Question

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

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

Step-by-step solution

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.

Simplify \(T^{2}\)

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

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.

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.

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 \(r_{2}=r\).