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I know that power function is defined as $\beta(\theta)=P(\mathbf{X}\in R\mid\theta)$, where $R$ is the rejection region. For a composite test, $\Theta_0^c$ is a set. How do we define "the power of a test" in this scenario? Do we use $\inf_{\theta\not\in\Theta_0}\beta(\theta)$ or $\sup_{\theta\not\in\Theta_0}\beta(\theta)$?

Thanks!

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The power of a test is $\mathbb{P}(reject \ H_0 | H_A \ is \ true) = \mathbb{P}(\vec{X} \in R | \theta \in \Theta_A) = \beta(\theta)$ for the same $\theta \in \Theta_A$. This means that the power of a test depends upon the specific $\theta$. It is not necessarily the case that the test has a single value for the power. You might be confusing the size or level of a test with its power.

We could look at the best possible power or the worst possible power by taking $\inf$s or $\sup$s over $\Theta_A$ like you are thinking about but I haven't seen this done much. I think part of the reason this isn't done much is that the results are often not very useful. Suppose we are testing something like $H_0: \theta \leq \theta_0$ vs $H_A: \theta > \theta_0$. Then we'll get the worst power for $\theta$ just a tiny bit greater than $\theta_0$ and the best power for $\theta \rightarrow \infty$. This isn't particularly insightful or helpful. What would be more practically useful is asking what power we get for a set of $\theta$'s so that we can choose our sample size and whatnot. This reduces to computing $\beta(\theta)$.

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  • $\begingroup$ I remember that I came across a scenario where the power is defined as the lower bound of the power function when the alternative hypothesis is true. But I forgot where I saw it. Your explanation is good. Whether or not defining the power as the lower bound does not make any fundamental difference. $\endgroup$ Apr 9, 2015 at 20:20
  • $\begingroup$ With the test $H_0: \theta \leq \theta_0$ vs $H_A: \theta > \theta_0$, the critical region $R$ can be found by finding $\sup_{\theta\in\Theta_0}\beta(\theta)$. Since the power function is defined for $\theta\in\Theta_A$, why is one taking $\sup$ over $\Theta_0$? $\endgroup$
    – psie
    Dec 1, 2020 at 12:53

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