Chapter 3.2, Problem 21E

Suppose that you read through this year’s issues of the New York Times and record each number that appears in a news article-the income of a CEO. the number of cases of wine produced by a winery, the total charitable contribution of a politician during the previous tax year, the age of a celebrity, and so on. Now focus on the leading digit of each number, which could be 1.2    8. or 9.

Your first thought might be that the leading digit X of a randomly selected number would be equally likely to be one of the nine possibilities (a discrete uniform distribution). However, much empirical evidence as well as some theoretical arguments suggest an alternative probability distribution called Benford’s law:

$p(x)\text{ = }P\text{(1st digit is }x{\text{) = log}}_{10}(\frac{x\text{ + 1}}{x})x\text{ = 1, 2,}\mathrm{...}\text{, 9 }$

1. a. Without computing individual probabilities from this formula, show that it specifies a legitimate pmf. Now compute the individual probabilities and compare to the corresponding discrete uniform distribution.
2. b. Obtain the cdf of X.
3. c. Using the cdf, what is the probability that the leading digit is at most 3? At least 5? [Note: Benford’s law is the basis for some auditing procedures used to detect fraud in financial reporting—for example, by the Internal Revenue Service.]

To determine

Show that the pmf is legitimate.

Explanation of Solution

Given info:

The income of a CEO, the number of cases of wine produced by a winery, the total charitable contribution of a politician during the previous tax year and other things were noted from a paper. After this, the focus is on the leading digit of each number like 1, 2, …9. The leading digit X is a randomly selected number would be equally likely to be one of the all possibilities.

The Benford’s law is given as

$\begin{array}{c}p\left(x\right)=P\left(\text{1st digit on x}\right)\\ ={\text{log}}_{10}\left(\frac{x+1}{x}\right),x=1,\mathrm{2...9}\end{array}$

Calculation:

For checking valid pmf the conditions are:

1. 1. $p\left(x\right)>0$
2. 2. ${\displaystyle \sum _{allx}p\left(x\right)}=1$

Here,

$\left(\frac{x+1}{x}\right)=\left(1+\frac{1}{x}\right)>1,\text{when }x=1,\mathrm{2..9}$

Thus, from the property of logarithm,

${\text{log}}_{10}\left(\frac{x+1}{x}\right)>0$

Hence, 1^st condition is satisfied.

Again,

$\begin{array}{c}{\displaystyle \sum _{x=1}^9{\text{log}}_{10}\left(\frac{x+1}{x}\right)}=\left\{\begin{array}{l}{\text{log}}_{10}\left(\frac{1+1}{1}\right)+{\text{log}}_{10}\left(\frac{2+1}{2}\right)+{\text{log}}_{10}\left(\frac{3+1}{3}\right)+\\ {\text{log}}_{10}\left(\frac{4+1}{4}\right)+{\text{log}}_{10}\left(\frac{5+1}{5}\right)+{\text{log}}_{10}\left(\frac{6+1}{6}\right)+\\ {\text{log}}_{10}\left(\frac{7+1}{7}\right)+{\text{log}}_{10}\left(\frac{8+1}{8}\right)+{\text{log}}_{10}\left(\frac{9+1}{9}\right)\end{array}\right\}\\ =\left\{\begin{array}{l}{\text{log}}_{10}\left(\frac{2}{1}\right)+{\text{log}}_{10}\left(\frac{3}{2}\right)+{\text{log}}_{10}\left(\frac{4}{3}\right)+\\ {\text{log}}_{10}\left(\frac{5}{4}\right)+{\text{log}}_{10}\left(\frac{6}{5}\right)+{\text{log}}_{10}\left(\frac{7}{6}\right)+\\ {\text{log}}_{10}\left(\frac{8}{7}\right)+{\text{log}}_{10}\left(\frac{9}{8}\right)+{\text{log}}_{10}\left(\frac{10}{9}\right)\end{array}\right\}\end{array}$

It is known that ${\displaystyle \sum _i\log x_i}=\text{log}\left({\displaystyle \prod _i{x}_i}\right)$.

Using this formula,

$\begin{array}{c}{\displaystyle \sum _{x=1}^9{\text{log}}_{10}\left(\frac{x+1}{x}\right)}={\text{log}}_{10}\left\{\begin{array}{l}\left(\frac{2}{1}\right)\times \left(\frac{3}{2}\right)\times \left(\frac{4}{3}\right)\times \left(\frac{5}{4}\right)\times \left(\frac{6}{5}\right)\\ \times \left(\frac{7}{6}\right)\times \left(\frac{8}{7}\right)\times \left(\frac{9}{8}\right)\times \left(\frac{10}{9}\right)\end{array}\right\}\\ ={\text{log}}_{10}10\\ =\frac{\text{log}10}{\text{log}10}\\ =1\end{array}$

Hence, 2^nd condition is also satisfied.

Thus, the pmf is a legitimate pmf.

To determine

Find the individual probabilities.

Compare the discrete uniform distribution.

Answer to Problem 21E

The probabilities are:

x	p(x)
1	0.30103
2	0.176
3	0.125
4	0.0969
5	0.0791
6	0.066
7	0.05799
8	0.05115
9	0.0457

Explanation of Solution

Calculation:

The probabilities:

For x = 1:

$\begin{array}{c}p\left(1\right)={\text{log}}_{10}\left(\frac{1+1}{1}\right)\\ ={\text{log}}_{10}2\\ =0.30103\end{array}$

For x = 2:

$\begin{array}{c}p\left(2\right)={\text{log}}_{10}\left(\frac{2+1}{2}\right)\\ ={\text{log}}_{10}\frac{3}{2}\\ =0.176\end{array}$

For x = 3:

$\begin{array}{c}p\left(3\right)={\text{log}}_{10}\left(\frac{3+1}{3}\right)\\ ={\text{log}}_{10}\frac{4}{3}\\ =0.125\end{array}$

For x = 4:

$\begin{array}{c}p\left(4\right)={\text{log}}_{10}\left(\frac{4+1}{4}\right)\\ ={\text{log}}_{10}\frac{5}{4}\\ =0.0969\end{array}$

For x = 5:

$\begin{array}{c}p\left(5\right)={\text{log}}_{10}\left(\frac{5+1}{5}\right)\\ ={\text{log}}_{10}\frac{6}{5}\\ =0.0791\end{array}$

For x = 6:

$\begin{array}{c}p\left(6\right)={\text{log}}_{10}\left(\frac{6+1}{6}\right)\\ ={\text{log}}_{10}\frac{7}{6}\\ =0.066\end{array}$

For x = 7:

$\begin{array}{c}p\left(7\right)={\text{log}}_{10}\left(\frac{7+1}{7}\right)\\ ={\text{log}}_{10}\frac{8}{7}\\ =0.05799\end{array}$

For x = 8:

$\begin{array}{c}p\left(8\right)={\text{log}}_{10}\left(\frac{8+1}{8}\right)\\ ={\text{log}}_{10}\frac{9}{8}\\ =0.05115\end{array}$

For x = 8:

$\begin{array}{c}p\left(9\right)={\text{log}}_{10}\left(\frac{9+1}{9}\right)\\ ={\text{log}}_{10}\frac{10}{9}\\ =0.0457\end{array}$

The probabilities:

x	p(x)
1	0.30103
2	0.176
3	0.125
4	0.0969
5	0.0791
6	0.066
7	0.05799
8	0.05115
9	0.0457

Comparison:

As the value of x increased, then the value of $p\left(x\right)$ decreased.

Hence, the distribution is not uniform distribution.

To determine

Find the cdf of X.

Answer to Problem 21E

The cdf is,

$F\left(x\right)=\{\begin{array}{l}0,x<0\\ 0.30103,1\le x<2\\ 0.477,2\le x<3\\ 0.602,3\le x<4\\ 0.699,4\le x<5\\ 0.778,5\le x<6\\ 0.844,6\le x<7\\ 0.90199,7\le x<8\\ 0.95314,8\le x<9\\ 1,x\ge 9\end{array}$

Explanation of Solution

Calculation:

Cumulative distribution function:

Here, the random variable X is a discrete random variables with values $x_1,x_2,\mathrm{...},x_n$. The cumulative distribution function of the random variable X is denoted as $F\left(x\right)$ and is given by:

$\begin{array}{c}F\left(x\right)=P\left(X\le x\right)\\ ={\displaystyle \sum _{x_i\le x}p\left(x_i\right)}\end{array}$

The distribution F(x) jumps on 0, 1,….8.

The cumulative probabilities are:

For x = 1:

$\begin{array}{c}F\left(1\right)=P\left(X\le 1\right)\\ =0.30103\end{array}$

For x = 2:

$\begin{array}{c}F\left(2\right)=P\left(X\le 2\right)\\ =P\left(X\le 1\right)+P\left(X=2\right)\\ =0.30103+0.176\\ =0.477\end{array}$

For x = 3:

$\begin{array}{c}F\left(3\right)=P\left(X\le 3\right)\\ =P\left(X\le 2\right)+P\left(X=3\right)\\ =0.477+0.125\\ =0.602\end{array}$

For x = 4:

$\begin{array}{c}F\left(4\right)=P\left(X\le 4\right)\\ =P\left(X\le 3\right)+P\left(X=4\right)\\ =0.602+0.0969\\ =0.699\end{array}$

For x = 5:

$\begin{array}{c}F\left(5\right)=P\left(X\le 5\right)\\ =P\left(X\le 4\right)+P\left(X=5\right)\\ =0.699+0.0791\\ =0.778\end{array}$

For x = 6:

$\begin{array}{c}F\left(6\right)=P\left(X\le 6\right)\\ =P\left(X\le 5\right)+P\left(X=6\right)\\ =0.778+0.066\\ =0.844\end{array}$

For x = 7:

$\begin{array}{c}F\left(7\right)=P\left(X\le 7\right)\\ =P\left(X\le 6\right)+P\left(X=7\right)\\ =0.844+0.05799\\ =0.90199\end{array}$

For x = 8:

$\begin{array}{c}F\left(8\right)=P\left(X\le 8\right)\\ =P\left(X\le 7\right)+P\left(X=8\right)\\ =0.90199+0.05115\\ =0.95314\end{array}$

For x = 9:

$\begin{array}{c}F\left(8\right)=P\left(X\le 9\right)\\ =P\left(X\le 8\right)+P\left(X=9\right)\\ =0.90199+0.0457\\ =1\end{array}$

Hence, the cdf is,

$F\left(x\right)=\{\begin{array}{l}0,x<0\\ 0.30103,1\le x<2\\ 0.477,2\le x<3\\ 0.602,3\le x<4\\ 0.699,4\le x<5\\ 0.778,5\le x<6\\ 0.844,6\le x<7\\ 0.90199,7\le x<8\\ 0.95314,8\le x<9\\ 1,x\ge 9\end{array}$

To determine

Find the probability that the leading digit is at most 3.

Find the probability that the leading digit is at least 5.

Answer to Problem 21E

The probability that the leading digit is at most 3 is 0.602 and the probability that the leading digit is at least 5 is 0.301.

Explanation of Solution

Calculation:

The probability that the leading digit is at most 3:

$\begin{array}{c}P\left(X\le 3\right)=F\left(3\right)\\ =0.602,\text{From the cdf}\end{array}$

The probability that the leading digit is at least 5:

$\begin{array}{c}P\left(X\ge 5\right)=1-P\left(X<5\right)\\ =1-P\left(X\le 4\right)\\ =1-F\left(4\right)\\ =1-0.699,\text{From the cdf}\end{array}$

$\text{=0}\text{.301}$

Hence, the probability that the leading digit is at most 3 is 0.602 and the probability that the leading digit is at least 5 is 0.301.