central limit theorem for binomial random variable - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T09:36:47Z http://mathoverflow.net/feeds/question/107894 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107894/central-limit-theorem-for-binomial-random-variable central limit theorem for binomial random variable metrics 2012-09-23T09:41:21Z 2012-09-23T10:26:33Z <p>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 $1/T^{1-\alpha}$ $0 &lt; \alpha &lt; 1$. 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 $S_n$?</p> http://mathoverflow.net/questions/107894/central-limit-theorem-for-binomial-random-variable/107896#107896 Answer by ansobol for central limit theorem for binomial random variable ansobol 2012-09-23T10:26:33Z 2012-09-23T10:26:33Z <p>In your example, unlike the CLT, you deal not with <em>one</em> 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.</p>