Why does the Central Limit Theorem have data tend towards a Normal Model?

Question

There are many functions that "look" like the normal model (which is $e^{-x^2}$) such as $\frac{1}{x^2+1}$ or $\operatorname{sech(x)}$. Why doesn't the Central Limit Theorem tend towards these functions? Why does it always tend toward the normal model?

Answer 1

Here is one way to intuitively see the Gaussian pop out (I don't claim any rigor). Consider a probability distribution $\psi:\mathbb R\to\mathbb R_{\geq0}$ with mean $0$ and variance $\sigma^2$. Let $\mathcal F$ denote the Fourier transform. Note that $$ (\mathcal F\psi)(0)=\int_{\mathbb R}\psi(x)\;dx=1, $$ $$ (\mathcal F\psi)'(0)=\int_{\mathbb R}ix\psi(x)\;dx=0, $$ $$ (\mathcal F\psi)''(0)=\int_{\mathbb R}-x^2\psi(x)\;dx=-\sigma^2. $$ Thus $(\mathcal F\psi)(t)=1-\sigma^2t^2/2+O(t^3)$. Now draw $X_1,\ldots,X_N$ independently from $\psi$ and let $\psi_n$ denote the probability distribution of $(X_1+\ldots+X_n)/\sqrt n$. Then $$ \psi_n(x)=\sqrt n(\psi*\psi*\ldots*\psi)(x\sqrt n) $$ where $*$ denotes convolution. Thus $$ (\mathcal F\psi_n)(t)=((\mathcal F\psi)(t/\sqrt n))^n \approx\left(1-\sigma^2t^2/(2n)\right)^n \approx\exp(-\sigma^2t^2/2). $$ Taking the inverse Fourier transform gives the expected result.