TITLE: Expected number of Pareto-optimal points
QUESTION [9 upvotes]: Suppose $S$ is a set of $n$ points in a plane.
A point is called maximal (or Pareto-optimal) if no other point in $S$ is both above and to the right of that point.
If each point in $S$ is chosen independently and uniformly at random from the unit square $[0,1]\times [1,0]$.  What is the exact expected number of Pareto-optimal points in $S$?

REPLY [6 votes]: Label the n points $p_1,…,p_n$ in descending order of their y coordinates.  $p_1$ is maximal since it is above all other points, and its expected contribution to the total number of maximal points is therefore 1.
$p_2$ is above points $p_3,…,p_n$ but below $p_1$, and is therefore maximal iff it is to the right of $p_1$.  Since the distribution of $p_1$ and $p_2$ in the x dimension is random and independent of that in the y dimension, it is equally likely that either one of them will be to the right of the other.  The probability that $p_2$ is maximal is therefore $\frac 1 2$, and its expected contribution to the total number of maximal points is $\frac 1 2$ x 1 = $\frac 1 2$.
Now consider any point $p_m$.  Generalising from the case of $p_2$, $p_m$ is above points $p_{m+1},…,p_n$ but below points $p_1,…,p_{m-1}$, and is therefore maximal iff it is to the right of all of  $p_1,…,p_{m-1}$.   Since the distribution of $p_1,…,p_m$ in the x dimension is random and independent of that in the y dimension, it is equally likely that any one of them will be to the right of all the others.  The probability that $p_m$ is maximal is therefore $\frac 1 m$, and its expected contribution to the total number of maximal points is $\frac 1 m$ x 1 = $\frac 1 m$.
The expected number of maximal points is therefore  1 + $\frac 1 2$ … + $\frac 1 n = H_n$, the $n$th Harmonic number.