## too many marginals This 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.

### 3 Responses to “too many marginals”

1. Gerard Letac Says:

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 $A=(1/2,0), B=(0,1/2),\ C=(1,0)$

and the three probabilities $m_{AB}(dx,dy), \ m_{BC}(dx,dy),\ m_{CA}(dx,dy)$

which are respectively the uniform distributions of the segments AB, BC, CA. Then $m=\frac{1}{3}\{m_{AB}+m_{BC}+m_{CA}\}.$

2. Gerard Letac Says:

Not very difficult. Correct, the set C of solutions is a convex set. Here is probably the simplest extremal point of C. Take $A^1=(0,1/2), A^2=(1/2,1), A^3=(1,0)$

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 $m=\frac{1}{3}(m^1+m^2+m^3)$

has the desired property…

3. […] 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 […]

This site uses Akismet to reduce spam. Learn how your comment data is processed.