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where b = s1/(µ1 - µ2). If, initially, one random variable is as likely as the other to be best, n = n* and the expected loss per trial is |
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(Given two arbitrary functions, Y(t) and Z(t), of the same variable t, " Y(t) ~ Z(t)" will be used to mean  while "Y(t) = Z(t)" means that under stated conditions the difference (Y(t) - Z(t)) is negligible.) |
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Proof: In order to select an n which minimizes the expected loss, it is necessary first to write q(N - n, n) as an explicit function of n. As defined above q(N - n, n) is the probability that x(2)(N) = x1. More carefully, given the observation, say, that x' = x(2)(N), we wish to determine the probability that x' = x1. That is, we wish to determine |
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as an explicit function of N - n and n. Bayes's theorem then gives us the equation |
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Letting q', q", and p designate Pr{x' = x(2) | x' = x1}, Pr{x' = x(2) | x' = x2}, and Pr{x' = x1}, respectively, and using the fact that x' must be x2 if it is not x1, this can be rewritten as |
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(If one random variable is as likely as the other to be best, then p = (1 - p) = ½.) |
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To derive q' let us assume that x' has received n trials out of the N total. Let  be the sum of the outcomes (payoffs) of N - n trials of x2 and let  be the corresponding sum for n trials of x1. Since q' has x' = x1 as a condition, q' is just the probability that  or, equivalently the probability that  . By the central limit theorem  approaches a normal distribution with mean µ2 and variance  ; similarly,  has mean µ1 and variance  . The distribution of  is given by the product (convolution) of the distributions of  and  ; by an elementary theorem (on the convolution of normal distributions) this is a normal distribution with mean µ1 - µ2 and variance  . Thus the probability  is the tail 1 - F(x0) of |
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