The lighthouse and the Cauchy distribution, Geometric Probability

Problem 55.1

A lighthouse stands at perpendicular distance $d$ from a straight shore. Its beam points in a uniform random direction on $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ (angles measured from the perpendicular to the shore). What is the distribution of the distance $X$ from the point at which the beam strikes the shore to the foot of the perpendicular from the lighthouse?

Solution. Let $\Theta$ be the random angle, uniform on $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ with density $1/\pi$ . Elementary trigonometry gives the distance along the shore: $X = d \tan \Theta.$

Since $\tan$ is monotone increasing on $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$ with $\tan(\pm \tfrac{\pi}{2}) = \pm \infty$ , $X$ takes every real value. To find its distribution, use the change-of-variable formula. For any $t \in \mathbb{R}$ , $\begin{align*} \mathbb{P}(X \leq t) &= \mathbb{P}(d \tan \Theta \leq t) = \mathbb{P}\!\left( \Theta \leq \arctan(t/d) \right) \\ &= \frac{\arctan(t/d) + \pi/2}{\pi}. \end{align*}$ Differentiating, the density of $X$ is $f_X(t) = \frac{1}{\pi} \cdot \frac{1/d}{1 + (t/d)^2} = \frac{d}{\pi (d^2 + t^2)}.$ This is the density of a Cauchy distribution centred at $0$ with scale $d$ . In summary, $\boxed{X \sim \operatorname{Cauchy}(0, d), \qquad f_X(t) = \frac{d}{\pi (d^2 + t^2)}.} \qedhere$

Lighthouse at perpendicular distance d from a straight shore. The beam, rotating through a uniform random angle \theta , strikes the shore at X = d \tan \theta . The resulting distribution of X is the Cauchy distribution with scale d . — Lighthouse at perpendicular distance $d$ from a straight shore. The beam, rotating through a uniform random angle $\theta$ , strikes the shore at $X = d \tan \theta$ . The resulting distribution of $X$ is the Cauchy distribution with scale $d$ .

The Cauchy distribution has the unusual property that its mean and variance are both undefined: $\int t \cdot f_X(t) \, dt$ diverges at both $\pm \infty$ , and the same for $\int t^2 f_X(t) \, dt$ . Thus the “average spot where the beam lands” is not well-defined, even though the problem looks perfectly symmetric. A famous consequence: the sample mean of $n$ iid Cauchy random variables is also Cauchy (not Gaussian), so the law of large numbers fails — averaging many observations does not reduce uncertainty.

This problem is a favourite of probability texts (Ross, A First Course in Probability) and a standard entry in early tutorials on the change of variables. The geometric interpretation — the beam of a uniformly rotating lighthouse — makes the otherwise pathological Cauchy law intuitive.

import numpy as np
d, N = 1.0, 10**7
theta = np.random.uniform(-np.pi/2, np.pi/2, N)
X = d * np.tan(theta)
# For d = 1, the CDF at t = 1 equals 1/2 + arctan(1)/pi = 3/4.
print(f"sim P(X<=1): {(X <= 1).mean():.5f}   exact: 0.75")
# sim P(X<=1): 0.75021   exact: 0.75