# There are d random variables. Given all k-D joint probability distributions with some k<d, what is the necessary and sufficient condition for these distributions to be feasible?

Consider $d$ random variables. For each set of $k$ variables, we are given a joint probability distribution. We want to know that whether these distributions correspond to a valid joint probability distribution of all $d$ variables. We can assume that each variable has a finite domain.

I think a necessary condition is that, all given distributions should agree with the same lower dimensional distributions when we integrates some variables out. But this seems not a sufficient condition.

Is there any simple necessary and sufficient condition? or can we find a simple but stronger necessary condition? or is the above necessary condition in fact sufficient? Thanks.

• Why is the condition you gave not sufficient? Do you have a counterexample? – John Wiltshire-Gordon Sep 28 '11 at 22:03
• I don't know. Since I can't prove it, I can't say that is sufficient. Do you know any proof? I think if the necessary condition is sufficed, then there exists a d-D joint distribution probably with some negative entries that are consistent with all given k-D joint distributions. But I don't know whether we can always find out a d-D joint distribution with all non-negative entries. – x10000year Sep 29 '11 at 5:14
• Try checking the simplest possible nontrivial case. d=3, k=2, all distributions are on $\{0,1|\]$. – Will Sawin Sep 29 '11 at 5:56
• For $k=2$ then you need the covariance matrix to be positive semidefinite. This is not guaranteed just by having the one dimensional distributions being consistent. – George Lowther Sep 29 '11 at 6:46
• To obtain a necessary and sufficient condition, I think you just need to apply the separating hyperplane (Hahn-Banach) theorem. – George Lowther Sep 29 '11 at 6:48

Let $X_1,Y_1,Y_2,Z_2,Z_3,X_3$, be six random variables having the same non-deterministic law such that: $X_1=Y_1$, $Y_2=Z_2$ and $(Z_3,X_3)$ are independent. Then there cannot exist $(X,Y,Z)$ such that $(X,Y)\sim (X_1,Y_1)$, $(Y,Z)\sim (Y_2,Z_2)$ and $(Z,X)\sim (Z_3,X_3)$.