# A question about the delta method in asymptotic distributions

I am reading up on the delta method from its Wikipedia page. Under the heading Univariate delta method the statement of the method is as follows:

If $$\sqrt{n}[X_n - \theta]\xrightarrow{\text{D}} \mathcal{N}(0,\sigma^2)$$ where $$X_n$$ is a sequence of random variables where $$\theta$$ and $$\sigma$$ are finite valued constants and $$\xrightarrow{D}$$ denotes convergence in distribution, then: $$\sqrt{n}[g(X_n) - g(\theta)]\xrightarrow{\text{D}} \mathcal{N}(0,[g^{'}(\theta)^2]\sigma^2)$$ Later on to prove this they ask us to note that $$X_n \xrightarrow{P}\theta$$ where $$\xrightarrow{P}$$ denotes convergence in probability

What justifies this claim? This seems to be sort of like a reverse central limit theorem and I feel like something very basic is in my blindspot. I need your help in figuring it out.

By Slutsky's theorem, since $$n^{-1/2} \to 0$$ and $$\sqrt{n} (X_n - \theta) \stackrel{D}{\to} N(0, \sigma^2)$$, we have $$\begin{equation*} n^{-1/2} \left\{ \sqrt{n} (X_n - \theta) \right\} \stackrel{D}{\to} 0 \times N(0, \sigma^2). \end{equation*}$$ Since this congergence in distribution to a constant implies convergence in probability to the same constant and since $$X_n = n^{-1/2} \left\{ \sqrt{n} (X_n - \theta) \right\} + \theta$$, $$\begin{equation*} X_n \stackrel{p}{\to} 0 + \theta = \theta. \end{equation*}$$
• Also worth noting that $X_n \xrightarrow{P}\theta$ means that $X_n$ is a (weakly) consistent estimator for $\theta$.