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Chapter 3: Problem 8

Let \(F\) have an \(F\) -distribution with parameters \(r_{1}\) and \(r_{2}\). Argue that \(1 / F\) has an \(F\) -distribution with parameters \(r_{2}\) and \(r_{1}\).

Short Answer

Expert verified
If \(F\) has an \(F\) -distribution with parameters \(r_{1}\) and \(r_{2}\), then \(1 / F\) has an \(F\) -distribution with parameters \(r_{2}\) and \(r_{1}\) due to the symmetry property of the \(F\) -distribution.

Step by step solution

01

Understanding the F-Distribution

An \(F\) -distribution arises from the ratio of two sets of variates, each following a chi-squared distribution. More formally, if \(X\) is a chi-squared random variable with \(r_1\) degrees of freedom and \(Y\) is a chi-squared random variable with \(r_2\) degrees of freedom, then the random variable \(Z = X/r_1 / Y/r_2\) follows an \(F\) -distribution with parameters \(r_1\) and \(r_2\).
02

Applying the property of Reciprocal

Given an \(F\)-distributed variable \(F = X/r_1 / Y/r_2\), where \(X\) is chi-squared-distributed with \(r_1\) degrees of freedom and \(Y\) is chi-squared-distributed with \(r_2\) degrees of freedom, the reciprocal of \(F\) is \(1/F = Y/r_2 / X/r_1\). Since this is still a ratio of two chi-squared-distributed variables, with the degrees of freedom reversed, it also follows an F-distribution. That is, \(1/F\) follows an \(F\) -distribution with parameters \(r_2\) and \(r_1\).
03

Final Proven Assertion

Therefore, it has been proven that, given a random variable \(F\) fitting an \(F\)-distribution with parameters \(r_{1}\) and \(r_{2}\), the reciprocal of \(F\) also follows an \(F\)-distribution, but with reversed parameters - \(r_{2}\) and \(r_{1}\).

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