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 \(T=W / \sqrt{V / r}\), where the independent variables \(W\) and \(V\) are, respectively, normal with mean zero and variance 1 and chi-square with \(r\) degrees of freedom. Show that \(T^{2}\) has an \(F\) -distribution with parameters \(r_{1}=1\) and \(r_{2}=r\). Hint: What is the distribution of the numerator of \(T^{2} ?\)

Short Answer

Expert verified
The variable \(T^{2}\) has an \(F\)-distribution with parameters \(r_{1}=1\) and \(r_{2}=r\).

Step by step solution

01

Compute \(T^{2}\)

The first step is to square \(T\) resulting in \(T^{2}= \frac{W^{2}}{V / r}\). This transforms the equation into a form where the independent variables \(W\) and \(V\) are squared.
02

Simplify \(T^{2}\)

Next, simplify the denominator to get \(T^{2} = \frac{W^{2}}{V / r} = \frac{W^{2}}{V} r\).
03

Understand Distributions for \(W^{2}\) and \(V\)

Given \(W\) is normally distributed with mean 0 and variance 1, \(W^{2}\) follows a chi-square distribution with 1 degree of freedom. \(V\) also represents a chi-square distribution with \(r\) degrees of freedom.
04

Illustrate the Ratio of Distributions

Since both the numerator and the denominator of \(T^{2}\) are chi-square distributed with 1 and \(r\) degrees of freedom respectively, and the denominator is multiplied by \(r\), the ratio \(T^{2} = \frac{W^{2}}{V} r\) gives an \(F\)-distribution.
05

Identifying the Parameters of the \(F\)-Distribution

An \(F\)-distribution is determined by its two parameters, namely the degrees of freedom of the numerator and the denominator. The degree of freedom of the numerator, given by the distribution of \(W^{2}\), is 1 (\(r_{1}=1\)). The degree of freedom of the denominator, given by the distribution of \(V\), is \(r\) (\(r_{2}=r\)). Thus, \(T^{2}\) follows an \(F\)-distribution with parameters \(r_{1}=1\) and \(r_{2}=r\).

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!

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

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