Suppose I have a measure $\mu$ over $\mathbb R_+$ given by its moments $\mu_0,...,\mu_n$, defined as :
$$\mu_k = \int x^{k} \partial\mu(x),\; k \in 1,...,n$$ Using Faà di Bruno's formula, I can obtain the corresponding cumulants $\kappa_0,...,\kappa_n$.
Say that there exists another measure $\nu$ that happens to have the set of moments $\kappa_0,...,\kappa_n$.
Is there some work somewhere about the relationship between $\mu$ and $\nu$ ?
$\newcommand\ka\kappa$$\newcommand\R{\mathbb R}$Without loss of generality, $\mu_0=1$, so that $\mu$ is a probability measure. Let $\ka$ be a measure on $\R$ with moments $\ka_0,\dots,\ka_n$ and $|\ka|:=\ka(\R)$.
Of course, only in exceptional cases (such as a case with $\mu_0=\mu_2=\mu_4=1$ and $\mu_1=0$) is a measure determined by finitely many of its moments. Therefore, usually the measure $\mu$ will not be determined by $\ka$ (nor $\ka$ will be determined by $\mu$).
However, for each measure $\ka$ with moments $\ka_0,\dots,\ka_n$ we can construct a measure $\nu$ with the same moments as $\mu$ for all moment orders from $0$ through $n$. Indeed, the accompanying (infinitely divisible) compound Poisson probability distribution $$\nu:=\nu_\ka:=e^{-|\ka|}e^{*\ka}=e^{-|\ka|}\sum_{j=0}^\infty\frac{\ka^{*j}}{j!}$$ will have the same moments as $\mu$ for all moment orders from $0$ through $n$; this is easily seen by considering the Fourier transform of $\nu_\ka$.
