Crofton’s expected chord of a disc, Geometric Probability

Problem 21.1

Pick a line uniformly at random from the space of lines that intersect the unit disc (“uniformly” with respect to the standard kinematic measure $d\phi \, dp$ on line-space). Let $\ell$ be the length of the chord cut out of the disc by this line. What is $\mathbb{E}[\ell]$ ?

Solution. Parametrise a line by $(\phi, p)$ where $\phi \in [0, \pi)$ is the angle of the line’s normal and $p \in \mathbb{R}$ is the signed distance from the origin. The kinematic measure $d\phi \, dp$ is the unique (up to scaling) measure on lines invariant under rigid motions.

A line $(\phi, p)$ intersects the unit disc iff $|p| < 1$ . Conditioning on this event, $\phi$ is uniform on $[0, \pi)$ and (by rotational symmetry) $p$ is uniform on $[-1, 1]$ .

The chord cut from the unit disc by such a line has length $\ell = 2 \sqrt{1 - p^2}$ (standard right-triangle formula). Hence $\mathbb{E}[\ell] = \frac{1}{2} \int_{-1}^{1} 2 \sqrt{1 - p^2} \, dp = \int_{-1}^{1} \sqrt{1 - p^2} \, dp = \frac{\pi}{2}.$ Therefore $\boxed{\mathbb{E}[\ell] = \frac{\pi}{2} \approx 1.571.} \qedhere$

Crofton’s general formula: for a convex body $K$ in the plane with area $A$ and perimeter $P$ , a uniform random line hitting $K$ has expected chord length $\mathbb{E}[\ell] = \frac{\pi A}{P}.$ For the unit disc, $A = \pi$ and $P = 2\pi$ , so $\mathbb{E}[\ell] = \pi \cdot \pi / (2 \pi) = \pi/2$ , matching. For the unit square, $A = 1$ and $P = 4$ , so $\mathbb{E}[\ell] = \pi/4 \approx 0.785$ .

Compare with chord-length sampling via uniform endpoints (Chapter 13): that gives $\mathbb{E}[\ell] = 4/\pi \approx 1.273$ , smaller than $\pi/2$ . With uniform midpoint in the disc (Chapter 20): $\mathbb{E}[\ell] = 4/3 \approx 1.333$ . Three different “uniform” rules yield three different averages — a reminder of Bertrand’s paradox.

import numpy as np
# Uniform random line: angle phi in [0, pi), signed distance p in [-1, 1].
phi = np.random.rand(10**7) * np.pi
p = np.random.uniform(-1, 1, 10**7)
chord = 2 * np.sqrt(1 - p*p)
print(f"sim: {chord.mean():.5f}   exact: {np.pi/2:.5f}")
# sim: 1.57026   exact: 1.57080