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.

This entry was posted on February 3, 2020 at 12:20 am and is filed under Kids, Statistics with tags blackboard, CEREMADE, compatible conditional distributions, Gibbs sampling, joint distribution, marginal density, mathematical puzzle, Université Paris Dauphine. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

3 Responses to “too many marginals”

Gerard Letac Says:
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

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

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

$m=\frac{1}{3}\{m_{AB}+m_{BC}+m_{CA}\}.$

Reply
Gerard Letac Says:
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

$A^1=(0,1/2), A^2=(1/2,1), A^3=(1,0)$

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…

Reply
Uniforms summing to a uniform – Robin Ryder's blog Says:
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 […]

Reply