Tag Archives: Benford

Benford law and lognormal distributions

Benford’s law is nowadays extremely popular (see e.g. http://en.wikipedia.org/…). It is usually claimed that, for a given set data set, changing units does not affect the distribution of the first digit. Thus, it should be related to scale invariant distributions. Heuristically, scale (or unit) invariance means that the density of the measure https://latex.codecogs.com/gif.latex?%20X (or probability function) https://latex.codecogs.com/gif.latex?f(x) should be proportional to https://latex.codecogs.com/gif.latex?f(kx). Thus, because densities integrate to 1, the proportionality coefficient has to be https://latex.codecogs.com/gif.latex?k^{-1}, and therefore, https://latex.codecogs.com/gif.latex?f should satisfy the following functional equation, https://latex.codecogs.com/gif.latex?%20kf(kx)=f(x), for all https://latex.codecogs.com/gif.latex?%20x in https://latex.codecogs.com/gif.latex?%20(1,\infty) and https://latex.codecogs.com/gif.latex?%20k in https://latex.codecogs.com/gif.latex?%20(0,\infty). The solution of this functional equation is https://latex.codecogs.com/gif.latex?%20f(x)=x^{-1}, I guess this can be proved easily solving ordinary differential equation


Now if https://latex.codecogs.com/gif.latex?%20D denotes the first digit of https://latex.codecogs.com/gif.latex?%20X, in base 10, then

https://latex.codecogs.com/gif.latex?%20\mathbb{P}(D=d)=\frac{\displaystyle{\int_d^{d+1}%20f(x)dx}}{{\displaystyle{\int_1^{10}%20f(x)dx}}}=\cdots=\frac{\displaystyle{\log\left(1+\frac{1}{d}\right)}}{\log(10)}Which is the so-called Benford’s law. So, this distribution looks like that

> (benford=log(1+1/(1:9))/log(10))
[1] 0.30103000 0.17609126 0.12493874 0.09691001 0.07918125 
[6] 0.06694679 0.05799195 0.05115252 0.04575749
> names(benford)=1:9
> sum(benford)
[1] 1
> barplot(benford,col="white",ylim=c(-.045,.3))
> abline(h=0)

To compute the empirical distribution from a sample, use the following function

> firstdigit=function(x){
+ if(x>=1){x=as.numeric(substr(as.character(x),1,1)); zero=FALSE}
+ if(x<1){zero=TRUE}
+ while(zero==TRUE){
+ x=x*10; zero=FALSE
+ if(trunc(x)==0){zero=TRUE}
+ }
+ return(trunc(x))
+ }

and then

> Xd=sapply(X,firstdigit)
> table(Xd)/1000

In Benford’s Law: An Empirical Investigation and a Novel Explanation, we can read

It is not a mathematical article, so do not expect any formal proof in this paper. At least, we can run monte carlo simulation, and see what’s going on if we generate samples from a lognormal distribution with variance https://latex.codecogs.com/gif.latex?%20\sigma^2. For instance, with a unit variance,

> set.seed(1)
> s=1
> X=rlnorm(n=1000,0,s)
> Xd=sapply(X,firstdigit)
> table(Xd)/1000
    1     2     3     4     5     6     7     8     9 
0.288 0.172 0.121 0.086 0.075 0.072 0.073 0.053 0.060 
> T=rbind(benford,-table(Xd)/1000)
> barplot(T,col=c("red","white"),ylim=c(-.045,.3))
> abline(h=0)

Clearly, it not far away from Benford’s law. Perhaps a more formal test can be considered, for instance Pearson’s https://latex.codecogs.com/gif.latex?%20\chi^2 (goodness of fit) test.

> chisq.test(T,p=benford)

	Chi-squared test for given probabilities

data:  T 
X-squared = 10.9976, df = 8, p-value = 0.2018

So yes, Benford’s law is admissible ! Now, if we consider the case where https://latex.codecogs.com/gif.latex?%20\sigma is smaller (say 0.9), it is a rather different story,

compared with the case where https://latex.codecogs.com/gif.latex?%20\sigma is larger (say 1.1)

It is possible to generate several samples (always the same size, here 1,000 observations), just change the variance parameter https://latex.codecogs.com/gif.latex?%20\sigma and compute the https://latex.codecogs.com/gif.latex?%20p-value of the test. There might be one tricky part: when generating samples from lognormal distributions with small variance, it might be possible that some digits do not appear at all. On that case, there is a problem with the test. So we just use here

> T=table(Xd)
> T=T[as.character(1:9)]
> T[is.na(T)]=0
> PVAL[i]=chisq.test(T,p=benford)$p.value

Boxplots of the https://latex.codecogs.com/gif.latex?%20p-value of the test are the following,

When https://latex.codecogs.com/gif.latex?%20\sigma is too small, it is clearly not Benford’s distribution: for half (or more) of our samples, the https://latex.codecogs.com/gif.latex?%20p-value is lower than 5%. On the other hand, when https://latex.codecogs.com/gif.latex?%20\sigma is large (enough), Benford’s distribution is the distribution of the first digit of lognormal samples, since 95% of our samples have  https://latex.codecogs.com/gif.latex?%20p-values higher than 5% (and the distribution of the https://latex.codecogs.com/gif.latex?%20p-value is almost uniform on the unit interval). Here is the proportion of samples where the https://latex.codecogs.com/gif.latex?%20p-value was lower than 5% (on 5,000 generations each time)

Note that it is also possible to compute the https://latex.codecogs.com/gif.latex?%20p-value of Komogorov-Smirnov test, testing if the https://latex.codecogs.com/gif.latex?%20p-value has a uniform distribution,

> ks.test(PVAL[,s], "punif")$p.value

Indeed, if https://latex.codecogs.com/gif.latex?%20\sigma is larger than 1.15 (around that value), it looks like Benford’s law is a suitable distribution for the first digit.