20065ee131A_1_EE131AF06HW3Solution

20065ee131A_1_EE131AF06HW3Solution - Solution of Assignment...

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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 0 ] = p , P [ A 1 ] = 1 - p , P [ B 0 | A 0 ] = 1 - ε , P [ B 1 | A 0 ] = ε , P [ B 0 | 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 ∈ { 0 , 1 } , the events A i and B j are independent. First we find the probabilities P [ B 0 ] and P [ B 1 ] . Since S = A 0 A 1 is a partition of the sample space, P [ B 0 ] = P [ B 0 A 0 ] + P [ B 0 A 1 ] = P [ A 0 ] · P [ B 0 | A 0 ] + P [ A 1 ] · P [ B 0 | A 1 ] = p · (1 - ε ) + (1 - p ) · ε = p + ε - 2 p ε ; and P [ B 1 ] = 1 - P [ B 0 ] = 1 - p - ε + 2 p ε. The independent requirements which must be satisfy are: P [ B 0 | A 0 ] = P [ B 0 ] , P [ B 0 | A 1 ] = P [ B 0 ] , P [ B 1 | A 0 ] = 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 ( + ) or negative ( - ) response. Suppose the test gives the correct answer 99% of the time. What is P [ -| H ] , the
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conditional probability that a person tests negative given that the person does have the HIV virus?
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