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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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In the sentence "hence there should be examples..." there seems to be missing something. Anyway: $m''(0)$ exists and is finite iff the second moment exists and is finite. This follows directly from the definition. – Tim van Beek Jul 8 '11 at 10:05
@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. – Ilya Jul 8 '11 at 10:19
up vote 5 down vote accepted

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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due to the Tim's comment, existence of $M_2$ implies existence and finitness of $m''(0)$ - but as I can see from your answer, and my counterexample, $m(t)$ need not to exist for $t>0$ if $M_2<\infty$. Do I miss anything? – Ilya Jul 8 '11 at 11:58
Gortaur: The lognormal distribution has finite moments of all orders, but $m(t) = \infty$ for any $t > 0$. – Shai Covo Jul 8 '11 at 12:11
Excellent, thanks. – Ilya Jul 8 '11 at 12:47
I've added the comment into the answer. – Shai Covo Jul 10 '11 at 8:39
if $\mathsf E \xi < 0$ and mgf exists in some neighborhood of $t=0$, does it mean that it will be $t'>0$ such that $m(t')=1$ (or $2$ or $100$)? I mean is it only possible for mgf to diverge to $+\infty$ - or it can not exist in some points due to the other reasons? – Ilya Jul 10 '11 at 16:11

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