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 $$\beta$$?

• Hint: On the assumption $H_0$ is correct, what is the probability $X_{(1)}\ge g$, as a function of $g$? May 10 at 9:59
• @Henry Since it's been a while, any chance you could post a full answer? May 21 at 5:43

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