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Expected values of a limiter

  1. Oct 22, 2011 #1
    For the limiter shown below, find the expected value for Y = g(X)
    dh6b6s.png

    attempt at solution:
    E[Y] = ∫g(x)f(x)dx, where f(x) is the probability density function with respect to x
    so...
    E[Y] = E[Y1] + E[Y2] + E[Y3]
    E[Y1] = ∫-af(x)dx where the limits of integration are from -∞ to -a
    so E[Y1] = -aFx(-a), where Fx is the cumulative distribution function
    E[Y2] = ∫xf(x)dx where the limits of integration are -a to a
    this is the part i'm having trouble with
    E[Y3] = ∫af(x)dx where limits of integration are a to ∞
    so E[Y3] = a(1-Fx(a))

    i'm having trouble simplifying E[Y2] into one expression because of the limits of integration. i know it would be just E[Y2] if the limits were -∞ to ∞, but that is not the case here
     
  2. jcsd
  3. Oct 23, 2011 #2
    I think that when the domain of integration is limited over a certain set A, the expected value becomes the conditional expected value of X given that X lies on A, times the probability that X lies on A.
     
  4. Oct 23, 2011 #3

    Ray Vickson

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    Unless you have a specific form for f(x) you have gone as far as you can: E(Y2) does not simplify further in any useful way. By the way: I would argue against your notation E(Y1), E(Y2), etc. You do not have three separate random variables Y1, Y2 and Y3; you just have three separate "pieces" of the same random variable Y.

    RGV
     
  5. Oct 23, 2011 #4
    hmm..ok
     
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