Expected maximum radius among n random points in a disc, Geometric Probability

Problem 37.1

Let $P_1, \ldots, P_n$ be iid uniform random points in the unit disc. What is the expected value of $\max_i |P_i|$ (the distance from the origin to the farthest point)?

Solution. The distance $R_i = |P_i|$ has CDF $\mathbb{P}(R_i \leq r) = \frac{\pi r^2}{\pi} = r^2, \qquad r \in [0, 1],$ with density $f(r) = 2r$ (from the polar decomposition).

For $M_n = \max_i R_i$ , the CDF is $\mathbb{P}(M_n \leq r) = r^{2n},$ so the density is $f_{M_n}(r) = 2 n r^{2n-1}$ . The expectation is $\mathbb{E}[M_n] = \int_0^1 r \cdot 2 n r^{2n-1} \, dr = \frac{2n}{2n+1}.$ Therefore $\boxed{\mathbb{E}\!\left[ \max_{1 \leq i \leq n} |P_i| \right] = \frac{2n}{2n+1}.} \qedhere$

Eight iid uniform random points in the unit disc. The farthest from the origin (copper, |P| \approx 0.806 ) is highlighted; the other seven (rose) lie closer in. For n = 8 , \mathbb{E}[\max |P_i|] = 16/17 \approx 0.941 . — Eight iid uniform random points in the unit disc. The farthest from the origin (copper, $|P| \approx 0.806$ ) is highlighted; the other seven (rose) lie closer in. For $n = 8$ , $\mathbb{E}[\max |P_i|] = 16/17 \approx 0.941$ .

Values: $n = 1$ gives $\tfrac23$ (the classical disc radius expectation); $n = 2$ gives $\tfrac45$ ; $n = 10$ gives $\tfrac{20}{21} \approx 0.952$ . As $n \to \infty$ , $\mathbb{E}[M_n] \to 1$ : a large sample almost surely has a point near the boundary. The convergence rate is $1 - \tfrac{1}{2n+1} \sim \tfrac{1}{2n}$ , characteristic of extreme-value statistics for samples bounded above.

import numpy as np
for n in [1, 2, 5, 10]:
    r = np.sqrt(np.random.rand(n, 10**6))
    print(f"n={n}: sim={r.max(axis=0).mean():.5f}  exact={2*n/(2*n+1):.5f}")
# n=1: sim=0.66658  exact=0.66667
# n=2: sim=0.79987  exact=0.80000
# n=5: sim=0.90919  exact=0.90909
# n=10: sim=0.95241  exact=0.95238