4

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.

x10000year
  • 43
  • 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 05:14
  • Try checking the simplest possible nontrivial case. d=3, k=2, all distributions are on ${0,1|]$. – Will Sawin Sep 29 '11 at 05:56
  • 1
    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 06:46
  • 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 06:48
  • Doing so, you get the necessary and sufficient condition, for all $f_1,\ldots,f_m\colon\mathbb{R}^d\to\mathbb{R}$ each just depending on $k$ variables, there exists $x\in\mathbb{R}^d$ with $\sum_i(f_i(x)-\mathbb{E}[f_i(X)])\ge0$. Could do with simplifying though. – George Lowther Sep 29 '11 at 07:06
  • Actually, this question is interesting from the point of view of quantum mechanics, where the k dimensional subspaces correspond to commuting sets of observables (so jointly observable). Therefore, Bell's inequality is one example of a necessary condition - and the fact that it is violated in some quantum systems demonstrates the necessity. – George Lowther Sep 29 '11 at 10:44

1 Answers1

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.

Benoît Kloeckner
  • 14,198