# Benford’s Law and Accelerated Growth

**Figure 1.**A snapshot of the distribution of leading digits in NASDAQ-100 stock prices.

The following example illustrates a mathematically rigorous deterministic (as opposed to probabilistic) counterpart of Benford’s law. Consider a geometric sequence, for instance

\[1, 2, 4, 8, 16, 32, 64, 128, 256,\ldots,\]

and extract from it the sequence of leading digits:

\[1, 2, 4, 8, 1, 3, 6, 1, 2, \ldots.\]

It turns out that the frequency \(p_k\) of digit \(k\) is well defined and given by

\[p_k=\lg (k+1) - \lg k, \tag1 \]

see [1]. In particular, the frequency decreases with \(k\):

\[p_1=\lg\frac{2}{1} > \lg \frac{3}{2} >\ldots > \lg\frac{10}{9}=p_9. \tag2 \]

In light of this example, if the price of a stock undergoes an exponential-like growth (in a loose analogy with the geometrical sequence), then the bias illustrated in Figure 1 may not be that surprising.

What is behind Benford’s frequency bias \((2)\) for the geometric series? A one-word answer is “acceleration.” To see why, consider a continuous counterpart of \(2^n\) — say, the exponential function \(e^t\), visualizing the point \(x=e^t \in {\mathbb R}\) as a particle moving with time along the \(x\)-axis.

Figure 2 shows the \(x\)-axis cut into segments \([10^j, 10^{j+1})\) and stacked on top of each other, all of them scaled (linearly) to the same length. This cutting and scaling allows us to see the leading digit of \(e^t\) at a glance. Now the reason for Benford’s law becomes clear: *since* \(e^t\) *accelerates, it passes the higher-digit segments faster and is thus less likely to be found there*.

**Figure 2.**The intuition behind Benford’s law.

Figure 2 makes it easy to compute the probability, i.e., the proportion of time, of observing the leading digit \(k\). To that end, we find the time spent having the leading digit \(k\) while traversing the \(j\)th row in Figure 2:

\[\ln [(k+1) 10^j]- \ln [k10^j] = \ln \frac{k+1}{k}.\]

We then divide it by the time of traversal \(\ln [10\cdot 10^j]-\ln 10^j= \ln 10\); both times are independent of \(j\), and thus the proportion of time spent with the leading digit \(k\) over time \([0,T]\) approaches

\[\frac{\ln \frac{k+1}{k} }{\ln 10} = \lg \frac{k+1}{k}\]

as \(T \rightarrow \infty\), the same as the discrete result \((1)\).

The figures in this article were provided by the author.

**References**

[1] Arnold, V.I. (1983). *Geometrical Methods in the Theory of Ordinary Differential Equations*. New York, NY: Springer-Verlag.

[2] Newcomb, S. (1881). Note on the frequency of use of the different digits in natural numbers. *American Journal of Mathematics, 4*(1), 39-40.