Let $X,Y,Z$ be real r.v. with $(X,Y)$, $(Y,Z)$ and $(Z,X)$ centered unit normal. How large can $\mathbb E (XYZ)$ be?
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2 Answers
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My entry is upper bound of $\sqrt{\frac 2 {\pi}}$ via $E(XYZ) \le E \frac { X^2|Y| + Z^2|Y|} 2 = E(X^2)E(|Y|)$ by the marginal independence of X and Y, and you can achieve $(\sqrt{\frac 2 {\pi}})^3$ by choosing $Z = sgn(X) sgn(Y) |W|$, W independent of everything, which works as e.g. conditional on X having some positive value the distribution of Z is the same as $sgn(Y) |W|$, which is easily seen to be normal.
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$\begingroup$ Very nice, thanks. Do you imply this could be the exact maximum? $\endgroup$ Commented Aug 16, 2016 at 15:19
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$\begingroup$ I don't think so but I don't know. $\endgroup$– user83457Commented Aug 16, 2016 at 15:50
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$\begingroup$ 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. $\endgroup$– user44143Commented Aug 17, 2016 at 12:57
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$\begingroup$ @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? $\endgroup$ Commented Aug 17, 2016 at 13:22
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1$\begingroup$ 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... $\endgroup$– fedjaCommented Aug 17, 2016 at 18:17
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For a discretized version of this problem, $E(XYZ)$ can be as high as $5/(4\sqrt{3})$, or ~ 0.72.
Suppose that $X$, $Y$, $Z$, are all constrained to have values in $\{\pm \sqrt{3}, \pm 1 / \sqrt{3}\}$. Let
\begin{align} P(X=x, Y=y, Z=z) = & 1/64 \text{ in the }\text{ 4 cases with } xyz = \sqrt{27}\\ & \ \ 0 \ \ \ \text{ in the 12 cases with } xyz = \sqrt{3}\\ & 3/64 \text{ in the 12 cases with } xyz = 1/\sqrt{3}\\ & 6/64 \text{ in the }\text{ 4 cases with } xyz = 1/\sqrt{27}\\ & \ \ 0 \ \ \ \text{ in the 32 cases with } xyz < 0\\ \end{align} which reduces the calculation of $E(XYZ)$ to multiplying the numbers on each row and adding up the results. The marginal distributions are \begin{align} P(X=x, Y=y) = & 1/64 \text{ in the 4 cases with } xy = \pm 3\\ & 3/64 \text{ in the 8 cases with } xy = \pm 1\\ & 9/64 \text{ in the 4 cases with } xy = \pm 1/3\\ \end{align} \begin{align} P(X=x) = & 1/8 \text{ in the 2 cases with } x = \pm \sqrt{3}\\ & 3/8 \text{ in the 2 cases with } x = \pm 1/\sqrt{3}\\ \end{align} So each of $X,Y,Z$ is a variable with mean 0 and variance 1, and they are pairwise independent.
I expect that there are similar continuous distributions with similar values for $E(XYZ)$.
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$\begingroup$ The discrete case is interesting too, though it doesn't address the question proper. Why not consider simply $X=\pm1$ with probability 1/2 each, where the solution is trivially $\frac14(\delta_{+++}+\delta_{+--}+\delta_{-+-}+\delta_{--+})$ ? That makes $XYZ=1$ with probability 1 ... $\endgroup$ Commented Aug 18, 2016 at 2:16
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$\begingroup$ I think the discrete approach can lead to a solution, eg: Start with X,Y,Z independently distributed as B(6,1/2), linearly transformed so each has mean 0 and variance 1. Restrict these to octants with XYZ>0. Then experiment over all prisms [a,b]x[c,d]x[e,f] to find where moving mass from one set of four vertices to the other will most increase E(XYZ). Repeat. The B(3,1/2) case leads to the solution above; from the B(6,1/2) case we could probably guess the answer with continuous setup easily enough. $\endgroup$– user44143Commented Aug 19, 2016 at 3:12