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The Wikipedia entry on the CLT states at one point: "For fixed large $n$ one can also say that the distribution of $S_n$ is close to the normal distribution with mean $\mu$ and variance $\frac1n\sigma^2$." $S_n = (\sum_{i=1}^nX_n) / n$ and $X_i$ are iids with mean $\mu$ and variance $\sigma^2$. I don't quite see how this follows from the other, more formal definitions. Is this statement true and what is the source/proof? |
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You seem to be asking about how the rescaling works. It utilizes the fact that linear transformations of a normal random variable are normal: if $$X\sim {\rm N}(m,s^2)$$ and $Y=\frac{X}{b}+a$ then $$Y\sim {\rm N}\Big(\frac{m}{b}+a,\frac{s^2}{b^2}\Big).$$ Now, if $$Z=\sqrt{n}\Big(\frac{1}{n}\sum_{i=1}^nX_n-\mu\Big)\approx {\rm N}(0,\sigma^2)$$ then consequently $$S_n=\frac{1}{n}\sum_{i=1}^nX_n=\frac{1}{\sqrt{n}}Z+\mu\approx{\rm N}\Big(\mu,\frac{\sigma^2}{n}\Big).$$ This corresponds to $m=0$, $s=\sigma$, $a=\mu$ and $b=\sqrt{n}$ in the general formulae above. |
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That statement seems to be inaccurate or misleading. When you have a limit, you don't automatically know that you are actually close for any particular fixed $n$. An additional result like that is called effective. One effective version of the Central Limit Theorem is the Berry-Esseen theorem, which bounds the difference between the CDF of $S_n$ and $\Phi$ in terms of the third central moment of $X$ and the standard deviation of $X$. To use this, you need the third central moment to exist, and you need a bound on it. The Central Limit Theorem does not require the third central moment to exist, though, so this does not work every time the Central Limit Theorem applies. In addition, since you don't have a uniform bound on the third central moments even when they exist, you can't say that $S_n$ resembles a normal distribution well. More precisely, if you choose $n$, I can pick a distribution for $X$ so that $S_n$ is far from normal, so that the convergence happens later. Let $X$ be $0$ with high probability ($\frac{n^2-1}{n^2}$), with equal small ($\frac{1}{2n^2}$) chances to be $n$ and $-n.$ This has mean $0$ and variance $1$, but the probability is over $\frac{n-1}{n}$ that $S_n = 0$. |
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