## too many marginals

**T**his week, the CEREMADE coffee room puzzle was about finding a joint distribution for (X,Y) such that (marginally) X and Y are both U(0,1), while X+Y is U(½,1+½). Beyond the peculiarity of the question, there is a larger scale problem, as to how many (if any) compatible marginals h¹(X,Y), h²(X,Y), h³(X,Y), …, need one constrains the distribution to reconstruct the joint. And wondering if any Gibbs-like scheme is available to simulate the joint.

February 5, 2020 at 3:57 pm

You correctly observe that the set C of suitable distributions of (X,Y)$is a compact convex set. Here is a simple extreme point m of C: define

and the three probabilities

which are respectively the uniform distributions of the segments AB, BC, CA. Then

February 4, 2020 at 10:02 pm

Not very difficult. Correct, the set C of solutions is a convex set. Here is probably the simplest extremal point of C. Take

and m³(dx dy) the uniform probability on the segment A¹A², m² the uniform probability on the segment A¹A³, m¹ the uniform probability on the segment A³A². Then

has the desired property…

February 4, 2020 at 4:58 pm

[…] for this. I wrote the problem in the lab coffee room, leading to nice discussions (see also Xian’s blog post). Here are two solutions to the […]