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What sequences of real-valued random variables $X_1,X_2,X_3,\ldots$ exist for which for all $n$ and all $k$ $$ \operatorname{cum}_k (X_1+\cdots+X_n) = \operatorname{cum}_k(X_1)+\cdots + \operatorname{cum}_k(X_n) $$ where $\operatorname{cum}_k$ is the $k$th cumulant functional, and $X_1,X_2,X_3,\ldots$ are not independent?

$$ \S $$

PS: The most usual definitions of "cumulant" that one comes across don't make clear any motivation behind the concept, so here is one that does: The $k$th cumulant $\operatorname{cum}_k(X)$ of (the probability distribution of) a random variable $X$ is the value of a certain polynomial in the first $k$ moments $\operatorname{E}(X^\ell),\ \ell=1,\ldots,k$. The polynomial is the unique one that is so chosen that

  • $\operatorname{cum}_k$ is shift-invariant for $k\ge 2$, i.e. $\operatorname{cum}_k(X+\text{constant}) = \operatorname{cum}_k(X)$ (and for $k=1$ it is shift-equivariant, i.e. $\operatorname{cum}_k(X+\text{constant}) = \operatorname{cum}_k(X)+\text{same constant}$); and
  • $\operatorname{cum}_k$ is homogeneous of degree $k$; and
  • $\operatorname{cum}_k$ is "cumulative", i.e. for independent random variables $X_1,X_2,X_3,\ldots$ one as for all $n$, $\operatorname{cum}_k(X_1+\cdots+X_n) = \operatorname{cum}_k(X_1) + \cdots + \operatorname{cum}_k(X_n)$.

(For example, the $4$th cumulant is the $4$th central moment minus $3$ times the square of the second central moment.)

(I suppose this characterization doesn't really explain why $k=1$ should be an exception in the first bulleted item above, where perhaps the more usual characterizations do.)

Michael Hardy
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  • For the record, I don't believe one should ever define cumulants in terms of Bell polynomials or any other polynomials for that matter. As far as I'm concerned, they should universally be introduced as follows: the mgf of the sum of independent random variables is the product of the mgf's of the factors, so sadly the higher moments are related by a polynomial mess (convolutions). However, taking the (formal) logarithm turns the product into a sum, sorting everything right out--sums go to sums. This definition for instance makes keej's reformulation of the original question obvious. – Joshua P. Swanson Dec 31 '18 at 08:00

1 Answers1

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If all of the cumulants are additive, that means the cumulant generating function is additive, meaning that $\log \textbf{E} e^{-itX_1} + \log \textbf{E} e^{-it X_2} = \log \textbf{E} e^{-it(X_1+X_2)}$.

Now your question reduces to A criterion for independence based on Characteristic function, and it turns out the variables do not have to be independent.

keej
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