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Suppose that I have two non-negative real valued random variables $x, y \in Z_+$ that always satisfy $$x+y \leq 1.$$ Also suppose that $E[x] = 1/2$ and $E[y] = 1/4$. What is the maximum possible value of $E[xy]$? Can it be larger than $1/8$?

More generally, is there a systematic way of analyzing $E[xy]$ for say other values of $E[x]$ and $E[y]$? (You can assume the assumptions $x, y \in Z_+$ and $x+y\leq 1$ continue to hold.)

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  • $\begingroup$ If $x$ is uniform $[0,1]$ and $y=x/2$, then $E(xy)=1/6\gt 1/8$ I believe. $\endgroup$ Apr 27, 2020 at 4:47
  • $\begingroup$ @BrendanMcKay But that wouldn't satisfy $x+y \leq 1$ since e.g. for $x=1$ you get $y=0.5$ and then $x+y = 1.5$. $\endgroup$
    – Mathman
    Apr 27, 2020 at 4:49
  • $\begingroup$ Right, I forgot that condition, sorry. $\endgroup$ Apr 27, 2020 at 4:52
  • $\begingroup$ Can you not introduce $z$ s.t. $y = (1-x) z$ and $x, z$ dependent random variables on $[0, 1]$, then $E[xy] = E[xz] - E[x^2z]$, and you get all relations from the expectations and covariances of $x, x^2, z$? $\endgroup$
    – user114668
    Apr 27, 2020 at 14:19
  • $\begingroup$ ... in which case you can likely get sharp bounds using Chebyshev, which are usually fulfilled by point masses, as below. $\endgroup$
    – user114668
    Apr 27, 2020 at 14:26

1 Answer 1

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It is possible to have $E[xy]=9/64$, by $$P\left[(x,y)=\left(\frac38,\frac18\right)\right]=\frac12$$ $$P\left[(x,y)=\left(\frac58,\frac38\right)\right]=\frac12$$

This can be guessed by knowing that optimal probability distributions are often normal, uniform, or concentrated at two points.

I don't have a proof that this is optimal, but here is a useful lemma: If $(x,y)$ and $(x',y')$ both have density at least $\epsilon$ in an optimal set-up, then the combination of $x<x'$ and $y>y'$ is impossible. The proof is that if those two inequalities both hold, we could increase $E[xy]$ by placing the density on $(x,y')$ and $(x',y)$ instead.

To actually prove that the above two-point solution is optimal, you'd probably apply arguments like the lemma to show that any optimal distribution must be concentrated on some curve with non-decreasing $x$ and non-decreasing $y$; then that it must be concentrated on some line segment; and then that it must be concentrated on two points.

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    $\begingroup$ Better: $P[(x,y)=(0,0)]=1/4$, $P[(x,y)=(2/3,1/3)]=3/4$, $E(xy)=1/6$. I agree that concentration at two points is likely to be the best option. $\endgroup$ Apr 27, 2020 at 10:02
  • $\begingroup$ With concentration at $(0,0)$ and one other point, the general optimum is $E(x)E(y)/E(x+y)$. I postulate this is globally the best. $\endgroup$ Apr 27, 2020 at 12:46
  • $\begingroup$ Thanks Matt and Brendan for your interesting examples. $\endgroup$
    – Mathman
    Apr 27, 2020 at 13:38
  • $\begingroup$ @BrendanMcKay, we can at least establish that the density should be concentrated at the origin and on the hypotenuse. Any weight of $\epsilon$ at $(t,u)$ with $0<t+u<1$ can be more optimally replaced with weight of $(t+u)\epsilon$ at $(t/(t+u), u/(t+u))$ and weight of $(1-t-u)\epsilon$ at the origin. This increases $E[xy]$ by $\epsilon tu/(t+u)-\epsilon tu$, which is always non-negative. $\endgroup$
    – user44143
    Apr 27, 2020 at 13:45
  • $\begingroup$ I wrote down a linear program to try all possible distributions. I can now guarantee that $1/6$ is the highest possible value for $E[xy]$ and more generally, it seems that @BrendanMcKay's conjecture that $E[xy] \leq E[x]E[y] / E[x+y]$ is correct. $\endgroup$
    – Mathman
    May 1, 2020 at 19:04

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