This can answered without any complicated maths.
It can be related to the following: Imagine you have $N$ marked cards in a pack of $m$ cards and shuffle them randomly. What is the probability that they are all at least distance $d$ apart?
Consider dealing the cards out, one by one, from the top of the pack. Every time you deal a marked card from the top of the deck, you then deal $d$ cards from the bottom (or just deal out the remainder if there's less than $d$ of them). Once all the cards are dealt out, they are still completely random. The dealt out cards will have distance at least d between all the marked cards if (and only if) none of the marked cards were originally in the bottom $(N-1)d$.
The probability that the marked cards are all distance d apart is the same as the probability that none are in the bottom $(N-1)d$.
The points uniformly distributed on a line segment is just the same (considering the limit as $m$$\rightarrow∞$). The probability that they are all at least a distance $d$ apart is the same as the probability that none are in the left section of length $(N-1)d$. This has probability $(1-\frac{(N-1)d}{L})^N$.
Integrating over $0$$\le$$d$$\le$$\frac{L}{(N-1)}$ gives the expected minimum distance of $\frac{L}{(N^2-1)}$.
I'm going to give a somewhat heuristic solution that nevertheless gives the right answers.
$E(D)$ is the distance from the origin to the closest of N points.
Let's replace $N$ with a random variable, namely the Poisson with mean $N$. Then your points form a Poisson process of rate $N$ on the unit cube.
Let's furthermore assume that $N$ is large enough that the closest point to the origin is at distance less than $0.5$ from the origin -- that is, $D < 0.5$ -- so we don't have to worry about the geometry of the cube. If there are a lot of points this is reasonable.
So, at least when $N$ is large, your question has approximately the same answer as: consider a Poisson process of rate $N$ in $R^3$. What's the distribution of the distance from the origin to the point closest to the origin?
But this scales with $N$ - a Poisson process of rate $N$ looks like a Poisson process of rate 1, shrunk by a factor of $N^{1/3}$. So it's enough to solve this problem when $N = 1$. We'll get $P(D < x) = F(x)$ when $N = 1$. For generic values of $N$ we'll get $P(D < x) = F(x N^{1/3})$.
So what's $P(D < x)$ when $N = 1$? It's easier to find $P(D > x)$; this is the probability that no point is within the sphere of radius $x$ centered at the origin. But this sphere has volume ${4 \over 3} \pi x^3$ and therefore this probability is $\exp(-4\pi x^3 /3)$. From this we find that the density of $D$ is $f(x) = 4\pi x^2 \exp(-4\pi x^3/3)$ and the expectation of $D$ is
$$ \int_0^\infty x f(x) \: dx = {\pi^{2/3} 2^{1/3} \over 3^{7/6} \Gamma(2/3)} = 0.5539602785 $$
by using Maple. I conjecture, therefore, that $\lim_{N \to \infty} E(D) N^{1/3}$ is equal to this constant; so $E(D) \approx 0.554 N^{-1/3}$.
To check this I simulated the case N = 1000 -- picking 1000 points in each of 1000 cubes, as follows in R:
cp = function() runif(3,-.5,.5)
dist = function(x) sqrt(sum(x^2))
x = rep(0, 10^3);
for (i in c(1:10^3)) {
m = 1;
for (j in c(1:10^3)) {
m = min(m,dist(cp())); x[i] = m;
}
}
The average distance turns out to be 0.05554898, very close to $0.554/10$. For picking 10000 points in each of 1000 cubes I get an average distance of 0.02573758
, very close to $0.554/10^{4/3}$.
Best Answer
Let $S$ be the length of the longest interval and let $X$ be the largest random variable among $n=N+1$ iid exponential random variables with mean 1. We have $E[S] = E[X]/n$, where $E[X]=\sum_{i=1}^n 1/i$.
From this paper:
L. Holst. On the lengths of the pieces of a stick broken at random. Journal of Applied Probability, 17(3):623–634, 1980.