In a footnote OP says: > I’m not yet good enough to intuitively get why the curvature of the probability-generating would be related to variance, but I’d be ha

In a footnote OP says: > I’m not yet good enough to intuitively get why the curvature of the probability-generating would be related to variance, but I’d be happy to receive pointers here. Here’s my intuition for this. The characteristic function is the Fourier transform of the density. If the density is in “t” units, the ch.f. is in f= 1/t units. It is the “inverse domain.” (I’m using “f” to suggest frequency, ie the Fourier coordinate.) Of course it is not a simple coordinate transformation! But some intuition does carry over. This is reflected in all sorts of ways. It’s one reason why the IFT formula is so functionally close to the FT formula. Anyway. Because of this, the behavior of the FT (ch.f.) very close to the origin (“f=0”) tells about the tails of the distribution (t = 1/f is large). In particular, high curvature around the origin tells you the tails are heavy. That’s the variance. This extends to the fourth moment. You can get even sharper curvature around the origin of the FT (ch.f. at f=0) with a large coefficient on the fourth order term. This corresponds to a large fourth moment of the pdf, or a high kurtosis. It’s useful to recall that, because of analytic continuation, knowing all the derivatives at the one point f=0 determines the ch.f. everywhere, and thereby determines the complete density. This corresponds to the fact that knowing all the moments determines the full density. So in a very real sense, you only need the ch.f. in a tight neighborhood of the origin! (Provided all moments are finite.)