In the usual proof of the central limit theorem via characteristic functions, we note that if $X_1, \dots, X_n$ are independent and identically distributed, with probability density function $f(x)$, and characteristic function $\hat f(t) = 1 - c t^2 + o(t^2)$, then $(X_1 + \dots + X_n) / \sqrt n$ has characteristic function $(\hat f(t/\sqrt n))^n = (1 - ct^2/n + o(t^2))^n \approx e^{-ct^2}$, a Gaussian. So the density function is the Fourier transform of a Gaussian, which one can check by a computation is again a Gaussian.

In real space, we have repeated convolution of the density function, and scaling. In frequency space, we have repeated pointwise multiplication of the characteristic function, and scaling. From this point of view, it seems like a coincidence that both of these processes result in the same thing.

Is there an explanation of this apparent coincidence? What does the fact that the Gaussian is self-dual have to do with its role in the central limit theorem? Is there a proof or generalization of the central limit theorem which exposes this symmetry between real space and frequency space?

One possible explanation is that a Gaussian is a very special object. For example, the Euclidean norm on $\mathbb R^n$ is special because it has the maximal amount of linear symmetries, and this means its dual norm also has that many symmetries and is therefore also Euclidean. The Gaussian on $\mathbb R^n$ is closely tied in to the Euclidean norm, so this argument has something to say about the Gaussian as well, but I haven't been able to articulate a concrete relationship to the CLT.

  • $\begingroup$ The fact that the characteristic function is the same is "because" the Gaussian, along with Hermite polynomials are eigenfunctions of the Fourier transform: en.m.wikipedia.org/wiki/Fourier_transform#Eigenfunctions $\endgroup$
    – Alex R.
    Jul 10, 2014 at 1:23
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    $\begingroup$ The generalized central limit theorem applies to stable distributions, which in general are not self dual. $\endgroup$
    – Alex R.
    Jul 10, 2014 at 1:38
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    $\begingroup$ Note that the sech density is also self dual, and is not a stable density. $\endgroup$ Nov 5, 2014 at 18:23


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