Problem 6. A certain binary PCM system transmits the two binary states X = +1, and X = -1 with equal probability. However, because of channel noise, the receiver makes recognition errors. Also, as a result of path distortion, the receiver may lose necessary signal strength to make any decision. Thus, there are three possible receiver states: Y = +1, Y = 0, and Y = -1. Y=0 corresponds to "loss of signal". Assume that: P(Y=-1|X = +1) = 0.1, P(Y=+1|X = -1) = 0.2, and P(Y=0|X = +1) = P(Y=0|X = -1) = 0.05 %3D %3D %3D %3D (a) Find the probability P(Y=+1), P(Y=-1), and P(Y=0). (b) Find the probability P(X=+1|Y = +1) and P(X=-1|Y = -1)

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Problem 6.
A certain binary PCM system transmits the two binary states X = +1, and X = -1 with equal probability.
However, because of channel noise, the receiver makes recognition errors. Also, as a result of path
distortion, the receiver may lose necessary signal strength to make any decision. Thus, there are three
possible receiver states: Y = +1, Y = 0, and Y = -1. Y=0 corresponds to "loss of signal". Assume that:
%D
P(Y=-1|X = +1) = 0.1, P(Y=+1|X = -1) = 0.2, and P(Y=0|X = +1) = P(Y=0|X = -1) = 0.05
%3D
%3D
(a) Find the probability P(Y=+1), P(Y=-1), and P(Y=0).
(b) Find the probability P(X=+1|Y = +1) and P(X=-1|Y = -1)
%3D
Transcribed Image Text:Problem 6. A certain binary PCM system transmits the two binary states X = +1, and X = -1 with equal probability. However, because of channel noise, the receiver makes recognition errors. Also, as a result of path distortion, the receiver may lose necessary signal strength to make any decision. Thus, there are three possible receiver states: Y = +1, Y = 0, and Y = -1. Y=0 corresponds to "loss of signal". Assume that: %D P(Y=-1|X = +1) = 0.1, P(Y=+1|X = -1) = 0.2, and P(Y=0|X = +1) = P(Y=0|X = -1) = 0.05 %3D %3D (a) Find the probability P(Y=+1), P(Y=-1), and P(Y=0). (b) Find the probability P(X=+1|Y = +1) and P(X=-1|Y = -1) %3D
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