the uniform distribution as input, and it is equal to the fraction of nonerased bits. Example 13.3: Noisy typewriter The transition matrix is shown in Eq. (12.7) for the simplified case of a six-character alphabet. Referring back to Fig. 12.4, we observe that each of the output characters, Z, is characterized by three nonzero conditional probabilities, e.g., for y = B: p(B|A) = p(B|B) = p(B|C) = 1/3. The corresponding probability is: ply = B) = p(B|A)p(A) + p(B|B)p(B) + p(B|C)p(C) (13.14) = [p(A)+p(B)+p(C)]/3 The joint probabilities are p(B, A) = p(B|A)p(A) = p(A)/3 p(B, B) = p(B|B)p(B) = p(B)/3 (13.15) p(B, C) = p(B|C)p(C) = p(C)/3. Using the above definitions, we obtain (13.16) As shown in Appendix K, the continuity of the solutions C and q stems from the limit 1/(1 + 2 W ) 22 a - n/2, where a +b=1 n and n 0. This limit yields q = 1/2, regardless of the value of a, and C = - log(a) 0 for a 1 13.1 Channel capacity where the sign means all possible circular permutations of the character triplets (i.e., BCD, DEF, XYZ, YZA). The mutual information is given by: H(Y|X) - [p(B, A) log p(B|A) + p(B, B) log p(B|B) + p(B, C) log p(B|C) log3. (13.17) The channel capacity is, therefore: C =max{H(Y)-H(YX)) = - (13.18) The expression between brackets is maximized when for each character triplet, A, B, C, we have p(A) = p(B) = p(C), which yields the optimal input distribution p(x) = 1/26. Thus, Eq. (13.18) becomes (13.19) = log log log 26 This result could have been obtained intuitively by considering that there exist 26/3 triplets of output characters that correspond to a unique input character.5 Thus 26/3 symbols can effectively be transmitted without errors as if the channel were noiseless. The fact that 26/3 is not an integer does not change this conclusion (an alphabet of 27 characters would yield exactly nine error-free symbols). Example 13.4: Asymmetric channel with nonoverlapping outputs The transition matrix is shown in Eq. (12.8). Since there is only one nonzero element in each row of the matrix, each output symbol 