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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.

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    $\begingroup$ Work out the variance. $\endgroup$ – J.G. Dec 2 '18 at 7:49
  • $\begingroup$ @J.G. but it should be just one version that is true.. $\endgroup$ – Arief Anbiya Dec 2 '18 at 7:57
  • $\begingroup$ 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). $\endgroup$ – Did Dec 2 '18 at 9:25
  • $\begingroup$ 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. $\endgroup$ – obscurans Dec 2 '18 at 23:59
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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.

Version 1 is correct, version 2 is wrong.

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