Expected minimum distance to the boundary of a square, Geometric Probability

Problem 10.1

A point $P = (X, Y)$ is chosen uniformly at random in the unit square $[0, 1]^2$ . What is the expected minimum distance from $P$ to the boundary of the square?

Solution. The minimum distance from $P$ to the boundary is $D = \min(X, \, 1 - X, \, Y, \, 1 - Y).$ Write $U = \min(X, 1 - X)$ and $V = \min(Y, 1 - Y)$ , so $D = \min(U, V)$ .

The random variable $X$ is uniform on $[0, 1]$ , and $U = \min(X, 1 - X)$ satisfies $U \leq u$ iff $X \leq u$ or $X \geq 1 - u$ , so $\mathbb{P}(U \leq u) = 2u$ for $u \in [0, \tfrac12]$ . Hence $U$ is uniform on $[0, \tfrac12]$ , with density $2$ . Similarly $V$ is uniform on $[0, \tfrac12]$ , independently of $U$ .

Now $D = \min(U, V)$ where $U, V$ are iid uniform on $[0, \tfrac12]$ . For iid uniforms on $[0, a]$ , $\mathbb{E}[\min(U, V)] = \int_0^a \mathbb{P}(\min > u) \, du = \int_0^a \left(1 - \tfrac{u}{a}\right)^2 du = \frac{a}{3}.$ With $a = \tfrac12$ : $\boxed{\mathbb{E}[D] = \frac{1}{6}.} \qedhere$

The same calculation in one dimension (distance from random point in $[0, 1]$ to its nearest endpoint) gives $\tfrac14$ ; in three dimensions (cube) it gives $\tfrac18$ ; in general dimension $d$ it is $\tfrac{1}{2(d+1)}$ . The factor $d+1$ arises because we take the minimum over $2d$ coordinate-distances, and folding by the midpoint makes each “half-coordinate” uniform on $[0, \tfrac12]$ .

import numpy as np
x, y = np.random.rand(2, 10**7)
d = np.minimum(np.minimum(x, 1-x), np.minimum(y, 1-y))
print(f"sim: {d.mean():.5f}   exact: {1/6:.5f}")
# sim: 0.16664   exact: 0.16667