The Exercise

A coin lands heads with probability $p$ and tails with probability $1-p$ . A second coin lands heads with probability $q$ and tails with probability $1-q$ . The exercise is to show that the trace distance between the two distributions $(p, 1-p)$ and $(q, 1-q)$ is $|p-q|$ .^✦

The Classical Trace Distance

The classical trace distance $D(p_x, q_x)$ of two real functions $p_x$ and $q_x$ on a finite set of outcomes $x$ , such as two probability distributions, is half the sum over the outcomes of the absolute differences of their values: $D(p_x, q_x) = \frac{1}{2} \sum_x |p_x - q_x| .$ (1)^✦

The Claim

On a set of two outcomes the sum in (1) has two terms, $|p-q|$ and $|(1-p)-(1-q)|$ , and they are equal.

Theorem 1. Let $p$ and $q$ be real numbers. Then the functions $(p, 1-p)$ and $(q, 1-q)$ on two outcomes satisfy $D\bigl((p, 1-p), (q, 1-q)\bigr) = |p - q| .$ (2)^✦

What Is Checked

The claim is stated over the classical trace distance (1), taken on a set of two outcomes and evaluated at $(p, 1-p)$ and $(q, 1-q)$ . The identity (2) holds for all real $p$ and $q$ , including those outside $[0,1]$ , where the two pairs are not probability distributions.