/*! 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 20 Let the joint pdf of \(X\) and \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let the joint pdf of \(X\) and \(Y\) be \(f(x, y)=\frac{12}{7} x(x+y), 0

Short Answer

Expert verified
The joint pdf of U and V, is given by \( f(u, v) = \frac{24}{7} u(v-u)\), for 0 < u < v < 1, and zero elsewhere.

Step by step solution

01

Identify the Joint pdf f(u, v)

We have \(U =\min (X, Y)\) and \(V =\max (X, Y)\). Then we consider the cases for \(u\) and \(v\). If \(0 < u < v < 1\), we have 2 cases, \(u = x\), \(v = y\) or \(u = y\), \(v = x\). Correspondingly, \(f(x, y) =\frac{12}{7} x(x+y)\) becomes \(f(u, v) = \frac{24}{7} u(v-u)\).
02

Find the Range

Since the given range for both X and Y is from 0 to 1, and \(v > u\), we therefore have 0 < u < v < 1. This is the range for the joint pdf of U and V.
03

Formulate the joint pdf of U, V

The joint pdf of U and V, therefore, is given by \( f(u, v) = \frac{24}{7} u(v-u)\), for 0 < u < v < 1, and zero elsewhere. The numeric factor has also changed to keep the pdf normalized, which can be checked by comparing the double integrals of the original and transformed pdfs over their respective domains.

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.

Joint Probability Density Function
The concept of a joint probability density function (pdf) is crucial when dealing with two or more random variables. It describes the likelihood of these variables taking on a specific set of values simultaneously.
In our given exercise, we have two random variables, \(X\) and \(Y\), with a joint pdf represented by \(f(x, y) = \frac{12}{7} x(x+y)\) over the range \(0 < x < 1\) and \(0 < y < 1\). Outside this range, the function is considered to be zero, which means the probability falls outside these bounds.
Understanding this joint pdf lets us predict the behavior and interaction of \(X\) and \(Y\) in the defined range. In practical terms, it helps us understand how changing one variable might influence another. This concept is not only pivotal in statistics but also in fields like economics or any domain where multi-variable systems are analyzed.
Random Variables Transformation
Transforming random variables is a common technique used to simplify complex problems. It involves changing the variables \(X\) and \(Y\) to new variables \(U\) and \(V\) where \(U = \min(X, Y)\) and \(V = \max(X, Y)\). This is useful for understanding relationships and dependencies in a simpler form.
In our exercise, this transformation is essential to derive the joint pdf of \(U\) and \(V\). We consider cases such as \(0 < u < v < 1\), with scenarios like \(u = x\) and \(v = y\) or \(u = y\) and \(v = x\). These cases help us express the joint pdf \(f(u, v)\) as \(\frac{24}{7}u(v-u)\).
Through this transformation, we can more easily analyze how the values of \(U\) and \(V\) distribute probabilities across different combinations of \(U\) and \(V\). Therefore, transformations can often make difficult problems more approachable by offering a new perspective on the distribution.
Probability Density Function Normalization
Normalization of a probability density function ensures that the total probability over the defined range equals one. This is a foundational principle in probability theory, as it ensures the correctness and logical consistency of a pdf.
In our given problem, when transforming from the joint pdf of \(X\) and \(Y\) to that of \(U\) and \(V\), we must adjust our pdf to maintain this normalization. The original function \(f(x, y)\) and our derived \(f(u, v)\) must have their total probabilities checked and matched.
This is done through the process of integration over the allowed ranges. By comparing the integrals of the original and new joint pdfs, we determine if they maintain normalized probability over their respective domains. Here, the normalization factor changed from \(\frac{12}{7}\) to \(\frac{24}{7}\) after transformation, ensuring the sum of probabilities stays consistent. Properly normalizing ensures we maintain unchanged necessary probabilities for accurate statistical analysis.

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

Frequently, the bootstrap percentile confidence interval can be improved if the estimator \(\widehat{\theta}\) is standardized by an estimate of scale. To illustrate this, consider a bootstrap for a confidence interval for the mean. Let \(x_{1}^{*}, x_{2}^{*}, \ldots, x_{n}^{*}\) be a bootstrap sample drawn from the sample \(x_{1}, x_{2}, \ldots, x_{n} .\) Consider the bootstrap pivot [analog of \((4.9 .12)]\) : $$ t^{*}=\frac{\bar{x}^{*}-\bar{x}}{s^{*} / \sqrt{n}} $$ where \(\bar{x}^{*}=n^{-1} \sum_{i=1}^{n} x_{i}^{*}\) and $$ s^{* 2}=(n-1)^{-1} \sum_{i=1}^{n}\left(x_{i}^{*}-\bar{x}^{*}\right)^{2} $$ (a) Rewrite the percentile bootstrap confidence interval algorithm using the mean and collecting \(t_{j}^{*}\) for \(j=1,2, \ldots, B .\) Form the interval $$ \left(\bar{x}-t^{*(1-\alpha / 2)} \frac{s}{\sqrt{n}}, \bar{x}-t^{*(\alpha / 2)} \frac{s}{\sqrt{n}}\right) $$ where \(t^{*(\gamma)}=t_{([\gamma * B])}^{*} ;\) that is, order the \(t_{j}^{*} \mathrm{~s}\) and pick off the quantiles.

Define the sets \(A_{1}=\\{x:-\infty

Assume that the weight of cereal in a "10-ounce box" is \(N\left(\mu, \sigma^{2}\right)\). To test \(H_{0}: \mu=10.1\) against \(H_{1}: \mu>10.1\), we take a random sample of size \(n=16\) and observe that \(\bar{x}=10.4\) and \(s=0.4\). (a) Do we accept or reject \(H_{0}\) at the \(5 \%\) significance level? (b) What is the approximate \(p\) -value of this test?

Let \(Y_{1}

Among the data collected for the World Health Organization air quality monitoring project is a measure of suspended particles in \(\mu \mathrm{g} / \mathrm{m}^{3}\). Let \(X\) and \(Y\) equal the concentration of suspended particles in \(\mu \mathrm{g} / \mathrm{m}^{3}\) in the city center (commercial district) for Melbourne and Houston, respectively. Using \(n=13\) observations of \(X\) and \(m=16\) observations of \(Y\), we test \(H_{0}: \mu_{X}=\mu_{Y}\) against \(H_{1}: \mu_{X}<\mu_{Y}\). (a) Define the test statistic and critical region, assuming that the unknown variances are equal. Let \(\alpha=0.05\). (b) If \(\bar{x}=72.9, s_{x}=25.6, \bar{y}=81.7\), and \(s_{y}=28.3\), calculate the value of the test statistic 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.