Weak Law of Large Numbers

Also known as Bernoulli's Theorem. Let , ..., be a sequence of independent and identically distributed random variables, each having a Mean $\left\langle{x_i}\right\rangle{} = \mu$ and Standard Deviation $\sigma$ . Define a new variable

$\begin{displaymath} x \equiv {x_1+\ldots +x_n\over n}. \end{displaymath}$

(1)

Then, as $n\to\infty$ , the sample mean $\left\langle{x}\right\rangle{}$ equals the population Mean $\mu$ of each variable.

$\begin{displaymath} \left\langle{x}\right\rangle{}=\left\langle{x_1+\ldots+x_n\o... ...+\ldots+\left\langle{x_n}\right\rangle{}) = {n\mu\over n}= \mu \end{displaymath}$

(2)

$\displaystyle \mathop{\rm var}\nolimits (x)$	$\textstyle =$	$\displaystyle \mathop{\rm var}\nolimits \left({x_1+\ldots +x_2\over n}\right)$
	$\textstyle =$	$\displaystyle \mathop{\rm var}\nolimits \left({x_1\over n}\right)+\ldots +\mathop{\rm var}\nolimits \left({x_n\over n}\right)$
	$\textstyle =$	$\displaystyle {\sigma^2\over n^2}+ \ldots + {\sigma^2\over n^2}= {\sigma^2\over n}.$	(3)

Therefore, by the Chebyshev Inequality, for all $\epsilon > 0$ ,

$\begin{displaymath} P(\vert x-\mu \vert \geq \epsilon) \leq {\mathop{\rm var}\nolimits (x)\over \epsilon^2}= {\sigma^2\over n\epsilon^2}. \end{displaymath}$

(4)

As $n\to\infty$ , it then follows that

$\begin{displaymath} \lim_{n\to \infty}P(\vert x-\mu \vert \geq \epsilon ) = 0 \end{displaymath}$

(5)

for $\epsilon$ arbitrarily small; i.e., as $n\to\infty$ , the sample Mean is the same as the population Mean.

Stated another way, if an event occurs times in Trials and if is the probability of success in a single Trial, then the probability that $\vert x/s-P\vert<\epsilon$ for $\epsilon$ an arbitrary Positive quantity approaches 1 as $s\to\infty$ .