Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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 $(1)$.

share|cite|improve this answer
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 $P(|f|<b)=P(0\leq f<b)+P(-b<f<0)$


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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.