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.

  • $\begingroup$ Why is the condition you gave not sufficient? Do you have a counterexample? $\endgroup$ Sep 28, 2011 at 22:03
  • $\begingroup$ 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. $\endgroup$
    – x10000year
    Sep 29, 2011 at 5:14
  • $\begingroup$ Try checking the simplest possible nontrivial case. d=3, k=2, all distributions are on $\{0,1|\]$. $\endgroup$
    – Will Sawin
    Sep 29, 2011 at 5:56
  • 1
    $\begingroup$ 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. $\endgroup$ Sep 29, 2011 at 6:46
  • $\begingroup$ To obtain a necessary and sufficient condition, I think you just need to apply the separating hyperplane (Hahn-Banach) theorem. $\endgroup$ Sep 29, 2011 at 6:48

1 Answer 1


I asked myself the very same question some time ago. First, let me show that the obvious necessary condition is not sufficient.

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 $(Z,X)\sim (Z_3,X_3)$.

Now, there is a condition that, added to the sub-dimensional joint law correspondence is sufficient for feasability. It is written in a short note http://www-fourier.ujf-grenoble.fr/~bkloeckn/papiers/compatibility.pdf on my web page in the case of three variables; the full condition is stated and proved by Hans G. Kellerer. in Verteilungsfunktionen mit gegebenen Marginalverteilungen. (Z. Wahrscheinlichkeitstheorie und Verw. Gebiete, 3:247–270 (1964), 1964.). Note if you do not recognize the name of the journal that its the former name of PTRF (Probability Theory and Related Fields). I do not know a reference in english, but you can have a look at the MR review of that paper.


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