Vtyjkyuk

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!

Joint density of the sample $(X_1,X_2,ldots,X_n)$ is

$$f_theta(x_1,ldots,x_n)=expleft(-sum_i=1^n(x_i-theta)right)mathbf1_x_(1)>thetaquad,,theta>0$$

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 $$varphi(x_1,ldots,x_n)=begincases1&,text if lambda(x_1,ldots,x_n)>k\0&,text if lambda(x_1,ldots,x_n)<kendcases$$

, where $$lambda(x_1,ldots,x_n)=fracf_theta_1(x_1,ldots,x_n)f_theta_0(x_1,ldots,x_n)$$

and $k(>0)$ is such that $$E_theta_0varphi(X_1,ldots,X_n)=alpha$$

Now,

beginalign
lambda(x_1,ldots,x_n)&=fracexpleft(-sum_i=1^n(x_i-theta_1)right)mathbf1_x_(1)>theta_1expleft(-sum_i=1^n(x_i-theta_0)right)mathbf1_x_(1)>theta_0
\\&=e^n(theta_1-theta_0)fracmathbf1_x_(1)>theta_1mathbf1_x_(1)>theta_0
\\&=begincasese^n(theta_1-theta_0)&,text if x_(1)>theta_1\0&,text if theta_0<x_(1)le theta_1endcases
endalign

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 $$E_theta_0varphi(X_1,ldots,X_n)=alpha$$

We thus have

$$varphi(x_1,ldots,x_n)=begincases1&,text if x_(1)>c\0&,text if x_(1)<cendcases$$

Again,

beginalign
E_theta_0varphi(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
endalign

So from the size condition we get $$c=theta_0-fraclnalphan$$

Finally, the test function is

$$varphi(x_1,ldots,x_n)=begincases1&,text if x_(1)>theta_0-fraclnalphan\0&,text if x_(1)<theta_0-fraclnalphanendcases$$

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
$$
clefrac
expn theta_1 - sum x_i
expn theta_0 - sum x_i
=
exp n(theta_1 - theta_0 ,
$$
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 $X_(1) sim mathcalExp_theta_1(n)$, the genral form of the MP is
$$
alpha = mathbbE_theta_1IX_(1) > c =mathbbP_theta_1(X_(1) > c) = expn(theta_1 - c),
$$
i.e., the MP is
$$
IX_(1) >theta_1-fracln alphan , .
$$
for $X_(1) ge theta_1$, and $0$ otherwise.

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.