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Amplification of stochastic advantage

Amplification of stochastic advantage. Biased : known with certainty one of the possible answer is always correct. Error can be reduced by repeat the algorithm. Unbiased example coin flip for p-correct “advantage” is p - 1/2. Let MC be a 3/4-correct unbiased

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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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