I posted this to mathoverflow and no one's saying anything.

$\newcommand{\cum}{\operatorname{cum}}$ Denote the joint cumulant of several random variables by $\cum(A,B,C,\ldots)$ (more precisely, the cumulant of the joint probability distribution). In particular, what is called the $n$th cumulant of (the probability distribution) of a random variable $A$ is $\cum(\ \underbrace{A,\ldots,A}_{\text{$n$}} \ )$.

Then Brillinger's law of total cumulance (Brillinger didn't use that term) says (in the concrete case of exactly four random variables) the following. There is one term for each partition of the set of random variables (so in the case of just four of them there are $15$ terms): $$ \begin{align} \cum(A,B,C,D) & = \cum(\cum(A,B,C,D\mid X)) \\[8pt] & \phantom{={}} + \underbrace{\cum(\cum(A,B,C\mid X),\cum(D\mid X)) + \cdots}_{\text{4 terms}} \\ & \phantom{={}} + \overbrace{\cum(\cum(A,B\mid X),\cum(C,D\mid X)) + \cdots}^{\text{3 terms}} \\[8pt] & \phantom{={}} + \underbrace{\cum(\cum(A,B\mid X),\cum(C\mid X),\cum(D\mid X))+\cdots}_{\text{6 terms}} \\[8pt] & \phantom{={}} + \cum(\cum(A\mid X),\cum(B\mid X),\cum(C\mid X),\cum(D\mid X)), \end{align} $$ and similarly for numbers of random variables other than $4$, so for example, for six random variables, there are $203$ terms since there are that many partitions of the set of six.


  • If $A,B,C,D$ are independent of $X$, then all terms vanish except the first, and the right and left sides of the equality become identical; and
  • If $A,B,C,D$ are completely determined by $X$, then all terms vanish except the last, which again simplifies to be identical to the expression to the left of "$=$".

So one can GUESS that as one moves from one extreme to the other, with varying degrees of dependence, the first term goes from being predominant to contributing little, and the last from contributing very little to predominating.

Statisticians have written innumerable volumes about the special case $\cum(A,A)$, where that is the same as $\operatorname{var}(A)$, so you're just talking about the "explained" and "unexplained" components of the variance, the the proportion that is explained is just the thing usually denoted $R^2$. The first of the $15$ terms above would be the component that's wholly unexplained by $X$ and the last would be the component that's wholly explained.

Is there some way to make that guess precise? Should one just regard the quotients from dividing each term divided by the total as a sort of summary measure of dependence?

In the case $\cum(A,A)$, every number in $[0,1]$ can occur as the quotient $$\frac{\text{explained component of variance}}{\text{total variance}},$$ and no numbers $>1$ or $<0$ can occur.

Which $15$-tuples of numbers summing to $1$ can occur as the tuple whose components are the terms on the right side divided by the one term on the left side? And more generally, for other numbers of random variables than $4$? Is that a hard problem comparable to moment problems?

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    $\begingroup$ As the help explains, cross-posting is discouraged (see the 3rd-last paragraph). Please don't cross post. The correct behaviour is to pick one site for your question, and if you decide later that it's better placed elsehwere, flag it for moderator attention and ask for it to be moved. $\endgroup$ – Glen_b Dec 16 '13 at 23:25
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    $\begingroup$ @Glen_b : It is customary that if no one answer on math.stackexchange.com, then one cross-posts with links to the older posting. Omitting the link and the statement that it's cross-posted is what is frowned on and draws comments. That that's different on stats.stackexchange.com is not something I suspected. $\endgroup$ – Michael Hardy Dec 17 '13 at 1:01
  • $\begingroup$ @Glen_b : The paragraph you cite says "cross-posting is not encouraged on SE sites". That is somewhat misinformed (as I explain above). $\endgroup$ – Michael Hardy Dec 17 '13 at 1:02
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    $\begingroup$ Custom is defined by what people do, and I can show plenty of precedent here for the usual behavior, which matches the help. I think it's interesting that you say that the help is misinformed, rather than consider the alternative, but I think we should take this to meta. $\endgroup$ – Glen_b Dec 17 '13 at 1:30
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    $\begingroup$ I think it's the same information as on the rest of the network, but if it's wrong, or it's being interpreted wrongly, we ought to discuss that. Anyway, I've posted a question to meta where I hope you take the opportunity to educate me on the error of my ways. I look forward to learning something it seems I've missed (no sarcasm, irony or any such intended; I really do hope to be educated, even if I might take a little convincing). $\endgroup$ – Glen_b Dec 17 '13 at 1:40

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