Likelihood Ratio Test: Uniform Distribution, Change of Inequality in Alternative Hypothesis

Question

Thanks, in advance. I would really appreciate any help here.

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Problem: Let $X_1, \ldots, X_n$ be a random sample from a $\operatorname{Uniform}(0,\theta)$ population. Let $$H_0 : \theta = \theta_0 \quad \text{and} \quad H_A : \theta < \theta_0.$$ The unrestricted MLE of $\theta$ is $X_{(n)}$ (the maximum).

(a) Show that for a likelihood ratio test of size $\alpha$ that the rejection region is $$\frac{X_{(n)}}{\theta_0} < \alpha^{1/n}.$$

Solution: This question I was able to do. We know that the rejection region for any GLRT is of the form $$RR = \{\Lambda(X_1, \ldots, X_n) < c\}.$$ So, $$\begin{align*} \Pr[\text{reject } H_0 \mid H_0 \text{ true}] &= \Pr[\Lambda(X_1, \ldots, X_n) < c \mid \theta = \theta_0] \\ &= \Pr[(X_{(n)}/\theta_0)^n < c \mid \theta = \theta_0 ] \\ &= \Pr[X_{(n)} < c^{1/n} \theta_0 \mid \theta = \theta_0 ] \\ &= \Pr[X_1 < c^{1/n} \theta_0 \mid \theta = \theta_0]^n \\ &= \biggl( \int_{x=0}^{c^{1/n} \theta_0} \frac{1}{\theta_0} \, dx \biggr)^{\! n} \\ &= (c^{1/n})^n = c \\ &= \alpha \end{align*}$$
Thus, $$RR = \{\Lambda(X_1, \ldots, X_n) < c\} = \left\{ \left(\frac{X_{(n)}}{\theta_0}\right)^{\! n} < \alpha \right\} = \left\{ \frac{X_{(n)}}{\theta_0} < \alpha^{1/n} \right\}.$$

(b) Now, let $$H_0 : \theta = \theta_0 \quad \text{and} \quad H_A : \theta > \theta_0.$$ Show that the LRT is simple in the sense that if $X_{(n)} < \theta_0$ we accept $H_0$ and if $X_{(n)} > \theta_0$ we reject $H_0$ for all $\alpha$.

Attempt at this question: So in this case would the rejection region of the LRT switch to this: $$RR = \{ \Lambda(X_1, \ldots, X_n) > c \}?$$ My reasoning is that if $\Lambda(X_1, \ldots, X_n) = (X_{(n)}/\theta_0)^n$, then $H_A$ implies that $\theta > \theta_0 = X_{(n)} > \theta_0$ and so we would expect $(X_{(n)}/\theta_0)^n$ to be a number larger than one if $H_A$ is true. And so, doing the same process as (a) for all $\alpha$ it would follow that if $X_{(n)}/\theta_0 > 1$ (as long as $X_{(n)} > \theta_0$) then we would always reject for all $\alpha$ since we end up with $(X_{(n)}/\theta_0)^n > \alpha \rightarrow X_{(n)}/\theta_0 > \alpha^{1/n}$ and $0 < \alpha \le 1$. Of course, the bound could be lower for different $\alpha$ that aren't one, but $X_{(n)}/\theta_0 > 1$ will always be a stronger condition.

Basically, this whole idea revolves around the RR being different from what I am used to seeing. What is wrong with my reasoning?

EDIT: I believe I have figured out this problem. Here was my solution in case anyone cares:

Since $\Lambda_r = (\frac {X_{(n)}}{\theta_0})^n$ , we know that if $H_A : \theta > \theta_0$ then this implies that $H_A : X_{(n)} > \theta_0$. So, we want to find c such that again,

$Pr[reject \ H_0 | H_0 \ is \ true] =Pr[\Lambda_r < c \ \ | \theta = \theta_0]$=$Pr[\ (\frac {X_{(n)}}{\theta_0})^n < c \ \ | \ \ \theta = \theta_0] = \alpha$.

If indeed $H_A : X_{(n)} > \theta_0$ is true then this would imply (as I originally stated) that $\Lambda_r > 1$. In other words, $RR = \{ \Lambda_r > 1\} $.

However, if $X \sim Uniform(0, \theta)$ distributed random variable then under $H_0$ it is impossible for the maximum $X_{(n)}$ to be greater than $\theta_0$ and so it is impossible for $\Lambda_r > 1$, since under the null $X \sim Uniform(0, \theta_0)$ and thus $Pr[ X_{(n)} > \theta_0 | \theta = \theta_0] = 0 $ .

Thus,

$Pr[reject \ H_0 | H_0 \ is \ true] = Pr[\ \Lambda_r > 1\ | \theta = \theta_0] = Pr[X_{(n)} > \theta_0 | \theta = \theta_0] = 0.$

Thus, it follows that if we observe a maximum value that is greater than $\theta_0$ that the null hypothesis cannot be true (ie. the alternative hypothesis must be true), since under the null hypothesis the probability of observing a maximum value greater than $\theta_0$ is zero, regardless of $\alpha$.