- #1
RedX
- 970
- 3
I want to show that the binomial distribution:
[tex]P(m)=\frac{n!}{(n-m)!m!}p^m(1-p)^{n-m} [/tex]
using Stirling's formula:
[tex]n!=n^n e^{-n} \sqrt{2\pi n} [/tex]
reduces to the normal distribution:
[tex]P(m)=\frac{1}{\sqrt{2 \pi n}} \frac{1}{\sqrt{p(1-p)}}
exp[-\frac{1}{2}\frac{(m-np)^2}{np(1-p)}]
[/tex]
Unfortunately, I keep on getting an extra term linear in m-np:
[tex] exp[\frac{m-np}{2n}\frac{2p-1}{(1-p)p}][/tex]
This term is zero if p=1/2, but I want to show that the binomial distribution reduces to the normal distribution for any probablity p.
Has anyone else had this problem, as I'm sure this derivation is fairly common?
[tex]P(m)=\frac{n!}{(n-m)!m!}p^m(1-p)^{n-m} [/tex]
using Stirling's formula:
[tex]n!=n^n e^{-n} \sqrt{2\pi n} [/tex]
reduces to the normal distribution:
[tex]P(m)=\frac{1}{\sqrt{2 \pi n}} \frac{1}{\sqrt{p(1-p)}}
exp[-\frac{1}{2}\frac{(m-np)^2}{np(1-p)}]
[/tex]
Unfortunately, I keep on getting an extra term linear in m-np:
[tex] exp[\frac{m-np}{2n}\frac{2p-1}{(1-p)p}][/tex]
This term is zero if p=1/2, but I want to show that the binomial distribution reduces to the normal distribution for any probablity p.
Has anyone else had this problem, as I'm sure this derivation is fairly common?