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Optimizing Error Reduction in Biased and Unbiased Algorithms through Amplification of Stochastic Advantage

This study explores error reduction in algorithms by amplifying stochastic advantage. Known with certainty, one possible answer is always correct. Error reduction is achieved through algorithm repetition. An unbiased example is the coin flip, with p-correct "advantage" being p - 1/2. When MC is 3/4-correct unbiased, the error probability of MC3 (majority vote) is 27/32-correct. The error probability of MC with advantage e > 0 in majority vote k times is analyzed. Repetition enhances error reduction and increases algorithm accuracy. To reduce error probability below a certain threshold, the number of repetitions is proportional to the advantage and logarithm of the threshold.

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Optimizing Error Reduction in Biased and Unbiased Algorithms through Amplification of Stochastic Advantage

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  1. Amplification of stochastic advantage Biased: known with certainty one of the possible answer is always correct. Error can be reduced by repeat the algorithm. Unbiasedexample coin flip for p-correct “advantage” is p - 1/2 Prabhas Chongstitvatana

  2. Let MC be a 3/4-correct unbiased What is the error prob. Of MC3 (mojority vote) ? MC3 is 27/32-correct ( > 84% ) Prabhas Chongstitvatana

  3. What is the error prob. of MC with advantage e > 0 in majority vote k times ? Let Xi = 1 if correct answer, 0 otherwise Pr[ Xi = 1 ] >= 1/2 + e assume for simplicity Pr[ Xi = 1 ] = 1/2 + e ; k is odd (no tie) E( Xi ) = 1/2 + e; Var(Xi ) = (1/2 + e) (1/2-e) = 1/4 - e2 Prabhas Chongstitvatana

  4. Let X is a random variable corresponds to the number of correct answer is k trials. E(X) = (1/2+e)k Var(X) = (1/4 - e2) k Prabhas Chongstitvatana

  5. error prob. Pr [ X <= k/2 ] can be calculated is normal distributed if k >= 30 Prabhas Chongstitvatana

  6. If need error < 5% Pr [ X < E(X) - 1.645 sqrt Var(X) ] ~ 5% (from the table of normal distribution) Pr [ X <= k/2 ] < 5% if k/2 < E(X) - 1.645 sqrt Var(X) k > 2.706 ( 1/(4e2) - 1 ) Prabhas Chongstitvatana

  7. Example e = 5%, which is 55%-correct unbiased Monte Carlo, k > 2.706 ( 1/(4e2) - 1 ) k > 267.894, majority vote repeat 269 times to obtain 95%-correct. Repetition turn 5% advantage into 5% error prob Prabhas Chongstitvatana

  8. It takes a large number of run for unbiased Compare to biased Run 55%-correct bias MC 4 times reduces the error prob. To 0.454 ~ 0.041 ( 4.1% ) Prabhas Chongstitvatana

  9. Not much more expensive to obtain more confidence. If we want 0.5% confidence (10 times more) Pr[X < E(X) - 2.576 sqrt Var(X) ] ~ 5% k > 6.636 (1/(4e2) - 1 ) This makes it 99.5%-correct with less than 2.5 times more expensive than 95%-correct. Prabhas Chongstitvatana

  10. To reduce error prob < del for an unbiased Monte Carlo with advantage e; the number of repetition is proportional to 1/e2, also to log 1/del Prabhas Chongstitvatana

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