Question

Let \(Y_{1}

Short answer

The complete sufficient statistic \(Y_{4}\) for \(\theta\) is indeed independent of the statistics \(Y_{1} / Y_{4}\) and \((Y_{1}+Y_{2}) / (Y_{3}+Y_{4})\) since we can express the joint density as a product of separate functions, each pertaining to the distinct statistics in question.

Step-by-step solution

Write Down the Joint Density of Order Statistics

The joint density of order statistics, when removed of the irrelevant constant, is given by: \( g(Y_{1}, Y_{2}, Y_{3}, Y_{4}; \theta) = Y_{4}^{-4}\) provided that \(0 < Y_{1} < Y_{2} < Y_{3} < Y_{4} < \theta\)

Express the Density as a Product of Two Functions

Now, it's essential to express this joint density as a product of two functions, one of \(Y_{4}\) and another of \(Y_{1} / Y_{4}\) and \((Y_{1}+Y_{2}) / (Y_{3}+Y_{4})\). Here's how: \( g(Y_{1}, Y_{2}, Y_{3}, Y_{4}; \theta) = h(Y_{4}; \theta).i[T(Y_{1}, Y_{2}, Y_{3}, Y_{4})]\) where \(h(Y_{4}; \theta) = Y_{4}^{-4}\) for \(0 < Y_{4} < \theta\), and \(i\), a function that indicates when the inequalities hold. In this case: \( i(T(Y_{1}, Y_{2}, Y_{3}, Y_{4})) = 1\) when \(0 < [Y_{1} / Y_{4}] < [Y_{2} / Y_{4}] < [Y_{3} / Y_{4}] < [Y_{4} / Y_{4}] < 1\), and zero otherwise

Prove the Independence

The joint density \(g(Y_{1}, Y_{2}, Y_{3}, Y_{4}; \theta)\) is seen to be independent of \(Y_{4}\), \(Y_{1} / Y_{4}\), and \((Y_{1}+Y_{2}) / (Y_{3}+Y_{4})\), because it can be separated into the product of two independent functions, \(h(Y_{4}; \theta)\) and \(i(T(Y_{1}, Y_{2}, Y_{3}, Y_{4}))\). Hence, we can conclude that \(Y_{4}\) is independent of \(Y_{1} / Y_{4}\) and \((Y_{1}+Y_{2}) / (Y_{3}+Y_{4})\)