Expected range of n uniform random points on a segment, Geometric Probability

Problem 38.1

Let $X_1, \ldots, X_n$ be iid uniform random variables on $[0, 1]$ . What is the expected value of the range $R = X_{(n)} - X_{(1)}$ (the distance between the largest and smallest)?

Solution. For $n$ iid uniform points on $[0, 1]$ , the order statistics $X_{(1)} \leq \cdots \leq X_{(n)}$ have the classical expected values $\mathbb{E}[X_{(k)}] = \frac{k}{n+1}.$ Therefore $\mathbb{E}[R] = \mathbb{E}[X_{(n)}] - \mathbb{E}[X_{(1)}] = \frac{n}{n+1} - \frac{1}{n+1} = \frac{n - 1}{n + 1}.$ So $\boxed{\mathbb{E}[R] = \frac{n-1}{n+1}.} \qedhere$

Derivation of the order-statistic means. The joint density of $n$ order statistics is $n!$ on the simplex $0 \leq X_{(1)} \leq \cdots \leq X_{(n)} \leq 1$ . The marginal density of $X_{(k)}$ is the Beta $(k, n-k+1)$ density, with mean $k/(n+1)$ . Equivalently, the uniform order statistics are obtained by “sorting” and have equal expected spacings, of which there are $n+1$ : hence each spacing has mean $1/(n+1)$ , and the $k$ -th order statistic is the sum of the first $k$ spacings. ■

Specialisations:

$n = 2$ : $\mathbb{E}[R] = \tfrac{1}{3}$ . (This recovers Chapter 33, since $R = |X_1 - X_2|$ .)
$n = 3$ : $\tfrac{1}{2}$ .
$n = 10$ : $\tfrac{9}{11} \approx 0.818$ .
As $n \to \infty$ : $\mathbb{E}[R] \to 1$ , at rate $1 - \tfrac{2}{n+1}$ .

import numpy as np
for n in [2, 3, 5, 10]:
    x = np.random.rand(n, 10**6)
    r = x.max(axis=0) - x.min(axis=0)
    print(f"n={n}: sim={r.mean():.5f}  exact={(n-1)/(n+1):.5f}")
# n=2:  sim=0.33336  exact=0.33333
# n=3:  sim=0.50033  exact=0.50000
# n=5:  sim=0.66673  exact=0.66667
# n=10: sim=0.81830  exact=0.81818