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Wikipedia and MathWorld[1] define the cumulant-generating function to be the log of the moment generating function, i.e. a Laplace transform, rather than a Four
by dbaupp 13y ago
Wikipedia and MathWorld[1] define the cumulant-generating function to be the log of the moment generating function, i.e. a Laplace transform, rather than a Fourier transform. This is (essentially) just notational since the proof still holds, but it does mean that the i's and -'s disappear, e.g. with the Laplace-CGF, c'(0) = μ, c''(0) = σ², c'''(0) = σ³ γ.
This proof also works with directly with moment generating functions and characteristic functions (which are the same as the two forms of CGF's without the logarithm), which avoids the slight issue of ln(cos(k)) when cos(k) = 0.
[1]: http://mathworld.wolfram.com/Moment-GeneratingFunction.html http://mathworld.wolfram.com/Moment-GeneratingFunction.html
- drostie 13y agoThat was in some sense intentional -- I prefer exp(i μ k) exp(-k² σ² / (2 n)) for the clarity of the expression in Fourier space. That issue is not avoided -- it becomes ln(cosh(k)), which still only has a finite radius of convergence due to branch points at ± i π/2, and a complicated branch-sheet structure. It is not dissimilar from 1/(1 + x^2) -- a perfectly smooth ordinary real function on the real line, very pleasant to work with, but the (real!) Taylor series at 0, 1 - x^2 + x^4 - x^6 + ... does not converge outside of the interval (-1, 1) because it must go to infinity at ± i on the complex plane.