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Old 10-07-2004, 05:45 AM
irchans irchans is offline
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Default normal distribution

jason1990 writes

[ QUOTE ]
Does anyone know, roughly, how many poker hands a person needs to play before their winnings become approximately normally distributed? For example, suppose someone has a winrate of 2BB/100 and a standard deviation of 15BB/100. They might think that if they play 100 hands, then they have a 95% chance of winning somewhere between -28BB and 32BB (a spread of 2 SD's). But this is only true if the result of playing 100 hands is (roughly) normally distributed. Is it?


[/ QUOTE ]

This brings up some interesting questions about the sum of identically distributed random variables.

Given identically distributed random variables x_i with 0 mean and standard deviation s. Let the random variable y_N be defined to be

y_N = (x_1 + x_2 + ... + x_N)/Sqrt(N).

By the central limit theorem, P( y_N > 2 s) converges to (1-Erf(Sqrt(2)))/2 = 0.02275 as N becomes large, but the theorem does not tell us how quickly it converges.

Question 1:
Is there a bound on P( y_N > 2 s), the probability that y_N will be more than 2 standard deviation above the mean?

Question 2: If we know that P( |x_i| > a) = 0 for some real number a, is there a sharper bound on P( y_N > 2 s) using a?
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Old 10-07-2004, 07:02 AM
pzhon pzhon is offline
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Default Re: normal distribution

[ QUOTE ]

Question 1:
Is there a bound on P( y_N > 2 s), the probability that y_N will be more than 2 standard deviation above the mean?


[/ QUOTE ]
The Chebyshev (multiple spellings) inequality is a quick, coarse estimate: The probability that a random variable is n standard deviations away from the mean is at most 1/n^2. A one-tailed version is slightly better: The probability that a random variable is n standard deviations above the mean is at most 1/(1+n^2). Here that is 1/5.

This is not sharp for n>1, but there isn't a bound depending on N and not the random variables that approaches the result for a normally distributed random variable. For example, consider a Poisson distribution such that 1 is 2 standard deviations above the mean. The mean is 3-2sqrt(2)=.1716. The probability it is 1 or greater is .1577, not too far below the .2 from the Chebyshev inequality. A Poisson distribution is divisible, so it is the sum of n IID random variables, in particular, n variables with a Poisson distributions of mean (3-2sqrt(2))/n.

[ QUOTE ]
Question 2: If we know that P( |x_i| > a) = 0 for some real number a, is there a sharper bound on P( y_N > 2 s) using a?

[/ QUOTE ]
Yes. That puts an upper bound on the higher moments, and these are used in effective versions of the Central Limit Theorem. I don't know them offhand.
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