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lecture9 notes - LECTURE 9 Last time Channel capacity...

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LECTURE 9 Last time: Channel capacity Binary symmetric channels Erasure channels Maximizing capacity Lecture outline Maximizing capacity: Arimoto-Blahut Convergence Examples
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Arimoto-Blahut Lemma 1: I ( X ; Y ) = max P X ( x ) P Y X ( y x ) P ± X | Y y ∈Y x ∈X | | log P ± X Y ( x | y ) P | X ( x ) Proof: I ( X ; Y ) = ² x ∈X ² x ∈Y P X | Y ( x | y ) P Y ( y ) log ³ P X | Y ( x | y ) ´ P X ( x ) Recall: y ) = P X ( x ) P Y X ( y | x ) P X | Y ( x | ² x ∈X P X ( x ) | P Y | X ( y | x ) and P Y ( y ) = ² x ∈X P X ( x ) P Y | X ( y | x )
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± ± ± Arimoto-Blahut I ( X ; Y ) y ∈Y P X P X ( x ) P Y X ( y x ) | | x ∈X Y ( x y ) | | P X ( x ) log I ( X ; Y ) y ∈Y P X P Y ( y ) P X Y ( x y ) | = | x ∈X Y ( x y ) | | P X ( x ) log P X Y ( x y ) | | P X P Y ( y ) P X Y ( x y ) log | = | Y ( x y ) | | y ∈Y x ∈X 1 x ) (using log( x ) 1 P Y ( y ) P X Y ( x y ) | | ∈Y ∈X y x y ∈Y x ∈X 0 P Y ( y ) P ± X Y ( x | y ) = |
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Arimoto-Blahut Capacity is C = max max P X ( x ) P Y X ( y x ) P X P ± X | Y y ∈Y x ∈X | | P X Y ( x y ) log ± P | X ( x ) | For fixed P X , RHS is maximized when P X Y ( x y ) = P X ( x ) P Y | X ( y | x ) ± | | ² x ∈X P X ( x ) P Y | X ( y | x ) fixed P ± X | Y , RHS is maximized when P Y X ( y x ) log( P ± X Y ( x y )) e ² y ∈Y | | | | P X ( x ) = ² x ∈X ³ ² y ∈Y P Y | X ( y | x ) log( P ± X | Y ( x |
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lecture9 notes - LECTURE 9 Last time Channel capacity...

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