Consider the following passage from the bottom of pg. 23 of Statistical Decision Theory and Bayesian Analysis:
Problem. I'm having trouble understanding the red highlighted; it seems that the textbook has computed the error not of rejecting but of either falsely rejecting or incorrectly accepting the null hypothesis.
Why I think this is so. First, the "0-1" loss can be displayed on the following table:
$$\begin{array}{c|c|c|} & \text{$H_0: θ = θ_0$ } & \text{$H_1: θ = θ_1$ } \\ \hline \text{$θ = θ_0$} & 0 & 1 \\ \hline \text{$θ = θ_1$ } & 1 & 0 \\ \hline \end{array}$$
From this table it is clear that
$$ R(θ_0, δ) = (P_{θ_0}(\text{Keep $H_0$}) ⋅ 0) + (P_{θ_0}(\text{Type I error}) ⋅ 1) = α_0 = P_{θ_0}(\text{Type I error}) $$
and
$$ R(θ_1, δ) = (P_{θ_1}(\text{Type II error}) ⋅ 1) + (P_{θ_1}(\text{Reject $H_0$}) ⋅ 0) = α_1 = P_{θ_1}(\text{Type II error}) $$
Now suppose, as in the passage, that $α_0 = 0.01$ and $α_1 = 0.99$. Then
$$ \frac{α_0 + α_1}{2} = \frac{0.01 + 0.99}{2} = 0.5 $$
But this doesn't seem to show that "half of all rejections of the null will actually be in error"; instead, it shows that half of all tests (which result in either a rejection or an acceptance of $H_0$) will lead to error.
What am I missing?
EDIT. I think I understand the solution. Basically, in order to compute the error rate when rejecting, one must calculate:
$$ \frac{P_{\theta_0}(H_1)}{P_{\theta_0}(H_1) + P_{\theta_1}(H_1)} = \frac{\alpha_0}{P_{\theta_0}(H_1) + P_{\theta_1}(H_1)} = {0.01 \over 0.01 + 0.01} = 0.5 $$
Is this the correct?