17, 2010 #2

    Hurkyl

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    1) If there was a restriction on a, the statement of the theorem should have said something. If the statement is right for positive a, then it's surely right for negative a as well.

    2) Why would you go through all the trouble of looking for the density distribution*, multiplying by x, and integrating that when you could just use that formula? :confused:


    *: The hypotheses cover the situation where the density cannot be written as a function
     
  4. May 17, 2010 #3
    The formula for general RV X is [tex]EX = -\int_{-\infty}^{0} F(x) dx + \int_{0}^{\infty} (1 - F(x)) dx[/tex]. This formula works in a much more general setting than you might expect. Some distributions don't have densities (singular distributions), for example http://en.wikipedia.org/wiki/Cantor_distribution but the formula still applies.
     
  5. May 18, 2010 #4
    I've seen this general formula. But does it imply that the "theorem" above is true for a<0 (e.g. a=-2, or a=-2.4) as well? I've done some manipulations and I think the theorem above is true for ANY a, but I would like someone to confirm this.

    Thanks!
     
  6. May 18, 2010 #5
    Try to prove it by manipulating the general formula I posted, it's not hard.
     
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