Is there a reason why we are used to write the CLT as $\sqrt{n}(\overline{X}_n-\mu)\stackrel{d}{\rightarrow}N(0,\sigma^2)$ and not as $\overline{X}_n\stackrel{d}{\rightarrow}N(\mu, \frac{\sigma^2}{n})$?

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    $\begingroup$ Because second is not true. Read this: stats.stackexchange.com/a/194331/90473 $\endgroup$
    – Neeraj
    Commented Jan 12, 2019 at 9:21
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    $\begingroup$ Have a look at the Wikipedia article on CLT>Remarks>Proof of classical CLT. At the bottom you will find exactly this writing $\endgroup$ Commented Jan 12, 2019 at 9:30
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    $\begingroup$ Opinions will differ on this, but I would consider the latter statement a legitimate abuse of notation. It is not formally correct, but it is a more useful intuitive statement of the CLT, and the underlying meaning of the statement seems to me to be obvious. Others will object to this statement, but I find it useful. $\endgroup$
    – Ben
    Commented Jan 12, 2019 at 9:34
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    $\begingroup$ Simply because the term on the right hand side is the limit as $n$ goes to $\infty$ and therefore cannot depend on $n$. $\endgroup$
    – Xi'an
    Commented Jan 12, 2019 at 10:32

1 Answer 1


The notation $\stackrel{d}{\rightarrow}$ in the CLT is shorthand for the formal limit statement:

$$\lim_{n \rightarrow \infty} \mathbb{P} \Big( \sqrt{n} (\bar{X}_n - \mu) \leqslant t \Big) = \Phi(t | 0, \sigma^2).$$

You will notice that the formal statement is a limit statement in $n$ and so all of the values of $n$ are on the left-hand-side of the equation. On the right-hand-side this value does not appear, since the limit operation removes it. The latter statement in your question is generally considered an acceptable shorthand, but it is a slight abuse of notation with respect this formal limit statement, since $n$ appears in the right-hand-side of the limiting symbol. So the reason it is not commonly used is that, formally speaking, the limit of a function taken with respect to $n$ cannot itself be a function of $n$.

  • $\begingroup$ Would it be correct then to state that the latter statement is wrong as statement of CLT but nevertheless a true property of $\overline{X}_n$, for example using the continuos mapping theorem and the first CLT statement $\stackrel{d}{\rightarrow}N(0,\sigma^2)$ ? $\endgroup$ Commented Jan 12, 2019 at 9:49
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    $\begingroup$ No. It is formally a false statement, and therefore not a true property of $\bar{X}_n$. Arguably it may be interpreted as an abuse of notation for the above formal property. $\endgroup$
    – Ben
    Commented Jan 12, 2019 at 9:51

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