Using NP lemma to find the most powerful test for uniform distribution

Question

I tried to use Neymann Pearson lemma and get the following, however, I do not know how to proceed from this point.

Answer 1

Here is my proof:

Consider $H_0: \theta = \theta_0$ v.s. $H_1 : \theta = \theta_1$

Under $H_0$, $X_{(n)} \sim Beta(n, 1)$. So it's easy to verify the test:

reject $H_0$ if $X_{(n)} \geq (1 - \alpha)^\frac{1}{n}$ is a level $\alpha$ test.

Then we want to prove it is a UMP test:

Knowing that $f(\vec x|\theta) = \theta^{-n}I_{(x_{n}\leq \theta)}$, then by choosing $k = \dfrac{\theta_0^n}{\theta_1^n}$,

we get $f(\vec x|\theta_1) > k\times f(\vec x|\theta_0)$ $\iff$ $\dfrac{1}{\theta_1^n}I_{(x_{n}\leq \theta_1)} > \dfrac{1}{\theta_1^n}I_{(x_{n}\leq \theta_0)} \iff \theta_0 < x_{(n)} < \theta_1$ .

Similarly, $f(\vec x|\theta_1) < k\times f(\vec x|\theta_0)$ $\iff$ $\dfrac{1}{\theta_1^n}I_{(x_{n}\leq \theta_1)} < \dfrac{1}{\theta_1^n}I_{(x_{n}\leq \theta_0)} \iff x_{(n)} \in \phi$ .

Denote $R = \{X_{(n)} \geq (1 - \alpha)^\frac{1}{n}\}$, then $\{\vec x|\theta_0 < x_{(n)} < \theta_1 \} \subseteq R$ and $\phi \subseteq R^c$.

Therefore, by NP - lemma, we can conclude that the test we derived is a UMP test.

Observe that the form of rejection region is independent of $\theta_1$, so it is also a UMP test for

$H_0: \theta = \theta_0$ v.s. $H_1 : \theta > \theta_0$