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Expected Value of Positive-Valued RV

  1. Jul 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that if X is a positive-valued RV, then E(X^k) ≥ E(X)^k for all k≥1

    3. The attempt at a solution
    Why do I feel like this is a counter-example:

    X = {1,2,4,8,16,...} (A positive-valued RV)
    m(X) = {1/2,1/4,1/16,1/32,...} (A distribution function that sums to one)

    Yet clearly,

    [tex]E[X] = \sum _{k=1} ^{∞} \frac{1}{2^k} {2}^{k-1} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ... = ∞[/tex]

    So the expected value diverges (ie. doesn't exist). So I can't do the proof because for this RV, the expectation DNE.
     
    Last edited: Jul 28, 2012
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  3. Jul 28, 2012 #2

    Ray Vickson

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    This is also an allowable case if you accept that ∞ ≥ ∞, and that EX = ∞ implies EX^k = ∞ for any k > 1.

    However, in the case that EX < ∞ and EX^k < ∞, can you do the proof then?

    RGV
     
  4. Jul 28, 2012 #3
    Why can't you? Both [itex]E[X^2][/itex] and [itex]E[X]^2[/itex] are infinite, right?
     
  5. Jul 28, 2012 #4
    So I went and got this from Wikipedia:

    Skjermbilde_2012_07_28_kl_4_39_15_PM.png

    I wouldn't have much of a problem accepting ∞ ≥ ∞ if I thought that the expectation of such an RV even existed. Since the expectation diverges in my example, it seems meaningless to me to discuss comparisons of its nonexistent expected value.
     
  6. Jul 28, 2012 #5

    Simon Bridge

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    I suspect you have over-complicated things... surely it is not true, in general, that the sum of powers is equal to the power of the sum - especially if all elements in the sum are positive? Consider if X is drawn from a set of two values for instance...

    That's if I've read it correctly that you have to show:
    [tex]E(X^k)=\frac{1}{N}\sum_{i=1}^{N}(x_i)^k = \bigg ( \frac{1}{N}\sum_{i=1}^{N}x_i \bigg )^k = \bigg ( E(X) \bigg ) ^k

    [/tex]
     
  7. Jul 28, 2012 #6

    Ray Vickson

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    Opinions vary, and not all books would agree with that Wiki article. Of course, the expectation does not exist (as a real number), but in such a case we sometimes write EX = ∞ and pretend that ∞ is in an extended real number field.

    Note that there are essentially two kinds of "EX does not exist": (1) EX = ∞; and (2) for X in all of ℝ, with X = X+ - X-, with X± ≥ 0 and EX+ = EX- = ∞ (that is, EX would be of the form ∞ - ∞).

    Anyway, if you don't like this, add finiteness statements to the hypotheses. That still leaves you with something to prove.

    RGV
     
  8. Jul 28, 2012 #7
    I don't think that's what I'm trying to show. I want to show a weak inequality, as opposed to a strict equality. Also, you seem to have assumed a uniform pdf (ie. the 1/N), which does not seem a fair assumption to me.
     
  9. Jul 28, 2012 #8
    Yes, I think I'll have to add convergence hypotheses. In which case, the proof is immediate from Jensen's Inequality.
     
  10. Jul 28, 2012 #9

    Ray Vickson

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    I think you also need to show that EX = ∞ implies EX^k =∞ for all k > 1; that is, if EX does not exist then neither does EX^k.

    RGV
     
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