How to arrive at well-know result of CLT from specific version of CLT?

Question

The version of the Central Limit Theorem I learned in class follows:

Given ${X}_{i},..., {X}_{n}$ iid RVs, such that $E[{X}_{1}] = 0 < \infty$ and $Var({X}_{1}) = {\sigma}^{2} < \infty,$ $\dfrac{{X}_{1} + ... +{X}_{n}}{\sqrt{n}}$ converges in distribution to $N(0, {\sigma}^{2})$ as n grows large.

Another common version of the CLT states that the sum of the ${X}_{i}$ is approximately distributed as $N(n\mu, n{\sigma}^{2})$ as n grows large (and a subsequent equivalence for the average).

I cannot figure out how the version I learned leads to the more common versions above. Any advice is appreciated.

Answer 1

The first can be rewritten to include a mean $\mu$ as

Given ${X}_{i},\ldots, {X}_{n}$ iid RVs, such that $E[{X}_{1}] = \mu$ (finite) and $Var({X}_{1}) = {\sigma}^{2}$ (finite), $\dfrac{({X}_{1}-\mu) + \cdots +({X}_{n} - \mu)}{\sqrt{n}}$ converges in distribution to $N(0, {\sigma}^{2})$ as n grows large.

This then suggests that for some $n$ large enough (how large will depend both on the distribution of $X_1$ and on how good you need the approximation to be)

• the distribution of $\dfrac{{X}_{1} + \cdots +{X}_{n} - n\mu}{\sqrt{n}}$ will be approximately that of $N(0, {\sigma}^{2})$
• the distribution of ${X}_{1} + \cdots +{X}_{n} - n\mu$ will be approximately that of $N(0, n{\sigma}^{2})$
• the distribution of ${X}_{1} + \cdots +{X}_{n}$ will be approximately that of $N(n\mu, n{\sigma}^{2})$
• the distribution of $\dfrac{{X}_{1} + \cdots +{X}_{n}}{n}$ will be approximately that of $N(\mu, \frac1n{\sigma}^{2})$

and the last two correspond to your second

the sum of the ${X}_{i}$ is approximately distributed as $N(n\mu, n{\sigma}^{2})$ as n grows large (and a subsequent equivalence for the average).