De Moivre Laplace theorem question

[noowutah]

Homework Statement

According to the de Moivre Laplace theorem

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

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):

\binom{n}{k}\approx{}2^{n}N(\frac{n}{2},\frac{n}{4})(k)

and therefore

\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

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

\sum_{k=0}^{m}k\binom{n}{k}

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

\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

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

Homework Equations

see above

The Attempt at a Solution

see above

 

[micromass]

Hi stlukits!

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

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

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

However, after that you use something like

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

For

f(x)=\left\{\begin{array}{c}x~\text{if}~x\leq m\\ 0~\text{if}~x>m\end{array}\right.
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

\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}

and thus

\sum_{k=1}^m{k\binom{n}{k}}=n\sum_{k=1}^m{\binom{n-1}{k-1}}

And now you can apply formula A again (which DOES work well!)

 

[noowutah]

Simple and brilliant. I'll try it. And thanks for explaining why (A) is legit.