Question

Let \(Y_{1}

Short answer

Yes, the variables \(Z_{1}=Y_{2}\) and \(Z_{2}=Y_{4}-Y_{2}\) are independent.

Step-by-step solution

Noting the Given Information

From the problem statement, it's known that \(Y_{1}<Y_{2}<Y_{3}<Y_{4}<Y_{5}\) are the order statistics of a random sample of size 5 drawn from a distribution with pdf \(f(x)=e^{-x}\), where \(0<x<\infty\). The task is to prove \(Z_{1}=Y_{2}\) and \(Z_{2}=Y_{4}-Y_{2}\) are independent.

Joint PDF of \(Y_{2}\) and \(Y_{4}\)

Calculate the joint pdf of \(Y_{2}\) and \(Y_{4}\). According to the principle of order statistics, the joint pdf of \(Y_{2}\) and \(Y_{4}\) is given by: \(f_{Y_{2}, Y_{4}}(y_{2}, y_{4})=20e^{-y_{2}}(1-e^{-(y_{4}-y_{2})})e^{-y_{4}}\), which focuses on the second and fourth order statistics and has a combination of 20 associated with it as there are \(\binom{5}{2}=20\) ways to choose 2 items out of 5.

Transformation of Variables

The next step is to transform the variables from \(Y_{2}\) and \(Y_{4}\) to \(Z_{1}\) and \(Z_{2}\). This involves calculating the Jacobian determinant for this transformation. It's found by differentiating \(Z_{1}=Y_{2}\) and \(Z_{2}=Y_{4}-Y_{2}\) with respect to \(Y_{2}\) and \(Y_{4}\), yielding a Jacobian determinant of 1.

Finding the Joint PDF of \(Z_{1}\) and \(Z_{2}\)

Proceeding with the transformation, the joint pdf of \(Z_{1}\) and \(Z_{2}\), \(f_{Z_{1}, Z_{2}}(z_{1}, z_{2})\), can now be computed. By substituting \(z_{1}\) for \(y_{2}\), and \(z_{1}+z_{2}\) for \(y_{4}\) into the joint pdf of \(Y_{2}, Y_{4}\) and also applying the Jacobian determinant, we get: \(f_{Z_{1}, Z_{2}}(z_{1}, z_{2})=20e^{-z_{1}}(1-e^{-z_{2}})e^{-(z_{1}+z_{2})}\).

Proving independence

Finally, we can show that \(Z_{1}=Y_{2}\) and \(Z_{2}=Y_{4}-Y_{2}\) are independent. Two random variables are independent if and only if their joint density function can be expressed as the product of their individual density functions. This is indeed the case here, as \(f_{Z_{1}, Z_{2}}(z_{1}, z_{2})=f_{Z_{1}}(z_{1})f_{Z_{2}}(z_{2})\), where \(f_{Z_{1}}(z_{1})=4e^{-2z_{1}}\) and \(f_{Z_{2}}(z_{2})=5e^{-z_{2}}(1-e^{-z_{2}})\).