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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 ).2012-06-04
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    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 ?. ;)2012-06-04
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    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.2012-06-04
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    @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)2012-06-04
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    it should be $p$ and not $1$.2012-06-04
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    When you multiply out the series, the coefficient for the $t^2$ term is: $-p^2/\sigma^2 + p^2/(2\sigma^2) + p/(2\sigma^2) = 1/(2n)$, since $\sigma^2 = (p - p^2)n$.2012-07-10
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    User @p.s. saw fit to deface the question Jul 9 '12 at 21:04. Why this edit was accepted Jul 9 '12 at 21:06 is a mystery since one cannot guess the question which the answer and comments addressed, from the modified version.2013-05-19

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