A random chord via a uniform midpoint, Geometric Probability

Problem 20.1

A chord of the unit disc is determined as follows: pick a point $M$ uniformly at random in the closed unit disc, and take the chord perpendicular to $\mathbf{OM}$ passing through $M$ (where $O$ is the disc’s centre). What is the expected length of this chord?

Solution. Let $r = |OM| \in [0, 1]$ denote the distance from the disc’s centre to the midpoint $M$ . A chord at perpendicular distance $r$ from the centre has length $\ell(r) = 2 \sqrt{1 - r^2}.$ For $M$ uniform in the unit disc, $\mathbb{P}(r \le t)$ is the area of the radius- $t$ disc over the area of the unit disc, namely $t^2$ ; differentiating, $r$ has density $f_r(t) = 2t$ on $[0, 1]$ .

Therefore $\mathbb{E}[\ell] = \int_0^1 2 \sqrt{1 - t^2} \cdot 2 t \, dt = 4 \int_0^1 t \sqrt{1 - t^2} \, dt.$ Substitute $u = 1 - t^2$ , $du = -2 t \, dt$ : $= 4 \int_0^1 \tfrac12 \sqrt{u} \, du = 2 \cdot \tfrac{2}{3} = \boxed{\tfrac{4}{3}.} \qedhere$

Two sample chords determined by uniform-in-disc midpoints M_1 and M_2 . Each chord is perpendicular to \mathbf{OM}_i at M_i , with length 2 \sqrt{1 - |M_i|^2} computed exactly. — Two sample chords determined by uniform-in-disc midpoints $M_1$ and $M_2$ . Each chord is perpendicular to $\mathbf{OM}_i$ at $M_i$ , with length $2 \sqrt{1 - |M_i|^2}$ computed exactly.

Compare with the “uniform endpoints” sampling of a chord (Chapter 13): there, two iid uniform points on the boundary circle give an expected chord length of $\tfrac{4}{\pi} \approx 1.27$ , smaller than the $\tfrac{4}{3} \approx 1.33$ obtained here. The two sampling schemes are both “uniform” in distinct senses — they are not the same probability measure on the space of chords.

This is the third sampling rule from Bertrand’s classical paradox, which asks a related question: what is the probability that a random chord of the unit circle exceeds $\sqrt 3$ ? The answer under “uniform midpoint in disc” sampling is $\mathbb{P}\left( 2 \sqrt{1 - r^2} > \sqrt 3 \right) = \mathbb{P}\left( r < \tfrac12 \right) = \frac{\pi (1/2)^2}{\pi} = \tfrac14,$ matching the classical Bertrand value under that sampling rule.

import numpy as np
# Sample midpoint M uniform in unit disc; r = |OM| has density 2r.
r = np.sqrt(np.random.rand(10**7))
chord = 2 * np.sqrt(1 - r*r)
print(f"sim E[chord]: {chord.mean():.5f}   exact: {4/3:.5f}")
# sim: 1.33349   exact: 1.33333