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 $Y$:

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where $\kappa _{r,s}$ denotes the various bivariate cumulants.

Here are the first 8 cumulants of $Z = X+Y$:

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The solution is immediately identifiable by induction as Pascal's Triangle / Binomial Theorem at play.

  • 2
    $\begingroup$ 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. $\endgroup$
    – becko
    Apr 16 '20 at 20:22

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