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In central limit theorems, we have the following conclusion $$ \sqrt{n}\bigg(\bigg(\frac{1}{n}\sum_{i=1}^n X_i\bigg) - \mu\bigg)\ \xrightarrow{d}\ \mathcal{N}(0,\;\sigma^2). $$

Some book says

Informally, it implies that the probabilistic rate at which $\frac{1}{n}\sum_{i=1}^n X_i$ approaches $\mu$ is $1/\sqrt{n}$. That is, $\frac{1}{n}\sum_{i=1}^n X_i$ must decay at a rate $1/\sqrt{n}$ to balance the $\sqrt{n}$ "blow-up" factor and yield a well-behaved random vector with the distribution $\mathcal{N}(0,\;\sigma^2)$ (i.e., well-behaved in the sense of being neither degenerate $0$ nor $\infty$ in magnitude).

I was wondering how the rate of convergence is formally stated in this probabilistic setting?

A related but more general question was asked a while ago.

Thanks and regards!

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    I think what the author may mean is that the scale parameter (standard deviation or interquartile range or whatever---take your pick) is $\mathbb{O}(\sqrt{n})$ as $n\to\infty$.2012-07-04

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