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Math Help - The Strong Law of Large Numbers

  1. #1
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    The 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.
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    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 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 f(A)=\frac{a}{n}.
    Then
     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)
     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
    P \left( \ |{(f(A)-p } | \geq \epsilon \right)\geq 1 - \frac{p\cdot (1-p)}{n\cdot \epsilon^2}

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

    that means:

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

    put into the inequality
    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
    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 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 = .95 and n = 125.

    Because
     \max_ {\subc{0 \leq p \leq 1}}{p(1-p)}=\frac{1}{4}
    the inequality could be transformed into:
    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
    \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
     \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!
    Attached Thumbnails Attached Thumbnails The Strong Law of Large Numbers-ipf-.gif   The Strong Law of Large Numbers-ges_grzahl.gif  
    Last edited by earboth; January 17th 2006 at 08:52 PM.
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  3. #3
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    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!
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    Thank you very much earboth, for your posting; it probably took a long time to post.
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