1
$\begingroup$

I have a two-sided Laplace-Stieltjes transform, $$ \int_{-\infty}^{+\infty} e^{-xt}d\mu(t) $$ that converges absolutely in $(a,b)$.

If the measure $\mu$ is finite,then $$ \int_{-\infty}^{+\infty}d\mu(t)=\mu(\mathbb{R})<\infty $$ can I conclude that $(a,b)$ MUST contain the origin?

In general, how changes the interval of convergence of a two-sided Laplace-Stieltjes transform with respect to $\mu$?

Thank you

$\endgroup$
0
$\begingroup$

Even if $\mu$ is not a positive measure, its total variation measure $|\mu|$ is, and "the integral converges absolutely" means $\int_{\mathbb R} e^{-xt}\ d|\mu|(t) < \infty$.
Now $\int_{A} e^{-xt}\ d|\mu|(t)$ is a convex function of $x$ for every measurable $A \subseteq \mathbb R$, so the set where this is finite is a convex set, i.e. an interval. If $\mu$ is a finite measure, that says $\int_{\mathbb R} e^{-xt}\ d|\mu|(t) < \infty$ for $x=0$, so $0$ is in the interval.

Note, however, that the interval doesn't have to be open. For example, take the measure with density $1/x^2$ for $x > 1$, $0$ elsewhere: this has interval $[0,\infty)$.

$\endgroup$
  • $\begingroup$ Do you know if is it possible to prove analyticity of the laplace-stieltjes transform for a $\mu$ non-finite? Eventually, do you have some references? $\endgroup$ – alemou Jun 24 '13 at 14:48
  • $\begingroup$ If $\int_{\mathbb R} e^{-xt}\ d|\mu|(t)$ converges for $x \in (a,b)$, then $\int_{\mathbb R} e^{-zt}\ d\mu(t)$ should be analytic in $a < \text{Re}(z) < b$, as this will be the limit of $\int_{-R}^R e^{-zt}\ d\mu(t)$ as $R \to \infty$, and convergence will be uniform on $c \le \text{Re}(z) \le d$ if $a < c < d < b$. $\endgroup$ – Robert Israel Jun 24 '13 at 15:07
0
$\begingroup$

In case $\mu$ is finite positive, let us consider its normalization $\mu'(\cdot):=\mu(\cdot)/\mu(\Bbb R)$ which is clearly a probability measure on reals. As a result, $$ f_\mu(x):=\int_\Bbb R\mathrm e^{-xt}\mu(\mathrm dt) = \mu(\Bbb R)\int_\Bbb R\mathrm e^{-xt}\mu'(\mathrm dt) = \mu(\Bbb R)f_{\mu'}(x) = \mu(\Bbb R)m_{\mu'}(-x). $$ where $m_{\mu'}$ is a moment-generating function (MGF) of the distribution $\mu'$. Clearly, the LHS is finite iff the RHS is, so that your question for positive finite $\mu$ is equivalent to studying the properties of MGF and in particular of heavy-tailed distributions. Note, the clearly $m_{\mu'}(0) = 1$ regardless of $\mu'$, but it does not have to exist anywhere else, e.g. if $\mu'$ is Cauchy distribution.

$\endgroup$
  • $\begingroup$ Thus, $\mu$ is positive and finite iff $(a,b)$ contain the origin? $\endgroup$ – alemou Jun 24 '13 at 12:48
  • $\begingroup$ @alexou No, $\mu$ is finite iff the integral converges for the origin (trivial). However, even if $\mu$ is finite and positive, the integral may not converge anywhere else. $\endgroup$ – Ilya Jun 24 '13 at 12:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.