Distance between two points on the boundary of a square

Solution. Parametrise a boundary point by arclength $s \in [0, 4)$ along the perimeter (starting from a corner, going counter-clockwise). Given two uniform points $S_1, S_2$ on $[0, 4)$, we condition on which sides the two points lie on.

By symmetry, the $16$ side-pair combinations fall into three categories:

• Same side (prob $\tfrac{4}{16} = \tfrac14$): distance is $|X - Y|$ for $X, Y$ iid uniform on $[0, 1]$. Expected value $\tfrac13$ (Chapter 33).

• Adjacent sides (prob $\tfrac{8}{16} = \tfrac12$): WLOG one point at $(X, 0)$ and the other at $(1, Y)$. Then with $U = 1 - X$, $$\mathbb{E}\!\left[\sqrt{U^2 + Y^2}\right] = \frac{\sqrt{2} + \ln(1 + \sqrt{2})}{3}.$$

• Opposite sides (prob $\tfrac{4}{16} = \tfrac14$): one point at $(X, 0)$ and the other at $(Y, 1)$, distance $\sqrt{(X - Y)^2 + 1}$. Then $$\mathbb{E}\!\left[\sqrt{(X-Y)^2 + 1}\right] = \frac{2 + 3 \ln(1+\sqrt 2) - \sqrt{2}}{3}.$$

Combining by the law of total expectation: $$\begin{align*} \mathbb{E}[\text{distance}] &= \tfrac14 \cdot \tfrac13 + \tfrac12 \cdot \tfrac{\sqrt 2 + \ln(1+\sqrt 2)}{3} \\ &\quad + \tfrac14 \cdot \tfrac{2 + 3\ln(1+\sqrt 2) - \sqrt 2}{3}. \end{align*}$$ A tidy algebraic simplification gives $$\boxed{\mathbb{E}[\text{distance}] = \frac{3 + \sqrt 2 + 5 \ln(1 + \sqrt 2)}{12} \approx 0.7351.} \qedhere$$

import numpy as np
u = np.random.rand(10**7) * 4  # position on perimeter
v = np.random.rand(10**7) * 4
def to_xy(s):
    x = np.where(s < 1, s,
        np.where(s < 2, 1,
        np.where(s < 3, 3 - s, 0)))
    y = np.where(s < 1, 0,
        np.where(s < 2, s - 1,
        np.where(s < 3, 1, 4 - s)))
    return x, y
x1, y1 = to_xy(u); x2, y2 = to_xy(v)
d = np.sqrt((x1-x2)**2 + (y1-y2)**2)
exact = (3 + np.sqrt(2) + 5*np.log(1 + np.sqrt(2))) / 12
print(f"sim: {d.mean():.5f}   exact: {exact:.5f}")
# sim: 0.73485   exact: 0.73509