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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

    [tex]\binom{n}{k}p^{k}q^{n-k}\approx\frac{1}{\sqrt{2\pi{}npq}}e^{-\frac{(k-np)^{2}}{2npq}}[/tex]

    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):

    [tex]\binom{n}{k}\approx{}2^{n}N(\frac{n}{2},\frac{n}{4})(k)[/tex]

    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

    [tex]\sum_{k=0}^{m}k\binom{n}{k}[/tex]

    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

    micromass

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    Hi stlukits! :smile:

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

    [tex]Y_n=\frac{X-n/2}{\sqrt{n/4}}[/tex]

    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]

    For

    [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

    [tex]\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}[/tex]

    and thus

    [tex]\sum_{k=1}^m{k\binom{n}{k}}=n\sum_{k=1}^m{\binom{n-1}{k-1}}[/tex]

    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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