Clarification of a statement in the paper

Question

I was going through the paper:

M. Raginsky, “Shannon meets Blackwell and Le Cam: Channels, codes, and statistical experiments,” IEEE ISIT Proceedings, pp. 1220–1224, 2011. (http://maxim.ece.illinois.edu/pubs/raginsky_ISIT11.pdf)

In page. 2, the author says that "..for any real-valued independent random variables $Z$ and $Z'$, if the law of $Z$ is a convolution factor of the law of $Z'$, then..."

What does it mean to say one p.d.f is a convolution factor of another p.d.f?

I tried going through the reference mention there. It was hard to digest.

Thanks for the help.

Answer 1

Based on the paragraph before that statement in the paper, and the comment above, it just says exactly what the comment says: $\mu$ is a convolution factor for $\nu$ if $\nu = \mu * \lambda$ for some measure $\lambda$.

In the context of the paper (in particular, that example), it simply asks whether $Z = Z' + Q$ for some r.v. $Q$. If that is the case, then you can "simulate" $W = Z + X$ as $W = (Q + Z') + X = Q + (Z' + X) = Q + W'$.