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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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    User @p.s. saw $f$it to de$f$ace 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, $f$rom the modi$f$ied version.2013-05-19

1 Answers 1

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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 $M_Y(0)=1$.