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Chapter 3: Problem 12

Show that $$ Y=\frac{1}{1+\left(r_{1} / r_{2}\right) W} $$ where \(W\) has an \(F\) -distribution with parameters \(r_{1}\) and \(r_{2}\), has a beta distribution.

Short Answer

Expert verified
The transformation \(Y = \frac{1}{1+\left(\frac{r_1}{r_2}\right) W}\) where \(W\) has an F-distribution with parameters \(r_1\) and \(r_2\) follows a Beta Distribution.

Step by step solution

01

- Expression of the PDF of F-Distribution

Firstly, recall the probability density function of the F-distribution. If \(W\) has an F-distribution with parameters \(r_1\) and \(r_2\), its PDF can be expressed as: \[ f(w|r_1,r_2) = \frac{\sqrt{\left(\frac{r_1 w^{r_1-2}}{(r_2+r_1 w)^{r_1+r_2}}\right) \frac{1}{w B(r_1/2,r_2/2)}}}{r_2^{r_2/2} r_1^{r_1/2}}, w > 0 \]
02

- Change of Variable

Next, we apply the change of variable to simplify our computations. Define \(Y = \frac{1}{1+\left(\frac{r_1}{r_2}\right) W}\). The derivative of \(Y\) with respect to \(W\) is \(-\frac{r_1}{r_2} \cdot \frac{1}{(1+\frac{r_1 W}{r_2})^2}\).
03

- Transforming the PDF

Substitute \(Y = \frac{1}{1+\left(\frac{r_1}{r_2}\right) W}\) into the PDF using the chain rule of differentiation. The transformed PDF will be: \[ f_Y(y|r1, r2) = f_W(w|r1, r2)|dw/dy| = \left[\sqrt{\left(\frac{(\frac{r_2 y - r_1 y^2}{(1-y)^2})^{r1-2}}{(\frac{r_2 y - r_1 y^2}{(1-y)} + r2)^{r1+r2}}\right) \frac{1}{(\frac{r_2 y - r_1 y^2}{(1-y)^2})B(r1/2,r2/2)}\right] \times \left[\frac{r1/r2}{(1+\frac{r1 W}{r2})^2}\right] \]
04

- Simplifying the PDF

After simplifying the equation, it comes to the form: \[ f_Y(y|r1, r2) = \frac{\sqrt{(r1 y^{r1-1}(1-y)^{r2-1})}}{B(r1/2,r2/2)}, 0 < y < 1 \]
05

- Recognising the Beta Distribution

This is exactly the PDF of a Beta Distribution with parameters \(r1\) and \(r2\). So we can conclude that the transformed variable \(Y\) follows a Beta Distribution.

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