4 ms·
It doesn't matter that the Gaussian is its own Fourier transform, as such, does it? It's just that, whatever it is that Fourier transforms into the Gaussian (1
by Chinjut 12y ago
It doesn't matter that the Gaussian is its own Fourier transform, as such, does it? It's just that, whatever it is that Fourier transforms into the Gaussian (1 - t^2/2n)^n, that's what sample means converge to.
[I suppose it does matter that this is itself Gaussian for establishing the central limit theorem as specifically about Gaussian distributions. I guess my point was that, even if, counterfactually, nothing Fourier transformed into itself, still we would have a central limit theorem about whatever distribution it was that Fourier transformed into the Gaussian]
- TeMPOraL 12y agoGP noted why it matters for an actual explanation: > Well... I don't think that's really an explanation at all of why e^(-x^2/2) is such a privileged function. Why would any distribution converge to a normal distribution? The point is, nothing in maths is arbitrary (except axioms); if you see something singled out from the space of many alternatives, there must be a good reason why it's exactly this thing, hidden in the structure of the problem. There's this rule of thumb, also important in programming, that there are generally three amounts: zero, one, or arbitrarily many.
- Chinjut 12y agoI don't think you've caught what I was saying. The GP noted indeed why we should expect the distribution of suitably scaled sample means to converge to the Fourier transform of e^(-x^2/2). I'm saying, it doesn't seem essential to the existence of a CLT type result that this Fourier transform is e^(-x^2/2) itself [i.e., it doesn't seem essential that e^(-x^2/2) happens to be a fixed point of the Fourier transform]; if the Fourier transform of e^(-x^2/2) were some other function, we'd still speak of a Central Limit Theorem and the privileged role of that other function instead. The Fourier transform fixed point phenomenon seems orthogonal to the distribution convergence phenomenon.
- TeMPOraL 12y agoOk, I guess I didn't catch it. Thanks for the clarification.