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The law of large numbers.

Topic closed. 11 replies. Last post 10 years ago by Amazing Grace.

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rainbow lake
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November 2, 2005
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 Posted: August 14, 2006, 7:50 pm - IP Logged

The weak law

The weak law of large numbers states that if X1, X2, X3, ... is an infinite sequence of random variables, where all the random variables have the same expected value μ and variance σ2; and are uncorrelated (i.e., the correlation between any two of them is zero), then the sample average

$\overline{X}_n=(X_1+\cdots+X_n)/n$

converges in probability to μ. Somewhat less tersely: For any positive number ε, no matter how small, we have

$\lim_{n\rightarrow\infty}\operatorname{P}\left(\left|\overline{X}_n-\mu\right|<\varepsilon\right)=1.$

Proof

Chebyshev's inequality is used to prove this result. Finite variance $\operatorname{Var} (X_i)=\sigma^2$ (for all i) and no correlation yield that

$\operatorname{Var}(\overline{X}_n) = \frac{n\sigma^2}{n^2} = \frac{\sigma^2}{n}.$

The common mean μ of the sequence is the mean of the sample average:

$E(\overline{X}_n) = \mu.$

Using Chebyshev's inequality on $\overline{X}_n$ results in

$\operatorname{P}( \left| \overline{X}_n-\mu \right| \geq \varepsilon) \leq \frac{\sigma^2}{{n\varepsilon^2}}.$

This may be used to obtain the following:

$\operatorname{P}( \left| \overline{X}_n-\mu \right| \leq \varepsilon) = 1 - \operatorname{P}( \left| \overline{X}_n-\mu \right| > \varepsilon) \geq 1 - \operatorname{P}( \left| \overline{X}_n-\mu \right| \geq \varepsilon) \geq 1 - \frac{\sigma^2}{\varepsilon^2 n}.$

As n approaches infinity, the expression approaches 1.

Proof ends here

The result holds also for the 'infinite variance' case, provided the Xi are mutually independent and their (finite) mean μ exists.

A consequence of the weak law of large numbers is the asymptotic equipartition property.

The strong law

The strong law of large numbers states that if X1, X2, X3, ... is an infinite sequence of random variables that are pairwise independent and identically distributed with E(|Xi|) < ∞   (and where the common expected value is μ), then

$\operatorname{P}\left(\lim_{n\rightarrow\infty}\overline{X}_n=\mu\right)=1,$

i.e., the sample average converges almost surely to μ.

If we replace the finite expectation condition with a finite second moment condition,  E(Xi2) < ∞ (which is the same as assuming that Xi has variance), then we obtain both almost sure convergence and convergence in mean square. In either case, these conditions also imply the consequent weak law of large numbers, since almost sure convergence implies convergence in probability (as, indeed, does convergence in mean square).

This law justifies the intuitive interpretation of the expected value of a random variable as the "long-term average when sampling repeatedly".

A weaker law and proof

Proofs of the above weak and strong laws of large numbers are rather involved. The consequent of the slightly weaker form below is implied by the weak law above (since convergence in distribution is implied by convergence in probability), but has a simpler proof.

Theorem. Let X1, X2, X3, ... be a sequence of random variables, independent and identically distributed with common mean μ < ∞, and define the partial sum Sn := X1 + X2 + ... +Xn. Then,  Sn / n converges in distribution to μ.

Proof. (See [1], p. 174) By Taylor's theorem for complex functions, the characteristic function of any random variable, X, with finite mean μ, can be written as

$\varphi(t) = 1 + it\mu + o(t), \quad t \rightarrow 0.$

Then, since the characteristic function of the sum of independent random variables is the product of their characteristic functions, the characteristic function of  Sn / n  is

$\left[\varphi\left({t \over n}\right)\right]^n = \left[1 + i\mu{t \over n} + o\left({t \over n}\right)\right]^n \, \rightarrow \, e^{it\mu}, \quad \textrm{as} \quad n \rightarrow \infty.$

The limit  eitμ  is the characteristic function of the constant random variable μ, and hence by the Lévy continuity theorem,  Sn / n converges in distribution to μ. Note that the proof of the central limit theorem, which tells us more about the convergence of the average to μ (when the variance σ 2 is finite), follows a very similar approach.

References

• Grimmett, G. R. and Stirzaker, D. R. (1992). Probability and Random Processes, 2nd Edition. Clarendon Press, Oxford. ISBN 0-19-853665-8.

rainbow lake
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 Posted: August 14, 2006, 7:51 pm - IP Logged

Which brings me to my next point.

To be continued.

United States
Member #39406
May 15, 2006
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 Posted: August 14, 2006, 8:06 pm - IP Logged

Can you illustrate this at a little more elementary level. I flunked math. No need in me trying to learn that stuff now is it.

rainbow lake
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 Posted: August 15, 2006, 8:30 am - IP Logged

Here is a series of numbers. These number have paid out over 200 million dollars in the last 3 months, What is interesting about these are, read up and down instead of across one way, take first 3 digits of each series of numbers can you predict what the next 3 numbers will be, only one rule 0 can not be a first digit,

3102123334443131528293347252131925323040626273341422413162142464983715384347322692043444681321222439413172133414238626303844451611142933424472212229343571119364146493835192034473122026293745174173439414626111420253136481011363846482618233234414649111214273448882021273648111114344042436915183435456115181929375151822314349269192024364810210111642432131736444548862830313448471172429344627
rainbow lake
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November 2, 2005
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 Posted: August 15, 2006, 8:31 am - IP Logged

3102123334443131528293347252131925323040626273341422413162142464983715384347322692043444681321222439413172133414238626303844451611142933424472212229343571119364146493835192034473122026293745174173439414626111420253136481011363846482618233234414649111214273448882021273648111114344042436915183435456115181929375151822314349269192024364810210111642432131736444548862830313448471172429344627

rainbow lake
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November 2, 2005
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 Posted: August 15, 2006, 8:32 am - IP Logged

this is not printing out the right way?

Solon, OH
United States
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January 7, 2004
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 Posted: August 15, 2006, 8:40 am - IP Logged

LOL I hear math is the worst even though i got a B+ in algebra and statistics lol.

rainbow lake
Member #25177
November 2, 2005
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 Posted: August 15, 2006, 8:56 am - IP Logged

3102123334443

13152829334725

2131925323040

6262733414224

1316214246498

371538434732

269204344446

813212224394

13172133414238

62630384444516

1114293342447

221222934357

11193641464938

351920344731

2202629374517

4173439414626

11142025313648

10113638464826

182332344144649

1112142734488

8202127364811

111434042436

915183435456

115181929375

15182231434926

9192024364810

2101116424321

317364445488

6283031344847

1172429344627

Copy and paste did not work

rainbow lake
Member #25177
November 2, 2005
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 Posted: August 15, 2006, 9:15 am - IP Logged

if  you read up and down you will notice only 8 of the first digits fall into the high number category.

and 19 of the numbers fall into the odd category.

that is first row up and down.

there are only 7 numbers 6 or above first row. reading up and down.

how about sums first 3 numbers across all the way down starts off as

Sum 4,5,6,14,5,11,17,12,5,14,3,5,3,9,4,12,3,2,13,3,10,3,15,7,7,19,3,11,16,9

so root sums 4,5,6,5,5,2,8,3,5,5,3,5,3,9,4,3,3,2,4,3,1,3,6,7,7,1,3,2,7,9

so i will print another series number in 2 days or less.

Can any one try and predict the next outcome.

you can look at these number any way you like but best way to look at them is as 4 digits then 4 digits then 4 or 6 digits going across in the series. so one could predict the following as a next number

12192535464849 although if you read each digit in its corresponding spot up and down this may not be a good choice.

if you care to try this I will tell you how much money you would have won if you played those numbers in a certain state,

if this is too complicated then dont attempt, but it is worth a try because when this post is done it will open your eyes on how to chose numbers in a certain lottery that we all play very wisely.

rainbow lake
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November 2, 2005
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 Posted: August 15, 2006, 9:59 pm - IP Logged

Ok looking in the first row my choice of first numbers would be , 2 or 5 or 8 or 4  or 9 for the next series draw. my second number will be a one so so far my combos are

21

51

81

41

91

but i will also have 10 combos starting with

10

11

12

13

14

15

16

17

18

19

now i will have to predict my 3rd row.

rainbow lake
Member #25177
November 2, 2005
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 Posted: August 15, 2006, 10:20 pm - IP Logged

One thing to keep in mind is on the first 3 digits that you chose will or can not be over sum of 18

so highest first 3 digits can be is 945  and 954 does not exist.

rainbow lake
Member #25177
November 2, 2005
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 Posted: August 17, 2006, 12:24 am - IP Logged

so i was off by a mile the next series number was

34825414822

Which  if looking at it normal way would be

3,4,8,25,42,48, Bonus 22

I did manage to pull off a 4 number win using this system on the western 6-49

I had all digits but not on same line only ended up with a 4 hit on western

i find its easier to look at the numbers drawn as a series such as

19141720388 and follow the columns up and down to predict next number to fall then move over to new column .

the only reason this is in the pick 3 forumn was to give it exposure maybe thinking it should have been in the pick 6 games.

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