What, if any, asymptotic arguments are used in moving between the various statements of the central limit theorem (e.g. in terms of sample means compared with standardised sample means)?


My question is somewhat trivial, and perhaps pedantic, but I would appreciate some concrete insight on it in any case.

In terms of the asymptotic normality of sample means, one version of the central limit theorem is

$$\overline{X}_n \overset{d}{\longrightarrow} \mathcal{N} \left(\mu, \frac{\sigma^2}{n} \right) \tag{1}$$

for i.i.d. $X_1, \dots, X_n$ with mean $\mu$ and variance $\sigma^2$. An equivalent formulation in terms of standardised sample means might be

$$\frac{ \sqrt{n}(\overline{X}_n - \mu)}{\sigma} \overset{d}{\longrightarrow} \mathcal{N}(0,1). \tag{2}$$

Most intermediate statistics references I have seen prove $(2)$ using some assumptions about the moment generating function, and Taylor approximations. I have noticed that pedagogically, they either present:

  • $(1)$ and $(2)$ as equivalent formulations of the same theorem.
  • Or they prove $(2)$ and assume that in asymptopia, we can move from $(2)$ to $(1)$. No further, explicit justification for how we move between $(2)$ and $(1)$ is presented, possibly because it is so fundamental and "standard".

In the latter case, could we view going from $(2)$ to $(1)$ as an instance of the continuous mapping theorem?


1 Answer 1


When you take a limit $n \to\infty$ then $n$ cannot occur on the right hand side, the limit expression cannot depend on $n$. So your $(1)$ is incorrect, while $(2)$ is the correct expression.

The way $(1)$ arises (but not as a limit, a an approximation) is that you decide that your $n$ is large enough to trust the CLT as an approximation. So you write $$ \frac{ \sqrt{n}(\overline{X}_n - \mu)}{\sigma} \overset{d}{\approx} \mathcal{N}(0,1). \tag{2} $$ where we have replaced $\overset{d}{\longrightarrow}$ by $\overset{d}{\approx} $, meaning approximately distributed as. Multiplying by $\sigma$ and dividing by $\sqrt{n}$ we then get $$ \overline{X}_n \overset{d}{\approx} \mathcal{N}(\mu,\frac{\sigma^2}n) $$ This does not involve any use of the continuous mapping theorem. As a fact, from the point that we decide to use the CLT result as an approximation, there is no asymptotics involved.

  • 1
    $\begingroup$ +1. Thank you for assisting in clarifying some of my erroneous preconceptions. $\endgroup$
    – microhaus
    Commented Jun 20, 2021 at 2:54

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