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Chapter 4: Problem 12

Let \(Y_{1}

Short Answer

Expert verified
The variables \(Z_{1}=Y_{1} / Y_{2}, Z_{2}=Y_{2} / Y_{3}\), and \(Z_{3}=Y_{3}\) are mutually independent because their joint pdf can be expressed as the product of their marginal pdf's.

Step by step solution

01

Identify the joint density of ordered statistics

For the given distribution with pdf \(f(x)=2x\), for \(0<x<1\), and zero elsewhere, the joint density function of order statistics will be \(f_{Y_1, Y_2, Y_3}(y_1, y_2, y_3) = 6y_1 y_2(1 - y_3)\), when \(0 < y_1 < y_2 < y_3 < 1\), and zero elsewhere.
02

Change variables

Let's consider the transformation of variables by the functions \(Z_{1}=Y_{1} / Y_{2}, Z_{2}=Y_{2} / Y_{3}\), and \(Z_{3}=Y_{3}\). The reverse transformation is obtained as \(Y_1 = Z_1Z_2Z_3\), \(Y_2 = Z_2Z_3\), and \(Y_3=Z_3\) The absolute value of the Jacobian determinant of this transformation is equal to \(Z_3^2\).
03

Find the transformed joint pdf

Substitute the reverse transformations into the joint pdf \(f_{Y_1, Y_2, Y_3}(y_1, y_2, y_3)\) and multiply by the absolute determinant to find the transformed joint pdf. \(f_{Z_1, Z_2, Z_3}(z_1, z_2, z_3) = 6z_1z_2^2z_3^3(1-z_3)^2 \), when \(0 < z_1 < 1, 0<z_2<1,0<z_3<1 \).
04

Prove independence

The joint pdf \(f_{Z_1, Z_2, Z_3}(z_1, z_2, z_3)\) can be expressed as product of marginal densities \(f_{Z_1}(z_1)\), \(f_{Z_2}(z_2)\) and \(f_{Z_3}(z_3)\). Hence, \(Z_1\), \(Z_2\) and \(Z_3\) are independent.

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Most popular questions from this chapter

Let \(X\) and \(Y\) denote independent random variables with respective probability density functions \(f(x)=2 x, 0

Define the sets \(A_{1}=\\{x:-\infty

Let the result of a random experiment be classified as one of the mutually exclusive and exhaustive ways \(A_{1}, A_{2}, A_{3}\) and also as one of the mutually exhaustive ways \(B_{1}, B_{2}, B_{3}, B_{4}\). Say that 180 independent trials of the experiment result in the following frequencies: \begin{tabular}{|c|c|c|c|c|} \hline & \(B_{1}\) & \(B_{2}\) & \(B_{3}\) & \(B_{4}\) \\ \hline\(A_{1}\) & \(15-3 k\) & \(15-k\) & \(15+k\) & \(15+3 k\) \\ \hline\(A_{2}\) & 15 & 15 & 15 & 15 \\ \hline\(A_{3}\) & \(15+3 k\) & \(15+k\) & \(15-k\) & \(15-3 k\) \\ \hline \end{tabular} where \(k\) is one of the integers \(0,1,2,3,4,5\). What is the smallest value of \(k\) that leads to the rejection of the independence of the \(A\) attribute and the \(B\) attribute at the \(\alpha=0.05\) significance level?

Similar to Exercise \(4.8 .2\) but now approximate \(\int_{0}^{1.96} \frac{1}{\sqrt{2 \pi}} \exp \left\\{-\frac{1}{2} t^{2}\right\\} d t\)

Let \(Y_{1}
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