Uniform maximum pivot interval

The sample sigma-algebra is the Borel sigma-algebra of ${\mathbb{R}}^n$ by [F7]. The maximum map $m(x)={\max }_i{x}_i$ satisfies $\mathrm{\mid }m(x)-m(y)\mathrm{\mid }\le {\max }_i\mathrm{\mid }{x}_i-y_i\mathrm{\mid }$, so it is continuous. The preimages of open sets are therefore open; since the Borel sigma-algebra is generated by open sets, preimages of every Borel set are Borel and $M=m(X)$ is measurable. Also $m\mathrm{/}\theta$ is continuous because $\theta >0$. The coordinate variables are independent by [F1], and [F2] gives their product joint law.

If $0\le u\le 1$, then $P_{\theta }(M\mathrm{/}\theta \le u)=P_{\theta }(X_1\le u\theta ,\dots ,X_n\le u\theta )={\prod }_{i=1}^{n}u=u^n$. For $u<0$ the probability is $0$, and for $u\ge 1$ it is $1$. This law does not depend on $\theta$, so $M\mathrm{/}\theta$ is a pivot by [F3].

The distribution function $u^n$ established in step 1.2 is continuous. Therefore $P_{\theta }({\alpha }^{1\mathrm{/}n}\le M\mathrm{/}\theta \le 1)=1-({\alpha }^{1\mathrm{/}n}{)}^n=1-\alpha$. By [F5], pivot inversion gives exact coverage.

Since $\theta >0$, the inequality ${\alpha }^{1\mathrm{/}n}\le M\mathrm{/}\theta \le 1$ is equivalent to $M\le \theta \le M\mathrm{/}{\alpha }^{1\mathrm{/}n}$. Thus the inverted set is $[M,M\mathrm{/}{\alpha }^{1\mathrm{/}n}]\cap (0,\mathrm{\infty })$, and [F4] identifies its coverage with the probability computed in step 2.1.