By definition, point estimator $\hat\theta_n(\mathbf{X})$ is asymptotically normal if $$\sqrt{n}(\hat\theta_n - \theta) \, \overset{d}{\longrightarrow} \, \mathcal{N}(0, \sigma^2(\theta)), \qquad \forall \theta \in \Theta$$ where $\sigma^2(\theta)$ is called asymptotic variance of $\hat\theta_n$.

Then the approximated distribution of the asymptotically normal estimator is the following: $$\hat\theta_n \, \overset{\mathrm{appr}}{\sim} \, \mathcal{N}\left(\theta, \frac{\sigma^2(\theta)}{n}\right). \qquad ~~~~ (1)$$

That's ok but in some sources I saw the following expression: $$\hat\theta_n \, \overset{\mathrm{appr}}{\sim} \, \mathcal{N}\left(\hat\theta_n, \frac{\sigma^2(\hat\theta_n)}{n}\right).\qquad (2)$$

This is confusing me because the parameters of a distribution must be constants (assuming frequentist inference).
So should I interpret expression (2) in frequentist statistics as $$\hat\theta_n(\mathbf{X}) \, \overset{\mathrm{appr}}{\sim} \, \mathcal{N}\left(\hat\theta_n(\mathbf{x}), \frac{\sigma^2(\hat\theta_n(\mathbf{x}))}{n}\right).\qquad (3)$$ where $\mathbf{x}$ is some fixed realization of a sample $\mathbf{X}$ ??

  • $\begingroup$ $\frac{1}{n}\sigma^2(\theta)$ is the large sample variance of $\hat\theta_n$; $\frac{1}{n}\sigma^2(\hat\theta_n)$ is the large sample estimated variance of $\hat\theta_n$. $\endgroup$ – StubbornAtom Mar 19 at 4:39

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