Following up on [Mean minimum distance for N random points on a one-dimensional line](https://mathoverflow.net/questions/1294/mean-minimum-distance-for-n-random-points-on-a-one-dimensional-line) and [Mean minimum distance for N random points on a unit square (plane)](https://mathoverflow.net/questions/124579/mean-minimum-distance-for-n-random-points-on-a-unit-square-plane/317409#317409), I have the following questions.


Given N random points in the unit square, what is the expected value of the Maximum distance between any two such points?


I know that the expected value of the distance is $\Big(2 + \sqrt{2} + 5 \log (1+ \sqrt{2}) \Big)$ and the (mysterious) distribution of the distances is given by the square line picking distribution (<http://mathworld.wolfram.com/SquareLinePicking.html>).


Thank you!



Mean maximum distance for N random points on a unit square
Here is a first approximation for large $n$.


The formulas are easier if we use the diamond whose whose corners are $(\pm 1,0)$, $(0,\pm 1)$. Then the pdf of the $x$-coordinate is just $1-|x|$.


The expected maximum distance is at least $2\, E[\max x]$. If $u>0$,
$$P(\max x < u) = P(\text{all }x < u)
= (1-(1-u)^2/2)^n \sim e^{-(1-u)^2 n/2}$$


So, ignoring the negligible possibility that the maximum might be negative,
\begin{align}
E[\max x] 
&= \int\_0^1 u(1-u)n\,e^{-(1-u)^2 n/2} du\\
&=1 - \sqrt{\frac{\pi}{2n}}\text{erf}\left(\sqrt{\frac n2}\right)\\\
&\sim 1-\sqrt{\frac{\pi}{2n}}
\end{align}


So the expected maximum distance on the diamond is at least $2-\sqrt{2\pi/n}$, and the expected maximum distance on the unit square is at least $\sqrt{2}-\sqrt{\pi/n}$.


Empirically the formula $\sqrt{2}-\sqrt{2/n}$ works well for $n=100$ or $n=1000$. So this rough estimation seems to get the right order of $n$, and better attention to using all four corners of the diamond would get a more accurate result.


