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!
Best Answer
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).