Central Limit Theorem and the Sample Sum This 27.1 The Theorem lists two equations for Z:
$$
Z=\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt n}}\\
Z = \frac{\sum_{i=1}^{n} X_i-n\mu}{\sqrt{n}\sigma}
$$
Is it correct to say that the second equation above is equivalent to the first but is simply expressed in terms of the sample sum rather than the sample mean, so every term is multiplied by $n$?
In other words, does what the Central Limit Theorem says about the sample mean also apply to the sample sum?
 A: You are correct that you can express $Z$ either of the ways that you wrote; that’s what it means to be “equal”. However, you have a misunderstanding about the central limit theorem, which explicitly concerns $Z$, not $\bar X$.
People often like thinking of the sample mean converging so that the following holds asymptotically, as it is an algebraic rearrangement of the central limit theorem.
$$
\bar X_n \sim N(\mu, \sigma^2/n)
$$
Such a notion is problematic, as it makes the convergence target a moving target, since $n$, therefore the variance, changes as the sample size increases. Further, if we have a distribution with bounded support, you are proposing that its sample mean could be off of the support, such as $\bar X=-1$ for an exponential distribution that does not give negative values. (A normal distribution has support on the entire real line.)
Further, the sum does not have to do anything close to converge. Consider $X_1,\dots,X_n\overset{iid}{\sim} U(1,2)$. That uniform distribution meets the assumptions of the classical central limit theorem. However, a sum of those values is going to diverge off to infinity, something like $1+1.1+1.7+1.2+\dots$ At least dividing by the sample size allows the mean to be controlled and kept from exploding off to infinity.
A: Yes, you are correct. Consider:
$$
\begin{aligned}
Z &= \frac{\sum_{i=1}^{n} X_i-n\mu}{\sqrt{n}\sigma}\\
&=\frac{n\bar{X}-n\mu}{n\frac{1}{\sqrt n}\sigma}\\
&=\frac{n}{n}\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt n}}\\
&=\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt n}}
\end{aligned}
$$
Although, as Dave points out, the CLT technically relates to a standardized or normalized sum of random variables and not the sample mean.
