How can one find $g$ so that the test will have size $\beta$ for the following uniform distribution case? I have the random sample $X_1, X_2, \dots, X_n$ drawn from the uniform distribution on $[\varphi, \varphi + 1]$. To test the null hypothesis $H_0 : \varphi = 0$ against the alternative hypothesis $H_1 : \varphi > 0$, we have the test
$$\text{Reject} \ H_0 \ \ \ \text{if} \ \ \ X_{(n)} \ge 1 \ \text{or} \ X_{(1)} \ge g,$$
where $g$ is a constant, $X_{(1)} = \min\{X_1, X_2, \dots, X_n\}, X_{(n)} = \max\{X_1, X_2, \dots, X_n\}$.
How can one find $g$ so that the test will some have size $\beta$?
 A: On the assumption $H_0$ is correct, the probability $X_{(n)} \ge 1$ is $0$, while the probability $X_{(1)} \ge g$ is the probability all the observations are in $(g,1]$ which is $(1-g)^n$.  So $\alpha = (1-g)^n$ and $g=1-\sqrt[n]{\alpha}$.
If you want the power of the test for some particular $\phi_1 > 0$, you do much the same calculation:

*

*if $1 \le \phi_1 $ then the probability $X_{(n)} \gt 1$ is $1$, while the probability  all the observations are in $[g,1]$ is $0$, so the power of the test is $1$


*if $g \le \phi_1 \le 1$ then the probability $X_{(n)} \gt 1$ is the probability not all the observations are in $[\phi_1,1]$ which is $1- (1-\phi_1)^n$,  while the probability  all the observations are in $[g,1]$ is the probability  all the observations are in $[\phi_1,1]$ which is $(1-\phi_1)^n$, so the power of the test is $1$


*if $0 \le \phi_1 \le g$ then the probability $X_{(n)} \gt 1$ is the probability not all the observations are in $[\phi_1,1]$  which is $1- (1-\phi_1)^n$, while the probability  all the observations are in $[g,1]$ is $(1-g)^n = \alpha$, so the power of the test is $$1- (1-\phi_1)^n +(1-g)^n = 1- (1-\phi_1)^n+\alpha$$
This last expression behaves as you might expect if the alternative hypothesis is in fact true and $\phi=\phi_1$:

*

*It is a continuous increasing function of $\phi_1$, being $\alpha$ when $\phi_1=0$ and $1$ when $\phi_1=g=1-\sqrt[n]{\alpha}$: the further the actual value of $\phi_1$ the more likely you are to reject the null hypothesis

*It is a continuous decreasing of $g$, being $1$ when $g=\phi_1$ and $1- (1-\phi_1)^n$ when $g=1$: the tighter you make the test the less likely you are to reject the null hypothesis

*It is a continuous increasing function of $\alpha$, being $1- (1-\phi_1)^n$ when $\alpha=0$ and $1$ when $\alpha=(1-\phi_1)^n$: the more willing you are to reject the null hypothesis the more likely you are to do so

