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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 $n\to\infty$. –  Michael Hardy Jul 4 '12 at 4:23

1 Answer 1

That the population mean $m=\frac{1}{n}\sum_i X_i$ "decays at a rate $1/\sqrt{n}$" does not make much sense to me, what "decays" is the difference $d=m - \mu$. A little more formally, in the context of CLT, the difference converges to zero in quadratic sense: its variance tends to zero, and hence also its standard deviation. And it's staightforward to show that the standard deviation tends to zero as $1/\sqrt{n}$ (which justifies the first sentence of your quote).

$$\sigma^2_d=\sigma^2_m= \frac{1}{n^2} \sum_i \sigma^2_{x_i}= \frac{\sigma_x^2}{n}$$

$$ \sigma_d = \frac{\sigma_x}{\sqrt{n}}$$ Then, we multiply $d$ by $\sqrt{n}$ so that its variance does not vanish (sort of a normalization).

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