Amplification of stochastic advantage

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
2
Let MC be a 3/4-correct unbiased What is the
error prob. Of MC3 (mojority vote) ?
MC3 is 27/32-correct ( gt 84 )
3
What is the error prob. of MC with advantage e gt
0 in majority vote k times ? Let Xi 1 if
correct answer, 0 otherwise Pr Xi 1 gt 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
4
Let
X is a random variable corresponds to the number
of correct answer is k trials.
E(X) (1/2e)k Var(X) (1/4 - e2) k
5
error prob. Pr X lt k/2 can be calculated
is normal distributed if k gt 30
6
If need error lt 5 Pr X lt E(X) - 1.645 sqrt
Var(X) 5 (from the table of normal
distribution) Pr X lt k/2 lt 5 if k/2 lt
E(X) - 1.645 sqrt Var(X) k gt 2.706 ( 1/(4e2) - 1
)
7
Example e 5, which is 55-correct unbiased
Monte Carlo, k gt 2.706 ( 1/(4e2) - 1 ) k gt
267.894, majority vote repeat 269 times to
obtain 95-correct. Repetition turn 5 advantage
into 5 error prob
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 )
9
Not much more expensive to obtain more
confidence. If we want 0.5 confidence (10 times
more) PrX lt E(X) - 2.576 sqrt Var(X) 5 k gt
6.636 (1/(4e2) - 1 ) This makes it 99.5-correct
with less than 2.5 times more expensive than
95-correct.
10
To reduce error prob lt del for an unbiased Monte
Carlo with advantage e the number of repetition
is proportional to 1/e2, also to log 1/del