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Expected values in Probability space

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data

    Let a probability space be [itex](Ω, \epsilon, P)[/itex]. A set of random variables X1,...,Xn

    Give an example where [itex]I_{p}(lim inf_{n -> ∞}X_{n}[/itex]) < [itex]lim inf_{n -> ∞}I_{p}(X_{n})[/itex]

    The attempt at a solution

    I know that [itex]I_{p}(lim inf_{n -> ∞}X_{n}[/itex])=[itex]E[lim inf_{n -> ∞}X_{n}[/itex]]
    and [itex]lim inf_{n -> ∞}I_{p}(X_{n})[/itex] = [itex]lim inf_{n -> ∞}E[X_{n}][/itex]

    I think I need to find a sequence of Xn such that lim inf Xn will have a smaller value than all the individual expected value, E[Xn].

    Am I on the correct path? I'm kind of stuck here and not sure how to proceed.

    Would be really really thankful for the help.
  2. jcsd
  3. Apr 30, 2013 #2


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    How is ##\displaystyle \liminf_{n \to \infty} X_n## defined? The limit of the "smallest value" of those random variables?
    In that case, I don't see how both can be equal apart from trivial Xi.
  4. Apr 30, 2013 #3
    Equality actually holds for surprisingly many cases. See for example, the dominated convergence theorem. So you'll need to look at sequences of random variables which fail that theorem.
    Without giving away too much, try to make a nonconvergent sequence ##X_n## such that ##E[X_n]## are all constant.
  5. Apr 30, 2013 #4


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    Staff: Mentor

    Okay, then I don't understand how to evaluate ##\displaystyle \liminf_{n \to \infty} X_n##.
  6. Apr 30, 2013 #5
    If ##X = \liminf_n X_n##, and ##\omega## is an outcome, then it is defined by
    [tex]X(\omega) = \liminf_n X_n(\omega)[/tex]

    So it's just the pointswise inferior limit.
  7. Apr 30, 2013 #6
    I saw this example....

    P(ω)=2, ω in Ω

    I can see that Xn is not convergent.

    But I'm not quite sure in computing the integrals.... I'm very bad with the lim inf concept >_< So, I have written my attempt on evaluating the integrals and reasoning that I've used.

    My attempt,
    lim inf Xn
    =lim inf [2nδω,n]
    Since this takes in values 0 or 2n

    = ∫2nδω,n2
    When ω=n, this reduces to ∫δω,n

    I'm not very sure about the way I evaluate the integral. Would be very helpful for some guidance.
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