from the form $P \left\{ \frac{S_n - n\mu}{\sigma\sqrt{n}} < \beta \right\} \to \mathfrak{N}(\beta)$ to the form $P \{ |S_n - n\mu| < \beta \sigma \sqrt{n} \} \approx \mathfrak{N(\beta)} - \mathfrak{N(-\beta)}$?
How to convert the Central Limit Theorem
2 Answers
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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0Nice 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. – 2012-09-15
Notice that $P(|f|
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.