This is a very nice comment, because it points out that the Gaussian "answer" is hidden in the structure of the question. You know the zero'th moment (the dist

This is a very nice comment, because it points out that the Gaussian "answer" is hidden in the structure of the question. You know the zero'th moment (the distribution has unit mass), and you have removed the first moment (by subtracting the mean), so the next is the second. By passing to a limit, you remove the higher moments from consideration, and isolate on the second-order behavior of the solution near the origin (in characteristic function coordinates). Only the second-order behavior matters, and the Gaussian is the distribution that has a characteristic function with a quadratic behavior at the origin (in the exponent). The real structure of the problem is that it assumes independence, or a similar condition (like mixing) strong enough to mimic independence. Independence is what allows you to pass to the limit and thereby focus on the second-order behavior around the origin. And in practice, you have just swept the bump under the rug, because you now have to justify your independence assumption -- and that's hard. My quibble is with the side comment about the Gaussian being its own Fourier transform. This property is required (i.e., the Gaussian must be a fixed point of the CLT), but the property does not provide any constraint on the class of possible solutions of the CLT. In fact, there is an infinite number of functions that transform into themselves. I can easily construct a new one from any Fourier-transformable function you give me. For example, I can make such a function that looks like a square wave with some wiggles in it. (Reference: http://www.systems.caltech.edu/dsp/ppv/papers/journal08post/PPVIETEeigenFT.pdf http://www.systems.caltech.edu/dsp/ppv/papers/journal08post/...)