Question

Exercises 21 and 22 examine how Benford’s law (Exercise 9) can be used to detect fraud.

Benford’s law and fraud A not-so-clever employee decided to fake his monthly expense report. He believed that the first digits of his expense amounts should be equally likely to be any of the numbers from 1 to 9. In that case, the first digit Yof a randomly selected expense amount would have the probability distribution shown in the histogram.

(a) What’s $P(Y<6)$? According to Benford’s law (see Exercise 9), what proportion of first digits in the employee’s expense amounts should be greater than 6? How could this information be used to detect a fake expense report?

(b) Explain why the mean of the random variable Yis located at the solid red line in the figure.

(c) According to Benford’s law, the expected value of the first digit is $\mu X=3.441$. Explain how this information could be used to detect a fake expense report.

Short answer

Part (a) $P\left(Y>6\right)=33.33\%$, Benford’s law:$P\left(Y>6\right)=15.5\%$

If the proportion of first digits greater than 6 is closer to 0.3333 than 0.155, the cost report is likely to be incorrect.

Part (b) The distribution is symmetric, and the mean for a symmetric distribution is exactly in the center of the distribution, implying that the mean must be at 5.

Part (c) If the average of the first digits in an expense report is less than 3.441, the expense report is likely to be fake.

Step-by-step solution

Part (a) Step 1. Given information.

The probability for each value is $\frac{1}{9}$.

Part (a) Step 2. Find PY>6.

First digit	$123456789$
Probability	$\frac{1}{9}\frac{1}{9}\frac{1}{9}\frac{1}{9}\frac{1}{9}\frac{1}{9}\frac{1}{9}\frac{1}{9}\frac{1}{9}$

The probability of each event is given as $\frac{1}{9}$.

$$\begin{gathered} P\left(Y=7\right)=\frac{1}{9} \\ P\left(Y=8\right)=\frac{1}{9} \\ P\left(Y=9\right)=\frac{1}{9} \end{gathered}$$

It can not be possible for the 1^st digit to be 2 different kinds of digits at the same time, therefore the above events are mutually exclusive.

Apply the addition rule for events of a mutually exclusive nature:

$$\begin{gathered} P\left(Y>6\right)=P\left(Y=7\right)+P\left(Y=8\right)+P\left(Y=9\right) \\ =\frac{1}{9}+\frac{1}{9}+\frac{1}{9} \\ =\frac{3}{9} \\ =\frac{1}{3}\approx 33.33\% \end{gathered}$$

Part (a) Step 3. Use Benford’s law.

Benford's Law:

First Digit	1	2	3	4	5	6	7	8	9
Probability	0.301	0.176	0.125	0.097	0.079	0.067	0.058	0.051	0.046

$$\begin{gathered} P\left(Y=7\right)=0.058 \\ P\left(Y=8\right)=0.051 \\ P\left(Y=9\right)=0.046 \end{gathered}$$

It can not be possible for the 1^st digit to be 2 different kinds of digits at the same time, therefore the above events are mutually exclusive.

Apply the addition rule for events of a mutually exclusive nature:

$$\begin{gathered} P\left(Y>6\right)=P\left(Y=7\right)+P\left(Y=8\right)+P\left(Y=9\right) \\ =0.058+0.051+0.046 \\ =0.155 \\ =15.5\% \end{gathered}$$

If the fraction of the first digit greater than 6 is closer to 0.3333 than 0.155, the cost report is likely to be fraudulent.

Part (b) Step 1. An explanation for the mean of the random variable Y is located at the solid red line in the figure.

$$\begin{gathered} \mu =\sum xP\left(X=x\right) \\ =1\times \frac{1}{9}+2\times \frac{1}{9}+3\times \frac{1}{9}+4\times \frac{1}{9}+5\times \frac{1}{9}+6\times \frac{1}{9}+7\times \frac{1}{9}+8\times \frac{1}{9}+9\times \frac{1}{9} \\ =\frac{1+2+3+4+5+6+7+8+9}{9} \\ =\frac{45}{9} \\ =5 \end{gathered}$$

In the following graphic, we can see that the anticipated value is 5 because the distribution is symmetric. The mean sits exactly in the middle of the distribution for a symmetric distribution, implying that the mean must be 5.

Part (c) Step 1. Explanation.

The uniform distribution's mean is 5, and Benford's law's mean is 3.441.

If the average of the first digits in an expense report is less than 3.441, the expense report is likely to be fake.

As a result, if we get an expense report with a mean of all initial digits that is closer to 5 than 3.441, we'll conclude it's a fake.