Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

. Let \(Y_{1}

Short Answer

Expert verified
The distribution function (CDF) of the smallest order statistic from a Weibull distribution, \(Y_{1}\), is given by \(F_{Y_1}(y_1)=1-e^{-n(y_1/\lambda)^{k}}\) and the probability density function (pdf) is given by \(f_{Y_1}(y_1) = n \frac{k}{\lambda } \left(\frac{y_1}{\lambda }\right) ^{k-1} e^{-n(y_1/\lambda)^{k}}\)

Step by step solution

01

Understand the properties of the Weibull distribution

The Weibull distribution has a pdf represented by \(f(y;\lambda,k) = \frac{k}{\lambda} \left(\frac{y}{\lambda}\right)^{k-1} e^{-(y/\lambda)^{k}} \text{ for } y\geq0, \lambda>0, k>0\). The cumulative distribution function (CDF) is represented by \(F(y;\lambda,k)=1-e^{-(y/\lambda)^{k}} \text{ for } y\geq0\).
02

Finding the joint distribution of order statistics

The joint distribution of ordered statistics of a random sample of size \(n\) from a population with pdf \(f(y)\) and cumulative distribution function (CDF) \(F(y)\) is \(f(y_1, y_2, ..., y_n)=n!f(y_1)f(y_2)...f(y_n)[F(y_1)...(F(y_2)-F(y_1))...(F(y_n)-F(y_{n-1}))...(1-F(y_n))]\). Here \(y_1<y_2<...<y_n\)
03

Derive the cumulative distribution function (CDF) of \(Y_{1}\)

The CDF of the minimum order statistic, \(Y_{1}\), is represented by \(F_{Y_1}(y_1)=1-(1-F(y_1))^n\), since \(1-F(y_1)\) represents the probability of a random observation exceeding \(y_1\) and \(1-(1-F(y_1))^n\) is the cumulative probability that at least one observation out of \(n\) does not exceed \(y_1\) (or at least one observation equals the minimum, \(y_1\)). For our specific case of the Weibull distribution, the CDF of \(Y_{1}\) becomes \(F_{Y_1}(y_1)=1-e^{-n(y_1/\lambda)^{k}}\)
04

Derive the probability density function (pdf) of \(Y_{1}\)

Once we have the CDF, we can find the pdf by taking the derivative of the CDF. In this case, the pdf of \(Y_{1}\) becomes \(f_{Y_1}(y_1) = \frac{d}{dy_1} F_{Y_1}(y_1) = n \frac{k}{\lambda } \left(\frac{y_1}{\lambda }\right) ^{k-1} e^{-n(y_1/\lambda)^{k}}\)

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 Vaia!

Key Concepts

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

Order Statistics
Order statistics are an important concept in probability and statistics. They involve arranging a set of random variables in increasing order. For instance, if you have a random sample of size \(n\) from a given distribution, the variables \(Y_{1}, Y_{2}, \dots, Y_{n}\) denote these variables ordered such that \(Y_{1} < Y_{2} < \cdots < Y_{n}\). Here, the smallest value is \(Y_{1}\) and the largest is \(Y_{n}\).
  • Order statistics help in finding extreme values, such as minimum or maximum.
  • They are widely used in reliability analysis and operations research.
  • Often utilized in financial risk modeling to identify worst-case scenarios.
In practice, you might be interested in the distribution of the smallest value (like \(Y_{1}\)), since it often holds importance in risk assessments and quality control sectors.
Cumulative Distribution Function
The cumulative distribution function (CDF) is a crucial tool for understanding probabilities. It describes the probability that a random variable takes a value less than or equal to a certain number. For a random variable \(Y\), the CDF is denoted as \(F(y)\) and is defined as:\[F(y) = P(Y \leq y)\]
  • The CDF starts from 0 and increases to 1 as you move along the distribution.
  • It provides a comprehensive description of all probability aspects of the distribution.
  • Useful in computing probabilities within a range and finding quantiles.
In the case of the Weibull distribution, the CDF helps you determine how likely it is for a variable to be less than or equal to a value \(y\). For instance, the CDF of the minimum order statistic, \(Y_{1}\), can be calculated as \(F_{Y_1}(y_1)=1-e^{-n(y_1/\lambda)^{k}}\). This formula reflects the cumulative probability of at least one observation being smaller or equal to a given \(y_1\) in a sample of size \(n\).
Probability Density Function
The probability density function (PDF) is another vital concept that complements the CDF. It provides the likelihood of a continuous random variable to take on a specific value. For a distribution's PDF, symbolized typically as \(f(y)\), it involves:\[f(y) = \lim_{\Delta y \to 0} \frac{P(y \leq Y < y + \Delta y)}{\Delta y}\]
  • The PDF is not a probability; it needs to integrate to 1 over the entire distribution.
  • For continuous distributions, the PDF at any particular point can be found by differentiating the CDF.
  • Useful in recognizing where values are most densely packed or spread out in a distribution.
In our special case, when it comes to the Weibull's smallest order statistic \(Y_{1}\), the process involves getting the probability density by differentiating its CDF. The resulting PDF is expressed as \(f_{Y_1}(y_1) = n \frac{k}{\lambda } \left(\frac{y_1}{\lambda }\right)^{k-1} e^{-n(y_1/\lambda)^{k}}\). This formula reveals how the probability density behaves, emphasizing occurrences of the smallest value in the sample.

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

Suppose \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample drawn from a \(N\left(\mu, \sigma^{2}\right)\) distribution. In this case, the pivot random variable for a confidence interval is $$ t=\frac{\bar{X}-\mu}{S / \sqrt{n}} $$ where \(\bar{X}\) and \(S\) are the sample mean and standard deviation, respectfully. Recall by Theorem \(3.6 .1\) that \(t\) has a Student \(t\) -distribution with \(n-1\) degrees of freedom; hence, its distribution is free of all parameters for this normal situation. In the notation of this section \(t_{n-1}^{(\gamma)}\) denotes the \(\gamma 100 \%\) percentile of a \(t\) -distribution with \(n-1\) degrees of freedom. Using this notation show that a \((1-\alpha) 100 \%\) confidence interval for \(\mu\) is $$ \left(\bar{x}-t^{(1-\alpha / 2)} \frac{s}{\sqrt{n}}, \bar{x}-t^{(\alpha / 2)} \frac{s}{\sqrt{n}}\right) $$

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a continuous type distribution. (a) Find \(P\left(X_{1} \leq X_{2}\right), P\left(X_{1} \leq X_{2}, X_{1} \leq X_{3}\right), \ldots, P\left(X_{1} \leq X_{i}, i=2,3, \ldots, n\right)\) (b) Suppose the sampling continues until \(X_{1}\) is no longer the smallest observation, (i.e., \(\left.X_{j}

Each of 51 golfers hit three golf balls of brand \(X\) and three golf balls of brand \(\mathrm{Y}\) in a random order. Let \(X_{i}\) and \(Y_{i}\) equal the averages of the distances traveled by the brand \(\mathrm{X}\) and brand \(\mathrm{Y}\) golf balls hit by the \(i\) th golfer, \(i=1,2, \ldots, 51\). Let \(W_{i}=X_{i}-Y_{i}, i=1,2, \ldots, 51 .\) To test \(H_{0}: \mu_{W}=0\) against \(H_{1}: \mu_{W}>0\), where \(\mu_{W}\) is the mean of the differences. If \(\bar{w}=2.07\) and \(s_{W}^{2}=84.63\), would \(H_{0}\) be accepted or rejected at an \(\alpha=0.05\) significance level? What is the \(p\) -value of this test?

Assume a binomial model for a certain random variable. If we desire a 90 percent confidence interval for \(p\) that is at most \(0.02\) in length, find \(n\). Hint: Note that \(\sqrt{(y / n)(1-y / n)} \leq \sqrt{\left(\frac{1}{2}\right)\left(1-\frac{1}{2}\right)}\).

. Let \(p\) equal the proportion of drivers who use a seat belt in a state that does not have a mandatory seat belt law. It was claimed that \(p=0.14 .\) An advertising campaign was conducted to increase this proportion. Two months after the campaign, \(y=104\) out of a random sample of \(n=590\) drivers were wearing their seat belts. Was the campaign successful? (a) Define the null and alternative hypotheses. (b) Define a critical region with an \(\alpha=0.01\) significance level. (c) Determine the approximate \(p\) -value and state your conclusion.

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.

Sign-up for free