I understand that in CLT, we have

$\sqrt{N} \left( \frac{\bar{X}_N - \mu}{\sigma} \right) \overset{d}\to N\left(0,1\right) \qquad (*)$

where $\bar{X}_N := \frac{1}{N} \sum_{i=1}^N X_i$ is the sample mean, and $\sqrt{N}$ inflates the $\frac{\bar{X}_N - \mu}{\sigma}$ to a $\mathcal{O}_p(1)$, while without the $\sqrt{N}$, $\frac{\bar{X}_N - \mu}{\sigma} \overset{p}\to 0$.

$\textbf{My Question is:}$ since $\sqrt{N} \left( \frac{\bar{X}_N - \mu}{\sigma} \right) = \frac{\bar{X}_N - \mu}{\frac{\sigma}{\sqrt{N}}} = \frac{\bar{X}_N - \mathbb{E}\left[\bar{X}_N\right]}{SD\left(\bar{X}_N\right)}$, is it correct to view CLT as $\frac{\bar{X}_N - \mathbb{E}\left[\bar{X}_N\right]}{SD\left(\bar{X}_N\right)} \overset{d}\to N\left(0,1\right)$? If yes, are we considering the $\sqrt{N}$ from expression $(*)$ as a part of the standard deviation of the sample mean, as opposed to an inflater, and does that mean as long as we can show that $SD\left(\bar{X}_N\right) \in \mathcal{O}_p(1)$, we do not need to consider about the rate of inflation (e.g. $\sqrt{N}$) separately?

  • $\begingroup$ What do you mean by $ SD\left(\bar{X}_N\right)$? $\endgroup$
    – Viktor
    Commented Dec 5, 2017 at 15:27

1 Answer 1


Formally speaking, the CLT is in principle, indeed concerned with the limiting behavior of standardized sums of random variables, say $S_n=(1/n)\sum X_i$,

$$\frac{S_n - \mathbb{E}\left[S_n\right]}{\sqrt{\text{Var}(S_n)}}$$

So indeed the $\sqrt n$ term comes from the standard deviation of the sum.

Still, thinking of it as an "inflater", as the OP says, is not without conceptual merits (although it is a "derived" inflater, not one we determine separately).


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