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For $n$ independent uniform points in a rectangle, condition on one point with normalized coordinates $(x,y)$. It is a maximal external point if all others avoid the upper-right rectangle of area $(1-x)(1-y)$, giving conditional probability $[1-(1-x)(1-y){]}^{n-1}$. Integrating gives $H_n/n$, where $H_n$ is a harmonic number. Linearity of expectation over the $n$ indicator random variables gives $\mathbb{E}{K}_n=H_n$. The same answer holds for independent continuous coordinate distributions, by transforming each coordinate with its distribution function.

Integrating the uniform density gives $\mathbb{E}{X}^k={\int }_0^1{x}^{k}dx=1/(k+1)$. By independence of random variables, $\mathbb{E}(XY{)}^k=\mathbb{E}{X}^k\mathbb{E}{Y}^k=1/(k+1{)}^2$. For the remaining expected value, integrate first in $x$:The value at $y=0$ is supplied by continuity. Therefore, with the harmonic number $H_m={\sum }_{j=1}^{m}1/j$,

Conditional on one sample point being $(x,y)$, it is a maximal external point precisely when none of the other $n-1$ points falls in the upper-right rectangle $(x,1]\times (y,1]$. Its area is $(1-x)(1-y)$, so its conditional probability of being maximal is $[1-(1-x)(1-y){]}^{n-1}$. Both $1-X$ and $1-Y$ are independent uniform random variables, hence a given point is maximal with probability $H_n/n$. Summing their indicator random variables and applying linearity of expectation gives the expected number of coordinatewise maxima:This grows only logarithmically with the sample size, even though all $n$ points are potential candidates.