Distribution of the $pn$ shortest edges out of $n$ uniform points, $p\to 0$ Suppose I sample $n$ points independently and uniformly at random in the unit square, and then I select the $pn$ shortest edges between all pairs of points, for fixed $0<p<1$.  For large $n$ and small $p$, what does the sum of the lengths of these edges look like (in a distributional sense)?  Geometric intuition says that the sum looks like $C(p)\sqrt{n}$ and that $C(p)\to0$ as $p\to0$, and I'm wondering what $C(p)$ looks like (say to first order)
 A: Here is a heuristic that agrees with the power proposed by @Bullet51 in the comments above, showing that $C(p)$ should grow like $p^{3/2}$. The sum should look like $pn$ times the typical order of the $(pn)$th smallest distance. 
To estimate that distance, consider a simpler problem, where the points are put in $Kn$ bins (each representing a sub-square of the unit square of side $(Kn)^{-1/2}$). Of course, this ignores the possibility that two points could be in neighbouring bins, and closer than points in their own bin, but I don't think it changes the order. 
If $n$ points are distributed between the $Kn$ bins, the distribution is approximately that each bin receives a Poisson number of points with mean $1/K$. Let $X_1,\ldots, X_{Kn}$ be the number of points in each bin, and assume these are independent Poisson random variables with mean $1/K$. Now the number of pairs of points lying in the same bin is $\sum_{i=1}^{Kn} \frac 12X_i(X_i-1)\sim Kn/2(\mathbb EX^2-\mathbb EX)$. For a Poisson with mean $1/K$, the term in parentheses is $1/K^2$, so the expected number of pairs of points lying in the same bin is $n/(2K)$. 
If we set this equal to $pn$, it tells you how big the bins have to be in order to have $pn$ pairs, that is $K=1/(2p)$. Hence our crude model suggests that the $(pn)$th smallest distance is of size roughly $\sqrt {1/(Kn)}\sim \sqrt{p/n}$. So the sum of the first $pn$ distances should grow like $p^{3/2}\sqrt n$. 
