# Confusion on the central limit theorem.

I have a confusion between two statements of the CLT.

1) Let $X_1,...$ be iid random variables with mean $\mu$ and finite variance $\sigma^2$ then $$\frac{X_1+X_2+...+X_n-\mu \color{red}{n}}{\sigma \sqrt n} \implies N(0,1)$$ where $\implies$ means convergence in distribution.

2) If $X_1,...$ are iid random variables with mean $\mu$ and finite variance $\sigma^2$ then $\bar{X}$ is approximately $N(\mu,\sigma^2)$ for large $n$. (Usually $n>30$).

From 2) We have for large $n$ $$\frac{X_1+X_2+...+X_n}{n}$$ is distributed $N(\mu, \sigma^2)$ then normalising gives

$$\frac{X_1+X_2+...+X_n-\mu}{n\times(\sigma/\sqrt n)}=\frac{X_1+X_2+...+X_n-\mu \color{red}{(1)}}{\sigma \sqrt n}$$ is approx $N(0,1)$.

There is a $n$ missing from the expression of 1) in what I get for 2) (indicated with a red $n$ and then a red $1$ where the $n$ should be).

What is going on here.

• In $2)$ $\bar{X_n}$ actually have $\mathcal{N}(\mu, \frac{\sigma^2}{n})$ distribution – Mesmerized student May 1 '17 at 12:06
• @Mesmerizedstudent No, the distribution of $\bar X$ is not normal in general. – Did May 1 '17 at 12:31
• @Did yes, you are right of course. – Mesmerized student May 1 '17 at 12:49

I'll write $Y_n$ for $(X_1+\cdots+X_n)/n$. Its mean is $\mu$ and its variance is $\sigma^2/n$. Its normalisation is $$\frac{Y_n-\mu}{\sigma/\sqrt n}=\frac{nY_n-n\mu}{\sqrt n\,\sigma}=\frac{X_1+\cdots+X_n-n\mu}{\sqrt n\,\sigma}.$$
The first expression in your question is evidently equivalent with:$$\frac{\overline{X}-\mu}{\sigma/\sqrt n}\implies N(0,1)$$ You will arrive there by dividing numerator and denominator of LHS by $n$.
On LHS we recognize the standardizing of $\overline{X}$
Btw, note that $\text{Var}\overline{X}=\sigma^2/n$ so that it is incorrect to state that the distribution of $\overline{X}$ is approximately $N(\mu,\sigma^2)$.