Proving a distribution convergs to a standard normal distribution

**redwings6** · March 29th 2010, 09:22 AM

Y has a binomial distribution based on n trials and success probability p. Then the point estimator (pn=Y/n) is an unbiased estimator of p. Prove that the distribution (pn-p)/sqr((pn*qn)/n) converges to a standard normal distribution.

I am supposed to the following theorem to prove it:
Suppose that Un has a distribution function that converges to a standard normal dist function as n-->∞. If Wn converges in probability to 1, then the dist. function Un/Wn converges to a Standard normal dist. function.

I am unsure of how to approach this so any help is greatly appreciated.

**Anonymous1** · March 29th 2010, 12:59 PM

Use $MGF.$ It suffices to show that Your $MGF \rightarrow MGF_{Standard-Normal}$ as $n\rightarrow \infty$

I actually recently did a similar problem, so if you show me what you have, I can work through it with you.

Start by deriving the $MGF$ of of the distribution in question.

[EDIT]Oops, I was thinking you were trying to prove something similar to the CLT.

**matheagle** · March 29th 2010, 02:05 PM

It seems that you just want to apply slutsky's theorem.

The CLT gives you ${P_n-p\over {\sqrt{pq\over n}}}\to N(0,1)$

All you need is the WLLN to finish...

${P_n\over p}\buildrel P \over \to 1$ and ${1-P_n\over 1-p}\buildrel P \over \to 1$

which gives you

$\sqrt{{P_n(1-P_n)\over p(1-p)}}\buildrel P \over \to 1$

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