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Notation for Expectation?

  1. Oct 17, 2013 #1
    How do I read, interpret the following definitions for the expectation of a random variable X?
    Assume the integral is over the entire relevant space for X.

    (1) E(X) = ∫ x dP
    (2) E(X) = ∫ x dF(X)

    If I asked you to calculate (1) or (2) for an arbitrary X, how does it look?
    My only other understanding of E(X) is to do pdf times x, integrate, plug in bounds, but that's assuming X is nice enough to have such a pdf. I appreciate any replies!
     
  2. jcsd
  3. Oct 17, 2013 #2

    Simon Bridge

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    http://en.wikipedia.org/wiki/Expected_value
    ... there are plenty of examples of expectation value calculations online - have you had a look?

    Note: none of your notations look right.
    The first looks looks like it's trying to be the Lebesgue Integral definition - which is more general, the second looks like it is trying to be the regular student definition while the second

    i.e.

    1. $$E[X]=\int_{\Omega}XdP$$

    2. $$E[X]=\int_{-\infty}^\infty xp(x)dx$$
     
  4. Oct 17, 2013 #3
    Okay I see. Would this be the other definition:

    E(X) = ∫ X dF(X) = ∫ x f(x) dx, where dF(X) = (dF/dx)dx = f(x)dx. But this is assuming dF/dx exists, or perhaps I should say this is assuming dF/dx is useful. What if F is a singular continuous distribution?
     
  5. Oct 17, 2013 #4

    mathman

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    E(X) = ∫xdF(x) is the most general form. When dF/dx exists, you can use it.
     
  6. Oct 17, 2013 #5

    Simon Bridge

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    i.e. dF/dx does not exist?
    Then you cannot use that definition.

    Does "expectation value" make sense in the absence of a probability density function?
     
  7. Oct 17, 2013 #6

    jbunniii

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    This is a Stieltjes integral. This integral is meaningful for any probability distribution, even when the cdf ##F## is not differentiable.

    In the case where ##F## is differentiable and the density function ##p(x) = dF/dx## integrates back to ##F(x)##, it is equivalent to ##E(X) = \int x p(x) dx##.

    In the case where ##F## is a "staircase" (the probability is all concentrated at discrete points), it is equivalent to ##E(X) = \sum_n a_n P(X = a_n)##, where ##a_n## are the x-coordinates of the jumps.
     
    Last edited: Oct 17, 2013
  8. Oct 17, 2013 #7
    How about for a d.f. F that is singularly continuous (neither absolutely continuous nor discrete)?
     
  9. Oct 17, 2013 #8

    jbunniii

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    ##E(X) = \int x dF(x)## is still meaningful in that case, but it may not be possible to write it in terms of a standard Riemann integral or a sum as in the two cases I noted above. Sometimes it can be written as a Riemann integral plus a sum, however (so called "mixed" continuous/discrete distribution). Do you have a specific distribution in mind?
     
  10. Oct 18, 2013 #9
    Yes, the Cantor distribution! Funny that you ask, I'm actually trying to work out a few problems based on the Cantor distribution right now. Wikipedia gives the Cantor distribution's expectation as 1/2 based on a symmetry argument. I was wondering if there is any other way to calculate it. I'm curious to know, because I am also asked to find ∫(x^2)dF(X) for the Cantor distribution, and I'm pretty stuck.
     
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