Let $X_1, \dots ,X_n$ be an i.i.d. sample from $\rm Unif(0, \theta)$. Consider testing $H_{0}: \theta = \theta_{0}$ against $H_{1}: \theta \neq \theta_{0}$. Show that an uniform most powerful (UMP) test is given by
\begin{align*} \delta(x) = \begin{cases} 1 & x_{(n)} > \theta_{0} \text{ or }x_{(n)} < \theta_{0}(\alpha)^{1/n}\\ 0 & \text{otherwise}\\ \end{cases} \end{align*}
where $x_{(n)}$ is the $n$-th order statistic.
First, $\mathbb{E}_{\theta_{0}}[\delta(X)] = \underbrace{P(x_{(n)} > \theta_{0})}_{=0} + P(x_{(n)} < \theta_{0}(\alpha)^{1/n}) = \left(\frac{\theta_{0}\alpha^{1/n}}{\theta_{0}}\right)^{n} = \alpha$, showing $\delta(x)$ is a level $\alpha$ test.
It remains to be shown that $\delta(x)$ has superior power over any other test $\phi(x)$ with level $\alpha$. In other words, $\mathbb{E}_{\theta_{1}}[\delta(x) - \phi(x)] \geq 0$.
We can partition by whether $x_{(n)}$ falls in acceptance or rejection regions
$$\mathbb{E}_{\theta_{1}}[\delta(x) - \phi(x)] = \mathbb{E}_{\theta_{1}}[(\delta(x) - \phi(x))1(\theta_{0}\alpha^{1/n} \leq x_{(n)} \leq \theta_{0})] + \mathbb{E}_{\theta_{1}}[(\delta(x) - \phi(x))1(x_{(n)} \in (-\infty, \theta_{0}\alpha^{1/n}) \cup (\theta_{0}, \infty))].$$
The second term is nonnegative because $\delta(x) = 1$ and $\phi(x) \leq 1$. The problem comes, in proving the first term is nonnegative. Below is an attempt:
\begin{align*} \mathbb{E}_{\theta_{1}}[(\delta(x) - \phi(x))1(\theta_{0}\alpha^{1/n} \leq x_{(n)} \leq \theta_{0})] = \underbrace{\frac{1}{\theta_{1}^{n}}}_{> 0}\underbrace{\int(\delta(x) - \phi(x))1(\theta_{0}\alpha^{1/n} \leq x_{(n)} \leq \theta_{0})dx}_{=h(\delta, \phi, \theta_{0})} \end{align*}
Thus, it is sufficent to show $h(\delta, \phi, \theta_{0}) > 0$. One trick strategy is to show that under $\theta_{0}$
\begin{align*} \frac{1}{\theta_{0}^{n}}h(\delta, \phi, \theta_{0}) &= \mathbb{E}_{\theta_{0}}[(\delta(x) - \phi(x))1(\theta_{0}\alpha^{1/n} \leq x_{(n)} \leq \theta_{0})]\\ &= \underbrace{\mathbb{E}_{\theta_{0}}[\delta(x)]}_{=\alpha} - \underbrace{\mathbb{E}_{\theta_{0}}[\phi(x)]}_{\leq \alpha}\\ &\geq 0 \end{align*}
However, the second equality above seems to be false, because by definition of $\delta(x)$, $\delta(x) = 0$ when $\theta_{0}\alpha^{1/n} \leq x_{(n)} \leq \theta_{0}$, so $\mathbb{E}_{\theta_{0}}[\delta(x)] = 0$ instead. Can anyone resolve this?
