Step 1

Alternatively prove that:

Theorem:

If x is a random variable with distribution B(n, p), then for sufficiently large n, the distribution of the

variablez \(=\frac{x\ -\ \mu}{\sigma}\ -\ N(0,\ 1)\)

where \(\mu=np\ \sigma^{2}=np(1\ -\ p)\)

Proof:

It can be prove using Moment generating function for binomial distribution. It's given as,

\(M_{x}(\theta)=(q\ +\ pe^{\theta})^{n}\)

where \(q = 1\ -\ p.\)

Step 2

By the linear transformation properties of the moment generating function.

\(M_{z}(\theta)=e\ -\ \frac{\mu\ \theta}{\sigma}M_{x}\left(\frac{\theta}{\sigma}\right)=e\ -\ \frac{\mu\ \theta}{\sigma}\left(q\ +\ pe\ \frac{\theta}{\sigma}\right)^{n}\)

Taking the natural log of both sides, and then expanding the power series of \(e^{\theta/\sigma}\)

Then, \(\ln\ M_{z}(\theta)=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\ \ln\left(q\ +\ pe\ \frac{\theta}{\sigma}\right)=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\ \ln\left(q\ +\ p\ \sum_{k=1}^{\infty}\frac{1}{k!}\left(\frac{\theta}{\sigma}\right)^{k}\right)\)

Since \(p\ +\ q = 1.\)

\(=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\ \ln\left(1\ +\ p\ \sum_{k=1}^{\infty}\frac{1}{k!}\left(\frac{\theta}{\sigma}\right)^{k}\right)\)

If n is made sufficiently large \(\sigma=\sqrt{npq}\) can be made large enough that for any fixed theta the absolute value of the sum above will be less than 1.

Let \(t=p\sum_{m=1}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\)

Thus for sufficiently large \(n,\ | t |\ <\ 1.\)</span>

The ln term in the previous expression is \(\ln (1\ +\ t)\ where\ | t |\ <\ 1,\)</span> and so we may expand this term as follows:

\(\ln(1\ +\ t)=\sum_{k=1}^{\infty}(-1)^{k\ -\ 1}\frac{t^{k}}{k}\)

Step 3

This means that

\(\ln\ M_{z}(\theta)=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\ \ln(1\ +\ t)=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\ \sum_{k=1}^{\infty}(-1)^{k_{1}}\)

\(\frac{t_{k}}{k}=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\left[p\ \sum_{m=1}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\ -\ p^{2}\left(\sum_{m=1}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{2}\right)+\ p^{3}\left(\sum_{m=1}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\right)^{3}\ -\ \cdots\right]=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\left[p\left[\frac{\theta}{\sigma}\ +\ \frac{1}{2}\left(\frac{\theta}{\sigma}\right)^{2}\ +\ \sum_{m=3}^{\infty}\ \frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\right]\ -\ p^{2}\left[\left(\frac{\theta}{\sigma}\right)^{2}\ +\ 2\left(\frac{\theta}{\sigma}\right)\sum_{m=2}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\ +\ \sum_{m=2}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\right]^{2}\right]\)

+ an infinite series of terms involving \(\left(\frac{\theta}{\sigma}\right)^{m}\ with\ m\ \geq\ 3]\)

By collecting terms in powers of \(\frac{\theta}{\sigma},\) we see that

\(\ln\ M_{z}(\theta)=(-\mu\ +\ np)\frac{\theta}{\sigma}\ +\ n(p\ -\ p^{2})\left(\frac{\theta}{\sigma}\right)^{2}\ +\ n\ \sum_{m=3}^{\infty}c_{m}\left(\frac{\theta}{\sigma}\right)^{m}\)

\(=\frac{np\ -\ \mu}{\sigma}\theta\ +\ \frac{np(1\ -\ p)}{\sigma^{2}}\theta^{2}\ +\ n\sum_{m=3}^{\infty}\frac{c_{m}}{\sigma_{m}}\theta^{m}\)

Here, the \(c_{k}\) terms don’t involve n, sigma or theta.

Since \mu=np\ and\ \sigma^{2}=mp(1\ -\ p)\) the coefficient of the theta term is 0 and the coefficient of the

\(\theta^{2}\) term is 1. Thus

\(\ln\ M_{z}(\theta)=\ \frac{\theta^{2}}{2}\ +\ \sum_{m=3}^{\infty}\ \frac{nc_{m}}{\sigma^{m}}\theta^{m}\)

Since the coefficient of each term in the sum has form

\(\frac{nc_{m}}{\sigma^{m}}=\frac{nc_{m}}{(npq)^{m/2}}= \frac{c_{m}}{(pq)^{m/2}}\ \cdot\ \frac{1}{n^{m/2\ -\ 1}}\ \rightarrow\ 0asn\ \rightarrow\ \infty\)

\(\lim_{n\ \rightarrow\ \infty}\ \ln\ M_{z}(\theta)=\frac{\theta^{2}}{2}\)

\(\lim_{n\ \rightarrow\ \infty}\ M_{z}(\theta)=e^{\theta2/2}\)

But note that by Property 3 of Normal Distribution the moment generating function for a random variable z with distribution N (0, 1) is

\(M_{z}(\theta)=e^{\theta^{2}/2}\)

Hence , Proved .

Alternatively prove that:

Theorem:

If x is a random variable with distribution B(n, p), then for sufficiently large n, the distribution of the

variablez \(=\frac{x\ -\ \mu}{\sigma}\ -\ N(0,\ 1)\)

where \(\mu=np\ \sigma^{2}=np(1\ -\ p)\)

Proof:

It can be prove using Moment generating function for binomial distribution. It's given as,

\(M_{x}(\theta)=(q\ +\ pe^{\theta})^{n}\)

where \(q = 1\ -\ p.\)

Step 2

By the linear transformation properties of the moment generating function.

\(M_{z}(\theta)=e\ -\ \frac{\mu\ \theta}{\sigma}M_{x}\left(\frac{\theta}{\sigma}\right)=e\ -\ \frac{\mu\ \theta}{\sigma}\left(q\ +\ pe\ \frac{\theta}{\sigma}\right)^{n}\)

Taking the natural log of both sides, and then expanding the power series of \(e^{\theta/\sigma}\)

Then, \(\ln\ M_{z}(\theta)=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\ \ln\left(q\ +\ pe\ \frac{\theta}{\sigma}\right)=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\ \ln\left(q\ +\ p\ \sum_{k=1}^{\infty}\frac{1}{k!}\left(\frac{\theta}{\sigma}\right)^{k}\right)\)

Since \(p\ +\ q = 1.\)

\(=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\ \ln\left(1\ +\ p\ \sum_{k=1}^{\infty}\frac{1}{k!}\left(\frac{\theta}{\sigma}\right)^{k}\right)\)

If n is made sufficiently large \(\sigma=\sqrt{npq}\) can be made large enough that for any fixed theta the absolute value of the sum above will be less than 1.

Let \(t=p\sum_{m=1}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\)

Thus for sufficiently large \(n,\ | t |\ <\ 1.\)</span>

The ln term in the previous expression is \(\ln (1\ +\ t)\ where\ | t |\ <\ 1,\)</span> and so we may expand this term as follows:

\(\ln(1\ +\ t)=\sum_{k=1}^{\infty}(-1)^{k\ -\ 1}\frac{t^{k}}{k}\)

Step 3

This means that

\(\ln\ M_{z}(\theta)=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\ \ln(1\ +\ t)=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\ \sum_{k=1}^{\infty}(-1)^{k_{1}}\)

\(\frac{t_{k}}{k}=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\left[p\ \sum_{m=1}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\ -\ p^{2}\left(\sum_{m=1}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{2}\right)+\ p^{3}\left(\sum_{m=1}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\right)^{3}\ -\ \cdots\right]=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\left[p\left[\frac{\theta}{\sigma}\ +\ \frac{1}{2}\left(\frac{\theta}{\sigma}\right)^{2}\ +\ \sum_{m=3}^{\infty}\ \frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\right]\ -\ p^{2}\left[\left(\frac{\theta}{\sigma}\right)^{2}\ +\ 2\left(\frac{\theta}{\sigma}\right)\sum_{m=2}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\ +\ \sum_{m=2}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\right]^{2}\right]\)

+ an infinite series of terms involving \(\left(\frac{\theta}{\sigma}\right)^{m}\ with\ m\ \geq\ 3]\)

By collecting terms in powers of \(\frac{\theta}{\sigma},\) we see that

\(\ln\ M_{z}(\theta)=(-\mu\ +\ np)\frac{\theta}{\sigma}\ +\ n(p\ -\ p^{2})\left(\frac{\theta}{\sigma}\right)^{2}\ +\ n\ \sum_{m=3}^{\infty}c_{m}\left(\frac{\theta}{\sigma}\right)^{m}\)

\(=\frac{np\ -\ \mu}{\sigma}\theta\ +\ \frac{np(1\ -\ p)}{\sigma^{2}}\theta^{2}\ +\ n\sum_{m=3}^{\infty}\frac{c_{m}}{\sigma_{m}}\theta^{m}\)

Here, the \(c_{k}\) terms don’t involve n, sigma or theta.

Since \mu=np\ and\ \sigma^{2}=mp(1\ -\ p)\) the coefficient of the theta term is 0 and the coefficient of the

\(\theta^{2}\) term is 1. Thus

\(\ln\ M_{z}(\theta)=\ \frac{\theta^{2}}{2}\ +\ \sum_{m=3}^{\infty}\ \frac{nc_{m}}{\sigma^{m}}\theta^{m}\)

Since the coefficient of each term in the sum has form

\(\frac{nc_{m}}{\sigma^{m}}=\frac{nc_{m}}{(npq)^{m/2}}= \frac{c_{m}}{(pq)^{m/2}}\ \cdot\ \frac{1}{n^{m/2\ -\ 1}}\ \rightarrow\ 0asn\ \rightarrow\ \infty\)

\(\lim_{n\ \rightarrow\ \infty}\ \ln\ M_{z}(\theta)=\frac{\theta^{2}}{2}\)

\(\lim_{n\ \rightarrow\ \infty}\ M_{z}(\theta)=e^{\theta2/2}\)

But note that by Property 3 of Normal Distribution the moment generating function for a random variable z with distribution N (0, 1) is

\(M_{z}(\theta)=e^{\theta^{2}/2}\)

Hence , Proved .