What's the distribution of the closest point from uniform samples? Suppose you have $N$ values $x_1, \ldots, x_N$ that are uniformly sampled in $[0; 1]$. For a random $x_k$ amongst the $(x_i)_i$ (with equiprobability), what is the expected value of the distance between $x_k$ and the closest $x_j\in (x_i)_i$ (with obviously $k\neq j$) ?
 A: For an analytic solution, you need to start by finding the distributions of differences
between adjacent order statistics. Then you need to deal with two cases: (a)
where $X_j$ is the max or min because those to order statistics have only
one neighbor, and (b) where $X_j$ is somewhere in the middle.
Intuitively, if you have $n$ observations, the order statistics are on average
spaced equally, so that the average distance between them is $\frac{1}{n+1}.$
That takes care of case (a). For case (b), the closest of two differences will be smaller.
By simulation, the average distance from the min $X_{1:n}$ to the next larger value
$X_{2:n}$ does indeed seem to be $\frac{1}{n+1}.$ By symmetry, the average
distance between $X_{(n-1):n}$ and $X_{n:n}$ is the same. Simulation if you happen to choose the smallest of $n=5:$
set.seed(725)  # for reproducibility
m = 10^6;  n = 5;  d = numeric(m)
for(i in 1:m) {
  x = sort(runif(n))
  d[i] = x[2] - x[1] }
mean(d)
[1] 0.1666103  # aprx 1/6

Simulation, if you happen to choose a value other
than the max or min. This seems to be half as large
as for the max or min.
set.seed(2019)
m = 10^5;  n = 5;  d = numeric(m)
for(i in 1:m) {
  x = sort(runif(n))
  d[i] =  min(diff(x[2:4])) }
mean(d)
[1] 0.08302884 

So it seems the answer for a randomly chosen one of $n = 5$ is $(2/5)(1/6) + (3/5)(1/12) = 7/60.$
