Evidently, ν and µ have identical marginal distributions: ν$_{X}$$_{i}$ = µ$_{X}$$_{i}$ , and also identical joint distributions of any consecutive pairs: ν$_{X}$$_{i}$$_{,X}$$_{i}$$_{+1}$ = µ$_{X}$$_{i}$$_{,X}$$_{i}$$_{+1}$ . Therefore and Since µ ∈ Λ( γ, q, c ) , we have ν ∈ Λ( γ, q, c ) . Let { X$_{i}$ } follow distribution µ and { Z$_{i}$ } follow distribution ν . Then where equality holds if and only if { X$_{i}$ } is a first-order Markov process. By (11) and (105), L ( ν ) ≥ L ( µ ) . So for any µ which maximizes L ( µ ) , ν can be generated from µ by (99) with L ( ν ) ≥ L ( µ ) . L ( µ ) must be maximized by a first-order Markov process. Property (c): Suppose ν is a stationary fist-order Markov process, sufficiently denote as ν = {X , P$_{X}$$_{2}$$_{|}$$_{X}$$_{1}$ } , where X is the state space of ν and P$_{X}$$_{2}$$_{|}$$_{X}$$_{1}$ is the transition probability distribution. Define a new first-order Markov process ¯ ν from ν as follows. Definition 1: Let ¯ ν , defined on the same state space X as ν , be a first-order Markov process denoted by ( X , P$_{Z}$$_{2}$$_{|}$$_{Z}$$_{1}$ ) , where P$_{Z}$$_{2}$$_{|}$$_{Z}$$_{1}$ ( z$_{2}$ | z$_{1}$ ) =            α z$_{1}$ = 0 z$_{2}$ = 0 , 1 − β z$_{1}$ = 0 z$_{2}$ = 0 , 1 − α η P$_{X}$ ( z$_{2}$ ) z$_{1}$ = 0 z$_{2}$ = 0 , β η P$_{X}$ ( z$_{2}$ ) z$_{1}$ = 0 z$_{2}$ = 0 , (106) 22 ν$_{X}$$_{i}$ ( { 0 } ) = µ$_{X}$$_{i}$ ( { 0 } ) (100) ν$_{X}$$_{i}$$_{,X}$$_{i}$$_{+1}$ ( { x$_{i}$ = 0 , x$_{i}$$_{+1}$ = 0 } ) = µ$_{X}$$_{i}$$_{,X}$$_{i}$$_{+1}$ ( { x$_{i}$ = 0 , x$_{i}$$_{+1}$ = 0 } . (101) I ( Z$_{1}$ ; Z ∞ 2 ) = I ( Z$_{1}$ ; Z$_{2}$ ) + I ( Z$_{1}$ ; Z ∞ 3 | Z$_{2}$ ) (102) = I ( Z$_{1}$ ; Z$_{2}$ ) (103) = I ( X$_{1}$ ; X$_{2}$ ) (104) ≤ I ( X$_{1}$ ; X ∞ 2 ) (105)