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3 votes
2 answers
278 views

The disintegration of the convolution of two probability measures

Let $G$ be a topological group with all the topological conditions in order that some form of the disintegration theorem be applicable (for instance, take $G$ metrizable). Let $N$ be normal and closed,...
Alex M.'s user avatar
  • 5,407
3 votes
1 answer
167 views

When are all the convolution roots of an infinitely divisible probability measure infinitely divisible?

Let $G$ be a topological group. Let us say that a probability measure $\mu$ on $G$ is strongly infinitely divisible (SID) if $\mu$ is infinitely divisible and any probability measure $\nu$ on $G$ ...
Iosif Pinelis's user avatar
3 votes
0 answers
741 views

Justification of the convolution operation of $L^1(G)$ functions where $G$ is a LCA group (measurability)

Suppose $G$ is a locally compact abelian Hausdorff group (LCA), and $\lambda$ is the Haar measure on it. We all know the convolution of two $L^1(\lambda)$ functions $f$ and $g$ on $G$ is defined as $$...
Hua Wang's user avatar
  • 960
2 votes
0 answers
80 views

Are the roots of an infinitely divisible probability infinitely divisible themselves?

Let $\mu$ be an infinitely divisible probability on a topological group $G$. If $\nu ^{* n} = \mu$ for some $n$, is $\nu$ an infinitely divisible probability too? A sufficient criterion would be to ...
Alex M.'s user avatar
  • 5,407