Asymptotic property of estimators I'm studying the asymptotic properties of estimators.
Let $\{ \hat{\theta}_T : T=1,2,3... \}$ be a sequence of estimators of the $p \times1$ vector $\theta \in \Theta $, and $T$ is the sample size. Then, by assuming that my consistent estimator has the following asymptotic distribution:
$\sqrt{T}(\hat{\theta}_T-\theta) \sim N(0,V)$
where $V$ is a $p \times p $ positive semidefinite matrix.
Then, we can assume $\hat{\theta}_T \sim N(\theta,V/T)$
I understand we have the factor $\sqrt{T}$ in the asymptotic distribution of the consistent estimator $\hat{\theta}$ because $(\hat{\theta}_T-\theta)$ has a degenerate distribution.
Can you show me why $(\hat{\theta}_T-\theta)$ has a degenerate distribution? That's what Marno Verbeek is saying


First of all, instead of stating "$\sqrt{T}(\hat{\theta}_T - \theta) \color{red}{\sim} N(0, V)$" for asymptotic normality of the estimator sequence $\{\hat{\theta}_T\}$, the correct notation should be $\sqrt{T}(\hat{\theta}_T - \theta) \color{red}{\to_d} N(0, V)$. In words, this means $\sqrt{T}(\hat{\theta}_T - \theta)$ converges to $N(0, V)$ in distribution (or in law/weakly). The "$\sim$" notation stands for the distributional form following it is exact, but not asymptotically.  By the same token, you should not say "$\hat{\theta}_T - \theta$" has a degenerate distribution" (I see your reference, which is not that accurate either and probably misled you), a more precise statement would be "$\hat{\theta}_T - \theta$ converges in distribution to a degenerate distribution as $T \to \infty$".
That said, what you are interested is an immediate consequence of the Slutsky's theorem.  Write
\begin{align}
\hat{\theta}_T - \theta = \frac{1}{\sqrt{T}} \times \sqrt{T}(\hat{\theta}_T - \theta),
\end{align}
$1/\sqrt{T} \to 0$ and $\sqrt{T}(\hat{\theta}_T - \theta) \to_d N(0, V)$ together imply that their product converges in distribution to the product of $0$ and $N(0, V)$, which is $0$.
