content_image/296734.jpg content_image/296731.jpg content_image/296732.jpg content_image/296733.jpg content_image/296728.jpg content_image/296727.jpg content_image/296729.jpg content_image/296730.jpg
where 

and 

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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 

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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$_{1}$ ) . 

