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!

Solution 1:

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

$$\varphi(x_1,\ldots,x_n)=\begin{cases}1&,\text{ if }x_{(1)}>\theta_0-\frac{\ln\alpha}{n}\\0&,\text{ if }x_{(1)}<\theta_0-\frac{\ln\alpha}{n}\end{cases}$$