Expected area of a random rectangle, Geometric Probability

Problem 32.1

Let $P_1 = (X_1, Y_1)$ and $P_2 = (X_2, Y_2)$ be two points chosen independently and uniformly in the unit square. Let $R$ be the axis-aligned rectangle with diagonally opposite corners $P_1$ and $P_2$ . What is the expected area of $R$ ?

Solution. The area of the axis-aligned rectangle is $|X_1 - X_2| \cdot |Y_1 - Y_2|$ . Since the pairs $(X_1, X_2)$ and $(Y_1, Y_2)$ are independent (the $x$ - and $y$ -coordinates of the two points are independent), so are $|X_1 - X_2|$ and $|Y_1 - Y_2|$ . Hence $\mathbb{E}[\operatorname{Area}(R)] = \mathbb{E}[|X_1 - X_2|] \cdot \mathbb{E}[|Y_1 - Y_2|].$

Each factor is the expected absolute difference of two uniform random variables on $[0, 1]$ . A direct calculation: for $X_1, X_2$ iid uniform on $[0,1]$ , the density of $|X_1 - X_2|$ is $2(1 - t)$ on $[0,1]$ , so $\mathbb{E}[|X_1 - X_2|] = \int_0^1 t \cdot 2(1 - t) \, dt = \left[ t^2 - \tfrac{2 t^3}{3} \right]_0^1 = 1 - \tfrac23 = \tfrac13.$ (Alternatively, by symmetry in the triangle $\{x_1 < x_2\}$ : $\mathbb{E}[x_2 - x_1] = \mathbb{E}[x_2] - \mathbb{E}[x_1] = \tfrac12 - \tfrac12 = 0$ unconditionally, but conditional on $x_1 < x_2$ , $\mathbb{E}[x_2 - x_1 \mid x_1 < x_2] = \tfrac13$ ; and the same value arises unconditionally for the absolute value.)

Hence $\boxed{\mathbb{E}[\operatorname{Area}(R)] = \tfrac13 \cdot \tfrac13 = \tfrac19.} \qedhere$

The factor $\tfrac13$ for $\mathbb{E}[|X_1 - X_2|]$ is such a frequent building block that it is worth knowing by heart; we meet it again in Chapter 33. The factor of $\tfrac19$ here is a clean demonstration of the multiplicative structure of product measures.

import numpy as np
x1, x2 = np.random.rand(2, 10**7)
y1, y2 = np.random.rand(2, 10**7)
area = np.abs(x1 - x2) * np.abs(y1 - y2)
print(f"sim: {area.mean():.5f}   exact: {1/9:.5f}")
# sim: 0.11114   exact: 0.11111