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Let $f(x)$ be a bilateral Laplace transform of a measure $\mu$: $$ f(x)=\int_{-\infty}^{+\infty} e^{-xt} d\mu(y). $$ Suppose that $f(x)$ converges absolutely in $(a,b)$, and $(a,b)$ do not contain the origin. It is always true that $f(x)$ is analytic in $(a,b)$? Or it is true just for finite measure $\mu$?

Moreover, if the measure $\mu$ is finite, then $(a,b)$ must contain the origin?

Thank you!

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The following references investigate the Moment Generating Function (finite measure):

There is a classical reference for the case where the interval $(a,b)$ contains the origin and a recent extension to the case of a general open interval $(a,b)$. This paper that gives a nice overview and proves the converse to Curtiss's theorem in the general case.

Refs:

Curtiss, J.H., 1942. A note on the theory of moment generating functions, 1, Ann. of Math. Statist, 13 (4) : 430-433.

Mukherjea, A., Rao, M. and Suen, S., 2006. A note on moment generating functions, Statistics & Probability Letters, 76 : 1185-1189.

P. Chareka. \The converse to Curtiss theorem for one-sided moment generating functions" (2008). arXiv: 0807.3392 [quant-ph].

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  • $\begingroup$ thank you! Then if the measure is finite a mgf is always analytic in the interval where converges absolutely, even if this interval do not contain the origin. Hence, is it possible that $\mu$ is finite and still the mgf not convergent in an interval containing the origin? $\endgroup$ – alemou Jun 22 '13 at 17:19
  • $\begingroup$ Yes, existence of mgf (in a nbhd of the origin) implies existence of all moments of the random variable. Hence mgf of any random variable that does not have all moments does not converge (in any nbhd of the origin). Here is a note I prepared on a related subject: dropbox.com/s/w6h7gjjomv5uohu/MGF_ConcIneq.pdf and an excellent post on mgfs: stats.stackexchange.com/questions/32706/… $\endgroup$ – mStudent Jun 22 '13 at 18:13
  • $\begingroup$ really helpfull! $\endgroup$ – alemou Jun 23 '13 at 13:40

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