Timeline for Bound on the joint distribution of three real random variables with given two dimensional marginals
Current License: CC BY-SA 3.0
13 events
when toggle format | what | by | license | comment | |
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Aug 24, 2016 at 20:05 | comment | added | Jean Duchon | An approach is sketched here: mathoverflow.net/questions/248205 | |
Aug 24, 2016 at 8:20 | comment | added | Jean Duchon | I now think a solution is not carried by a piecewise smooth set but by a fractal 2-dimensional set. | |
Aug 18, 2016 at 4:20 | comment | added | user44143 | @JeanDuchon, your counting seems off: the plane orthogonal to (1,1,-1) is just x+y=z, so it has no sector in the --+ octant. I don't think 9 angular sectors will give the right marginals. | |
Aug 18, 2016 at 2:04 | comment | added | Jean Duchon | Pieces of Gaussians on the 12 angular sectors (of the 3 planes othogonal to $(1,\pm1,\pm1)$ except $(1,1,1)$ ) contained in the 4 octants $+++$, $+\,-\,-$, $-\,+\,-$ and $-\,-\,+$ should do the trick, I think. Proving it is optimal is beyond me, I'm afraid. | |
Aug 17, 2016 at 18:17 | comment | added | fedja | I would rather look at the surfaces $\Phi(x)+\Phi(y)+\Phi(z)=c$ (in the positive octant; the other 3 are the same with minuses). At least, the surface area on these ones projects correctly (proportionally to the planar bivariate Gaussian measure) to all 3 planes at each point... | |
Aug 17, 2016 at 15:22 | comment | added | Jean Duchon | Then Gaussians restricted to quarter planes in sectors defined by the signs of $x,y$ and $z$ (a variant of michael's example, but singular) ?? | |
Aug 17, 2016 at 15:11 | comment | added | fedja | Linear planes are not terribly attractive because if you cube the equation $ax+by+cz=0$, you'll be able to express $xyz$ as a linear combination of products of at most $2$ variables, so in this case $E(XYZ)=0$ | |
Aug 17, 2016 at 13:41 | comment | added | Jean Duchon | ... or maybe a sum of Gaussians on some of the planes $x\pm y\pm z=0$ ?? (excluding $x+y+z=0$, probably) | |
Aug 17, 2016 at 13:22 | comment | added | Jean Duchon | @MattF. At first I find this surprising. But why not? Minimizing the Kullback-Leibler divergence of the distribution of $(X,Y,Z)$ with respect to the unit Gaussian, with some given positive value of $\mathbb E(XYZ)$ (less than the maximum) must have this entropy diverge when the value approaches the maximum. Then we should search for a maximizing distribution (of $(X,Y,Z)$) carried by a two dimensional set. Couldn't it be a Gaussian on some plane? | |
Aug 17, 2016 at 12:57 | comment | added | user44143 | It seems that no continuous distribution is maximal. Eg if [a,b]^3 has positive mass, move epsilon of mass from each of the corners with odd numbers of a's to each of the corners with odd numbers of b's. This keeps the marginals constant and increases E(XYZ) by (b-a)^3 epsilon. | |
Aug 16, 2016 at 15:50 | comment | added | user83457 | I don't think so but I don't know. | |
Aug 16, 2016 at 15:19 | comment | added | Jean Duchon | Very nice, thanks. Do you imply this could be the exact maximum? | |
Aug 16, 2016 at 13:35 | history | answered | user83457 | CC BY-SA 3.0 |