Question

Benford’s law and fraud

(a) Using the graph from Exercise 21, calculate the standard deviation σ_Y. This gives us an idea of how much variation we’d expect in the employee’s expense records if he assumed that first digits from 1 to 9 were equally likely.

(b) The standard deviation of the first digits of randomly selected expense amounts that follow Benford’s law is ${\sigma }_X=2.46$. Would using standard deviations be a good way to detect fraud? Explain your answer.

Short answer

Part (a) The standard deviation is 2.5820.

Part (b) No

Step-by-step solution

Part (a) Step 1. Given information.

The given information is:

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

Part (a) Step 2. Find the standard deviation.

The expected value is:

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

The standard deviation is:

$$\begin{gathered} {\sigma }^2=\sum {\left(x-\mu \right)}^{2}P\left(x\right) \\ ={\left(1-5\right)}^2\times \frac{1}{9}+{\left(2-5\right)}^2\times \frac{1}{9}+{\left(3-5\right)}^2\times \frac{1}{9}+{\left(4-5\right)}^2\times \frac{1}{9}+{\left(5-5\right)}^2\times \frac{1}{9}+ \\ {\left(6-5\right)}^2\times \frac{1}{9}+{\left(7-5\right)}^2\times \frac{1}{9}+{\left(8-5\right)}^2\times \frac{1}{9}+{\left(9-5\right)}^2\times \frac{1}{9} \\ =\frac{20}{3} \\ \sigma =\sqrt{{\sigma }^2} \\ =\sqrt{\frac{20}{3}} \\ =\frac{2\sqrt{15}}{3}\approx 2.5820 \end{gathered}$$

Part (b) Step 1. Explanation.

The uniform distribution's standard deviation is 2.5820, while Benford's law's standard deviation is 2.46.

The two standard deviations are quite similar, so using the standard deviation to detect fraud is not a good idea because the two distributions have about the same standard deviation (the two distributions are difficult to distinguish by their standard deviation).