Question

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 $S_n$?

Answer 1

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.