Binary digits of uniform random variables

… are independent fair coin tosses.

Let \(X\) be a random number in the unit interval \([0,\,1]\), and let \(Z \equiv (Z_k)_{k\in \mathbb N}\) represent the sequence of its binary digits, so that \(Z_k \in \{0,\,1\}\) for all \(k\) and:

\[ X = \sum _{k = 1} ^\infty Z_k \cdot 2^{-k} \] Notice that the representation \(Z\) is unique for all \(X\) outside of a countable subset of the unit interval.¹

The cool theorem proved below is that \(X\) is uniformly distributed in \([0,\,1]\) if and only if all \(Z_k\)’s are independent and \(\text{Pr}(Z_k = 1) = \text{Pr}(Z_k = 0) = \frac{1}{2}\). That is to say, the binary representation \(Z\) of a random variable \(X\sim \text{Unif}(0,\,1)\) amounts to a sequence of independent fair coin tosses.

We fix \(n \in \mathbb N\) and decompose the unit interval as follows:

\[ [0,1) = \cup _{j = 1} ^{2^n} I^n_j,\quad I^n_j = [(j-1) \cdot2^{-n},j\cdot2^{-n}) \] Each interval \(I^n_j\) corresponds to a specific set of values for the first \(n\) digits \(Z_1,\,Z_2,\,\dots,\,Z_n\), that is \(X\in I^n _j\) if and only if \(Z_1 = z_1,\,Z_2 =z_2,\,\dots,\,Z_n=z_n\) for some \(z_1,\,z_2,\,\dots,\,z_n\) that depend on the interval \(I^n _j\). Therefore:

\[ \text{Pr}(X\in I^n _j) = \text{Pr}(Z_1 = z_1,\,Z_2 = z_2,\,\dots,\,Z_n = z_n) \] Now, \(X\) is uniformly distributed if and only if the left hand side of this equation equals \(2^{-n}\) for all \(n\in \mathbb N\) and \(1\leq j \leq 2^{n}\) ². Furthermore, the \(2^{n}\) possible values of \(j\) correspond to the \(2^{n}\) possible values of \(z_1,\,z_2,\,\dots,\,z_n\) in the right hand side. Therefore, \(X\) is uniform if and only if:

\[ \text{Pr}(Z_1 = z_1,\,Z_2 = z_2,\,\dots,\,Z_n = z_n) = 2^{-n} \] for all \(z_1,\,z_2,\,\dots,\,z_n \in \{0,\,1\}\). More generally, this implies that, for any \(k \in \mathbb N \cup \{0\}\) we have:

\[ \text{Pr}(Z_{k+1} = z_1,\,Z_{k+2} = z_2,\,\dots,\,Z_{k+n} = z_n) = 2^{-n} = \prod _{i=1}^{n}\text{Pr}(Z_{k+i} = z_i), \]

where the second equality follows from the special case \(n=1\). This is equivalent to saying that all \(Z_k\)’s are independent, each having \(\text{Pr}(Z_k = 1) = \text{Pr}(Z_k = 0) = \frac{1}{2}\).

---

1. That is, the set of numbers that have a finite expansion \(X = \sum _{k = 1} ^N Z_k \cdot 2^{-k}\) for some finite \(N\), with \(Z_N = 1\). These numbers also have the equivalent infinite expansion \(X = \sum _{k = 1} ^{N-1} Z_k \cdot 2^{-k} + \sum _{k = N+1} ^{\infty}2^{-k}\). For these numbers we can make the convention of using the first (finite) representation.↩︎

2. That this is sufficient follows from the fact that any interval of the real line can be obtained by taking countable unions and intersections of intervals of the form \(I^n _j\).↩︎