## More on Benford’s Law

**I**n connection with an earlier post on Benford’s Law, i.e. the probability that the first digit of a random variable *X* isis approximately—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, iffollows Benford’s Law, thenalso follows Benford’s Law for *any* random variablethat is independent from… Now, the funny thing is that, if you take a normal sampleand check whether or not Benford’s Law applies to this sample, it does not. But if you take a second normal sampleand consider the product sample, 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:

x=abs(rnorm(10^6)) b=trunc(log10(x)) -(log(x)<0) plot(hist(trunc(x/10^b),breaks=(0:9)+.5)$den,log10((2:10)/(1:9)), xlab="Frequency",ylab="Benford's Law",pch=19,col="steelblue") abline(a=0,b=1,col="tomato",lwd=2) x=abs(rnorm(10^6)*x) b=trunc(log10(x)) -(log(x)<0) points(hist(trunc(x/10^b),breaks=(0:9)+.5,plot=F)$den,log10((2:10)/(1:9)), pch=19,col="steelblue2") x=abs(rnorm(10^6)*x) b=trunc(log10(x)) -(log(x)<0) points(hist(trunc(x/10^b),breaks=(0:9)+.5,plot=F)$den,log10((2:10)/(1:9)), pch=19,col="steelblue3")

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

May 29, 2011 at 12:13 am

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