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Unformatted text preview: Lec+nr¢Z' 5’: $59 Annoumh:
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number Lg,‘£ r>o i Xilvh'c‘ 0/. ' (a. Sh»: ECefx] = [0%) ”5]“ Performance of Chemov bound and Markov bound Consider transmitting an N —bit binary string through a binary symmetric channel with bit—ﬂipping probability
(i.e., biterror probability) 6. Deﬁne random variable X to be the number of bit errors in the received binary string.
We are interested in the tail distributions of X, i.e., P,{X 2 k} for k = 0, 1, 2, . . . ,N. The exact value of the
tail distribution is easy to write but involves a summation that is not easy to understand analytically. Thus, simple upper bounds are often useful. Here we compare the performance of the Chemov and Markov bounds to the exact
values of the tail distribution P,‘{X 2 k}: Exact value: P1.{X 2 k} = ilk (1;) 6i(].— 6)N_i, 0 S k g N
Markov bound: P,{X 2 k} 3 Mil %, 1 g k g N Performance of Chemov and Markov bound (N=100. 2:001) 10 m Exact value 4, — — — Chemov bound
“" Markov bound i Pr (X >= k) (log scale)
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 Fall '08
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