国家重点实验室 。信道总容量保持不变，但是被不均等分配了 o Conservation: 2C(W)=I(X X2;YY)=I(UU,;YY) =I(U;YY)+1(U,;YY,IU) =I(U;YY)+I(U2;YY,U)=C(W-)+C(W+) o Extremization: C(W-)≤C(W)≤C(W+) with equality iff C(W)=0 or 1. 14信道总容量保持不变，但是被不均等分配了 Conservation: Extremization: with equality iff C(W) = 0 or 1. 14 ( ) ( ) 1 2 12 1 2 12 1 12 2 12 1 1 12 2 12 1 2( ) ( ; ) ( ; ) ( ; ) ; | ( ; ) ; ( ) ( ) C W I X X YY I UU YY I U YY I U YY U I U YY I U YYU C W C W − + = = = + = + =+ CW CW CW ( ) () ( ) − + ≤ ≤