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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 $\beta$?

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