Expected distance on a segment, Geometric Probability

Problem 33.1

Let $X, Y$ be two points chosen independently and uniformly on the unit segment $[0,1]$ . What is the expected distance $\mathbb{E}[|X - Y|]$ ?

Solution. By symmetry, condition on $X < Y$ (the event has probability $\tfrac12$ ). In this case, $|X - Y| = Y - X$ . Compute directly: $\mathbb{E}[Y - X \mid X < Y] = \frac{1}{\tfrac12} \int_0^1 \int_0^y (y - x) \, dx \, dy = 2 \int_0^1 \frac{y^2}{2} \, dy = \frac{1}{3}.$ By symmetry between $X$ and $Y$ , the conditional expectation is the same given $X > Y$ , so $\boxed{\mathbb{E}[|X - Y|] = \tfrac13.} \qedhere$

Alternatively, think of the three pieces of a stick snapped at two uniform points: they have lengths $X_{(1)}, X_{(2)} - X_{(1)}, 1 - X_{(2)}$ , where $X_{(1)} = \min(X,Y)$ and $X_{(2)} = \max(X,Y)$ . The middle piece has length $|X - Y|$ , and by symmetry among the three order statistics each piece has expected length $\tfrac13$ . The answer $\tfrac13$ is therefore just the equality of the three expected part-lengths.

This constant turns up in many places in geometric probability. We have already seen it in Chapter 32 as a factor of $\tfrac19$ , and the higher moments $\mathbb{E}[|X-Y|^k]$ play a role in other classical computations.

import numpy as np
x, y = np.random.rand(2, 10**7)
print(f"sim: {np.abs(x-y).mean():.5f}   exact: {1/3:.5f}")
# sim: 0.33333   exact: 0.33333