where and The process ¯ ν is described by ( X , α, β, P$_{X}$ ) . It is easy to prove that the stationary distribution P$_{Z}$ of ¯ ν is equal to P$_{X}$ of ν , ¯ ν ∈ Λ( γ, q, c ) . Moreover, ¯ ν satisfies the same power and duty cycle constraint ν satisfies, i.e., ¯ ν ∈ Λ( γ, q, c ) . Furthermore let B$_{i}$ = 1$_{{}$$_{X}$$_{i}$$_{̸}$$_{=0}$$_{}}$ , then Let b$_{i}$ = 1$_{{}$$_{z}$$_{i}$$_{̸}$$_{=0}$$_{}}$ . Since Z$_{i}$ and Z$_{i}$$_{+1}$ are independent given B$_{i}$ = 1$_{{}$$_{Z}$$_{i}$$_{̸}$$_{=0}$$_{}}$ and B$_{i}$$_{+1}$ = 1$_{{}$$_{Z}$$_{i}$$_{+1}$$_{̸}$$_{=0}$$_{}}$ . Based on (106) to (113), it is easy to see that The inequality in (117) follows since X$_{1}$ → X$_{2}$ → B$_{2}$ forms a Markov chain then I ( X$_{1}$ ; B$_{2}$ ) ≤ I ( X$_{1}$ ; X$_{2}$ ) [21] and B$_{2}$ → X$_{1}$ → B$_{1}$ also forms a Markov chain then I ( B$_{2}$ ; B$_{1}$ ) ≤ I ( B$_{2}$ ; X$_{1}$ ) . 23 I ( Z$_{1}$ ; Z$_{2}$ ) = E$^{{}$ log P$_{Z}$$_{2}$$_{|}$$_{Z}$$_{1}$ ( Z$_{2}$ | Z$_{1}$ ) P$_{Z}$$_{2}$ ( Z$_{2}$ ) } (114) = E$^{{}$ log P$_{B}$$_{2}$$_{|}$$_{B}$$_{1}$ ( B$_{2}$ | B$_{1}$ ) P$_{B}$$_{2}$ ( B$_{2}$ ) } (115) = I ( B$_{1}$ ; B$_{2}$ ) (116) ≤ I ( X$_{1}$ ; X$_{2}$ ) . (117) P$_{Z}$$_{2}$$_{|}$$_{Z}$$_{1}$ ( z$_{2}$ | z$_{1}$ ) = P$_{B}$$_{2}$$_{|}$$_{B}$$_{1}$ ( b$_{1}$ | b$_{2}$ ) P$_{X}$ ( z$_{2}$ ) P$_{B}$$_{2}$ ( b$_{2}$ ) , (113) P$_{B}$$_{2}$$_{|}$$_{B}$$_{1}$ (0 | 0) = α (111) P$_{B}$$_{2}$$_{|}$$_{B}$$_{1}$ (1 | 1) = β. (112) α = P$_{X}$$_{2}$$_{|}$$_{X}$$_{1}$ (0 | 0) (108) β = P ( X$_{2}$ ∈ S$_{1}$ | X$_{1}$ ∈ S$_{1}$ ) (109) η = P ( X ∈ S$_{1}$ ) . (110) S$_{1}$ = X \ { 0 } (107)