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Local limit theorem

  1. Jun 14, 2008 #1
    1. The problem statement, all variables and given/known data
    This theorem is from Shiryaev. Can someone PLEASE explain how they are using the little o notation here. It makes no sense to me how they say that a FIXED real number is little o of something. I thought f(n) = o(g(n)) mean that for ever c>0 there exists an n_o such that when n => n_0, |f(n)|<c|g(n)|.

    2. Relevant equations



    3. The attempt at a solution
     

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  2. jcsd
  3. Jun 14, 2008 #2

    Astronuc

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    Staff: Mentor

  4. Jun 14, 2008 #3
  5. Jun 14, 2008 #4

    Hurkyl

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    Where did they say that?
     
  6. Jun 14, 2008 #5
    It says for all [tex]x \in R^1[/tex]
     
  7. Jun 14, 2008 #6

    Hurkyl

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    That doesn't sound like a FIXED real number to me. And besides, they clarify what they mean immediately afterwards.
     
    Last edited: Jun 14, 2008
  8. Jun 14, 2008 #7
    They say for all [tex]x \in R^1[/tex] such that [tex]x = o(npq)^{1/6}[/tex]. To me that means that you can take any real number that you want and see if it is [tex]o(npq)^{1/6}[/tex]. I don't know how else to interpret it.
     
  9. Jun 14, 2008 #8

    Hurkyl

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    How about exactly how they interpreted it in the attachment you posted?
     
  10. Jun 14, 2008 #9
    That is precisely my question: How did they interpret it?
     
  11. Jun 14, 2008 #10

    Hurkyl

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    as [itex]n \rightarrow \infty[/itex],
    [tex]
    \sup_{\left\{ x : |x| \leq \psi(n) \right\}}
    \left|
    \frac{P_n(np + x \sqrt{npq})}{
    \frac{1}{\sqrt{2\pi npq}} e^{-x^2 / 2}
    } - 1
    \right| \rightarrow 0,
    [/tex]​
    where [itex]\psi(n) = o(npq)^{1/6}[/itex].
     
  12. Jun 14, 2008 #11
    OK. I will have to think about that. I am kind of confused about the sup above. Is \psi(n) fixed? If not, I don't understand how the sup is taken. Is it taken over all x AND over all \psi(n) that are [tex]o(npq)^{1/6}[/tex]?
     
  13. Jun 15, 2008 #12
    anyone?
     
  14. Jul 8, 2008 #13
    I FINALLY figured this out! Hurky! was right that they clarify what they mean-that just wasn't clicking for me.
     
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