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.
Unbiased example coin flip
for p-correct “advantage” is p - 1/2
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2. Prabhas Chongstitvatana
Let MC be a 3/4-correct unbiased
What is the error prob. Of MC3 (mojority vote) ?
MC3 is 27/32-correct
( > 84% )
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3. Prabhas Chongstitvatana
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
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4. Prabhas Chongstitvatana
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
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5. Prabhas Chongstitvatana
error prob. Pr [ X <= k/2 ] can be calculated
is normal distributed if k >= 30
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6. Prabhas Chongstitvatana
If need error < 5%
Pr [ X < E(X) sqrt Var(X) ] ~ 5%
(from the table of normal distribution)
Pr [ X <= k/2 ] < 5% if
k/2 < E(X) sqrt Var(X)
k > ( 1/(4e2) - 1 )
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7. Prabhas Chongstitvatana
Example e = 5%, which is 55%-correct unbiased Monte Carlo,
k > ( 1/(4e2) - 1 )
k > ,
majority vote repeat 269 times to obtain 95%-correct.
Repetition turn 5% advantage into 5% error prob
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8. Prabhas Chongstitvatana
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 ~ ( 4.1% )
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9. Prabhas Chongstitvatana
Not much more expensive to obtain more confidence.
If we want 0.5% confidence (10 times more)
Pr[X < E(X) sqrt Var(X) ] ~ 5%
k > (1/(4e2) - 1 )
This makes it 99.5%-correct with less than 2.5 times more expensive than 95%-correct.
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10. Prabhas Chongstitvatana
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
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