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 […]

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