Theoretical Statistics and Mathematics Unit, ISI Delhi

January 21, 2015 (Wednesday) ,
3:30 PM at Webinar

Speaker:
Christian Berg,
University of Copenhagen, Denmark

Title:
On the determinacy/indeterminacy of the Stieltjes moment sequence $(n!)^c,c>0$

Abstract of Talk

Let $f$ be a non-zero Bernstein function, i.e.,
$$
f(s)=a+bs+\int_0^\infty \left(1-e^{-xs}\right)\,d\nu(x),\;a,b\ge 0,\nu\ge 0.
$$
There exists a uniquely determined
product convolution semigroup $(\rho_c)_c>0$ on $(0,\infty)$ such that
\begin{equation}\label{eq:1}
\int_0^\infty x^n\,d\rho_c(x)=(f(1)\cdots\ldots\cdot f(n))^c,\quad c>0,n=0,1,\ldots,
\end{equation}
see \cite{B1}. The Stieltjes moment sequence in \eqref{eq:1} is always
determinate when $c\le 2$ as an easy consequence of the Carleman criterion. However, for $c>2$,
it can be determinate or indeterminate depending on $f$. In fact, in the case $f(s)=s$, where
the moment sequence is $(n!)^c$,
it was proved in \cite{B1} that the moment sequence is indeterminate. In this case $\rho_c=e_c(t)dm(t)$, where $m$ is Lebesgue measure on the half-line and
\begin{equation}\label{eq:2}
e_c(t)=\frac{1}{2\pi}\int_{-\infty}^\infty t^{ix-1}\Gamma(1-ix)^c\,dx,\quad t>0.
\end{equation}
The proof of the indetermincy was quite delicate based on asymptotic formulas for stable distributions due to Skorokhod.
In recent work with Jos\'e L\'opez, Spain, \cite{B:L} we have found the asymptotic behaviour of $e_c$ at infinity, viz.
\begin{equation}\label{eq:3}
e_c(t)\sim \frac{(2\pi)^{(c-1)/2}}{\sqrt{c}}\frac{e^{-ct^{1/c}}}{t^{(c-1)/(2c)}}.
\end{equation}
From \eqref{eq:3} it is easy to derive the indeterminacy of $e_c$ from a criterion of Krein.
In the lecture I will give the necessary background for the above.
\begin{thebibliography}{abc}
\bibitem{B1} Berg, C., On powers of Stieltjes moment sequences, I. {\it J. Theor. Prob.} {\bf 18} (2005), 871--889.
\bibitem{B:L} Berg, C., L{\'o}pez, J. L., Asymptotic behaviour of the Urbanik semigroup. To appear
in J. Approx. Theory 2015.
\end{thebibliography}