Which variance estimate to use for a Wald test?

Question

I have seen the following justification for the Wald test of the null hypothesis $H_0: \theta = \theta_0$ for a scalar parameter $\theta$. When $\hat{\theta}_n$ is the MLE for $\theta$ estimated from an independent sample of size $n$, under the null hypothesis we have $\sqrt{n}\left(\hat{\theta}_n - \theta_0\right) \rightarrow N\left(0, \dfrac{1}{i(\theta_0)}\right)$ in distribution as $n\rightarrow \infty$, where $i(\theta_0)$ is the expected information for a single observation, evaluated at $\theta_0$. So it seems to me that we should use the test statistic

$ \dfrac{\sqrt{n}\left(\hat{\theta}_n - \theta_0\right)}{\sqrt{\dfrac{1}{i(\theta_0)}}}$

which will be approximately $N(0,1)$ for large $n$. However, it seems to be more common to write the Wald statistic as

$ \dfrac{\sqrt{n}\left(\hat{\theta}_n - \theta_0\right)}{\sqrt{\dfrac{1}{i\left(\hat{\theta}\right)}}},$

i.e., to evaluate the expected information at $\hat{\theta}$ rather than at $\theta_0$. My question is, considering that we need the distribution of the test statistic under the null to perform our hypothesis test, doesn't it make more sense to try and estimate the standard error under the null, i.e., to estimate $s.e.\left(\hat{\theta}\right)$ by $\sqrt{\dfrac{1}{i\left(\theta_0 \right)}}$?

Answer 1

Either approach is legitimate, with both leading to the same asymptotic null distribution of the statistic.

$\sqrt{n}(\hat{\theta}_n - \theta_0) \rightarrow_d N(0, i(\theta_0)^{-1})$ implies that $\hat{\theta}_n \to_p \theta_0$ so that the continuous mapping theorem (CMT) yields that $i(\hat{\theta}_n) \to_p i(\theta_0)$, provided, as is the case in regular problems, that $i$ is continuous. Then, again by the CMT, $$ \sqrt{\dfrac{1}{i(\hat{\theta}_n)}}\to_p\sqrt{\dfrac{1}{i(\theta_0 )}} $$ and Slutzky's theorem yields that $$ \dfrac{\sqrt{n}(\hat{\theta}_n - \theta_0)}{\sqrt{\dfrac{1}{i(\hat{\theta})}}}\to_dN(0,1)$$ under $H_0$ as well.