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$$x\sim \frac{x-np}{\sqrt{npq}} \overset{d}{\to} N(0,1)$$

I want to show that normalised binomial distribution converges in distribution to standard normal distribution.

Note that: Convergence in distribution mean

$F$ is cumulative distribution function

If $F_n(x) \to F(x)$ as $n \to \infty $ then $x_n \overset{d}{\to} x$

Also note that

$$M_{x_n}(t) \to M_x(t) \Rightarrow F_n(x) \to F(x) \Rightarrow x_n \overset{d}{\to} x$$

where $M$ is called as moment generating function.

so I need to show that

$$\lim_{n\to \infty} M_{x_n}(t)=M_x(t)$$

For binomial distribution, the moment generating function is $[pe^t +(1-p)]^n$

And for standard normal distribution, the moment generating function is $e^{t^2/2}$

pe-perry
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user3911
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  • It is the standardized binomial $Z_n = \frac{X - np}{\sqrt{np(1-p)}}$ that converges to standard normal Z. Were you asked to use MGFs or is that your idea? The Wikipedia article on the DeMoivre_Laplace CLT does the proof using Stirling's formula and PDFs. – BruceET Dec 21 '15 at 23:30
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    @BruceET http://math.stackexchange.com/questions/648243 – BCLC Dec 22 '15 at 00:16

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