# Finding the uniformly most powerful test

Let $X_1,X_2,...,X_n$ denote a random sample from density,
$$f(x;\theta)={1\over 2\theta}, \quad 0<x<2\theta.$$ Find the uniformly most powerful test for testing $H_0:\theta \le \theta_0$ vs $H_1:\theta>\theta_0$.

Let $H_1:\theta=\theta_1$ where $\theta_1>\theta_0$. Then from Neyman-Pearson Lemma I get $\Delta={L_1\over L_0}={\theta_0^n\over \theta_1^n}$.
Then the rejection criteria is reject $H_0$ if $\Delta>k$.

Question: But how can I find $k$ and show that it is independent of $\theta_1$?

This is how I derived the rejection region $\Delta={L1\over L_0}$.
$$L(\theta)=\prod\left({1\over 2\theta}\right)=\left({1\over 2\theta}\right)^n.$$
So: $$\Delta={\left({1\over 2\theta_1}\right)^n \over \left({1\over 2\theta_0}\right)^n}={\theta_0^n \over \theta_1^n}.$$

$L_1(\theta)=\left({1\over 2\theta_1}\right)^n$ if $X_{(n)}<2\theta_1$ and $0$ otherwise.

$L_0(\theta)=\left({1\over 2\theta_0}\right)^n$ if $X_{(n)}<2\theta_0$ and $0$ otherwise.

So $\Delta={\theta_0^n \over \theta_1^n}$ if $X_{(n)}<\min(2\theta_0,2\theta_1)$ and $0$ otherwise.

• Your likelihood ratio appears not to depend on the data $x_1, \ldots, x_n$. Suggest you show how you derived it. Dec 1, 2015 at 14:38
• @Scortchi I have now included how I derived it in my post Dec 1, 2015 at 14:53
• That can't be right: if, say, $x_3=10$ then you know that $2\theta \geq 10$, & values of $\theta \leq 5$ must therefore have zero likelihood. Think about expressing this using an indicator function. Dec 1, 2015 at 14:57
• @Scortchi I don't know how to handle indicator functions. I edited the post with what I know.Is this how to use indicator function? Dec 1, 2015 at 15:37
• Indicator functions are convenient but inessential. Check the null & alternative hypotheses: they're $\theta\leq\theta_0$ & $\theta>\theta_0$, so you want to use a generalized likelihood-ratio test rather than the LRT for simple hypotheses. And you know $\min(2\theta_0,2\theta_1)$. Dec 1, 2015 at 16:34