Does the Central Limit Theorem concern the sum or the average of iid random variables?

Question

I have seen the central limit theorem explained in two ways:

1. If $X_{1}, \dots , X_{n}$ are i.i.d. random variables, then their sum $Y$, when standardized, will converge in distribution to the standard normal distribution as $n$ goes to infinity.
2. If $X_{1}, \dots , X_{n}$ are i.i.d. random variables, then their arithmetic mean $\overline{Y}$, when standardized, will converge in distribution to the standard normal distribution as $n$ goes to infinity.

Which is the correct interpretation, as the sum of idd rvs or as the arithmetic mean? Or are they both variants of some overarching CLT?

Answer 1

Both are true.

The key word here is 'standardized'. Both $Y$ and $\bar{Y}$ are standardized using different means and variances so that the limiting distribution is identical.

ie. Suppose $X_i$ are i.i.d with mean $\mu$ and variance $\sigma^2$.

Then it follows that $E[Y] = E[\sum X_i] = \sum E[X_i] = n\mu$

Likewise, $E[\bar{Y}] = \frac{1}{n}E[Y] = \mu$

Doing the same with variance and using $Var(cX) = c^2 Var(X)$ it is straightforward to see that:

$Var(Y) = n\sigma^2$ and $Var(\bar{Y}) = \frac{\sigma^2}{n}$

Hence when we standardize we get:

$$Y: Z = \frac{Y - n\mu}{\sqrt n\sigma} = \frac{Y/\sqrt n-\sqrt n \mu}{\sigma}$$

$$\bar{Y}: Z = \frac{\bar{Y} - \mu}{\sigma/\sqrt n} = \frac{Y/n - \mu}{\sigma/\sqrt n} = \frac{Y/\sqrt n - \sqrt n\mu}{\sigma}$$

So that standardizing either gives the same result, $Z$.