Suppose you take a bunch of differentiable functions, all of which have a global maximum at 0, and add them pointwise.
Usually you will get a single peak at 0 towering above the rest. The only special case is if ∃x≠0:∀i:fi(x)=fi(0) In the neighbourhood of 0, the function is approximately parabolic. (Its differentiable.) You take the exponent, this squashes everything but the highest peak down to nearly 0. (In relative terms). The highest peak turns into a sharp spiky Gaussian . You take the inverse Fourier transform and get a shallow Gaussian.
Even if you are unlucky enough to start with several equally high peaks in your fi’s then you still get something thats kind of a Gaussian. This is the case of a perfectly multimodal distribution, something 0 except on exact multiples of a number. The number of heads in a million coin flips forms a Gaussian out of dirac deltas at the integers.
But the condition of having a maximum at 0 in the Fourier transform is weaker than always being positive. If ∀x:f(x)≥0 then ∫f(x)dx≥∫f(x)e2πiθxdx
Suppose you take a bunch of differentiable functions, all of which have a global maximum at 0, and add them pointwise.
Usually you will get a single peak at 0 towering above the rest. The only special case is if ∃x≠0:∀i:fi(x)=fi(0) In the neighbourhood of 0, the function is approximately parabolic. (Its differentiable.) You take the exponent, this squashes everything but the highest peak down to nearly 0. (In relative terms). The highest peak turns into a sharp spiky Gaussian . You take the inverse Fourier transform and get a shallow Gaussian.
Even if you are unlucky enough to start with several equally high peaks in your fi’s then you still get something thats kind of a Gaussian. This is the case of a perfectly multimodal distribution, something 0 except on exact multiples of a number. The number of heads in a million coin flips forms a Gaussian out of dirac deltas at the integers.
But the condition of having a maximum at 0 in the Fourier transform is weaker than always being positive. If ∀x:f(x)≥0 then ∫f(x)dx≥∫f(x)e2πiθxdx