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Possible Duplicate:
Existence of the moment generating function and variance

Given that there is an interval $-h < t < h$ where MGF exists, does it imply that the distribution's $E(X^n)$ will be defined for all $n$?

kjetil b halvorsen
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  • I don't think this is an *exact* duplicate; this one is about all the moments, not just the variance – Peter Flom Oct 03 '12 at 10:16
  • @Peter: That's true, but there is no essential difference in this case and this is discussed in the answer there. There was no option to close as an *essentially* exact duplicate. :-) – cardinal Oct 03 '12 at 10:22
  • OK, @Cardinal then I will add my close vote – Peter Flom Oct 03 '12 at 10:25

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Yes. The moment generating function need not exist and a distribution does not necessarily have any finite moments but if the momnet generating function exists in an open interval then quoting from Wikipedia '

Calculations of moments: The moment-generating function is so called because if it exists on an open interval around t = 0, then it is the exponential generating function of the moments of the probability distribution:

E(X$^n$)=d$^n$/dt$^n$ M$_x$(0) where M$_x$(t) is the moment generating function of the random variable X

n should be nonnegative.

Michael R. Chernick
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    But note that the opposite is untrue: A distribution may have all moments finite, but still the moment-generating function may not exist! An example is the lognormal distribution. This is the basis for an example of two different distributions having all moments equal: One of the distributions in the example is lognormal. This could not happen with a distribution havingn a mgf: it characterizes the distribution. – kjetil b halvorsen Oct 02 '12 at 22:48
  • @kjetilbhalvorsen Thanks for the add on contribution in comments. – Michael R. Chernick Oct 02 '12 at 23:22