20065ee131A_1_EE131AF06HW3Solution

20065ee131A_1_EE131AF06HW3Solution - Solution of Assignment...

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Unformatted text preview: Solution of Assignment 3, EE 131A Due: Wednesday October 25, 2006 Problem 1. Consider a binary communication channel where the probabilities of inputs “0” and “1” into the system are p and 1- p , respectively. Suppose that the transmission errors occur at random with probability ε . For i = 0 , 1 , let A i be the event “input was i ,” and B i be the event “output was i .” Thus P [ A ] = p , P [ A 1 ] = 1- p , P [ B | A ] = 1- ε , P [ B 1 | A ] = ε , P [ B | A 1 ] = ε , and P [ B 1 | A 1 ] = 1- ε . Find the value of ε for which the input of the channel is independent of the output of the channel. Solution. We have to find the value of ε such that, for i, j ∈ { , 1 } , the events A i and B j are independent. First we find the probabilities P [ B ] and P [ B 1 ] . Since S = A ∪ A 1 is a partition of the sample space, P [ B ] = P [ B ∩ A ] + P [ B ∩ A 1 ] = P [ A ] · P [ B | A ] + P [ A 1 ] · P [ B | A 1 ] = p · (1- ε ) + (1- p ) · ε = p + ε- 2 p ε ; and P [ B 1 ] = 1- P [ B ] = 1- p- ε + 2 p ε. The independent requirements which must be satisfy are: P [ B | A ] = P [ B ] , P [ B | A 1 ] = P [ B ] , P [ B 1 | A ] = P [ B 1 ] , and P [ B 1 | A 1 ] = P [ B 1 ] . Therefore, the following identities must satisfy: 1- ε = p + ε- 2 p ε ε = p + ε- 2 p ε ε = 1- p- ε + 2 p ε 1- ε = 1- p- ε + 2 p ε. Any of these identities implies ε = 1 2 . Problem 2. Suppose that for the general population, 1 in 5000 people carries the human immun- odeficiency virus (HIV). A test for the presence of HIV yields either a positive ( +...
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This note was uploaded on 06/09/2010 for the course EE 131A taught by Professor Lorenzelli during the Spring '08 term at UCLA.

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20065ee131A_1_EE131AF06HW3Solution - Solution of Assignment...

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