Question

Show that the \(n\) th order statistic of a random sample of size \(n\) from the uniform distribution having pdf \(f(x ; \theta)=1 / \theta, 0

Short answer

The \(n\)th order statistic of a random sample of size \(n\) either from a uniform distribution or from a more general class of distributions \(f(x ; \theta) = Q(\theta) M(x), 0<x<\theta\), is a sufficient statistic for \(\theta\).

Step-by-step solution

Solving for the Uniform Distribution

To demonstrate this, start by preparing the joint pdf of \(n\) random variables \(X_1, X_2, ..., X_n\) which are uniformly distributed as \[f(x_1, ..., x_n|\theta) = \frac{1}{\theta^n}, 0 x_{(n)}\), where \(x_{(n)}\) is the \(n\)th (i.e. the maximum) order statistic. Otherwise, the likelihood ratio is zero. We can see that the likelihood ratio does not depend on the specific values of the data other than the \(n\)th order statistic. Thus, using the likelihood ratio theorem for sufficiency, \(x_{(n)}\) is sufficient for \(\theta\).

Generalizing the Result

To generalize the result for the case of a pdf defined as \(f(x ; \theta) = Q(\theta) M(x), 0<x<\theta, 0<\theta<\infty\), observe that under the conditions given in the problem, \(Q(\theta)\) functions as a scaling factor to the function \(M(x)\) such that the integral of \(M(x)\) over the range of possible \(x\) values is \(\frac{1}{Q(\theta)}\). In this case, prepare the likelihood ratio as in the uniform distribution case and observe that the likelihood comparison also does not depend on the individual data values, other than \(x_{(n)}\). Thus \(x_{(n)}\) is sufficient for \(\theta\) in this more general case.