The expectation of two sides of rectangle is equal. Can we deduce that in the expectation the rectangle is not very far from being a square?

Question

Let $T$ be a set of $n\ge 3$ points in the plane such that not all of them lie in a common line. Pick two distinct points $\{a=\left( \begin{array}{c} a_{1} \\a_{2} \end{array} \right) ,b=\left( \begin{array}{c} b_{1} \\b_{2} \end{array} \right)\}$ uniformly at random from $T$. Let $R$ be the rectangle whose sides are horizontal and vertical, and whose two diagonally opposite vertices are $\{a,b\}$. Let $X$ be the length of horizontal side of $R$ and $Y$ be the length of its vertical side (i.e. $X=|a_{1}-b_{1}|$ and $Y=|a_{2}-b_{2}|$).

Assuming $\mathbb{E}(X)=\mathbb{E}(Y)$, how small can be expected value of random variable $Z := \min(X,Y)$? What lower bounds can one get for $\mathbb{E}(Z)$?

Answer 1

Since you can scale everything linearly, let's suppose $\mathbb E(X) = \mathbb E(Y) = 1$.

I'm not sure this is optimal, but I think it must be close. Consider a case where all but two of the points are at the origin, with one point (call it $c$) at $(n/2, 0)$ and one ($d$) at $(0, n/2)$. (You actually want $n$ distinct points, so make all but two points very close to the origin, and insert the word "approximately" as needed below). Then $X = 0$ unless one of your two points is $c$ (which has probability $2/n$ and makes $X = n/2)$, so $\mathbb E(X) = 1$, and similarly $Y = 0$ unless one of your points is $d$. $Z = 0$ unless both $c$ and $d$ are chosen, which has probability $2/(n(n-1))$, so $\mathbb E(Z) = 1/(n-1)$.