wo points is always below the maximum). Therefore, the maximum of H(X; Y) > 0 is given by the root of the derivative dH(X;Y)/dq. Thus, we must solve dH(X; f(b)} (K10) Using the definition of f in Eq. (K8) and going through elementary calculations yields the following solution for q: (K11) Capacity of binary communication channels (K12) Concerning the continuity of the above solution in the case a + b = 1, see the note at the end of this appendix. The optimal distribution defined in Eqs. (K11) and (K12) can now be substituted into Eq. (K10). After elementary calculation, we obtain: C = max H(X;Y) (K13) log(1 Define (K14) and substitute U in Eq. (K13) to get the channel capacity: C = log(1+2) + = U (K15) (K16) The optimal probability distribution q = p(x1) = 1 - p(x2) defined in Eqs. (K11) and (K12) seemingly has a pole in a + b = 1. I show herewith that it is actually defined over the whole plane a, b E [0, 1], namely, that it is analytically defined in the limiting case a+b = 1. I shall establish this by first setting a +b=1 - E in Eq. (K11), then using the Taylor expansion of the function f(a+e) to the second order, i.e., which gives By substituting this result in Eq. (K11), and expanding the exponential term, one easily finds 1/(1 +2 W 21 a - E/2, which gives the limit q = 1/2, corresponding to Appendix K the uniform distribution. However, such a distribution does not maximize the mutual information H(Y; X), since we have seen that in the limiting case a + b = 1 we have H(Y;X) = 0. I show next that the channel capacity is also defined over the plane a,be [0, 1], including the limiting case a +b = 1. Indeed, using the same Taylor expansion as previously, we easily obtain ~loga. Substituting this result and 1/(1 + 2 a - E/2 2 a into Eq. (K15) yields C = log(1+2") ++~~ - log a + log a = 0, which is the expected channel capacity in the limiting case a + b = 1. The function V is also found to take the limit V 2 log(1 - a), which, from Eq. (K15), also gives C = log(2U + 2r) 22 0. Appendix L (Chapter 13) Converse proo