Cumulant of sum of correlated random variables?

Let $$X,Y$$ be two random variables. We denote by $$[X^k]$$ and $$[Y^k]$$ the $$k$$'th order cumulants of $$X$$ and $$Y$$, respectively. I'm interested in computing the $$k$$'th order cumulant of $$Z = X+Y$$.

If $$X,Y$$ were independent, then a well-known property implies that $$[Z^k] = [X^k] + [Y^k]$$.

Now suppose $$X,Y$$ are not independent. Can we write an expression for the cumulants of $$Z$$, from the cumulants of $$X$$, $$Y$$, plus additional terms related to the dependence between $$X$$ and $$Y$$ (e.g., the cumulants of products $$XY$$?

Update: wolfie's post almost answers this question. This expansion feels like something that should be known in the literature. So I am adding the reference tag here, in case anyone can suggest relevant papers.

There might be a number of ways to tackle this. To get a feel for the problem, my first thought was which tool / function could be used to check it out. In the mathStatica package for Mathematica, there is a general function that can find cumulants of power sums $$s_{r,t}=\sum _{i=1}^n X_i^r Y_i^t$$. For example, $$s_{1,0} = \sum _{i=1}^n X_i$$ and $$s_{0,1} =\sum _{i=1}^n Y_i$$.

Then, in a bivariate dependent world, $$Z = X+Y$$ can be written as $$s_{1,0} + s_{0,1}$$ where we are considering the very simple case of $$n = 1$$. Then, the problem at hand is to express the $$r^\text{th}$$ cumulant of Z, in terms of cumulants of the $$X$$ and $$Y$$. This can be done using the CumulantMomentToCumulant function.

Here, for example, is the 3rd cumulant of Z expressed in terms of the bivariate cumulants of $$X$$ and $$Y$$: where $$\kappa _{r,s}$$ denotes the various bivariate cumulants.

Here are the first 8 cumulants of $$Z = X+Y$$: The solution is immediately identifiable by induction as Pascal's Triangle / Binomial Theorem at play.

• This is very cool. I did not expect to get a binomial rule here. I tried to prove it but it gets very messy and I did not manage to get it done. Apr 16 '20 at 20:22