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Consider two RV's $X_1$ and $X_2$, independently and uniformly distributed over $[0,1]$.

I want to find the largest possible area $A\subset [0,1]^2$ while keeping the average of $X_1$ and $X_2$ inside of $A$ above a certain level, say $y\in(1/2,1)$. For $y\leq 1/2$, it is easy to see that $A=[0,1]^2$ solves the problem, and that's why I excluded $y$'s below $1/2$.

So, the programming problem should be

$$\max_{A\in[0,1]^2}\int_Adx_1dx_2$$ $$s.t. E[X_1|(X_1,X_2)\in A]\geq y~\textrm{and}\\E[X_2|(X_1,X_2)\in A]\geq y.$$

Is there a general solution to this problem or any reference I can have a look to find similar types of problems?

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1 Answer 1

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Adding the two constraints gives $E[X_1 + X_2 \mid (X_1, X_2) ∈ A] ≥ 2y$. The optimal $A$ under only this weaker constraint is clearly the region $x_1 + x_2 ≥ c$ for appropriately chosen $c$. Specifically,

  • if $\frac12 ≤ y ≤ \frac23$, then $c$ is the root of $3 - c^3 = 3(2 - c^2)y$ with $0 ≤ c ≤ 1$;
  • if $\frac23 ≤ y ≤ 1$, then $c = 3y - 1$.

But since this $A$ is symmetric in $x_1$ and $x_2$, it happens to satisfy the original stronger constraints—so it’s optimal under them as well.

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  • $\begingroup$ In a parallel way, if we may consider another weaker constraint, say $E[\alpha X_1+(1-\alpha)X_2|(X_1,X_2)\in A]\geq y$, $\alpha\in(0,1)$. If this is the case, can we say $\alpha x_1+(1-\alpha)x_2\geq c$ for an appropriately chosen $c$ can also be a solution to the problem? $\endgroup$
    – Andeanlll
    Sep 20, 2022 at 5:41
  • $\begingroup$ @Andeanlll No: if $a ≠ \frac12$, that solution won’t be symmetric in $x_1$ and $x_2$, so it will fail to satisfy the original stronger constraints. $\endgroup$
    – Anders Kaseorg
    Sep 20, 2022 at 6:19
  • $\begingroup$ Got it. Thanks. $\endgroup$
    – Andeanlll
    Sep 22, 2022 at 5:07

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