# Given pairwise marginal distributions, what can we say about the full joint?

From reading previous posts, I understand that, if we have pairwise marginals, say $$P(A,B)$$, $$P(B,C)$$, and $$P(C,A)$$, this doesn't allow us to reconstruct the full joint $$P(A,B,C)$$. But does it allow us to say whether or not a valid probability distribution $$P(A,B,C)$$ exists or not?

For instance, suppose $$A,B,C$$ are binary, and that the pairwise marginals are such that the events that each configuration where the variables take different values are equally likely. So $$P(A=1,B=0) = P(A=0,B=1) = 0.5$$, and likewise for the other two marginals. Do these pairwise marginals correspond to a valid probability distribution over all three variables? Intuitively I would say no, because the only events that occur are clearly when "adjacent" variables take different values, but having $$A \neq B, B \neq C$$, and $$C \neq A$$ induces a contradiction clearly given the binary nature of the variables, but I have no idea how one might go about proving this (if my conclusion about this distribution is even correct, that is) in a less "handwavey" way, since I don't think we can simply determine the probability of each of the 8 configurations of the 3 variables and come to the conclusion based on whether or not they sum to 1.

So my question is: If they don't allow us to construct the joint, what information do the marginals give us about the full joint? Is it simply whether or not it exists, or is it more than that? And, in the specific example above, what is the conclusion?

• Commented Apr 30 at 9:27

If there exists a joint distribution, we have $$p_0(a,b,c)=p_1(a,b)p_2(c|a,b)=p_3(b,c)p_4(a|b,c)=p_5(c,a)p_6(b|c,a)$$ hence $$p_2(c|a,b)=\dfrac{p_3(b,c)p_4(a|b,c)}{\int p_3(b,\chi)p_4(a|b,\chi)\text d\chi}=\dfrac{p_5(c,a)p_6(b|c,a)}{\int p_5(\chi,a)p_6(b|\chi,a)\text d\chi}$$ and $$p_4(a|b,c) %\dfrac{p_3(b,c)p_4(a|b,c)}{\int p_3(b,\chi)p_4(a|b,\chi)\text d\chi} =\dfrac{p_5(c,a)p_6(b|c,a)}{\int p_5(c,\alpha)p_6(b|c,\alpha)\text d\alpha}$$ which implies that the fixed point functional equation (in the conditional density $$\color{red}{p_6}$$) $$\require{amsmath} \dfrac{p_3(b,c)}{{\Large \int} p_3(b,\chi)\dfrac{p_5(\chi,a)\color{red}{p_6}(b|\chi,a)}{\int p_5(\chi,\alpha){\color{red}{p_6}}(b|\chi,\alpha)\,\text d\alpha}\,\text d\chi}=\dfrac{\int p_5(c,\alpha)\color{red}{p_6}(b|c,\alpha)\,\text d\alpha}{\int p_5(\chi,a)\color{red}{p_6}(b|\chi,a)\,\text d\chi}$$ must have a solution and the same for both other pairs $$(p_3,p_1)$$ and $$(p_1,p_3)$$.