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

Show that the square of a noncentral \(T\) random variable is a noncentral \(F\) random variable.

Short Answer

Expert verified
The square of a noncentral \(T\) random variable is a noncentral \(F\) random variable. This is due to the underlying mathematical definitions and properties of these distributions.

Step by step solution

01

Define a Noncentral \(T\) Distribution

A noncentral \(T\) variable may be defined as the ratio \(T = X / \sqrt{Y}\), where \(X\) follows a normal distribution with mean \(\mu\) and standard deviation \(1\), and \(Y\) follows a chi-square distribution with one degree of freedom. It's essential to note that \(X\) and \(Y\) are independent random variables.
02

Square the Noncentral \(T\) Distribution

Squaring the \(T\) random variable yields \(T^2 = X^2 / Y\). The \(X^2\) term is a chi-square distribution with one degree of freedom, and \(Y\) is a chi-square distribution with one degree of freedom, based on the definition from Step 1.
03

Define a Noncentral \(F\) Distribution

A noncentral \(F\) distribution is defined as the ratio of two independent chi-square random variables each divided by their respective degrees of freedom. In other words, let \(Z_1\) and \(Z_2\) be independent random variables each following chi-square distribution with degrees of freedom \(d_1\) and \(d_2\) respectively, then the noncentral \(F\) distribution can be defined as \(F = Z_1/d_1 / Z_2/d_2 = Z_1 / Z_2\) if \(d_1 = d_2 = 1\).
04

Compare \(T^2\) and \(F\) Definitions

Compare the \(T^2\) and \(F\) distribution definitions. \(T^2 = X^2 / Y\) matches the definition of a noncentral \(F\) distribution if both \(X^2\) and \(Y\) are chi-square with one degree of freedom. Hence, this shows that the square of the noncentral \(T\) random variable is a noncentral \(F\) random variable.

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

Using the notation of Section 9.2, assume that the means \(\mu_{j}\) satisfy a linear function of \(j\), nanely \(\mu_{j}=c+d[j-(b+1) / 2] .\) Let independent random samples of size \(a\) be taken from the \(b\) normal distributions having means \(\mu_{1}, \mu_{2}, \ldots, \mu_{b}\), respectively, and common unknown variance \(\sigma^{2}\). (a) Show that the maximum likelihood estimators of \(c\) and \(d\) are, respectively, \(\hat{c}=\bar{X}_{. .}\) and $$ \hat{d}=\frac{\sum_{j=1}^{b}[j-(b-1) / 2]\left(\bar{X}_{. j}-\bar{X}_{. .}\right)}{\sum_{j=1}^{b}[j-(b+1) / 2]^{2}} $$ (b) Show that $$ \begin{aligned} \sum_{i=1}^{a} \sum_{j=1}^{b}\left(X_{i j}-\bar{X}_{. .}\right)^{2}=\sum_{i=1}^{a} \sum_{j=1}^{b}\left[X_{i j}-\bar{X}_{. .}-\hat{d}\left(j-\frac{b+1}{2}\right)\right]^{2} \\ &+\hat{d}^{2} \sum_{j=1}^{b} a\left(j-\frac{b+1}{2}\right)^{2} \end{aligned} $$ (c) Argue that the two terms in the right-hand member of Part (b), once divided by \(\sigma^{2}\), are independent random variables with \(\chi^{2}\) distributions provided that \(d=0\) (d) What \(F\) -statistic would be used to test the equality of the means, that is, \(H_{0}: d=0 ?\)

If \(A_{1}, A_{2}, \ldots, A_{k}\) are events, prove, by induction, Boole's inequality $$ P\left(A_{1} \cup A_{2} \cup \cdots \cup A_{k}\right) \leq \sum_{1}^{k} P\left(A_{i}\right) $$ Then show that $$ P\left(A_{1}^{c} \cap A_{2}^{c} \cap \cdots \cap A_{k}^{c}\right) \geq 1-\sum_{1}^{b} P\left(A_{i}\right) $$

Suppose \(\mathbf{A}\) is a real symmetric matrix. If the eigenvalues of \(\mathbf{A}\) are only 0 's and 1 's then prove that \(\mathbf{A}\) is idempotent.

Student's scores on the mathematics portion of the ACT examination, \(x\), and on the final examination in the first-semester calculus ( 200 points possible), \(y\), are given. (a) Calculate the least squares regression line for these data. (b) Plot the points and the least squares regression line on the same graph. (c) Find point estimates for \(\alpha, \beta\), and \(\sigma^{2}\). (d) Find 95 percent confidence intervals for \(\alpha\) and \(\beta\) under the usual assumptions. $$ \begin{array}{cc|cc} \hline \mathrm{x} & \mathrm{y} & \mathrm{x} & \mathrm{y} \\ \hline 25 & 138 & 20 & 100 \\ 20 & 84 & 25 & 143 \\ 26 & 104 & 26 & 141 \\ 26 & 112 & 28 & 161 \\ 28 & 88 & 25 & 124 \\ 28 & 132 & 31 & 118 \\ 29 & 90 & 30 & 168 \\ 32 & 183 & & \\ \hline \end{array} $$

Let the \(4 \times 1\) matrix \(\boldsymbol{Y}\) be multivariate normal \(N\left(\boldsymbol{X} \boldsymbol{\beta}, \sigma^{2} \boldsymbol{I}\right)\), where the \(4 \times 3\) matrix \(\boldsymbol{X}\) equals $$ \boldsymbol{X}=\left[\begin{array}{rrr} 1 & 1 & 2 \\ 1 & -1 & 2 \\ 1 & 0 & -3 \\ 1 & 0 & -1 \end{array}\right] $$ and \(\beta\) is the \(3 \times 1\) regression coeffient matrix. (a) Find the mean matrix and the covariance matrix of \(\hat{\boldsymbol{\beta}}=\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1} \boldsymbol{X}^{\prime} \boldsymbol{Y}\). (b) If we observe \(\boldsymbol{Y}^{\prime}\) to be equal to \((6,1,11,3)\), compute \(\hat{\boldsymbol{\beta}}\).

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