Can Stirling's Formula Derive the Normal Distribution from the Binomial?

[RedX]

I want to show that the binomial distribution:

$$P(m)=\frac{n!}{(n-m)!m!}p^m(1-p)^{n-m}$$

using Stirling's formula:

$$n!=n^n e^{-n} \sqrt{2\pi n}$$

reduces to the normal distribution:

$$P(m)=\frac{1}{\sqrt{2 \pi n}} \frac{1}{\sqrt{p(1-p)}}<br /> <br /> exp[-\frac{1}{2}\frac{(m-np)^2}{np(1-p)}]$$

Unfortunately, I keep on getting an extra term linear in m-np:

$$exp[\frac{m-np}{2n}\frac{2p-1}{(1-p)p}]$$

This term is zero if p=1/2, but I want to show that the binomial distribution reduces to the normal distribution for any probability p.

Has anyone else had this problem, as I'm sure this derivation is fairly common?

 

[mathman]

It would help to show your derivation in detail. In any case the central limit theorem could be used.

 

[RedX]

It would help to show your derivation in detail. In any case the central limit theorem could be used.

Sure. It's a little bit lengthy though, so it might take some work to read it:

$$P(m)=\frac{\sqrt{2 \pi n}n^ne^{-n}}{\sqrt{2 \pi m}m^me^{-m}*\sqrt{2 \pi (n-m)}(n-m)^{(n-m)}e^{-(n-m)}}p^m(1-p)^{n-m}=<br /> <br /> \frac{n^{n+1}}{\sqrt{2 \pi n}*m^{m+\frac{1}{2}}*(n-m)^{(n-m)+\frac{1}{2}}}p^m(1-p)^{n-m}$$

$$\frac{n^{n+1}}{\sqrt{2 \pi n}*m^{m+\frac{1}{2}}*(n-m)^{(n-m)+\frac{1}{2}}}p^m(1-p)^{n-m}=<br /> <br /> \frac{1}{\sqrt{2\pi n}} \exp[\log[\frac{n^{n+1}}{m^{m+\frac{1}{2}}*(n-m)^{(n-m)+\frac{1}{2}}}p^m(1-p)^{n-m}]]$$


$$\frac{1}{\sqrt{2\pi n}} \exp(\log[\frac{n^{n+1}}{m^{m+\frac{1}{2}}*(n-m)^{(n-m)+\frac{1}{2}}}p^m(1-p)^{n-m}]]=<br /> \frac{1}{\sqrt{2\pi n}}\exp[(n+1) \log n -(m+^1/_2) \log m$$

$$-(n-m+^1/_2) \log (n-m)+m\log p + (n-m) \log[1-p] ]$$

Now I set m=pn+k in the expression, where pn is the average number of successes, and k is the difference from the average, and use that k<<<pn to expand the logarithms in power series up to second order (i.e., ln(1+x)=x-.5x^2).

So for example:

$$(m+^1/_2) \log m=(pn+k+^1/_2)\log (pn+k)=<br /> (pn+k+^1/_2)[\log (pn)+\log (1+k/(pn))]=<br /> (pn+k+^1/_2)[\log (pn)+k/(pn)-.5 k^2/(pn)^2]<br />$$

After keeping all terms up to second order, I get:

$$P(m)=\frac{1}{\sqrt{2 \pi n}}\frac{1}{\sqrt{p(1-p)}}\exp[-\frac{1}{2}\frac{k^2}{np(1-p)}] \exp[\frac{k}{2n} \frac{2p-1}{(1-p)p}]$$

and that last factor doesn't make sense, but I've checked my results three times and it keeps coming out. But if p=1/2, then it becomes 1, and I do get the normal distribution.

 

[Stephen Tashi]

Just a thought: Stirlings approximation for n! doesn't converge to n! as n approaches infinity. However the ratio of the approximation to n! converges to 1. Does anyone know the situation whe we use Stirings approximations for $C_r^n$ ?

 

[mathman]

http://en.wikipedia.org/wiki/De_Moivre–Laplace_theorem

Above is the derivation that you are looking for. It should be able to help you resolve the problem.