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Problem: Let $X_n \sim \operatorname{Bin}(n,p_n) $ where $p_n \xrightarrow{} 0$ and $np_n \xrightarrow{} \infty$. What I need to show is that

$$\frac{X_n - np_n}{\sqrt{np_n}} \xrightarrow{d} N(0,1) \text{ as } n\xrightarrow{} \infty.$$ My thoughts: My first thought was to set $$Y_n=\frac{X_n - np_n}{\sqrt{np_n}}$$ and investigate $P(Y_n=k) = p_{X_n}(\sqrt{np_n}k + np_n) $ as $n$ goes to infinity, but this led to very messy calculations so I don't know if it is the right approach. My second attempt was with Central Limit Theorem, rewriting $X_n$ as a sum of Bernoulli random variables, $X_n = Z_1 + \dots + Z_n$, $Z_n \sim \operatorname{Be}(p_n)$. The expectation and variance of each $Z_i$ is dependent on $n$ in that case. Will that violate any assumptions in CLT?

I prefer working out these kind of questions from definition rather than using a theorem and previous results, so if anyone can show me that I would be very grateful!

Thanks.

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Your second attempt is definitely on the right track. For the first one, do you mean $k$ to be an integer? Because obviously that's not right. Plus, even if you allow $k$ to be any real number, you still get into trouble because the probability goes to zero. You need to be careful in going from a discrete to a continuous distribution. –  Raskolnikov Jul 3 '13 at 19:14
    
Hm.. true, didn't think of that. How would one approach this question if one was not allowed to use CLT to solve it? –  Lotus3000 Jul 3 '13 at 19:21
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The reason why this doesn't yield instantly to the most usual form of the central limit theorem is that these are not identically distributed, since $p_n$ depends on $n$. And the fact that $np_n\to\infty$ matters. One approach might be to try to prove this the same way the central limit theorem is proved, by looking at the limit of the sequence of characteristic functions. –  Michael Hardy Jul 3 '13 at 19:24
    
Thanks for the comment. But for every $n$, all $Z_i$ are indentically distributed? –  Lotus3000 Jul 3 '13 at 19:32
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I would do this in two steps. First take $n\to \infty$ and $p\to 0$ while keeping $np=m$ for some finite integer $m$. That brings you from the Binomial distribution to the Poisson distribution with parameter $m$ because $$ P(X_n=k)=\binom{n}{k}p^k(1-p)^{n-k}\approx\frac{1}{k!}\left(\frac{np}{1-p}\right)^k(1-p)^n \rightarrow\frac{m^k}{k!}e^{-m}. $$ Then view this Poisson distribution with parameter $m$ as the result of summing $m$ independent Poisson random variables, each with parameter $1$. You can then use the central limit theorem to take $m\to\infty$.

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