Functional Invariance of the MLE

Question

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

1. $\hat{\theta}$ is asymptotically normally distributed about $\theta$;
2. $\hat{\alpha} = 1/\hat{\theta}$ 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.

Answer 1

It's possible situation because of asymptotics.

When the $\hat \theta$ is close enough to $\theta$, $\hat \alpha= \frac{1}{\hat \theta}= \frac{1}{\theta} \left(1+\frac{\hat \theta -\theta}{\theta} \right)^{-1} \simeq \frac{1}{\theta} \left(1-\frac{\hat \theta -\theta}{\theta} \right)= \frac{1}{\theta}-\frac{1}{\theta^2}(\hat \theta -\theta)=\alpha-\frac{1}{\theta^2}(\hat \theta -\theta)$

$\hat \alpha -\alpha\simeq -\frac{1}{\theta^2}(\hat \theta -\theta)$

I guess this equation doesn't look to be a contradiction.

And actually $-\frac{1}{\theta^2}$ is $g'(\theta)$. This result can be generally applied.

I am not a native English speaker and also a self-study learner. I hope this answer is helpful.