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Chapter 9: Problem 4

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.

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Most popular questions from this chapter

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. $$ \begin{array}{cc|cc} \hline \mathrm{x} & \mathrm{y} & \mathrm{x} & \mathrm{y} \\ \hline 25 & 138 & 20 & 100 \\ 20 & 84 & 25 & 143 \\ 26 & 104 & 26 & 141 \\ 26 & 112 & 28 & 161 \\ 28 & 88 & 25 & 124 \\ 28 & 132 & 31 & 118 \\ 29 & 90 & 30 & 168 \\ 32 & 183 & & \\ \hline \end{array} $$

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Often in regression the mean of the random variable \(Y\) is a linear function of \(p\) -values \(x_{1}, x_{2}, \ldots, x_{p}\), say \(\beta_{1} x_{1}+\beta_{2} x_{2}+\cdots+\beta_{p} x_{p}\), where \(\boldsymbol{\beta}^{\prime}=\left(\beta_{1}, \beta_{2}, \ldots, \beta_{p}\right)\) are the regression coefficients. Suppose that \(n\) values, \(\boldsymbol{Y}^{\prime}=\left(Y_{1}, Y_{2}, \ldots, Y_{n}\right)\) are observed for the \(x\) -values in \(\boldsymbol{X}=\left[x_{i j}\right]\), where \(\boldsymbol{X}\) is an \(n \times p\) design matrix and its ith row is associated with \(Y_{i}, i=1,2, \ldots, n .\) Assume that \(Y\) is multivariate normal with mean \(\boldsymbol{X} \boldsymbol{\beta}\) and variance-covariance matrix \(\sigma^{2} \boldsymbol{I}\), where \(\boldsymbol{I}\) is the \(n \times n\) identity matrix. (a) Note that \(Y_{1}, Y_{2}, \ldots, Y_{n}\) are independent. Why? (b) Since \(\boldsymbol{Y}\) should approximately equal its mean \(\boldsymbol{X} \boldsymbol{\beta}\), we estimate \(\boldsymbol{\beta}\) by solving the normal equations \(\boldsymbol{X}^{\prime} \boldsymbol{Y}=\boldsymbol{X}^{\prime} \boldsymbol{X} \boldsymbol{\beta}\) for \(\boldsymbol{\beta}\). Assuming that \(\boldsymbol{X}^{\prime} \boldsymbol{X}\) is non- singular, solve the equations to get \(\hat{\boldsymbol{\beta}}=\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1} \boldsymbol{X}^{\prime} \boldsymbol{Y}\). Show that \(\hat{\boldsymbol{\beta}}\) has a multivariate normal distribution with mean \(\boldsymbol{\beta}\) and variance-covariance matrix $$ \sigma^{2}\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1} $$ (c) Show that $$ (\boldsymbol{Y}-\boldsymbol{X} \boldsymbol{\beta})^{\prime}(\boldsymbol{Y}-\boldsymbol{X} \boldsymbol{\beta})=(\hat{\boldsymbol{\beta}}-\boldsymbol{\beta})^{\prime}\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)(\hat{\boldsymbol{\beta}}-\boldsymbol{\beta})+(\boldsymbol{Y}-\boldsymbol{X} \hat{\boldsymbol{\beta}})^{\prime}(\boldsymbol{Y}-\boldsymbol{X} \hat{\boldsymbol{\beta}}) $$ say \(Q=Q_{1}+Q_{2}\) for convenience. (d) Show that \(Q_{1} / \sigma^{2}\) is \(\chi^{2}(p)\). (e) Show that \(Q_{1}\) and \(Q_{2}\) are independent. (f) Argue that \(Q_{2} / \sigma^{2}\) is \(\chi^{2}(n-p)\). (g) Find \(c\) so that \(c Q_{1} / Q_{2}\) has an \(F\) -distribution. (h) The fact that a value \(d\) can be found so that \(P\left(c Q_{1} / Q_{2} \leq d\right)=1-\alpha\) could be used to find a \(100(1-\alpha)\) percent confidence ellipsoid for \(\beta\). Explain.
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