# Coverage probability for Uniform$(0, \theta)$

Let $$X_1 \dots X_n$$ denote a random sample from a uniform $$(0, \theta$$) distribution.

PROBLEM: Compute the coverage probability for the CI: $$\left(\frac{X_{(n)}}{0.95}, \frac{X_{(n)}}{0.25}\right)$$ using $$W = X_{(n)}/\theta$$ which has density $$F_w = w^n$$

My Work $$\begin{split} CI &= \left(\frac{X_{(n)}}{0.95} \leq \theta \leq \frac{X_{(n)}}{0.25}\right)\\ &\iff \left((\frac{X_{(n)}}{0.95} \leq X_{(n)}/w \leq \frac{X_{(n)}}{0.25}\right)\\ &\iff \left(\frac{1}{0.95} \leq 1/w \leq \frac{1}{0.25}\right) \end{split}$$ And this is just $$P\big(w \in [0.25, 0.95]\big) = F_w(0.95) - F_w(0.25) = 0.95^n - 0.25^n$$.

The coverage probability is the probability that $$\theta\in \left(\frac{X_{(n)}}{0.95},\frac{X_{(n)}}{0.25}\right)$$.
That is, because $$\theta>0$$,
\begin{align} P\left[\frac{X_{(n)}}{0.95}<\theta<\frac{X_{(n)}}{0.25}\right]&=P\left[0.25<\frac{X_{(n)}}{\theta}<0.95\right] \\&=P\left[\frac{X_{(n)}}{\theta}<0.95\right]-P\left[\frac{X_{(n)}}{\theta}<0.25\right] \\&=0.95^n-0.25^n \end{align}