Likelihood ratio test rejection region for uniform parameters Let $Y_1...Y_n$ be a sample from a uniform distribution over the interval $[0, \theta]$. For $H_0 = \theta_0$, $H_a = \theta_a$, $\theta_a \gt \theta_0$ we know that the most powerful $\alpha$-level test has a rejection region {$y_{max} \gt k$} for some constant k. Find the constant k.
First I tried to write out the likelihood ratio test:
$$\frac{\prod_{i=1}^{n} \frac{1}{\theta_0}}{\prod_{i=1}^{n} \frac{1}{\theta_a}} \lt k$$
$$\left(\frac{\theta_a}{\theta_0}\right)^n \lt k$$
$$\frac{\theta_a}{\theta_0} \lt k, \text{    (k absorbed the root)}$$
$$\theta_a \lt k\theta_0 $$
$$\theta_0 \gt \frac {\theta_a} {k} $$
...and I'm stuck here. My best attempt is as follows:
The MLE of $\theta_0$  is $y_{max}$ and so 
$$y_{max} \gt \frac {\theta_a} {k} $$
What we are looking for is k such that $P(y_{max} > \frac{\theta_a}{k}) \le \alpha$.
Since these are uniform random variables, if $\frac{\theta_a}{k} = 1 - \alpha$ then $P(y_{max} > \frac{\theta_a}{k}) \le \alpha$ and so $k = \frac {\theta_a}{1 - \alpha}$
Intuitively thought I feel like it should be just {$y_{max} \gt \theta_a$} since as soon as the max is bigger than your null hypothesis your null hypothesis becomes false.
 A: You're given the test-statistic and the decision rule for the test, which is reject $H_0$ if $y_{max} > k$. Now you want to find $k$ such that the test has size $\alpha$; that is, you want to find the minimum $k$ such that the probability of rejecting the null if it is true is less than $\alpha$. 
In math, you want to find the smallest $k$ such that
$$
P(y_{max} > k  \, | \, \theta = \theta_0) \le \alpha.
$$
The probability above is
$$
P(y_{max} > k  \, | \, \theta = \theta_0)  = 1 - P(y_{max} \le k \, | \, \theta = \theta_0) = 1 - P(Y_1 \le k)^n = 1 - \left( \frac{k}{\theta_0} \right)^n.
$$
Notice that the second equality follows because the maximum of a random sample is less than $k$ if and only if all of the $Y_i$ are less than $k$, and since they're all independent this is just $P(Y_1 \le k)^n$.
Alright, so now you can solve for $k$ and get
$$
k \ge \theta_0 (1-\alpha)^{1/n}.
$$
This inequality gives all the possible values of $k$ such that the test has size $\alpha$. You actually wanted the $k$ that maximizes the power. That $k$ should be $\theta_0 (1-\alpha)^{1/n}$ (since it is the smallest $k$ such that the test has size $\alpha$).
