# Chernoff Bounds - Chernoff Bounds [Ref. Wozencraft and...

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Chernoff Bounds [Ref. Wozencraft and Jacobs] Consider a random variable X with known mean X . Form a sample mean 1 1 ˆ n nj j XX n = = . Then 0 () ˆ Pr{ } ˆ Pr{ } n n Xd n X d dX e X d λ ≥>    ≤< where 0 is found from the relation 0 0 [] X X E Xe d Ee = . X d ˆ n X fx

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Homework PROOF: We will prove the case dX > . 12 For notational simplicity, let . n YX X X = + ++ { } { } ( ) ˆ Then Pr Pr n X d Y nd u Y nd ≥= = , where () u is the unit step function. Define ( ) for any 0 Y nd gY e λ = > . Then ( ) ( ) for any g Y u Y nd Y −∞< <∞ . Therefore Pr{ } ( ) Y nd g Y ≥≤ . nX nd
Now { } ( ) { } () n n n Xd Y n d nd Y nd X d X g Y e ee e e e λ λλ −− = = = = = Therefore ( ) { } ˆ Pr{ } 0 n n X d e for any ≥≤ > The bound is minimized when ( ) ( ) ( ) ( ) ( ) 00 0 XX e X de Xe de Xe = −≡ ⇒=

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Example. Consider a memoryless binary symmetric channel with BER p=0.1 . We repeat the source symbol N times for error correction purpose. Assume N is an odd integer. Find an upper bound to the probability of an error. Let { } 0,1 n X denote the channel error at the n -th repeat. Then 1 ˆ [ ] Pr{ } Pr{ 1/2} 2 N nN n N PE X X = = ≥= Since 0.1 Xp = = and 1/2 dX = > , we can use the rhs chernoff bound:

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## This note was uploaded on 11/23/2010 for the course EE EE528 taught by Professor Majungsoo during the Spring '10 term at Korea Advanced Institute of Science and Technology.

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Chernoff Bounds - Chernoff Bounds [Ref. Wozencraft and...

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