> Well... I don't think that's really an explanation at all of why e^(-x^2) is such a privileged function. Why would any distribution converge to a normal distr

> Well... I don't think that's really an explanation at all of why e^(-x^2) is such a privileged function. Why would any distribution converge to a normal distribution? > It essentially boils down to the Fourier transform Precisely. One motivation for taking a fourier transform is that the the p.d.f. of the sum of several random variables is the convolution of corresponding p.d.fs, and that convolution in the original domain is equivalent to multiplication in the fourier domain (by the convolution theorem).