Relation between different versions of the central limit theorem, i.e.$ X\sim N(0,1)$ and $X\sim N(0,\sigma^{2})$

130 Views Asked by At

I do not understand what is the difference between the following two versions of the central limit theorem:

  1. $S_{n}=\frac{\sum X_{i}-\mu}{\sigma\sqrt{n}}\stackrel{D}{\rightarrow}X\sim N(0,1)$

  2. $S_{n}=\frac{\sum X_{i}-\mu}{\sqrt{n}}\stackrel{D}{\rightarrow}X\sim N(0,\sigma^{2})$

Are they different theorems that just happen to look really similar? I think that one implies the other. Is there any other connection?

Thank you very much in advance

1

There are 1 best solutions below

0
On BEST ANSWER

They are the same thing. You can prove that they are equivalent using the continuous mapping theorem, but, as @Did said, that is really unnecessary, as you can also prove it easily by the definition:

Suppose that $S_{n}'=\frac{\sum X_{i}-\mu}{\sigma\sqrt{n}}\stackrel{D}{\rightarrow}Z\sim N(0,1)$, and $S_n =\frac{\sum X_{i}-\mu}{\sqrt{n}}$. We proceed by definition $$ P(S_n \leq t) = P(\sigma S_n' \leq t)=P(S_n' \leq \frac t\sigma) \rightarrow P(Z \leq \frac t\sigma) = P(\sigma Z\leq t) $$ and we know that $\sigma Z \sim N(0,\sigma^2)$. Hence, $S_{n}=\frac{\sum X_{i}-\mu}{\sqrt{n}}\stackrel{D}{\rightarrow}X\sim N(0,\sigma^{2})$.

The converse can be proven in the same manner.