In All of Statistics, chapter 11 (pg. 183), Larry Wasserman states in his description of the Wald Test:
We are testing the null hypothesis $ \hat{\theta} = \theta_0 $ versus the alternative hypothesis $ \theta \neq \theta_0 $.
He then says we should assume that $ \hat{\theta} $ is asymptotically normal, i.e. converges in distribution to a standard normal: $$ \frac{\sqrt{n}(\hat{\theta} - \theta_0)}{\hat{\text{se}}} \rightsquigarrow N(0, 1). $$
(Wasserman uses $ \rightsquigarrow $ to denote converging in distribution.)
My impression is that this is a reasonable assumption because of the Central Limit Theorem.
But then, when proving the following theorem:
Asymptotically the Wald test has size $ \alpha $, that is, $$ \mathbb{P}_{\theta_0}\left(\lvert Z \rvert > z_{\alpha/2}\right) \rightarrow \alpha $$ as $ n \rightarrow \infty $,
He says:
Under $ \theta = \theta_0 $, $ (\hat{\theta} - \theta_0)/\hat{se} \rightsquigarrow N(0, 1). $
I don't see how $ (\hat{\theta} - \theta_0)/\hat{se} \rightsquigarrow N(0, 1) $ follows from the fact that $$ \frac{\sqrt{n}(\hat{\theta} - \theta_0)}{\hat{\text{se}}} \rightsquigarrow N(0, 1). $$
Doesn't this imply that $$ \lim_{n \rightarrow \infty} P\left(\frac{\sqrt{n}(\hat{\theta} - \theta_0)}{\hat{\text{se}}} \leq z\right) = \lim_{n \rightarrow \infty} P\left((\hat{\theta} - \theta_0)/\hat{se} \leq z\right), $$ which is not true?
Is this a mistake or am I missing something?
