## An entropy-related derivation of Benford’s law

In this thread of posts from 2012, the possibility of drawing a natural number at random was discussed. A solution offering relative chances was given, and stated to be in accordance with Benford’s law.

Then, due to unforeseen circumstances, the thread remained unfinished. In fact it stopped precisely at the point for which I had started the thread in the first place :-).

I wish to return to this thread, but it seems worthwhile to repeat the result mentioned above in a different wording. Why? Well, searching on “Benford’s law” I didn’t find any comparable entropy-related derivation (perhaps I should say motivation) in the literature/internet. And perhaps more telling, Theodore Hill (when deriving Benford’s law from considering a random’ mix of distributions, see A Statistical Derivation of the Significant-Digit Law) explicitly mentions the difficulty of drawing a natural number at random, if the sum probability must be 1, which would seem to exclude the density function $\frac{1}{x}$.

So the trick to turn to relative chances $\frac{P(n)}{P(m)}$ seems new [except it isn’t, see the next post], and it does yield Benford’s law. The entropy-related motivation for these relative chances also seems to be new. The (relative) chances involved in drawing a natural number at random will play a role in the discussion to come (on Laplacian determinism, digital physics and foundations of probability theory). But first let us explicitly derive Benford’s law from these chances:

Solution to drawing a natural number at random’:

* We can only assign relative chances, and the role of the natural number $0$ remains mysterious.

* For $1\leq n,m \in \mathbb{N}$ let’s denote the relative chance of drawing $n$ vs. drawing $m$ by: $\frac{P(n)}{P(m)}$.

* For $1\leq n,m \in \mathbb{N}$, we find that $\frac{P(n)}{P(m)} = \frac{\log{\frac{n+1}{n}}}{\log{\frac{m+1}{m}}}$

(* An alternative discrete’ or Zipfian’ case $P_{\rm discrete}$ can perhaps be formulated, yielding: for $1\leq n,m \in \mathbb{N}$, we find that $\frac{P_{\rm discrete}(n)}{P_{\rm discrete}(m)} = \frac{m}{n}$.)

The entropy-related motivation for these chances can be found in this thread of posts from 2012.

Now to arrive at Benford’s law, we consider the first digit of a number $N$ in base 10 (the argument is base-independent though) to be drawn at random according to our entropy-related chances. We then see that the relative chance of drawing a 1 compared to drawing a 2 equals $\frac{\log 2}{\log\frac{3}{2}}$ which in turn equals $\frac{^{10}\log 2}{^{10}\log\frac{3}{2}}$. Since the sum of the probabilities of drawing 1, 2,…,9 equals 1, one finds that the chance of drawing $i$ as first digit for $N$ equals $^{10}\log\frac{i+1}{i}$.

A second-digit law can be derived similarly, for example adding the relative chances for 11, 21, 31, 41, 51, 61, 71, 81, 91 vs. the sum of the relative chances for 12, 22, 32, 42, 52, 62, 72, 82, 92 to arrive at the relative chance of drawing a 1 as second digit vs. the chance of drawing a 2 as second digit. And so on for third-digit laws etc.

This shows that the second-digit distribution is more uniform than the first-digit distribution, and that each next-digit law is more uniform than its predecessor, which fits with our entropy argument.

In the next post the thread on information and probability will be continued.