Proof of the power of the wald test

Question

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?

Answer 1

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.

Answer 2

I am fortunate to have come across this thread while revising! Just to complete the proof, here's my working.

I am using Wasserman's notation of $z_{\alpha}$ where $z_{\alpha} = \Phi^{-1} (1-\alpha)$, which means that $z_{\alpha}$ is positive for $\alpha < 0.5$.

Continuing from math and heropup's answer:

$$\beta (\theta_*) = 1 - \mathbb{P}\left(-z_{\alpha/2} < \frac{\hat{\theta} - \theta_*}{\hat{se}} < z_{\alpha/2} \right) $$ $$= 1 - \mathbb{P}\left(-z_{\alpha/2} < \frac{\hat{\theta} - \theta_0}{\hat{se}} + \frac{\theta_0 - \theta_*}{\hat{se}} < z_{\alpha/2} \right)$$

Using the assumption of asymptotic normality, $\frac{\hat{\theta} - \theta_0}{\hat{se}} \sim N(0,1)$

Denote the standard normal as $Z$

$$\beta (\theta_*) = 1 - \mathbb{P}\left(-z_{\alpha/2} < Z + \frac{\theta_0 - \theta_*}{\hat{se}} < z_{\alpha/2} \right) $$ $$ = 1 - \mathbb{P}\left(-z_{\alpha/2} - \frac{\theta_0 - \theta_*}{\hat{se}} < Z < z_{\alpha/2} - \frac{\theta_0 - \theta_*}{\hat{se}} \right)$$

$$ = 1 - \Phi\left(z_{\alpha/2} - \frac{\theta_0 - \theta_*}{\hat{se}}\right) + \Phi\left( -z_{\alpha/2} - \frac{\theta_0 - \theta_*}{\hat{se}} \right)$$

By some additional manipulation using the symmetry property of the distribution (I guess for aesthetics?):

$$= 1 - \Phi\left( \frac{\theta_0 - \theta_*}{\hat{se}} + z_{\alpha/2}\right) + \Phi\left(\frac{\theta_0 - \theta_*}{\hat{se}} - z_{\alpha/2}\right)$$