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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

Let \(Q=X_{1} X_{2}-X_{3} X_{4}\), where \(X_{1}, X_{2}, X_{3}, X_{4}\) is a random sample of size 4 from a distribution which is \(N\left(0, \sigma^{2}\right) .\) Show that \(Q / \sigma^{2}\) does not have a chi-square distribution. Find the mgf of \(Q / \sigma^{2}\).

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 \(\gamma^{2}\).

Using the background of the two-way classification with one observation per cell, show that the maximum likelihood estimator of \(\alpha_{i}, \beta_{j}\), and \(\mu\) are \(\hat{\alpha}_{i}=\bar{X}_{i .}-\bar{X}_{. .}\) \(\hat{\beta}_{j}=\bar{X}_{. j}-\bar{X}_{. .}\), and \(\hat{\mu}=\bar{X}_{. .}\), respectively. Show that these are unbiased estimators of their respective parameters and compute \(\operatorname{var}\left(\hat{\alpha}_{i}\right), \operatorname{var}\left(\hat{\beta}_{j}\right)\), and \(\operatorname{var}(\hat{\mu})\).

Let \(X_{1 j}, X_{2 j}, \ldots, X_{a_{f} j}\) represent independent random samples of sizes \(a_{j}\) from a normal distribution with means \(\mu_{j}\) and variances \(\sigma^{2}, j=1,2, \ldots, b\). Show that $$ \sum_{j=1}^{b} \sum_{i=1}^{a_{j}}\left(X_{i j}-\bar{X}_{. .}\right)^{2}=\sum_{j=1}^{b} \sum_{i=1}^{a_{j}}\left(X_{i j}-\bar{X}_{. j}\right)^{2}+\sum_{j=1}^{b} a_{j}\left(\bar{X}_{. j}-\bar{X}_{. .}\right)^{2} $$ or \(Q^{\prime}=Q_{3}^{\prime}+Q_{4}^{\prime} .\) Here \(\bar{X}_{. .}=\sum_{j=1}^{b} \sum_{i=1}^{a_{j}} X_{i j} / \sum_{j=1}^{b} a_{j}\) and \(\bar{X}_{. j}=\sum_{i=1}^{a_{j}} X_{i j} / a_{j} .\) If \(\mu_{1}=\mu_{2}=\) \(\cdots=\mu_{b}\), show that \(Q^{\prime} / \sigma^{2}\) and \(Q_{3}^{\prime} / \sigma^{2}\) have chi-square distributions. Prove that \(Q_{3}^{\prime}\) and \(Q_{4}^{\prime}\) are independent, and hence \(Q_{4}^{\prime} / \sigma^{2}\) also has a chi-square distribution. If the likelihood ratio \(\Lambda\) is used to test \(H_{0}: \mu_{1}=\mu_{2}=\cdots=\mu_{b}=\mu, \mu\) unspecified and \(\sigma^{2}\) unknown against all possible alternatives, show that \(\Lambda \leq \lambda_{0}\) is equivalent to the computed \(F \geq c\), where $$ F=\frac{\left(\sum_{j=1}^{b} a_{j}-b\right) Q_{4}^{\prime}}{(b-1) Q_{3}^{\prime}} $$ What is the distribution of \(F\) when \(H_{0}\) is true?

Show that \(\sum_{j=1}^{b} \sum_{i=1}^{a}\left(X_{i j}-\bar{X}_{i .}\right)^{2}=\sum_{j=1}^{b} \sum_{i=1}^{a}\left(X_{i j}-\bar{X}_{i}-\bar{X}_{. j}+\bar{X}_{. .}\right)^{2}+a \sum_{j=1}^{b}\left(\bar{X}_{. j}-\bar{X}_{. .}\right)^{2} .\)
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