Benford's law – formula

Question

May I ask what does the "$k$" represent in Benford's law?

The formula I'm struggling with is in here – $$\mathbb P(d) =\sum\limits_{k=10^{n-2}}^{10^{n-1}-1} \log_{10}\left(1+\frac1{10k+d}\right)$$

Thanks in advance.

Answer 1

A quick look at the Wikipedia page of the Benford's law gives a good definition.

Answer 2

$$\mathbb P(d) =\sum\limits_{k=10^{n-2}}^{10^{n-1}-1} \log_{10}\left(1+\frac1{10k+d}\right)$$ is an expression for the probability under Benford's law that the $n$th digit being considered is $d$, with $d$ in $\{1,2,3,4,5,6,7,8,9\}$.

For this sum to to be meaningful, you need $n\ge 2$; if instead you are considering the first digit then you have $\mathbb P(d) = \log_{10}\left(1+\frac1{d}\right)$.

As an example, suppose $d=7$ and $n=3$ so you are considering the probability that the third digit is $7$ with Benford's law. You want the probability that numbers start $107\ldots$ or $117\ldots$ or $127\ldots$ and so on up to $997\ldots$.

That will be $\log_{10}\left(1+\frac1{107}\right)+\log_{10}\left(1+\frac1{117}\right)+\log_{10}\left(1+\frac1{127}\right) +\cdots+\log_{10}\left(1+\frac1{997}\right)$ which could be written as $\sum\limits_{k=10}^{99} \log_{10}\left(1+\frac1{10k+7}\right)$, i.e. the original expression with $d=7$ and $n=3$.