# How to convert the Central Limit Theorem

from the form $P \left\{ \frac{S_n - n\mu}{\sigma\sqrt{n}} < \beta \right\} \to \mathfrak{N}(\beta)$ to the form $P \{ |S_n - n\mu| < \beta \sigma \sqrt{n} \} \approx \mathfrak{N(\beta)} - \mathfrak{N(-\beta)}$?

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Accepting an answer seems in order here. – Did Oct 14 '12 at 10:59

The central limit theorem says that for every real number $\beta$, $$\lim_{n\to\infty}\Pr\left(\frac{S_n - n\mu}{\sigma\sqrt{n}} \le \beta \right) = \Phi(\beta),$$ where $S_n$ is the mean of an i.i.d. sample of size $n$ from a population with mean $\mu$ and finite variance $\sigma^2$. Now consider $$\Pr\left(-\beta\le\frac{S_n - n\mu}{\sigma\sqrt{n}} \le \beta \right).$$ Recall that if $A,B$ are mutually exclusive events, then $$\Pr(A\text{ or }B)=\Pr(A)+\Pr(B).\tag{1}$$ Apply this to the case where \begin{align} A & = \left[\frac{S_n - n\mu}{\sigma\sqrt{n}} \le -\beta\right], \\[10pt] B & = \left[-\beta\le\frac{S_n - n\mu}{\sigma\sqrt{n}} \le -\beta\right], \end{align} So that $$[A\text{ or }B] =\left[\frac{S_n - n\mu}{\sigma\sqrt{n}}\le\beta\right].$$ Then the event $[A\text{ or }B]$ is $\displaystyle\left[\frac{S_n - n\mu}{\sigma\sqrt{n}} \le \beta\right]$. Now apply $(1)$.

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I am accustomed to using $\Phi$ for the c.d.f. of the standard normal distribution, and $N$ or $\mathcal{N}$ or the like to refer to the distribution itself, so that, for example, $X\sim N(\mu,\sigma^2)$ means $X$ has a normal distribution with expectation $\mu$ and variance $\sigma^2$. – Michael Hardy Sep 14 '12 at 4:46
Nice solution. I was using $\sqrt{x^2}$ to construct the absolute value. And didn't realise union of interval could solve it too. Thank you. – RHS Sep 15 '12 at 0:12

Notice that $P(|f|<b)=P(0\leq f<b)+P(-b<f<0)$

so,

$P\{|S_n-n\mu|\leq \beta\sigma\sqrt{n}\}=N(\beta)-N(0)-N(-\beta)+N(0)=N(\beta)-N(-\beta)$

where we used the fact that $N(x)$ is continuous in $x$ to resolve the $\leq,<$ part.

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