# central limit theorem for binomial random variable

I'm confused about applying central limit theorem to Bernoulli random variables. Let $X_i=\frac{n}{\sqrt{n-1}}(Z_i - \frac{1}{n})$ where $Z_i$ is iid Bernoulli($\frac{1}{n}$). Then $E[X_i]=0$ and $Var(X_i) = 1$. Thus, it seems that standard Lindberg-Levy CLT can be applied to $S_n = \frac{1}{\sqrt{n}} \sum_{i-1}^n X_i$ which is a linear function of a binomial random variable. But the moment generating function of $S_n$ doesn't converge to that of standard normal, and the convergency works only when the probability parameter of the Bernoulli function is $1/T^{1-\alpha}$ $0 < \alpha < 1$. I read a couple of textbook and couldn't find if any further condition is required to apply CLT to $iid$ variables with finite mean and variance. What's wrong with applying CLT to $S_n$?

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I think this is a nice question for you to work through, but too localized for this site. Here is a simpler example to consider: $X_{i,n} = n, -n, 0$ with probabilities $\frac{1}{2n^2}, \frac{1}{2n^2}, \frac{n^2-1}{n^2}$ respectively. $\sum_{i=1}^n X_{i,n}$ is $0$ with probability at least $1 - 1/n$ so this converges in distribution to the constant $0.$ The CLT is not effective without additional assumptions. You can't replace $\forall X \forall \epsilon \exists N$ with $\forall \epsilon \exists N \forall X.$ –  Douglas Zare Sep 23 '12 at 17:51

In your example, unlike the CLT, you deal not with one sequence of independent random variables $X_i$, but with a two-parameter family $X^{(n)}_i$, where for example $X_1^{(n)}$ and $X_1^{(m)}$ have different statistics for $m \neq n$ and cannot be both denoted simply by $X_1$. Therefore as $n$ grows, the sum $S_n$ does not merely grow term by term as in the standard CLT, but is replaced with an entirely new sum for each $n$. No surprise that this particular scaling gives a Poisson distribution; other scalings of $X^{(n)}_i$ may give yet other non-normal distributions.