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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

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample of size \(n\) from a distribution which is \(N\left(0, \sigma^{2}\right)\). Prove that \(\sum_{1}^{n} X_{i}^{2}\) and every quadratic form, which is nonidentically zero in \(X_{1}, X_{2}, \ldots, X_{n}\), are dependent.

Let \(\mathbf{X}^{\prime}=\left[X_{1}, X_{2}\right]\) be bivariate normal with matrix of means \(\boldsymbol{\mu}^{\prime}=\left[\mu_{1}, \mu_{2}\right]\) and positive definite covariance matrix \(\mathbf{\Sigma}\). Let $$ Q_{1}=\frac{X_{1}^{2}}{\sigma_{1}^{2}\left(1-\rho^{2}\right)}-2 \rho \frac{X_{1} X_{2}}{\sigma_{1} \sigma_{2}\left(1-\rho^{2}\right)}+\frac{X_{2}^{2}}{\sigma_{2}^{2}\left(1-\rho^{2}\right)} $$ Show that \(Q_{1}\) is \(\chi^{2}(r, \theta)\) and find \(r\) and \(\theta\). When and only when does \(Q_{1}\) have a central chi-square distribution?

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.

Here \(Q_{1}\) and \(Q_{2}\) are quadratic forms in observations of a random sample from \(N(0,1) .\) If \(Q_{1}\) and \(Q_{2}\) are independent and if \(Q_{1}+Q_{2}\) has a chi-square distribution, prove that \(Q_{1}\) and \(Q_{2}\) are chi-square variables.

Let the independent random variables \(Y_{1}, Y_{2}, \ldots, Y_{n}\) have, respectively, the probability density functions \(N\left(\beta x_{i}, \gamma^{2} x_{i}^{2}\right), i=1,2, \ldots, n\), where the given numbers \(x_{1}, x_{2}, \ldots, x_{n}\) are not all equal and no one is zero. Find the maximum likelihood estimators of \(\beta\) and \(\gamma^{2}\).

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