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?

  • 4
    $\begingroup$ Explicitly, the central limit theorem does not concern convergence of the sample mean. The central limit theorem says what it says it says: the $Z$ random variable is asymptotically standard normal. $\endgroup$
    – Dave
    Dec 22, 2021 at 18:08
  • $\begingroup$ Thanks @Dave "the Z random variable is asymptotically standard normal" ... noted. $\endgroup$ Dec 23, 2021 at 6:05

2 Answers 2


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.

  • $\begingroup$ While the CLT is explicitly about Z it is implicitly about $\bar{X}$ or $\sum X_i$. The theorem is central to statistics that deals with repeated experiments or combinations of many (small) sources of error. The point of the theorem (the history of the theorem and why it has been developed) is that such statistics have a normal distribution as limit, bit more importantly can often be approximated with a normal distribution. The explicit theorem is never strictly relevant (the limit is never truly achieved, there is no infinity). $\endgroup$ Dec 26, 2021 at 18:40

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.

  • $\begingroup$ I am not Dave, but I think you might have gotten him wrong. The point is that CLT holds for $Z$ regardless of how you represent it (as a sum, as an average or what not). $\endgroup$ Dec 22, 2021 at 18:36
  • $\begingroup$ Yes. But to the best of my knowledge, which may well be flawed, $Z$ is almost always referred to as the standardized sum of IID variables in the classic case, and the standardized sums of i but not necessarily id variables in the more exotic Lyapunov/Lipschutz formulations. $\endgroup$
    – Avraham
    Dec 22, 2021 at 18:38
  • 1
    $\begingroup$ @Richard The CLT does not apply to the sum. The sum almost surely does not converge. $\endgroup$
    – whuber
    Dec 22, 2021 at 18:47
  • 1
    $\begingroup$ @whuber, I agree with the essence. My point was that equivalent representations of the same object do not behave differently. $\endgroup$ Dec 22, 2021 at 21:12
  • $\begingroup$ @Richard I agree. This question is merely one of how to do arithmetic with fractions. It isn't a statistical issue. $\endgroup$
    – whuber
    Dec 22, 2021 at 21:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.