I never studied statistics and read that The Strong Law of Large Numbers is a famous theorem. Could someone explain it to me without any concepts of statistics for I have never studied them.

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- Nov 12th 2005, 06:13 PMThePerfectHackerThe Strong Law of Large Numbers
I never studied statistics and read that The Strong Law of Large Numbers is a famous theorem. Could someone explain it to me without any concepts of statistics for I have never studied them.

- Jan 16th 2006, 09:55 AMearboth
Hello,

we've got something in common: I never studied statistics too. Therefore my explanations are probably(!) a little bit simple. A further remark on the following text: I live in Germany and think and write in German; to write to you I've to translate my text into English. And I know that there are missing a lot of specific words. So sorry.

But maybe you get a hint where you can start to study.

Empiric (Statistical) Law of Large Number

If you have independent experiments and you count the desired outcome A exactly a times, than the relative frequency is $\displaystyle f(A)=\frac{a}{n}$.

If you repeat the experiments under exactly the same conditions, you'll find, that the relative frequency will stabilize(?) in a narrowing interval. The relative frequency could be considered as a estimated value of p(A) (the probability of A)

The Weak Law of large Numbers (Bernoulli's Law)

If you have a Bernoulli-chain(?) (a Bernoulli-experiment is repeated very often, this is called in German a chain) with the probability p for the desired outcome A and the relative frequency of the observed outcomes A $\displaystyle f(A)=\frac{a}{n}$.

Then

$\displaystyle P \left( l {(f(A)-p } | \geq \epsilon \right)\leq \frac{p\cdot (1-p)}{n\cdot \epsilon^2}$

where ε is the relative error (aberration?).

The Strong Law of large Numbers (Borel's Law)

$\displaystyle P \left( \lim _{ \subc{n\rightarrow\infty}}{f(A)}=p \right)=1$

where p is the value of the probability of the desired outcome A.

The inequality above can be transformed into

$\displaystyle P \left( \ |{(f(A)-p } | \geq \epsilon \right)\geq 1 - \frac{p\cdot (1-p)}{n\cdot \epsilon^2}$

let $\displaystyle \beta:= 1 - \frac{p\cdot (1-p)}{n\cdot \epsilon^2}$

that means:

$\displaystyle \epsilon = \sqrt{\frac{p(1-p)}{(1-\beta)n}}$

put into the inequality

$\displaystyle P \left( |{(f(A)-p } | \geq \sqrt{\frac{p(1-p)}{(1-\beta)n}} \right)\geq \beta $

For a fix β and a fix n you'll get for every p an interval

$\displaystyle I_p := \left[ p - \sqrt{\frac{p(1-p)}{(1-\beta)n}} ; p + \sqrt{\frac{p(1-p)}{(1-\beta)n}} \right]$

where you can find with the probability of at least ß the value of f(A).

If you consider the pairs (p, f(A)) as coordinates of points you'll get an ellipse. I've attached below such an ellipse for ß = .95 and n = 125.

Because

$\displaystyle \max_ {\subc{0 \leq p \leq 1}}{p(1-p)}=\frac{1}{4}$

the inequality could be transformed into:

$\displaystyle P\left( \ | {(f(A)-p } | \geq \epsilon \right)\leq \frac{1}{4 \cdot n\cdot \epsilon^2}$

Notice that the RHS of the inequality is independent from p.

Let for instance be

$\displaystyle \frac{1}{4 \cdot n\cdot \epsilon^2}= 1 \%$

This equation describes the probability (by a given n) with which the relative frequency f(A) will be found in the interval

$\displaystyle \left[ p - \epsilon; p+\epsilon \right]$

In the example:

n: 10 1000 10000 100000

ε: .5 .158 .05 .016

You'll get a so called 99%-funnel. I've attached below the graph of this funnel.

I hope this text could help a little bit further on.

Greetings! - Jan 17th 2006, 08:50 PMearboth
Hello,

here I'm again.

In additon to my posting: If you've got the opportunity to get hold of:

Marek Fisz, Probability Theory and Mathematical Statistic

then catch it!

It was published 1963 (I believe) in Poland and was re-edited in 2003(?) by Krieger Publishing. It's the very best book about those topics(?) (items (?)) I've ever read: many examples, and even I can understand the calculations.

Perhaps you should look at www.krieger-publishing.com.

Best wishes! - Jan 18th 2006, 01:48 PMThePerfectHacker
Thank you very much earboth, for your posting; it probably took a long time to post.