I have a small confusion over describing the cutoff point for the critical region in a likelihood ratio test when the null hypothesis is composite. Take this exercise in particular:
Let $(X_1,X_2,\ldots,X_n)$ be a random sample from a shifted exponential distribution with density $$f_{\theta}(x)=e^{-(x-\theta)}\mathbf1_{x\ge\theta}\quad,\,\theta\in\mathbb R$$
I am to derive the likelihood ratio test of size $\alpha$ for testing $$H_0:\theta\le \theta_0\quad\text{ against }\quad H_1:\theta>\theta_0,$$ where $\theta_0$ is a specified value of $\theta$.
Given the sample $(x_1,x_2,\ldots,x_n)$, the likelihood function is
$$L(\theta\mid x_1,\ldots,x_n)=\prod_{i=1}^n f_{\theta}(x_i)=\exp\left[-\sum_{i=1}^n (x_i-\theta)\right]\mathbf1_{x_{(1)}\ge\theta}\quad,\,\theta\in\mathbb R$$
Unrestricted MLE of $\theta$ is clearly $$\hat\theta=X_{(1)}$$
And I think the restricted MLE of $\theta$ when $\theta\le \theta_0$ is $$\hat{\hat\theta}=\begin{cases}\hat\theta&,\text{ if }\hat\theta\le\theta_0 \\ \theta_0&,\text{ if }\hat\theta>\theta_0 \end{cases}$$
So my LR test statistic is
\begin{align} \Lambda(x_1,\ldots,x_n)&=\frac{\sup_{\theta\le\theta_0}L(\theta\mid x_1,\ldots,x_n)}{\sup_{\theta}L(\theta\mid x_1,\ldots,x_n)} \\\\&=\frac{L(\hat{\hat\theta}\mid x_1,\ldots,x_n)}{L(\hat\theta\mid x_1,\ldots,x_n)} \\\\&=\begin{cases}1&,\text{ if }\hat\theta\le\theta_0\\\\\frac{L(\theta_0\mid x_1,\ldots,x_n)}{L(\hat\theta\mid x_1,\ldots,x_n)}&,\text{ if }\hat\theta>\theta_0\end{cases} \end{align}
If $\hat\theta\le\theta_0$, we trivially accept $H_0$.
Now when $\hat\theta>\theta_0$,
$$\Lambda(x_1,\ldots,x_n)=\frac{L(\theta_0\mid x_1,\ldots,x_n)}{L(\hat\theta\mid x_1,\ldots,x_n)}=e^{n(\theta_0-x_{(1)})}$$
Therefore, when $x_{(1)}>\theta_0$,
\begin{align} \Lambda(x_1,\ldots,x_n)<\text{ constant }&\implies x_{(1)}-\theta_0>\text{ constant } \end{align}
My confusion is whether I should rewrite that last line as $x_{(1)}>\text{ constant }$ or keep it as it is.
We know that an appropriate test statistic here is $$2n(X_{(1)}-\theta)\sim \chi^2_2$$
But under $H_0$, how is $\theta<\theta_0$ reflected in this statistic (if it was a simple null it would have been fine)? Is it correct to say, that under $H_0$, $2n(X_{(1)}-\theta_0)\sim \chi^2_2$?
If I describe the critical region as $x_{(1)}>c$, then I have to find $c$ subject to $$P_{H_0}(X_{(1)}>c)=\alpha$$
I can do this by finding the probability directly. But suppose I want to find the cutoff point in terms of the $\chi^2_2$ fractile. Then I get for some $k (=c-\theta_0)$, $$P_{H_0}(2n(X_{(1)}-\theta_0)>2n k)=\alpha,$$ which gives me $$2nk=\chi^2_{2;\alpha}\implies k=\frac{1}{2n}\chi^2_{2;\alpha},$$ where $\chi^2_{2,\alpha}$ is the $(1-\alpha)$th fractile of a $\chi^2_2$ distribution.
Then my decision rule for $\hat\theta>\theta_0$ would be "Reject $H_0$ at size $\alpha$ if $X_{(1)}>\frac{1}{2n}\chi^2_{2;\alpha}$".
Is there anything wrong with the last line? Or should I write the cutoff point in terms of $c$ , that is $c=\frac{1}{2n}\chi^2_{2;\alpha}+\theta_0$ ?
In other words, should I write the cutoff point including or excluding $\theta_0$?
This question arises because I have been told that the LRT coincides with the UMP test whenever the latter exists. And for the UMP test (UMP test exists because $f_{\theta}$ has the MLR property), the cutoff point for the critical region is expressed in terms of $\theta_0$.
