I'm reading through Larry Wasserman Book "All of Statistics". Currently I'm in chapter hypothesis testing, the chapter can be found here. I would like to prove theorem 10.6, I quote:
Suppose the true value of $\theta$ is $\theta_* \neq \theta_0$. The power $\beta(\theta_*)$ - the probability of correctly rejecting the null hypothesis - is given (approximately) by $$ 1- \Phi(\frac{\theta_0-\theta_*}{\hat{se}}+z_{\alpha/2})+\Phi(\frac{\theta_0-\theta_*}{\hat{se}}-z_{\alpha/2})$$
We assume that under $H_0: \theta = \theta_0$ the quantity $\frac{\hat{\theta}-\theta_0}{\hat{se}}$ is asymptotically standard normal. We reject $H_0$ if $|W|>z_{\alpha/2}$, where $W:=\frac{\hat{\theta}-\theta_0}{\hat{se}}$
My work so far:
By definition: $$\beta(\theta_*)= P(|W|>z_{\alpha/2})=P(z_{\alpha/2}<W<-z_{\alpha/2})=P(z_{\alpha/2}<\frac{\hat{\theta}-\theta_*}{\hat{se}}<-z_{\alpha/2})=P(z_{\alpha/2}<\frac{\hat{\theta}-\theta_0}{\hat{se}}+\frac{\theta_0-\theta_*}{\hat{se}}<-z_{\alpha/2})=P(z_{\alpha/2}-\frac{\theta_0-\theta_*}{\hat{se}}< Z <-z_{\alpha/2}-\frac{\theta_0-\theta_*}{\hat{se}})=P(z_{\alpha/2}+\frac{\theta_0-\theta_*}{\hat{se}}< Z <-z_{\alpha/2}+\frac{\theta_0-\theta_*}{\hat{se}})$$ where $Z$ is a standard normal and we used that $Z$ is symmetric. The last expression is equal $$P(z_{\alpha/2}+\frac{\theta_0-\theta_*}{\hat{se}}< Z <-z_{\alpha/2}+\frac{\theta_0-\theta_*}{\hat{se}})\approx\Phi(-z_{\alpha/2}+\frac{\theta_0-\theta_*}{\hat{se}})-\Phi(z_{\alpha/2}+\frac{\theta_0-\theta_*}{\hat{se}})$$
however I'm somehow missing the $+1$. Where is my mistake?
I find the way you write compound inequalities to be remarkably and disturbingly strange. If I were to write $$3 < x < -4$$ when what I really mean is $$x > 3 \quad \text{or} \quad x < -4,$$ the former expression would technically mean that $x$ has no solution: there are no real numbers that are simultaneously greater than $3$ and less than $-4$.
In the same vein, then, when you go from $$\Pr[|W| > z_{\alpha/2}] = \Pr[z_{\alpha/2} < W < -z_{\alpha/2}],$$ I can only assume that what you really mean is to write $$\Pr[|W| > z_{\alpha/2}] = 1 - \Pr[|W| \le z_{\alpha/2}] = 1 - \Pr[-z_{\alpha/2} \le W \le z_{\alpha/2}].$$ I think that you will find that this will fix your problems.