Consider the binary communication channel. The channel input symbol X may assume the state 0 or the state 1, and, similarly, the channel output symbol Y may assume either the state 0 or the state 1. Because of the channel noise, an input 0 may convert to an output 1 and vice versa. The channel is characterized by the channel transition probabilities $p_0, q_0, p_1$, and $q_1$, defined by -

$p_0$ & = & $P(y_1|x_0) $ \ $p_1$ &=& $P(y_0|x_1)$ \ $q_0$ &=& $P(x_1|y_0)$ \ $q_1$ &=& $P(x_0|y_1)$

where $x_0$, and $x_1$, denote the events (X = 0) and (X = 1), respectively, and $y_0$ and $y_1$, denote the events (Y = 0) and (Y = 1), respectively.

And $p_0 +q_0$ = 1 = $p_1 +q_1$ , $P(x_0)$ = 0.5 , $p_0$ = 0.1 and $p_1$ = 0.2

(a) Find P($y_0$) and P($y_1$).

(b) If a 0 was observed at the output, what is the probability that a 0 was the input state i.e. $P(x_0 | y_0)$ = ?

(c) If a 1 was observed at the output, what is the probability that a 1 was the input state i.e. $P(x_1 | y_1)$ = ?

(d) Calculate the probability of error $P_e$.

(a) Find P($y_0$) and P($y_1$) $$q_0 = 1 - p_0 = 1 - 0.1= 0.9 $$ $$q_1 = 1 - p_1 = 1 - 0.8 = 0.8 $$

$$ P(y_0) =P(y_0 | x_0 ) P(x_0) + P(y_0 | x_1) P(x_1) $$ $$ = 0.9(0.5) + 0.2(0.5) $$ $$ = 0.55 $$ $$ P(y_1) = P(y_1 | x_0 ) P(x_0) + P(y_1 | x_1) P(x_1) $$ $$= 0.1(0.5) + 0.8(0.5)$$ $$ = 0.45 $$ $$\mathbf{ P(y_0) = 0.55 \; \; and \; \; P(y_1) = 0.45 }$$

(b) If a 0 was observed at the output, what is the probability that a 0 was the input state i.e. $P(x_0 | y_0)$ = ? } $$ P(x_0 | y_0) = \frac{ P(y_0 | x_0) P(x_0) } {P(y_0)} $$ $$ = \frac {0.9(0.5)}{0.55} $$ $$ = 0.818 $$ $$\mathbf{ P(x_0 | y_0) = 0.818 }$$

(c) If a 1 was observed at the output, what is the probability that a 1 was the input state i.e. $P(x_1 | y_1)$ = ? $$ P(x_1 | y_1) = \frac{ P(y_1 | x_1) P(x_1) } {P(y_1)} $$ $$ = \frac {0.8(0.5)}{0.45} $$ $$ = 0.889 $$ $$\mathbf{ P(x_1 | y_1) = 0.889 }$$ (d) Calculate the probability of error $P_e$ $$ P_e = P(y_1|x_0)P(x_0) + P(y_0|x_1) P(x_1) $$ $$ = 0.1(0.5) + 0.2(0.5) $$ $$ = 0.15 $$ $$\mathbf{ P_e = 0.15 }$$