lecture9 notes

Y x e x x x x px x note also that xx px x

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Unformatted text preview: n � � P (y |x) log(PX |Y (x|y )) e y∈Y Y |X �� � PX (x) = � � ) log(P �X |Y (x�|y)) PY |X (y |x y ∈Y x� ∈X e � � � PY |X (y |x) PY |X (y |x) log � y ∈Y PX (x� )PY |X (y |x� ) x� ∈X PX (x)e ⎛ = � �⎞ � PY |X (y |x) PY |X (y |x� ) log � ⎜ ⎟ y ∈Y � PX (x� )PY |X (y |x� ) � ) ⎜e ⎟ x� ∈X x� ∈X PX (x ⎝ ⎠ Note also that � x∈X PX (x) = 1. This may b e very hard to solve. Arimoto-Blahut Proof: The first two statements follow immedi­ ately from our lemma For any value of x where PX |Y (x|y ) = 0, PX (x) should b e set to 0 to obtain the max­ imum. To find the maximum over the PMF PX , let us first ignore the constraint of p ositiv­ ity and use a Lagrange multiplier for the � x PX (x) = 1 Then ∂ � � { x∈X y∈Y PX (x)PY |X (y |x) log ∂ PX (x) � λ ( x∈X PX (x) − 1)} = 0 � � �X |Y (x|y ) P + PX (x) Arimoto-Blahut Equivalently � � � � − log(PX (x))−1+ y∈Y PY |X (y |x) lo...
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This document was uploaded on 03/19/2014 for the course EECS 6.441 at MIT.

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