/*! 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 10 A family of pdf's that has been ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A family of pdf's that has been used to approximate the distribution of income, city population size, and size of firms is the Pareto family. The family has two parameters, \(k\) and \(\theta\), both \(>0\), and the pdf is $$ f(x ; k, \theta)=\left\\{\begin{array}{cl} \frac{k \cdot \theta^{k}}{x^{k+1}} & x \geq \theta \\ 0 & x<\theta \end{array}\right. $$ a. Sketch the graph of \(f(x ; k, \theta)\). b. Verify that the total area under the graph equals 1 . c. If the rv \(X\) has pdf \(f(x ; k, \theta)\), for any fixed \(b>\theta\), obtain an expression for \(P(X \leq b)\). d. For \(\theta

Short Answer

Expert verified
a) Sketch a decreasing curve starting from x=θ. b) Total area under the curve = 1. c) P(X ≤ b) = 1 - (θ/b)^k for b > θ. d) P(a ≤ X ≤ b) = (θ/a)^k - (θ/b)^k for θ < a < b.

Step by step solution

01

Understand the PDF Definition

The given probability density function (pdf) is for a Pareto distribution with parameters \( k \) and \( \theta \). The pdf is only positive for \( x \geq \theta \). Mathematically, it is defined as: $$ f(x; k, \theta) = \frac{k \cdot \theta^k}{x^{k+1}} \text{ for } x \geq \theta $$ and \( f(x; k, \theta) = 0 \) for \( x < \theta \).
02

Sketch the PDF Graph

To sketch \( f(x; k, \theta) \), note that the function is only defined for \( x \geq \theta \) and follows a power-law distribution, decreasing as \( x \) increases. Start the graph at \( x = \theta \) with a value of \( \frac{k \cdot \theta^k}{\theta^{k+1}} = \frac{k}{\theta} \), then let it decline asymptotically towards zero as \( x \to \infty \).
03

Verify Total Area Equals 1

Calculate the area under the curve from \( \theta \) to \( \infty \). This is an improper integral: $$ \int_{\theta}^{\infty} \frac{k \cdot \theta^k}{x^{k+1}} \, dx = \left[ -\frac{\theta^k}{x^k} \right]_{\theta}^{\infty} = 1 $$ The calculation verifies that the total area under the pdf is 1, confirming a valid probability distribution.
04

Find P(X ≤ b) for b > θ

To find \( P(X \leq b) \), calculate the integral from \( \theta \) to \( b \): $$ \int_{\theta}^{b} \frac{k \cdot \theta^k}{x^{k+1}} \, dx = 1 - \left( \frac{\theta}{b} \right)^k $$ This is the cumulative distribution function (CDF) evaluated at \( b \).
05

Find P(a ≤ X ≤ b) for θ < a < b

Using the CDF found previously, calculate \( P(a \leq X \leq b) = P(X \leq b) - P(X < a) \). Thus, $$ P(a \leq X \leq b) = \left[1 - \left( \frac{\theta}{b} \right)^k\right] - \left[1 - \left( \frac{\theta}{a} \right)^k\right] = \left( \frac{\theta}{a} \right)^k - \left( \frac{\theta}{b} \right)^k $$

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 Density Function
The concept of a probability density function (PDF) revolves around understanding how probabilities are distributed over a continuous range of possible outcomes. For the Pareto distribution, its PDF helps in modeling phenomena like income or population sizes across various categories. Key characteristics of the PDF include the parameters \( k \) and \( \theta \), where \( k \) influences the tail behavior of the distribution and \( \theta \) sets the threshold above which the distribution applies. In mathematical terms, the PDF for a Pareto distribution is given by \[ f(x ; k, \theta) = \frac{k \cdot \theta^k}{x^{k+1}} \quad \text{for } x \geq \theta \] This means that as \( x \) increases, the value of \( f(x ; k, \theta) \) decreases. This defines the power-law nature of the distribution, showcasing a rapid fall in probability density with increasing \( x \). Notably, the function is zero for any \( x < \theta \), emphasizing that the distribution only applies to values above the parameter \( \theta \). Understanding a PDF is crucial, as it provides insights into how concentrated or spread out the outcomes are over the threshold. It's all about seeing the likelihood of a system exhibiting specific behavior based on the value of \( x \).
Cumulative Distribution Function
The cumulative distribution function (CDF) offers a complementary perspective to the PDF by quantifying the probability that a random variable \( X \) will take a value less than or equal to a specified point \( b \). For the Pareto distribution, calculating the CDF involves integrating the PDF from \( \theta \) to \( b \). The expression for the CDF in this context is given by: \[ P(X \leq b) = 1 - \left( \frac{\theta}{b} \right)^k \] Here, the subtraction from 1 denotes the probability of obtaining a value greater than \( b \), making it intuitive by flipping the domain consideration of the PDF. This CDF clearly shows how probabilities accumulate as we progress along the \( x \)-axis from \( \theta \) to \( b \), explaining the distribution's inclination to assign lower probabilities to higher \( x \) values. Remember, the CDF is a useful tool for providing a full probability accounting, making it easier to understand the cumulative likelihood of system command under different conditions. It bridges the specific point outcomes of the PDF into a broader range of predictions.
Probability Distribution Verification
Verifying a probability distribution means confirming that the total probability across the entire space is precisely 1, as any probability distribution on a continuous interval must completely cover the unit interval. For the Pareto distribution, this involves computing the integral of the PDF across its support. To demonstrate this for the Pareto, you calculate: \[ \int_{\theta}^{\infty} \frac{k \cdot \theta^k}{x^{k+1}} \, dx = 1 \] This verifies using improper integral evaluations, showing that all possible values from \( \theta \) to \( \infty \) account for a total probability of 1. Essentially, it checks that the distribution density sums up correctly over its domain, ensuring isolated regions of improbably high density cannot skew overall probability. By conducting this verification, it assures that the behavior of the Pareto distribution conforms to the fundamental law of probability. This process is essential in validating models, as it guarantees they are mathematically sound and applicable in real-world statistical scenarios.

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

Find the following percentiles for the standard normal distribution. Interpolate where appropriate. a. \(91 \mathrm{st}\) b. 9 th c. 75 th d. 25 th e. 6 th

Suppose Appendix Table A.3 contained \(\Phi(z)\) only for \(z \geq 0\). Explain how you could still compute a. \(P(-1.72 \leq Z \leq-.55)\) b. \(P(-1.72 \leq Z \leq .55)\) Is it necessary to tabulate \(\Phi(z)\) for \(z\) negative? What property of the standard normal curve justifies your answer?

Suppose that when a transistor of a certain type is subjected to an accelerated life test, the lifetime \(X\) (in weeks) has a gamma distribution with mean 24 weeks and standard deviation 12 weeks. a. What is the probability that a transistor will last between 12 and 24 weeks? b. What is the probability that a transistor will last at most 24 weeks? Is the median of the lifetime distribution less than 24 ? Why or why not? c. What is the 99th percentile of the lifetime distribution? d. Suppose the test will actually be terminated after \(t\) weeks. What value of \(t\) is such that only \(.5 \%\) of all transistors would still be operating at termination?

Vehicle speed on a particular bridge in China can be modeled as normally distributed ("Fatigue Reliability Assessment for Long-Span Bridges under Combined Dynamic Loads from Winds and Vehicles" \(J\). of Bridge Engr., 2013: 735-747). a. If \(5 \%\) of all vehicles travel less than \(39.12 \mathrm{~m} / \mathrm{h}\) and \(10 \%\) travel more than \(73.24 \mathrm{~m} / \mathrm{h}\), what are the mean and standard deviation of vehicle speed? [Note: The resulting values should agree with those given in the cited article.] b. What is the probability that a randomly selected vehicle's speed is between 50 and \(65 \mathrm{~m} / \mathrm{h}\) ? c. What is the probability that a randomly selected vehicle's speed exceeds the speed limit of \(70 \mathrm{~m} / \mathrm{h}\) ?

The article "A Model of Pedestrians" Waiting Times for Street Crossings at Signalized Intersections" (Transportation Research, 2013: 17-28) suggested that under some circumstances the distribution of waiting time \(X\) could be modeled with the following pdf: $$ f(x ; \theta, \tau)=\left\\{\begin{array}{cl} \frac{\theta}{\tau}(1-x / \tau)^{\theta-1} & 0 \leq x<\tau \\ 0 & \text { otherwise } \end{array}\right. $$ a. Graph \(f(x ; \theta, 80)\) for the three cases \(\theta=4,1\), and .5 (these graphs appear in the cited article) and comment on their shapes. b. Obtain the cumulative distribution function of \(X\). c. Obtain an expression for the median of the waiting time distribution. d. For the case \(\theta=4, \tau=80\), calculate \(P(50 \leq X \leq 70)\) without at this point doing any additional integration.

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.