Expected squared distance in a disc, Geometric Probability

Problem 31.1

Let $P_1, P_2$ be points chosen independently and uniformly in the closed disc of radius $r$ . What is the expected squared distance $\mathbb{E}[|P_1 - P_2|^2]$ ?

Solution. Treat $P_1, P_2$ as two-dimensional vectors. Expanding the squared distance, $|P_1 - P_2|^2 = |P_1|^2 + |P_2|^2 - 2 \, P_1 \cdot P_2.$

Take expectations. By independence, $\mathbb{E}[P_1 \cdot P_2] = \mathbb{E}[P_1] \cdot \mathbb{E}[P_2]$ . By the rotational symmetry of the disc, $\mathbb{E}[P_i] = 0$ (each component has zero mean). Hence $\mathbb{E}[|P_1 - P_2|^2] = 2 \, \mathbb{E}[|P|^2],$ where $P$ is a single uniform random point in the disc.

Compute in polar coordinates with density $1/(\pi r^2)$ : $\begin{align*} \mathbb{E}[|P|^2] &= \frac{1}{\pi r^2} \int_0^{2\pi} \!\! \int_0^r s^2 \cdot s \, ds \, d\theta \\ &= \frac{1}{\pi r^2} \cdot 2\pi \cdot \frac{r^4}{4} = \frac{r^2}{2}. \end{align*}$ Therefore $\boxed{\mathbb{E}[|P_1 - P_2|^2] = 2 \cdot \tfrac{r^2}{2} = r^2.} \qedhere$

The appeal of this result is its total absence of elliptic integrals. The first moment $\mathbb{E}[|P_1 - P_2|] = \tfrac{128}{45\pi} \cdot r \approx 0.905\,r$ (“disc line picking”) is much harder to compute; it reduces to an elliptic integral via the law of cosines. The squared distance admits a much cleaner analysis thanks to the bilinearity of the inner product and the vanishing of the mean vector.

import numpy as np
# Uniform on unit disc via (sqrt(u), theta) polar
r = np.sqrt(np.random.rand(2, 10**7))
theta = np.random.rand(2, 10**7) * 2*np.pi
x, y = r*np.cos(theta), r*np.sin(theta)
d2 = (x[0]-x[1])**2 + (y[0]-y[1])**2
print(f"sim: {d2.mean():.5f}   exact: 1.0")
# sim: 1.00012   exact: 1.0