MP test construction for shifted exponential distribution For the pdf $f_{\theta}(x)=e^{-(x-\theta)} , x \ge \theta$, find a most powerful test of size $\alpha$, using Neyman Pearson Lemma to test $\theta=\theta_{0}$ against $\theta=\theta_1(> \theta_0)$, based on a sample of size $n$.
I am facing difficulty as the parameter here is range dependent 
However, if $X_{(1)}>\theta_1$, then $f_1(x)>\lambda f_0(x)$ if $e^{n(\theta_1- \theta_0)}> \lambda$ would mean rejection of null hypothesis.
But how will I make this test a size $\alpha$ test? The ratio is coming to be constant.
Please help!
 A: Joint density of the sample $(X_1,X_2,\ldots,X_n)$ is 
$$f_{\theta}(x_1,\ldots,x_n)=\exp\left(-\sum_{i=1}^n(x_i-\theta)\right)\mathbf1_{x_{(1)}>\theta}\quad,\,\theta>0$$
By N-P lemma, a most powerful test of size $\alpha$ for testing $H_0:\theta=\theta_0$ against $H_1:\theta=\theta_1(>\theta_0)$ is given by $$\varphi(x_1,\ldots,x_n)=\begin{cases}1&,\text{ if }\lambda(x_1,\ldots,x_n)>k\\0&,\text{ if }\lambda(x_1,\ldots,x_n)<k\end{cases}$$
, where $$\lambda(x_1,\ldots,x_n)=\frac{f_{\theta_1}(x_1,\ldots,x_n)}{f_{\theta_0}(x_1,\ldots,x_n)}$$
and $k(>0)$ is such that $$E_{\theta_0}\varphi(X_1,\ldots,X_n)=\alpha$$
Now,
\begin{align}
\lambda(x_1,\ldots,x_n)&=\frac{\exp\left(-\sum_{i=1}^n(x_i-\theta_1)\right)\mathbf1_{x_{(1)}>\theta_1}}{\exp\left(-\sum_{i=1}^n(x_i-\theta_0)\right)\mathbf1_{x_{(1)}>\theta_0}}
\\\\&=e^{n(\theta_1-\theta_0)}\frac{\mathbf1_{x_{(1)}>\theta_1}}{\mathbf1_{x_{(1)}>\theta_0}}
\\\\&=\begin{cases}e^{n(\theta_1-\theta_0)}&,\text{ if }x_{(1)}>\theta_1\\0&,\text{ if }\theta_0<x_{(1)}\le \theta_1\end{cases}
\end{align}
So $\lambda(x_1,\ldots,x_n)$ is a monotone non-decreasing function of $x_{(1)}$, which means
$$\lambda(x_1,\ldots,x_n)\gtrless k \iff x_{(1)}\gtrless c$$, for some $c$ such that $$E_{\theta_0}\varphi(X_1,\ldots,X_n)=\alpha$$
We thus have
$$\varphi(x_1,\ldots,x_n)=\begin{cases}1&,\text{ if }x_{(1)}>c\\0&,\text{ if }x_{(1)}<c\end{cases}$$
Again,
\begin{align}
E_{\theta_0}\varphi(X_1,\ldots,X_n)&=P_{\theta_0}(X_{(1)}>c)
\\&=\left(P_{\theta_0}(X_1>c)\right)^n
\\&=e^{n(\theta_0-c)}\quad,\,c>\theta_0
\end{align}
So from the size condition we get $$c=\theta_0-\frac{\ln\alpha}{n}$$
Finally, the test function is
$$\varphi(x_1,\ldots,x_n)=\begin{cases}1&,\text{ if }x_{(1)}>\theta_0-\frac{\ln\alpha}{n}\\0&,\text{ if }x_{(1)}<\theta_0-\frac{\ln\alpha}{n}\end{cases}$$
A: Comment: This is a tricky problem--pretty much for the reason you mention.
It may help to consider the case $n = 1$ for $\theta_0 = 1,\,\theta_1 = 5.$
Then plots of the PDF are shown below.  Suppose we agree to Reject $H_0: \theta = 1$
against $H_a: \theta= 5$ when the single observation (also the smallest) $X > 5,$ otherwise fail to reject.
Then it is easy to see that the significance level of the test is $\alpha \approx 0.0025.$
Can you write the LR in this case? When you understand the problem for $n = 1,$
then go on the the general case.

A: If $X_{(1)} \in (\theta_0, \theta_1)$, then there is no uncertainty and you sure that $H_0$ right. If $X_{(1)} \ge \theta_1$, then the MP test of size $\alpha$ is: reject $H_0$ if 
$$
c\le\frac
{\exp\{n \theta_1 - \sum x_i \}}
{\exp\{n \theta_0 - \sum x_i \} }
=
\exp\{ n(\theta_1 - \theta_0 \},
$$
which is clearly not helpful as it constant for evey $n$. However, note that the LR is monotone increasing function of $\theta_1$, hence using the fact that $X_{(1)} \sim \mathcal{E}xp_{\theta_1}(n)$, the genral form of the MP is 
$$
\alpha = \mathbb{E}_{\theta_1}I\{X_{(1)}  > c \}=\mathbb{P}_{\theta_1}(X_{(1)}  > c) = \exp\{n(\theta_1 - c)\},
$$
i.e., the MP is
$$
I\{X_{(1)} >\theta_1-\frac{\ln \alpha}{n}\} \, .
$$
for $X_{(1)} \ge \theta_1$, and $0$ otherwise.
