Expected distance between two points on a circle, Geometric Probability

Problem 13.1

Let $P$ and $Q$ be two points chosen independently and uniformly on a circle of radius $r$ . What is the expected distance $\mathbb{E}[|PQ|]$ ?

Solution. Parametrise $P$ and $Q$ by angles $\alpha, \beta$ independently uniform on $[0, 2\pi)$ . The chord length between them is $\begin{align*} |PQ| &= \sqrt{(r\cos\alpha - r\cos\beta)^2 + (r\sin\alpha - r\sin\beta)^2} \\ &= r \sqrt{2 - 2\cos(\alpha - \beta)} = 2 r \left| \sin \tfrac{\alpha - \beta}{2} \right|. \end{align*}$

Let $\phi = \alpha - \beta \pmod{2\pi}$ ; by the standard change-of-variable for circular uniform random variables, $\phi$ is uniform on $[0, 2\pi)$ . Then $\mathbb{E}[|PQ|] = \mathbb{E}\!\left[2 r \left|\sin \tfrac{\phi}{2}\right|\right] = \frac{2r}{2\pi} \int_0^{2\pi} \sin \tfrac{\phi}{2} \, d\phi,$ using $\sin(\phi/2) \geq 0$ on $[0, 2\pi]$ . The integral equals $[-2 \cos(\phi/2)]_0^{2\pi} = 2 - (-2) = 4$ , so $\boxed{\mathbb{E}[|PQ|] = \frac{2r}{2\pi} \cdot 4 = \frac{4r}{\pi}.} \qedhere$

For the unit circle ( $r = 1$ ) this gives $\mathbb{E}[|PQ|] = 4/\pi \approx 1.273$ . This is the expected length of a “random chord” under the convention that both endpoints are chosen uniformly on the circle’s boundary.

import numpy as np
theta = np.random.rand(2, 10**7) * 2*np.pi
chord = 2 * np.abs(np.sin((theta[0] - theta[1]) / 2))
print(f"sim: {chord.mean():.5f}   exact: {4/np.pi:.5f}")
# sim: 1.27298   exact: 1.27324