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I found some difficulties in here. We know that if $X$ has Binomial distribution with $1$ trial and $p$ success, or what we called $X$~$Bin(1,p)$, we have $\mu=p$ and $\sigma^2=p(1-p)=pq$. From that, we have $\bar{X_n}$ stochastically convergent to $p$ with variance equals $\frac{pq}{n}$.

Then, we know that $\frac{\bar{X_n}-p}{\sqrt{\frac{pq}{n}}}$ has limiting distribution to $N(0,1)$. But, what if the denumerator is $\sigma^2=\frac{pq}{n}$ instead of $\sigma=\sqrt{\frac{pq}{n}}$? Here, we can't use the central limit theorem to find the limiting distribution of $\frac{\bar{X_n}-p}{\frac{pq}{n}}$, I guess.

I've tried to find with limiting Moment-Generating Function, and what I found is very complicated. It says that the limiting MGF is equal to $e^{-t(\frac{p}{pq/n})}((1-p)+pe^{\frac{t}{pq}})^n$. So, I can't try to find the distribution with such MGF.

So, is there any simpler method to find the limiting distribution of $\frac{\bar{X_n}-p}{\frac{pq}{n}}$?

Thank you.

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  • $\begingroup$ How can it have success probability n? $\endgroup$ – dsaxton Jun 10 '16 at 21:45
  • $\begingroup$ @dsaxton do u mean p? $\endgroup$ – MusMeong Jun 10 '16 at 21:51
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    $\begingroup$ You are trying to find the limiting distribution of $\sqrt{n/(pq)}$ times a sequence that converges to a constant. That $\sqrt{n}$ term makes it completely obvious that your sequence diverges. $\endgroup$ – whuber Jun 10 '16 at 22:27
  • $\begingroup$ So, we don't have any limiting distribution of it? @whuber $\endgroup$ – MusMeong Jun 11 '16 at 0:30

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