# Interval of convergence of a Laplace-Stieltjes transform

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

## 2 Answers

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 $[0,\infty)$.

• 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? – alemou Jun 24 '13 at 14:48
• 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 $a < c < d < b$. – Robert Israel Jun 24 '13 at 15:07

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.

• Thus, $\mu$ is positive and finite iff $(a,b)$ contain the origin? – alemou Jun 24 '13 at 12:48
• @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. – Ilya Jun 24 '13 at 12:56