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# Benford’s Law and Accelerated Growth

Figure 1. A snapshot of the distribution of leading digits in NASDAQ-100 stock prices.
Benford’s law, discovered by Simon Newcomb [2], is an empirical observation that in many sets of numbers arising from real-life data, the leading digit of a number is more likely to be 1 than 2, which in turn is more likely than 3, and so on. Figure 1 shows a count of leading digits from the 107 NASDAQ-100 prices, sampled around noon on June 14, 2017. Although the monotonicity is clearly violated, there is a bias in favor of low leading digits.

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 $$(1)$$.