I'm confused about applying central limit theorem to Bernoulli random variables. Let $X_i=\frac{n}{\sqrt{n1}}(Z_i  \frac{1}{n})$ where $Z_i$ is iid Bernoulli($\frac{1}{n}$). Then $E[X_i]=0$ and $Var(X_i) = 1 $. Thus, it seems that standard LindbergLevy CLT can be applied to $ S_n = \frac{1}{\sqrt{n}} \sum_{i1}^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 $S_n$?

1$\begingroup$ 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 $\frac{1}{2n^2}, \frac{1}{2n^2}, \frac{n^21}{n^2}$ 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 $\forall \epsilon \exists N \forall X.$ $\endgroup$– Douglas ZareCommented Sep 23, 2012 at 17:51
1 Answer
In your example, unlike the CLT, you deal not with one sequence of independent random variables $X_i$, but with a twoparameter 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 nonnormal distributions.

1$\begingroup$ Well, there are plenty of valid CLTs for twoparameter families. The point is that they have additional conditions which this example does not meet. $\endgroup$ Commented Sep 23, 2012 at 22:32