More on Benford’s Law

In connection with an earlier post on Benford’s Law, i.e. the probability that the first digit of a random variable X is1\le k\le 9is approximately\log\{(k+1)/k\}—you can easily check that the sum of those probabilities is 1—, I want to signal a recent entry on Terry Tiao’s impressive blog. Terry points out that Benford’s Law is the Haar measure in that setting, but he also highlights a very peculiar absorbing property which is that, ifXfollows Benford’s Law, thenXYalso follows Benford’s Law for any random variableYthat is independent fromX… Now, the funny thing is that, if you take a normal samplex_1,\ldots,x_nand check whether or not Benford’s Law applies to this sample, it does not. But if you take a second normal sampley_1,\ldots,y_nand consider the product samplex_1\times y_1,\ldots,x_n\times y_n, then Benford’s Law applies almost exactly. If you repeat the process one more time, it is difficult to spot the difference. Here is the [rudimentary—there must be a more elegant way to get the first significant digit!] R code to check this:

b=trunc(log10(x)) -(log(x)<0)
    xlab="Frequency",ylab="Benford's Law",pch=19,col="steelblue")
b=trunc(log10(x)) -(log(x)<0)
b=trunc(log10(x)) -(log(x)<0)

Even better, if you change rnorm to another generator like rcauchy or rexp at any of the three stages, the same pattern occurs.

3 Responses to “More on Benford’s Law”

  1. […] uninformative interview with Le Monde, Xavier Gabaix focused on the Zipf laws (connected with the Benford law I mentioned a while ago about the Iranian […]

  2. […] probability introduction] Feller’s Introduction to Probability Theory (volume 2) gets Benford’s Law “wrong”. While my interest in Benford’s Law is rather superficial, I find the […]

  3. […] evidenced by the analyses of the Iranian election. Christian Robert, in his wonderful blog, has an entry describing a fascinating property of Benford’s law. In some ways it calls into question the […]

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