Expected squared distance in a unit square, Geometric Probability

Problem 40.1

Let $P_1, P_2$ be two points chosen independently and uniformly in the unit square $[0, 1]^2$ . What is the expected squared distance $\mathbb{E}[|P_1 - P_2|^2]$ ?

Solution. Write $P_i = (X_i, Y_i)$ . The squared distance decomposes as $|P_1 - P_2|^2 = (X_1 - X_2)^2 + (Y_1 - Y_2)^2.$ By linearity, $\mathbb{E}[|P_1 - P_2|^2] = \mathbb{E}[(X_1 - X_2)^2] + \mathbb{E}[(Y_1 - Y_2)^2].$ For $X_1, X_2$ iid uniform on $[0, 1]$ , $\mathbb{E}[(X_1 - X_2)^2] = \operatorname{Var}(X_1 - X_2) = 2 \operatorname{Var}(X_1) = \tfrac{2}{12} = \tfrac{1}{6},$ using $\operatorname{Var}(X_1) = 1/12$ for uniform on $[0, 1]$ . Similarly for the $Y$ -terms.

Therefore $\boxed{\mathbb{E}[|P_1 - P_2|^2] = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}.} \qedhere$

Compare the analogous constant for the unit disc (Chapter 31): $\mathbb{E}[|P_1 - P_2|^2] = 1$ for radius $1$ . The two constants reveal a dimensional structure: for a planar region $K$ with centroid at the origin, $\mathbb{E}[|P_1 - P_2|^2] = 2 \cdot \mathbb{E}[|P|^2],$ so the ratio between disc and square values reflects the moment of inertia per unit area. For the unit disc, $\mathbb{E}[|P|^2] = 1/2$ ; for the unit square centred at the centroid, $\mathbb{E}[|P|^2] = 1/6$ , giving $\mathbb{E}[|P_1 - P_2|^2] = 2 \cdot 1/6 = 1/3$ .

import numpy as np
p = np.random.rand(2, 2, 10**7)
d2 = np.sum((p[:,0] - p[:,1])**2, axis=0)
print(f"sim: {d2.mean():.5f}   exact: {1/3:.5f}")
# sim: 0.33311   exact: 0.33333