Two important properties of the maximum likelihood estimator (MLE) are functional invariance and asymptotic normality.
Functional invariance: If $\hat{\theta}$ is the MLE for $\theta$, and if $g(\theta)$ is any transformation of $\theta$, then the MLE for $\alpha = g(\theta)$ is $\hat{\alpha} = g(\hat{\theta})$.
Asymptotic normality: If $\hat{\theta}$ is the MLE for $\theta$ then $\sqrt{n}(\hat{\theta} - \theta) \xrightarrow[]{d} \mathcal{N}(0, I^{-1})$, where $I$ is the Fisher information matrix.
My question is what happens when $\alpha = 1/\theta$?
If $\hat{\theta}$ is the MLE for $\theta$, then by the functional invariance property $\hat{\alpha} = 1/\hat{\theta}$. The asymptotic normality property means that
- $\hat{\theta}$ is asymptotically normally distributed about $\theta$;
- $\hat{\alpha} = 1/\hat{\theta}$ is asymptotically normally distributed about $1/\theta$.
To me this looks to be a contradiction since the reciprocal of a normal random variable is not itself normally distributed.