4 ms·
Why is it important for the existence of a CLT type result that the Gaussian be a fixed point of the Fourier transform? Regardless of whether it were or not, we
by Chinjut 12y ago
Why is it important for the existence of a CLT type result that the Gaussian be a fixed point of the Fourier transform? Regardless of whether it were or not, we would find that characteristic functions of (suitably scaled) sample means converged to e^(-x^2/2), and thus the Fourier transform of e^(-x^2/2) would be their limiting distribution. That this Fourier transform is e^(-x^2/2) again does not seem essential.
- mturmon 12y agoHmm, you know, I think you have a point. Clearly any CLT solution must be a fixed point of convolution (as the Gaussian is). The other property ("Gaussian is its own Fourier transform") seemed important, but I'm unable to formalize the argument, and don't think that "fixed point of convolution" requires it. Thanks for keeping me honest. Unless I'm missing something, your observation would render the closing statement in the GP comment ("because the Gaussian is its own Fourier transform") completely irrelevant.