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If $X_n$ is Binomial (n,p) then show that $\frac{X_n-np}{\sqrt{npq}}$ is asymptotically Normal $(0,1)$.

Do I need to calculate the moments of each $\frac{X_n-np}{\sqrt{npq}}$ and then show that the moment sequence converge to the moments of Normal distribution.

Any answer or reference is highly appreciated.

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You can directly prove that the MGF $M_{Z_n}(t)$ of the standardised binomial variate $Z_n=\frac{X_n-np}{\sqrt{npq}}$ tends to the MGF of the $\mathcal{N}(0,1)$ distribution for large $n$. I briefly sketch the outline of the proof.

You have $M_{Z_n}(t)=\mathbb E(e^{tZ_n})=\exp\left(\frac{-tnp}{\sqrt{npq}}\right)\left(q+p\exp\left(\frac{t}{\sqrt{npq}}\right)\right)^n$

$\implies\ln M_{Z_n}(t)=\frac{-tnp}{\sqrt{npq}}+n\ln\left(1+p\left(\exp\left(\frac{t}{\sqrt{npq}}\right)-1\right)\right)$

$\qquad\qquad\qquad\quad=\frac{-tnp}{\sqrt{npq}}+n\ln(1+y)$, where $y=p\left(\left(\frac{t}{\sqrt{npq}}+\frac{t^2}{2npq}+\cdots\right)-1\right)$

Now expand the $\log$ series (assuming $y\in(-1,1]$ of course) to get

$\ln M_{Z_n}(t)=\frac{-tnp}{\sqrt{npq}}+\left(\left(\frac{tnp}{\sqrt{npq}}+\frac{t^2}{2q}+\cdots\right)-\left(\frac{pt^2}{2q}+\cdots\right)+\cdots\right)$

$\qquad\qquad\to \frac{t^2}{2}$ as $n\to\infty$

This indeed shows that $Z_n\stackrel{a}{\sim}\mathcal{N}(0,1)$.

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Check that $\mathbb{E}(X_n)=np$ and $\text{Var}(X_n)=np(1-p)$ and use CLT

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  • $\begingroup$ I got you @Martin. If I want to calculate exactly without using the proof style of CLT, then is there any other way around? $\endgroup$
    – TRUSKI
    Dec 25 '17 at 4:51
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Just to give more justification for Martin's answer. To directly apply the classical CLT, you must identify that $X_n$ has the same distribution as the sum of $n$ i.i.d. Bernoulli random variables, i.e., $$X_n \overset{d} = Y_1 + Y_2 + \cdots + Y_n,$$ where $Y_1, \ldots, Y_n \text{ i.i.d.} \sim \text{Bin}(1, p)$.

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