2
$\begingroup$

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.

$\endgroup$
1
3
$\begingroup$

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 $Y$:

enter image description here

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

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

enter image description here

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

$\endgroup$
1
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.