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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.

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This basically follows directly from Slutsky's theorem. Below are the details.

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*}

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  • $\begingroup$ Also worth noting that $X_n \xrightarrow{P}\theta$ means that $X_n$ is a (weakly) consistent estimator for $\theta$. $\endgroup$
    – Ben
    Commented Jul 31, 2022 at 21:53
  • $\begingroup$ Wonderful! Thank you. Beats me why they'd assume it to be obvious and not even hint at the proof. $\endgroup$ Commented Aug 1, 2022 at 4:05

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