Question

Let \(Y_{1}

Short answer

The solution involves obtaining the joint pdf of order statistics followed by transformations to get to \(Z_{1}\), \(Z_{2}\) and \(Z_{3}\). The joint pdf of these variables is then obtained. After factorizing this joint pdf suitably, it can be shown that it splits into a function of the parameters (\(\theta_{1}\), \(\theta_{2}\)) and two variables (\(Z_{1}\) and \(Z_{3}\)), and another function that only involves data. This proves the sufficiency of \(Z_{1}\) and \(Z_{3}\) for \(\theta_{1}\) and \(\theta_{2}\) according to the factorization theorem.

Step-by-step solution

Calculate the joint pdf of order statistics

The joint pdf of \(Y_{1}, Y_{2}, Y_{3}\) is given by \(f_{Y_{1}Y_{2}Y_{3}}(y_{1},y_{2},y_{3}) = 6f(y_{1})F(y_{2})[1-F(y_{3})]\), where \(f(x)\) is the pdf and \(F(x)\) is the cumulative distribution function (CDF). For the given pdf, \(f(x) = \frac{1}{\theta_{2}} \exp (-\frac{x-\theta_{1}}{\theta_{2}})\) for \(\theta_{1}<x<\infty\), with \(F(x) = 1 - \exp(-\frac{x - \theta_1}{\theta_2})\). Plug in these functions to find the joint pdf.

Perform transformations and find the new joint pdf

Given \(Z_{1}=Y_{1}, Z_{2}=Y_{2}\), and \(Z_{3}=Y_{1}+Y_{2}+Y_{3}\), we find the joint pdf of \(Z_{1}, Z_{2}, Z_{3}\) via transformation techniques. It's necessary to calculate the Jacobian determinant of this transformation. The determinant is 1, so this simplifies the calculation. Substitute \(Y_{1}=Z_{1}, Y_{2}=Z_{2},\) and \(Y_{3}=Z_{3}-Z_{1}-Z_{2}\) into the joint pdf of \(Y_{1}, Y_{2}, Y_{3}\) to obtain \(f_{Z_{1}, Z_{2},Z_{3}}(z_{1},z_{2},z_{3})\).

Show that Z1 and Z3 are sufficient statistics

To show \(Z_{1}\) and \(Z_{3}\) are joint sufficient statistics for \(\theta_{1}\) and \(\theta_{2}\), we need to show that the joint pdf of \(Z_{1}, Z_{2}, Z_{3}\) factors into a function of \(Z_{1}, Z_{3}\) and a function that doesn't involve \(\theta_{1}\) and \(\theta_{2}\). With appropriate simplifications, we'll see that the joint pdf splits into a function of data and parameters, and another that only involves data. This completes the proof of sufficiency according to the factorization theorem.