A T F B B T 92 If the transformations are clear from context then we simply say

# A t f b b t 92 if the transformations are clear from

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∈ A T , F ∈ B B T . (9.2) If the transformations are clear from context then we simply say that the chan- nel is stationary. Intuitively, a right shift of an output event yields the same probability as the left shift of an input event. The different shifts are required because in general only T A x and not T - 1 A x exists since the shift may not be invertible and in general only T - 1 B F and not T B F exists for the same reason. If the shifts are invertible, e.g., the processes are two-sided, then the definition is equivalent to ν T A x ( T B F ) = ν T - 1 A x ( T - 1 B F ) = ν x ( F ) , all x A T , F ∈ B B T (9.3) that is, shifting the input sequence and output event in the same direction does not change the probability. The fundamental importance of the stationarity of a channel is contained in the following lemma. Lemma 9.3.1 If a source [ A, μ ] , stationary with respect to T A , is connected to channel [ A, ν, B ] , stationary with respect to T A and T B , then the resulting hookup μν is also stationary (with respect to the cartesian product shift T = T A × B = T A × T B defined by T ( x, y ) = ( T A x, T B y ) ). Proof: We have that μν ( T - 1 F ) = Z ( x ) ν x (( T - 1 F ) x ) . Now ( T - 1 F ) x = { y : T ( x, y ) F } = { y : ( T A x, T B y ) F } = { y : T B y F T A x } = T - 1 B F T A x 9.3. STATIONARITY PROPERTIES OF CHANNELS 165 and hence μν ( T - 1 F ) = Z ( x ) ν x ( T - 1 B F T A x ) . Since the channel is stationary, however, this becomes μν ( T - 1 F ) = Z ( x ) ν T A x ( F T A x ) = Z dμT - 1 A ( x ) ν x ( F x ) , where we have used the change of variables formula. Since μ is stationary, however, the right hand side is Z ( x ) ν x ( F ) , which proves the lemma. 2 Suppose next that we are told that a hookup μν is stationary. Does it then follow that the source μ and channel ν are necessarily stationary? The source must be since μ ( T - 1 A F ) = μν (( T A × T B ) - 1 ( F × B T )) = μν ( F × B T ) = μ ( F ) . The channel need not be stationary, however, since, for example, the stationarity could be violated on a set of μ measure 0 without affecting the proof of the above lemma. This suggests a somewhat weaker notion of stationarity which is more directly related to the stationarity of the hookup. We say that a channel [ A, ν, B ] is stationary with respect to a source [ A, μ ] if μν is stationary. We also state that a channel is stationary μ -a.e. if it satisfies (9.2) for all x in a set of μ -probability one. If a channel is stationary μ -a.e. and μ is stationary, then the channel is also stationary with respect to μ . Clearly a stationary channel is stationary with respect to all stationary sources. The reason for this more general view is that we wish to extend the definition of stationary channels to asymptotically mean stationary channels. The general definition extends; the classical definition of stationary channels does not. Observe that the various definitions of stationarity of channels immediately extend to block shifts since they hold for any shifts defined on the input and output sequence spaces, e.g., a channel stationary with respect to T N A and T K B could be a reasonable model for a channel or code that puts out K symbols  #### You've reached the end of your free preview.

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