# Joint distribution from marginals

I have a question about a joint distribution calculated in a paper I am reading.

There are three random variable a, b and c such that $$a,b,c \in \{+1,-1\}$$ and then the joint distribution is given by: $$p(a,b,c) = \frac{1}{8}(1 + aE_{A} + bE_{B} + cE_{C} + abE_{AB} + acE_{AC} + bcE_{BC} + abcE_{ABC})$$ where $$E_{A}$$, $$E_{B}$$ and $$E_{C}$$ are the single-party marginals, $$E_{AB}$$, $$E_{BC}$$ and $$E_{AC}$$ the two-party marginals, and $$E_{ABC}$$ is the three-body correlator.

I feel like this should be something I should have seen in an introductory course on probability but I can't seem to prove it. Also if I convince myself that it's just adding up all the possible cases, I think 1 should be part of the other expected values (i.e. it is already considered in them) so I don't see the point in adding it separately. Also would we have the same expression if the possible values were $$\{+1,0\}$$ (or any other set of size 2) ?

I am used to the notation $$E$$ as the expected value, but this is totally unrelated to that and it is about marginals. Am I correct?

I would be pleased to see a complete proof or a link to study this fact.

Note that the positivity of $$p(a,b,c)$$ implies constraints on marginals, in particular $$p(+ + +) + p(−−−) ≥ 0$$ implies $$E_{AB} + E_{AC} + E_{BC} ≥ −1.$$

which I don't understand.

• Can you add a link to the paper? Nov 19, 2021 at 12:07
• I think I heard the term 'Rademacher chaos' but not sure how deeply it is related to your question. Anyway, the expansion is more or less the result of the fact that monomials $\prod_{x\in A}x$ for $A\subseteq \{a,b,c\}$ spans the space of functions of the form $\{-1,1\}^3\to\mathbb{R}$. Nov 19, 2021 at 12:10
• arxiv.org/abs/1906.06495
– Pegi
Nov 19, 2021 at 12:37
• It is on the second page.
– Pegi
Nov 19, 2021 at 12:38
• The equations $(3)$-$(10)$ in the paper indicate that $E_A,E_B,E_C,E_{AB},E_{AC},E_{BC}$ and $E_{ABC}$ does indeed represent the expected values $E[A],E[B],E[C],E[AB],E[AC],E[BC]$ and $E[ABC]$. Nov 19, 2021 at 14:27