/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Suppose that you read through th... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(1,2, \ldots, 8\), 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: $$ \begin{aligned} p(x) &=P(1 \text { st digit is } x)=\log _{10}\left(\frac{x+1}{x}\right), \\ x &=1,2, \ldots, 9 \end{aligned} $$ a. Without computing individual probabilities from this formula, show that it specifies a legitimate pmf. b. Now compute the individual probabilities and compare to the corresponding discrete uniform distribution. c. Obtain the cdf of \(X\). d. Using the odf, 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.]

Short Answer

Expert verified
Sum of Benford's probabilities is 1, confirming a valid PMF. Benford's probabilities differ significantly from uniform. CDF is the cumulative sum of Benford's probabilities. Probability of digit at most 3 is approximately 0.477 and at least 5 is approximately 0.327.

Step by step solution

01

Define a Legitimate Probability Mass Function (PMF)

A probability mass function (PMF) is legitimate if the sum of probabilities for all possible outcomes equals 1. Hence, we need to show that \( \sum_{x=1}^{9} \log_{10} \left( \frac{x+1}{x} \right) = 1 \).
02

Compute Individual Probabilities Using Benford's Law

For each \( x \) from 1 to 9, calculate the probability using the formula \( p(x) = \log_{10}\left(\frac{x+1}{x}\right) \). For example, \( p(1) = \log_{10}\left(\frac{2}{1}\right) \), \( p(2) = \log_{10}\left(\frac{3}{2}\right) \), and so on. This will give us probabilities: \( p(1) \approx 0.301 \), \( p(2) \approx 0.176 \), etc.
03

Compare to a Discrete Uniform Distribution

In a uniform distribution, the probability for each digit is \( \frac{1}{9} \approx 0.111 \). Compare these constant probabilities to those found from Benford's Law, noting the stark differences.
04

Obtain the Cumulative Distribution Function (CDF)

Calculate the cumulative probabilities \( F(x) = P(X \leq x) = \sum_{i=1}^{x} p(i) \) for each \( x \) from 1 to 9. This means, for example, \( F(1) = p(1) \), \( F(2) = p(1) + p(2) \), etc.
05

Calculate the Probabilities Using the CDF

For at most 3, compute \( F(3) = P(X \leq 3) = p(1) + p(2) + p(3) \). For at least 5, use the complement rule: \( P(X \geq 5) = 1 - F(4) \), where \( F(4) \) is the cumulative probability up to 4.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Mass Function
A probability mass function (PMF) is a key concept in statistics used to specify the probability of a discrete random variable taking on particular values. For a PMF to be valid, the probabilities for all possible outcomes must sum to one. In the case of Benford's Law, the PMF is given for leading digits in numerical data. The formula \[ p(x) = \log_{10}\left(\frac{x+1}{x}\right) \] defines these probabilities for the digits 1 through 9. To verify this as a legitimate PMF, we can sum the probabilities and check if they equal 1. This ensures it follows the basic rule of PMFs, making it a valid probability distribution.
Discrete Uniform Distribution
The discrete uniform distribution is a simple distribution where each outcome is equally likely. For example, if you're rolling a fair six-sided die, each number from 1 to 6 has an equal probability of 1/6. For the case of leading digits from 1 to 9, in a uniform distribution, each digit would have a probability of \( 1/9 \approx 0.111 \).
In contrast, Benford's Law offers a different perspective, as it assigns different probabilities to each leading digit based on historical frequencies found in data sets. Comparing this with the discrete uniform distribution highlights how Benford's Law deviates from simple equal probabilities, showing us real-world data complexities.
Cumulative Distribution Function
The cumulative distribution function (CDF) helps us determine the probability that a random variable takes a value less than or equal to a certain point. For Benford's Law, the CDF is built from its PMF.
The cumulative probability for a digit \( x \) is found by summing the probabilities for all digits up to \( x \), using \[ F(x) = \sum_{i=1}^{x} p(i) \].
This means, for example, \( F(3) \) gives the probability that the leading digit is 1, 2, or 3. By calculating cumulative probabilities, we can answer questions like the likelihood of a digit being less than or equal to 3 or at least 5, providing deeper insights into the data distribution.
Auditing Procedures
Auditing procedures often utilize Benford's Law because it uncovers irregularities in financial data reporting. If financial figures are manipulated, they might not follow the expected distribution of leading digits as prescribed by Benford's Law.
This makes it a powerful tool in detecting fraud or anomalies. For example, auditors can compare the observed frequency of leading digits in a company's financial records against the expected frequencies. Significant deviations can signal the need for deeper investigation, making Benford's Law an instrumental part of modern auditing procedures, particularly in domains like tax audits and forensic accounting.
Leading Digit Distributions
The study of leading digit distributions is a fascinating field, with Benford's Law at its core. It predicts that in many naturally occurring data sets, the number 1 will be the most common leading digit, followed by 2, and so on, with 9 being the least common.
This is contrary to the intuitive assumption that all digits would appear with equal likelihood.
The distribution indicates how real-world numerical datasets, like financial reports or population numbers, tend to skew. Understanding these distributions provides valuable insights into data patterns, helping professionals across different fields, such as forensic analytics, fraud detection, and data science, to leverage these unexpected patterns in practical applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A chemical supply company currently has in stock \(100 \mathrm{lb}\) of a chemical, which it sells to customers in 5-lb containers. Let \(X=\) the number of containers ordered by a randomly chosen customer, and suppose that \(X\) has pmf \begin{tabular}{l|llll} \(x\) & 1 & 2 & 3 & 4 \\ \hline\(p(x)\) & \(.2\) & \(.4\) & \(.3\) & \(.1\) \end{tabular} Compute \(E(X)\) and \(V(X)\). Then compute the expected number of pounds left after the next customer's order is shipped and the variance of the number of pounds left. [Hint: The number of pounds left is a linear function of \(X\).]

Suppose small aircraft arrive at an airport according to a Poisson process with rate \(\alpha=8 / \mathrm{h}\), so that the number of arrivals during a time period of \(t\) hours is a Poisson rv with parameter \(\hat{\lambda}=8 t\). a. What is the probability that exactly 6 small aircraft arrive during a 1 -h period? At least 6 ? At least 10 ? b. What are the expected value and standard deviation of the number of small aircraft that arrive during a 90 -min period? c. What is the probability that at least 20 small aircraft arrive during a \(2 \frac{1}{2} h\) period? That at most 10 arrive during this period?

The mode of a discrete random variable \(X\) with pmf \(p(x)\) is that value \(x\) * for which \(p(x)\) is largest (the most probable \(x\) value). a. Let \(X-\operatorname{Bin}(n, p)\). By considering the ratio \(b(x+1 ; n, p) / b(x, n, p)\), show that \(b(x, n, p)\) increases with \(x\) as long as \(x

The number of requests for assistance received by a towing service is a Poisson process with rate \(\alpha=4 / \mathrm{h}\). a. Compute the probability that exactly ten requests are received during a particular 2-h period. b. If the operators of the towing service take a \(30-\) min break for lunch, what is the probability that they do not miss any calls for assistance? c. How many calls would you expect during their break?

Automobiles arrive at a vehicle equipment inspection station according to a Poisson process with rate \(\alpha=10 / \mathrm{h}\). Suppose that with probability .5 an arriving vehicle will have no equipment violations. a. What is the probability that exactly ten arrive during the hour and all ten have no violations? b. For any fixed \(y \geq 10\), what is the probability that \(y\) arrive during the hour, of which ten have no violations? c. What is the probability that ten "no-violation" cars arrive during the next hour? [Hint: Sum the probabilities in part (b) from \(y=10\) to \(\infty\).]

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.