Convergence of power function is hypothesis testing

Question

Let $\widehat{\theta}$ be the MLE of a parameter $\theta$ and let $\widehat{\text{se}}=\{nI(\widehat{\theta})\}^{-\frac12}$ where $I(\theta)$ is the Fisher information. Consider testing$$ H_0:\theta=\theta_0\,\,\text{versus}\,\,H_1:\theta\neq \theta_0. $$ Consider the Wald test with rejection region $R=\{(x_1,...,x_n):|Z|>z_{\alpha/2}\}$ where $Z=(\widehat{\theta}-\theta_0)/\widehat{\text{se}}$. Let $\theta_1>\theta_0$ be some alternative. Show that $\beta(\theta_1)\to 1$ as $n\to\infty$, where $\beta$ is the power function.

My approach: I tried to use the definition to get$$ \beta(\theta)=\mathbb{P}_\theta(X\in R)=\mathbb{P}_\theta(|\widehat{\theta}-\theta_0|/\widehat{\text{se}}>z_{\alpha/2}) $$ given that the true value of $\theta$ is $\theta_1$, but I'm stuck and not even sure if this is correct for $\theta_1$. Any ideas?

Answer 1

In the meantime, with the guidance of Ceph in the comment section, I believe I found an answer.

Let $H_1:\theta=\theta_1$. Under $H_1$, we define $W=(\widehat{\theta}-\theta_1)/\widehat{\text{se}} \rightsquigarrow N(0,1)$. Hence,\begin{align*} \beta(\theta_1)&=\mathbb{P}_{\theta_1}(|Z|>z_{\alpha/2})\\ &=\mathbb{P}_{\theta_1}(Z>z_{\alpha/2})+\mathbb{P}_{\theta_1}(Z<-z_{\alpha/2})\\ &=\mathbb{P}_{\theta_1}\left(\frac{\widehat{\theta}-\theta_0}{\widehat{\text{se}}}>z_{\alpha/2} \right)+\mathbb{P}_{\theta_1}\left(\frac{\widehat{\theta}-\theta_0}{\widehat{\text{se}}}<-z_{\alpha/2} \right)\\ &=\mathbb{P}_{\theta_1}(\widehat{\theta}>\theta_0+\widehat{\text{se}}\,z_{\alpha/2})+\mathbb{P}_{\theta_1}(\widehat{\theta}<\theta_0-\widehat{\text{se}}\,z_{\alpha/2})\\ &=\mathbb{P}_{\theta_1}\left( \frac{\widehat{\theta}-\theta_1}{\widehat{\text{se}}}>\frac{\theta_0-\theta_1}{\widehat{\text{se}}}+z_{\alpha/2} \right)+\mathbb{P}_{\theta_1}\left( \frac{\widehat{\theta}-\theta_1}{\widehat{\text{se}}}<\frac{\theta_0-\theta_1}{\widehat{\text{se}}}-z_{\alpha/2} \right)\\ &=\mathbb{P}\left(W> \frac{\theta_0-\theta_1}{\widehat{\text{se}}}+z_{\alpha/2} \right)+\mathbb{P}\left(W< \frac{\theta_0-\theta_1}{\widehat{\text{se}}}-z_{\alpha/2} \right)\\ &\geq \mathbb{P}\left(W> \frac{\theta_0-\theta_1}{\widehat{\text{se}}}+z_{\alpha/2} \right). \end{align*} As $n\to\infty$, $\widehat{\text{se}}\to 0$ and since $\theta_1>\theta_0$, $(\theta_0-\theta_1)/\widehat{\text{se}}\to -\infty$ and thus $\beta(\theta_1)\to 1$.