Maximal entropy distribution on three variables knowing its marginals on any two

Question

Observation 0: Given a finite set $X$, the probability distribution on $X$ with highest entropy is the uniform one. This is well known.

Observation 1: Given two finite sets $X,Y$ and two probability distributions, $p$ on $X$ and $q$ on $Y$, there always exists a probability distribution on $X\times Y$ which has these given distributions as marginals, namely the one $p(x)\,q(y)$ whose projections are independent, and it is the one with highest entropy. (See here and here.)

Question: Given three finite sets $X,Y,Z$ and three probability distributions, $p$ on $X\times Y$, $q$ on $X\times Z$ and $r$ on $Y\times Z$:

• how can I decide (in function of $p,q,r$) whether there is a probability distribution on $X\times Y\times Z$ which has these given distributions as pairwise marginals? (I mean, is there a simpler way to state this than by writing the obvious linear feasibility problem expressing it?)

• and more importantly, assuming there is one, how can I compute (in function of $p,q,r$) the one with highest entropy? does this distribution (or this problem) at least have a name?

I am interested both in what can theoretically be said about this problem/distribution, and about computing it in practice. We can assume $|X|=|Y|=|Z|=2$ if this is of help.

(A similar question was asked here, but it's not quite the same setup, and not as symmetric.)

Answer 1

This is only about $|X|=|Y|=|Z|=2$ case: let $X=Y=Z=\{0,1\}$. And $f$ be a "seed" joint distribution on $X\times Y\times Z$. Then if another distribution has same pairwise marginals, there difference is one dimensional: $\mu$ on $(0,0,0),(1,1,0),(1,0,1),(0,1,1)$ and $-\mu$ on compliment.

Then maximizing entropy is equivalent to $(f(0,0,0)+\mu)(f(1,1,0)+\mu)(f(1,0,1)+\mu)(f(0,1,1)+\mu)=(f(1,0,0)-\mu)(f(0,1,0)-\mu)(f(0,0,1)-\mu)(f(1,1,1)-\mu)$. Which seems no simple expression of roots are available.

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Some updates on general case: for every 2-subset of $X,Y$ and $Z$, there is such pertubations and if $f$ is the distribution maximizing entropy, it should satisfy $f(x_0,y_0,z_0)f(x_1,y_1,z_0)f(x_1,y_0,z_1)f(x_0,y_1,z_1)=f(x_1,y_0,z_0)f(x_0,y_1,z_0)f(x_0,y_0,z_1)f(x_1,y_1,z_1)$. Which indicates $f(x,y,z)=\exp(u(x,y)+v(y,z)+w(x,z))$ for some $u,v,w$. It's not hard to show it's also a sufficient condition.