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De Moivre Laplace theorem question

  1. Jul 1, 2011 #1
    1. The problem statement, all variables and given/known data

    According to the de Moivre Laplace theorem


    For p=q=1/2 this translates nicely into an approximation for the binomial distribution by the normal distribution (the +1/2 is a continuity correction):


    and therefore

    [tex]\mbox{(A)}\quad \sum_{k=0}^{m}\binom{n}{k}\approx{}2^{n}\int_{-\infty}^{m+\frac{1}{2}}N(\frac{n}{2},\frac{n}{4})(x)dx[/tex]

    This appears to be correct. I am trying to solve a problem for which I need


    and I am wondering what would keep me from reasoning in analogy to (A) so that

    [tex]\mbox{(B)}\quad \sum_{k=0}^{m}k\binom{n}{k}\approx{}2^{n}\int_{-\infty}^{m+\frac{1}{2}}xN(\frac{n}{2},\frac{n}{4})(x)dx[/tex]

    Unfortunately, when I use (B) I get some very counter-intuitive results (too complicated to expand on them here).

    2. Relevant equations

    see above

    3. The attempt at a solution

    see above
    Last edited: Jul 1, 2011
  2. jcsd
  3. Jul 1, 2011 #2
    Hi stlukits! :smile:

    Let X have the binomial(n,1/2) distribution, and let Z have the normal(0,1) distribution. Let


    What you are using is that Yn converges in distribution to Z. This is correct and yields the good results you're getting.

    However, after that you use something like

    [tex]E[f(Y_n)]\rightarrow E[f(Z)][/tex]


    [tex]f(x)=\left\{\begin{array}{c}x~\text{if}~x\leq m\\ 0~\text{if}~x>m\end{array}\right.[/tex]
    However, this does not need to hold if you only know convergence in distribution. By Portmanteau theorem, this only holds for functions f with certain properties (like boundedness, which this function f is not).

    However, not all is lost. You can do


    and thus


    And now you can apply formula A again (which DOES work well!)
  4. Jul 2, 2011 #3
    Simple and brilliant. I'll try it. And thanks for explaining why (A) is legit.
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