By J. Wolfowitz (auth.)

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3) must Since R(n) has a maximum at n = be a constant, say K -1, for all indices k such that nk> 0, and must be < K - 1 for all indices k such that nk = O. Define BI = {k In" > O} B2 = Hence EZ* (k) {k Ink = O}. 5) Now a C= 2: ii" EZ*(k). "=1 Hence K = C. 9. c. 1. c. 5) holds with K = C. 8. Suppose given a code (n, N, Ä) for this channel. Suppose that the i th word of the code is to be sent (i = 1, ... , N), that the first (j - 1) letters of the word have already been sent (j = 1, ...

1. 8 satisfies N < 2nc +KY -;;, where K depends on A but not on n. 1) 52 4. Compound Channels We shall first need a preliminary lemma. c. 1. 12). Let Z(k), k = 1, ... , a, be a chance variable with the distribution P{Z(k) = j} = w(j I k), j = 1, ... , a. , instead of writing Z(k) as a subscript to n' we write it as an argument for typographical simplicity). Now R(n) = - i~j~ n;w(j Ii) log ;~ njw(j I i) a a a a ( a ) + 2: 2: n; w (j I i) log w (j I i) . ;=1 j=1 Differentiate R(n) with respect to nk' k to the fact that 2:ni = 1.

2. 1 by a more careful examination of the constants involved. Since the latter depend continuously on w (. 2. c. let a = 2, w (1 1 1) > T' and Iw(11 1) -w(21 2) 1 be sufficiently small. Then, if 0 ::;:; A. < ; and n is sufficiently large, there does not exist a code (n, 2nC - K ,V;-, A), where K o is a positive constant which depends on w (. ) and A.. 5. The finite-state channel with state calculable by both sender and receivet. 's, {w (. 1 . 1 s), s = 1, ... , t}. f. is called a "state" of the channel.