Substituting x r0 for n r pr sn exp x 0 r0 x

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Unformatted text preview: µX (a) = γX (ro) − roa for a < X , r < 0 γX (ro) ￿ slope = γX (ro) = a Pr {Sn ≥ na} ≤ exp(nµX (a)) The Chernoff b ound, optimized over r, is essentially ex­ ponentially tight; i.e., Pr {Sn ≥ na} ≥ exp(n(µX (a) − ￿)) for large enough n. 12 In looking at threshold problems, we want to find the probability that Pr {Sn ≥ α} for any n. Thus we want a b ound that focuses on variable n for a fixed α, i.e., on when the threshold is crossed if it is crossed. We want a b ound of the form Pr {Sn ≥ α} ≤ exp αf (n) Start with the b ound Pr {Sn ≥ na} ≤ exp(n[γX (r0) − r0a]), ￿ with α = an and r0 such that γX (r0) = α/n. Substituting ￿ α/γX (r0) for n, ￿￿ γ (r ) Pr {Sn ≥ α} ≤ exp α X 0 − r0 ￿ γX (r0) ￿￿ 13 ￿￿ γ (r ) Pr {Sn ≥ α} ≤ exp α X 0 − r0 ￿ γX (r0) 0 r ro ❅ ❅ ❅ ❅ ❅ ❅ slope = X r∗ ￿￿ ro − γ (ro)/γ ￿(r0) slope = γ ￿(ro) = α/n γ (ro) ￿ When n is very large, the slope γX (r0) is close to 0 and the horizontal intercept (the negative exponent) is very large. As n decreases, the intercept decreases to r∗ and then increases...
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This note was uploaded on 01/13/2012 for the course ELECTRICAL 6.262 taught by Professor Staff during the Fall '11 term at MIT.

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