$X_1,X_2$ are iid uniform ($\theta, \theta+1$). We consider the rejection region given by
{(x1, x2): x1 + x2 > C}$\cup$
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{(x1, x2): x1 > 1}$\cup$
[
{(x1, x2): x2 > 1}.
What is the power function corresponding to this rejection region? This is from George Casella question 8.13 part (d). The question is Let $X_1, X_2$ be iid uniform ($\theta, \theta+1$). For testing $H_0:\theta=0$ versus $H_1:\theta>0$. We have a test: $\phi(X_1, X_2)$: Reject $H_0$ if $X_1 + X_2 > C$. Then the author claims that if we change the rejection region to the above (2 more union sets), this new test will have the same size but is more powerful than the current.
The solution is attached below. I don't know how to analyze this new test is more powerful than the original rejection region $x_1 + x_2 > C$. If deriving the exact power function of the new rejection region is too hard, just comparison is also helpful. I don't know how to compare.
Best Answer
You have not told us the question, but just from what you have told us, we might conclude
$\{(x_1, x_2): x_1 + x_2 > C\} \cup \{(x_1, x_2): x_1 > 1\} \cup \{(x_1, x_2): x_2 > 1\}$ contains $\{(x_1, x_2): x_1 + x_2 > C\} $ as a subset, so has either greater probability, or the same probability.
If under the null hypothesis $\mathbb P(X_1 > 1)=0$ and $\mathbb P(X_2 > 1)=0$, then the value of $C$ for a test of given statistical significance will not change for the larger rejection region
If under the alternative hypothesis $\mathbb P((X_1 > 1) \cap (X_1 + X_2 \le C))> 0$ or $\mathbb P((X_2 > 1) \cap (X_1 + X_2 \le C))> 0$ then the larger rejection region will lead to a more powerful test