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To prove CLT of binomial distribution,

$$X \sim \mbox{bin}(n,p)$$ $M_X(t)=(p e^t+q)^n$ where $M$ is mgf.

Let $Z=\frac{X-np}{ \sqrt{npq}}$, $\sigma =\sqrt{npq}$, then $$ \begin{align} M_Z(t)&=e^{-\frac{npt}{\sigma}} (p e^{t/\sigma} + q)^n\\ &=\left[\left(1- \frac{pt}{\sigma}+\frac{p^2t^2}{2\sigma^2}+\ldots\right) \left(1 \mbox{?}+ \frac{pt}{\sigma}+\frac{pt^2}{2σ^2}+\ldots\right)\right]^n\\ &=\left(1+t^2/2n+d(n)/n\right)^n \end{align} $$

where $\lim_{n \rightarrow \infty} d(n)=0$, so $\lim_{n \rightarrow \infty} M_Z(t)=e^{\frac{t^2}{2}}$

In here, I can't understand the results of taylor expansion.

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Dear noname, You need to write up in Latex. Or else it looks like mess. Its a good practice to expand the acronyms you are using ( eg: mgf in this case ). –  Iyengar Jun 4 '12 at 17:27
    
Dear noname, I tried editing the answer with all my patients. The brackets you used are completely redundant and are not letting me keeping the terms in the bracket. I stopped editing as a single mis-placing of bracket, may completely change the output. Remember the butterfly effect of Chaotic theory ?. ;) –  Iyengar Jun 4 '12 at 17:34
    
thanks for your comment. I tried to use Latex but failed. sorry for that. let me shorten my question. taylor expansion of p*e^(t/σ) = (1+p*(t/σ)+(p/2!)*(t/σ)^2+...) and I can't understand why the first term is 1 not p. –  noname Jun 4 '12 at 17:45
    
@noname LaTeX is often not that difficult! 1. Surround mathematical expressions by: $\$$'signs (use $\$\$$'signs to get centred display style). 2. Use surround indices and exponents by curly braces: $\{\}$. 3. Put backslash $\backslash$ in front of standard mathematical functions... (check out the links on my user page to learn more) –  AD. Jun 4 '12 at 19:06
2  
it should be $p$ and not $1$. –  robjohn Jun 4 '12 at 21:16

1 Answer 1

The first coefficient is definitely $1$.

Consider that $pe^t+q=p\left(1+t+\frac12t^2+\cdots\right)+q=1+pt+\frac12pt^2+\cdots$ since $p+q=1$.

As an aside, note that for any random variable $Y$, $M_Y(t)=\mathrm E(\mathrm e^{tY})$ yields $M_Y(0)=1$.

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