# Binary digits of uniform random variables

… are independent fair coin tosses.

Valerio Gherardi https://vgherard.github.io
2024-01-29

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 $$I^n _j$$.↩︎

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Text and figures are licensed under Creative Commons Attribution CC BY-SA 4.0. Source code is available at https://github.com/vgherard/vgherard.github.io/, unless otherwise noted. The figures that have been reused from other sources don't fall under this license and can be recognized by a note in their caption: "Figure from ...".

### Citation

Gherardi (2024, Jan. 29). vgherard: Binary digits of uniform random variables. Retrieved from https://vgherard.github.io/posts/2024-01-29-binary-digits-of-uniform-random-variables/
@misc{gherardi2024binary,
}