Transmission of Binary Signals Let T_0, T_1, R_0, R_1 designate the events T_0 0

Question

Transmission of Binary Signals Let T_0, T_1, R_0, R_1 designate the events T_0 0 is transmitted T_1 1 is transmitted R_0 0 is received R_1 1 is received The transmission channel is imperfect in that there is a non-zero probability of a transmitted 1 being received as 0. The channel is thus characterized by the set of conditional probabilities P(R_1 |R_1) P(R_0|R_1) P(R_1|T_0) P(R_0|T_0) These and input and output probabilities are conveniently displayed in the diagram below. Note that P(R_1|T_1) + P(R_0 |T_1) = 1 and P(R_1|T_0) + P(R_0|T_0) = 1 P(R_1) = P(R_1|T_1) P(T_1) + P(R_1|T_0) P(T_0) P(R_0) = P(R_0|T_1) P(T_1) + P(R_0|T_0) P(T_0) Given: P(_1) = 0.4 (Source data) P(R_1|T_1) = 0.8.; P(R_0|T_0) = 0.9 (Channel data) Find: P(T_1|R_1) = Prob. that a 1 was transmitted given that 2 1 was received P(T_0|T_0) = Prob. that 2 0 was transmitted given that 2 0 was received P(T_1|R_1) = 0.842 P(T0|R_0) = 0.871

Explanation / Answer

Solution

Back-up Theory

Since total probability must be 1, given P(T1), P(T0) = 1 – P(T1) ………………(1)

P(A/B) = {P(B/A)P(A)}/P(B) …………………………………………….(2)

Now, to work out the solution,

Given: P(T1) = 0.4 ……………………………………………………….(3)

P(R1/T1) = 0.8 …………………………………………………………..(4)  

P(R0/T0) = 0.9 …………………………………………………………..(5)

P(R1/T1) + P(R0/T1) = 1 ……………………………………………….(6)

P(R1/T0) + P(R0/T0) = 1 ……………………………………………….(7)

P(R1) = P(R1/T1)P(T1) + P(R1/T0)P(T0)………………………………(8)

P(R0) = P(R0/T1)P(T1) + P(R0/T0)P(T0)………………………………(9)

(1) and (3) => P(T0) = 0.6 ………………………………………… … .(10)

Part (1)

P(T1/R1) = P(R1/T1)P(T1)/P(R1) by (2) ………………………………(11)

By (8), P(R1) = (0.8x0.4)[given] + (1 – 0.9)(0.6)[from (7)(5)(10)] = 0.38

So, (11) => (0.8x0.4)/0.38 = 0.842 ANSWER

Part (2)

P(T0/R0) = P(R0/T0)P(T0)/P(R0) by (2) ………………………………(12)

By (9), P(R0) = (0.2x0.4)[from(6) (4)(3)] + (0.9)(0.6)[from (5)(10)] = 0.62

So, (12) => (0.9x0.6)/0.52 = 0.871 ANSWER

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