$X: \mathcal{U}(0,\theta), \theta>0.$ Prove that $W=(\frac{Y_n}{\theta})^n, Y_n=\max_{1 \leq k \leq n} X_k$ is $\mathcal{U}(0,1)$. Find $P\{a-\gamma<W<a\}$, where $0<\gamma<a\leq 1$. Using this find the $100 \% \gamma$ confidence interval for unknown parameter $\theta.$ Find $a$ for which this interval is shortest.
My attempt: I have proven that $W:\mathcal{U}(0,1)$ because I found that $F_{W}(t)=t,0<t<1.$ Also $P\{a-\gamma<W<a\}=F_{W}(a)-F_W(a-\gamma)=\gamma.$ The problem arises when I need to figure out this confidence interval, becasue I have not done this except with standard normal distributions and student distributions. How is this done with uniform distributions?
If $a,b\in(0,1)$ are so chosen that $\Pr(a<W<b) = b-a = \gamma$, then you have \begin{align} \Pr\left( a < \left( \frac Y \theta \right)^n < b \right) & = \gamma \\[10pt] \Pr\left( \frac Y {b^{1/n}} < \theta < \frac Y {a^{1/n}} \right) & = \gamma \end{align} If you want the shortest confidence interval with that confidence level, you need to find the value of $a$ that minimizes the length $$ \frac Y {a^{1/n}} - \frac Y {b^{1/n}} = \frac Y {a^{1/n}} - \frac Y {(a+\gamma)^{1/n}}. $$