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Math Help - gambling problem/ standard deviation

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    Quote Originally Posted by CrazyAsian View Post
    Can someone explain why you multiply the estimate of SD by the square root of the sample size?
    We have to start at the beginning and along the way we'll see why there are two ways to calculate the SD.

    The total of the actual payoffs is a sum of independent random variables \sum_i R_i. The standard deviation S of this sum is the square root of the variance V of the sum. The variance of a sum of independent random variables is the sum of the variances V_i of each R_i , so

    V = \sum_i V_i .

    There are two ways to calculate the V_i . The most accurate is to use the known pot sizes P_i and the win probabilities w_i to calculate the actual variance. Since w_i P_i is the mean of the payoff R_i , the actual variance is

    V_i = w_i (P_i - w_i P_i)^2 + (1-w_i)(0 - w_i P_i)^2 .

    Then the standard deviation of \sum_i R_i is

    S = \sqrt{V} = \sqrt{\sum V_i} .

    The 95% confidence interval around the expected total payoff E = \sum{w_i P_i} is

    [ E- 1.96 S, E+ 1.96 S ].

    There is no multiplication by \sqrt{n} of the sample size. Where did that go? Well, there is a second way to get the variance V and that is to estimate it ignoring the w_i. We assume the random variables R_i have the same variance v' and estimate that as

    v' = \sum_i (R_i - M)^2 / (n-1)

    where M = \sum_i R_i /n is the mean actual payoff.

    Then the variance V is estimated as

    V' = \sum_i v' = n v' .

    V' is an estimate of the actual variance V which is not exact because it treats the variation of w_i as unknown.

    The estimated standard deviation of \sum R_i is

    S' = \sqrt{V'} = \sqrt{n v'} = \sqrt{v'}\sqrt{n} = s' \sqrt{n},

    where

    s' = \sqrt{v'} = \sqrt{\sum_i (R_i - M)^2 / (n-1)}

    is the estimated common standard deviation of the R_i .

    Then the 95% confidence interval is

    [ E- 1.96 S', E+ 1.96 S' ] = [ E- 1.96 s' \sqrt{n}, E+ 1.96 s' \sqrt{n} ],

    which is the familiar one.
    Last edited by JakeD; May 26th 2007 at 09:34 PM.
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