# Central Limit Theorem different versions

Which one is correct about the Central Limit Theorem:

1. If $$X_{1}, ... , X_{n}$$ are i.i.d continuous random variables with mean $$\mu$$ and variance $$\sigma^2$$, then as $$n \rightarrow \infty$$, $$\sqrt{n} \frac{\bar{X} - \mu}{\sigma}$$ will have the standard normal distribution

or

1. If $$X_{1}, ... , X_{n}$$ are i.i.d continuous random variables with mean $$\mu$$ and variance $$\sigma^2$$, then as $$n \rightarrow \infty$$, $$\sqrt{n} \frac{\bar{X} - \mu}{\sigma}$$ will have normal distribution $$N(0, \sigma^2)$$

According to a book: "Introduction to Mathematical Statistics", by Hogg and Craig, the first one is true. But I also see from other sources that the second one is true.

Thanks.

• Work out the variance. – J.G. Dec 2 '18 at 7:49
• @J.G. but it should be just one version that is true.. – Arief Anbiya Dec 2 '18 at 7:57
• Both versions are incorrect (for example, if the distribution of every $X_i$ is discrete, then none of the random variables $\sqrt n(\bar X_n-\mu)/\sigma$ is normally distributed). – Did Dec 2 '18 at 9:25
• The above statement is itself incorrect. The conditions of this version of CLT is simply that $\mathrm{var}(X)<\infty$. CLT claims nothing about finite $\sqrt{n}(\bar{X}_n-\mu)/\sigma$ (you need the Berry-Esseen theorem for that), it simply says this sequence converges in distribution to a normal. – obscurans Dec 2 '18 at 23:59

Since $$X_i$$ are iid, we know that $$\mathrm{var}\left(\sqrt{n}\frac{\bar{X}-\mu}{\sigma}\right)=\frac{n}{\sigma^2}\mathrm{var}\left(\bar{X}-\mu\right)=\frac{n}{\sigma^2}\mathrm{var}\left(\frac{1}{n}\sum_{i=1}^nX_i\right)=\frac{n}{\sigma^2}\frac{1}{n^2}(n\sigma^2)=1$$ so the limit is a standard normal.