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For a real-valued r.v. $\xi$ we suppose and existence of its moment generating function $m(t) = \mathsf E\mathrm e^{t\xi}$ for all $t\in (-h,h)$ where $h>0$. I wonder how many moments $M_n = \mathsf E\xi^n$ are finite.

I know that using the differentiation under the Lebesgue integral, local existence of $m(t)$ implies an existence of $M_1 = \mathsf E\xi$. On the other hand for the 2nd moment it seems that the finiteness of $M_2$ is equivalent to the statement that $m''(0)$ exists, hence there should be examples when $m(t)$ in the neighborhood of $0$ while $M_2 = \infty$.

For an example when $M_2$ does exist without an existence of $m(t)$ for $t>0$ we can consider a r.v. with a density $ f(x) = \frac{2}{\pi(1+x^2)^2}. $

Could you provide such an example when $m(t)$ exists in the neighborhood of $t=0$ but $M_2 = \infty$? Maybe you can also refer me to the literature since this question is not covered in my book on the probability.

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    @Tim: how does it follow *directly* from the definition? With regards to the example - I meant the case when $m(t)$ exists, but does not have a second derivative at $t=0$. Edited for the clarity.2011-07-08

1 Answers 1

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As is well known, if the moment-generating function (mgf) exists in some open interval containing $0$, then all moments are finite. Indeed, suppose that $\xi$ has a finite mgf in some open interval containing $0$. Then, there exists a $t \neq 0$ such that $ \int_{( - \infty ,0)} {e^{ (- |t|)x} F(dx)} < \infty $ and $ \int_{[0,\infty)} {e^{|t|x} F(dx)} < \infty , $ where $F$ is the distribution of $\xi$. Given $n \in \mathbb{N}$, this implies that $ \int_{( - \infty ,0)} {|x|^n F(dx)} < \infty $ and $ \int_{[0 ,\infty)} {x^n F(dx)} < \infty , $ respectively (note that $\int_{[ - M,M]} {|x|^n F(dx)} < \infty $, for any $M > 0$ fixed). Hence $ \int_{( - \infty ,\infty)} {|x|^n F(dx)} < \infty, $ so the $n$th moment of $\xi$ is finite.

EDIT (two well-known facts):

1) The converse is not true: the lognormal distribution has finite moments of all orders, but $m(t)=\infty$ for any $t>0$.

2) In case $\xi$ has a finite mgf $m(\cdot)$ in some open interval containing $0$, then it holds $ m^{(n)}(0) = {\rm E}(\xi^n), $ and $m(\cdot)$ may be expanded in a power series about $0$ as $ m(t) = \sum\limits_{n = 0}^\infty {\frac{{m^{(n)} (0)}}{{n!}}t^n } = \sum\limits_{n = 0}^\infty {\frac{{{\rm E}(\xi ^n )}}{{n!}}t^n } . $

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    Thanks, that is what I thought.2011-07-10