Random walk from the centre of a unit square and cube, Geometric Probability

Problem 42.1

You start at the centre of the unit square and pick a random direction to move in, with all directions equally likely. You move along this direction until you reach a point on the perimeter of the square. On average, how far can you expect to have travelled?

Problem 42.2

You start at the centre of the unit cube. You pick a random direction uniformly on the sphere and move in that direction until reaching the surface of the cube. On average, how far can you expect to have travelled?

Solution, part (a): the square. Place the square with corners at $(0, 0)$ , $(1, 0)$ , $(1, 1)$ , $(0, 1)$ . By symmetry the mean length over all directions equals the mean over the triangle with vertices $(1/2, 1/2)$ , $(1, 1/2)$ , $(1, 1)$ . Pick a direction $\theta \in [0, \pi/4]$ ; the ray hits the right edge at distance $d(\theta) \;=\; \frac{1}{2 \cos \theta}.$ Weighting by the fraction of angles in the fundamental sector, $4/\pi$ per radian of $\theta$ , $\mathbb{E}[d] \;=\; \frac{4}{\pi} \int_0^{\pi/4} \frac{1}{2 \cos \theta} \, d\theta \;=\; \frac{2}{\pi} \int_0^{\pi/4} \sec \theta \, d\theta \;=\; \frac{2}{\pi} \, \ln\!\bigl(\sec \theta + \tan \theta\bigr) \Big|_0^{\pi/4}.$ At $\theta = \pi/4$ the bracketed quantity is $\sqrt 2 + 1$ ; at $\theta = 0$ it is $1$ . Therefore $\boxed{\mathbb{E}[d_{\text{square}}] \;=\; \frac{2}{\pi} \, \ln\!\bigl(\sqrt 2 + 1\bigr) \;=\; \frac{\ln(3 + 2 \sqrt 2)}{\pi} \;\approx\; 0.5611.}$ ■

The integral of $\sec \theta$ admits four classical derivations worth recording.

Method 1. Multiply the integrand by $(\sec x + \tan x)/(\sec x + \tan x) = 1$ . The numerator becomes $\sec^2 x + \sec x \tan x$ , which is exactly the derivative of $\sec x + \tan x$ . Setting $u = \sec x + \tan x$ reduces the integral to $\int du/u = \ln|u| + C = \ln|\sec x + \tan x| + C$ .

Method 2 (Weierstrass substitution). With $t = \tan(x/2)$ , the half-angle formulas give $\cos x = (1 - t^2)/(1 + t^2)$ and $dx = 2 \, dt/(1 + t^2)$ , so $\int \sec x \, dx \;=\; \int \frac{2}{1 - t^2} \, dt \;=\; \ln \!\left|\frac{1 + t}{1 - t}\right| + C \;=\; \ln \!\left|\tan\!\bigl(\tfrac{\pi}{4} + \tfrac{x}{2}\bigr)\right| + C,$ the last step by the tangent addition formula.

Method 3 (complex exponentials). With $\cos x = (e^{ix} + e^{-ix})/2$ and the substitution $u = e^{ix}$ , $\int \sec x \, dx \;=\; \frac{2}{i} \int \frac{du}{u^2 + 1} \;=\; -2 i \, \arctan(e^{ix}) + C,$ which, on the real line, reduces to $\ln|\sec x + \tan x| + C$ .

Method 4 (hyperbolic). With $x = i t$ , $\cos(i t) = \cosh t$ , so $\int \sec x \, dx \;=\; i \int \operatorname{sech} t \, dt \;=\; 2 i \arctan\!\bigl(\tanh(t/2)\bigr) + C,$ again equivalent to $\ln|\sec x + \tan x| + C$ after substituting back.

Solution, part (b): the cube. Place the cube with corners at $(0, 0, 0)$ and $(1, 1, 1)$ . A random direction on the sphere is $\vec v \;=\; (\sin \varphi \cos \theta, \, \sin \varphi \sin \theta, \, \cos \varphi),$ with uniform measure $(4\pi)^{-1} \sin \varphi \, d\varphi \, d\theta$ . From the centre the distance to the boundary is $d \;=\; \frac{1}{2 \, m}, \qquad m \;=\; \max\bigl(|\sin \varphi \cos \theta|, \, |\sin \varphi \sin \theta|, \, |\cos \varphi|\bigr).$

The magnitude $m = \max(|x|, |y|, |z|)$ is unchanged by the $48$ symmetries of the cube, so it suffices to average over a single fundamental region: the directions in the first octant whose $x$ -component is largest in magnitude. There are $24$ congruent such regions (eight octants, each subdivided by which of the three coordinates is largest), each of solid angle $4\pi/24$ . In it, $\theta \in \left[0, \tfrac{\pi}{4}\right], \qquad \varphi \in \left[\arctan(\sec \theta), \, \tfrac{\pi}{2}\right].$ Here $d = 1/(2 \sin \varphi \cos \theta)$ , and the $\sin \varphi$ in the spherical measure $\sin \varphi \, d\varphi \, d\theta$ cancels the $\sin \varphi$ in the denominator of $d$ . Averaging over the region multiplies the integral by $24/(4\pi)$ ; together with the factor $\tfrac12$ in $d = 1/(2m)$ this gives the prefactor $3/\pi$ : $\mathbb{E}[d] \;=\; \frac{3}{\pi} \int_0^{\pi/4} \frac{\tfrac{\pi}{2} - \arctan(\sec \theta)}{\cos \theta} \, d\theta \;=\; \frac{3}{\pi} \int_0^{\pi/4} \frac{\arctan(\cos \theta)}{\cos \theta} \, d\theta,$ where the last step uses $\arctan(x) + \arctan(1/x) = \pi/2$ for $x > 0$ . The integral does not collapse to elementary constants. Numerically, $\int_0^{\pi/4} \frac{\arctan(\cos \theta)}{\cos \theta} \, d\theta \;\approx\; 0.63951,$ giving $\boxed{\mathbb{E}[d_{\text{cube}}] \;\approx\; 0.6107.} \qedhere$

The expected distance rises from $0.561$ ( $2$ D) to $0.611$ ( $3$ D) by just under $9\%$ . An extra dimension opens longer body-diagonal directions (up to $\sqrt 3 / 2 \approx 0.866$ ), but the cube’s six faces cut most rays off quickly, so the uniform-direction average lifts only modestly above the inradius $\tfrac12$ .

import numpy as np
np.random.seed(0); n = 10**6

# 2D: uniform theta, distance from centre of unit square to boundary.
theta = np.random.uniform(0, 2*np.pi, n)
dx, dy = np.cos(theta), np.sin(theta)
t = np.minimum(
    np.minimum(np.where(dx>0, 0.5/dx, np.inf), np.where(dx<0, -0.5/dx, np.inf)),
    np.minimum(np.where(dy>0, 0.5/dy, np.inf), np.where(dy<0, -0.5/dy, np.inf)))
exact_2d = (2/np.pi) * np.log(1 + np.sqrt(2))
print(f"2D sim: {t.mean():.5f}   exact: {exact_2d:.5f}")
# 2D sim: 0.56113   exact: 0.56110

# 3D: uniform direction on sphere, distance from centre of unit cube to boundary.
theta = np.random.uniform(0, 2*np.pi, n)
u = np.random.uniform(-1, 1, n)
s = np.sqrt(1 - u*u)
dx, dy, dz = s*np.cos(theta), s*np.sin(theta), u
t = np.minimum.reduce([
    np.where(dx>0, 0.5/dx, np.inf), np.where(dx<0, -0.5/dx, np.inf),
    np.where(dy>0, 0.5/dy, np.inf), np.where(dy<0, -0.5/dy, np.inf),
    np.where(dz>0, 0.5/dz, np.inf), np.where(dz<0, -0.5/dz, np.inf)])
print(f"3D sim: {t.mean():.5f}   integral target: 0.61069")
# 3D sim: 0.61070