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.
Best Answer
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 $m\to\infty$.