What is the correct definition of a Power Function?

Question

In Casella Berger's Statistical Inference, they define a power function of a hypothesis test with rejection region $R$ to be the function of $\theta$ define by $\beta(\theta) = P_\theta(X\in R)$ for some data $X$. Suppose that $H_0: \theta\in \Theta_0$ and $H_1: \theta \in \Theta_0^c$.

Furthermore, they state that:

$$ P_\theta(X\in R) = \begin{cases} \text{probability of a Type 1 error} &\mbox{if } \theta\in \Theta_0\\ \text{one minus the probability of a Type 2 error} & \mbox{if } \theta\in \Theta_0^c\end{cases} $$

However, my understand is always that the power function is the probability of rejecting the null, given that the null is false. This doesn't match the above. What is wrong here? Thanks!

Answer 1

Consider if you have a simple null, like $\mu=\mu_0$ against a two-sided alternative. Then your power function has a "hole" at $\mu_0$.

The usual definition of power function fills in the hole, making the power function defined for all possible values of $\theta$.

Sure, at that point it's not power, but calling it a "rejection rate function" just because you defined the function at one point where it isn't measuring power is a little clumsy.

Answer 2

Power is the probability that the observation is in the rejection region when some value in the parameter space of the alternative is correct (falsely rejecting the null hypothesis). But when the two distributions are identical, the rejection region for the null hypothesis also corresponds to the non-rejection region for the alternative, so $\alpha =1-\beta$. Think of the case of two univariate normal distributions with variance 1 and mean 0 under the null hypothesis and a one-sided alternative mean >0. Then as the alternative mean gets closer to zero, the power drops all the way down to $\alpha$. A drawing showing the critical region with the standard normal and the normal shift to the right of a mean $\mu>0$ should make this clear.