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math-Student
math-Student
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The answer is as follows. We need to show that

$$I(Z; X)-I(X; Y)\leq H(Z|Y).$$ The rightleft hand side of this is simplified as $H(X|Y)-H(X|Z)$, so we need to show that $H(X|Y)-H(X|Z)\leq H(Z|Y)$. Since conditioning reduces the entropy we have $$H(X|Y)-H(X|Z)\leq H(X|Y)-H(X|Y,Z)=I(X;Z|Y)\leq H(Z|Y).$$

The answer is as follows. We need to show that

$$I(Z; X)-I(X; Y)\leq H(Z|Y).$$ The right side of this is simplified as $H(X|Y)-H(X|Z)$, so we need to show that $H(X|Y)-H(X|Z)\leq H(Z|Y)$. Since conditioning reduces the entropy we have $$H(X|Y)-H(X|Z)\leq H(X|Y)-H(X|Y,Z)=I(X;Z|Y)\leq H(Z|Y).$$

The answer is as follows. We need to show that

$$I(Z; X)-I(X; Y)\leq H(Z|Y).$$ The left hand side of this is simplified as $H(X|Y)-H(X|Z)$, so we need to show that $H(X|Y)-H(X|Z)\leq H(Z|Y)$. Since conditioning reduces the entropy we have $$H(X|Y)-H(X|Z)\leq H(X|Y)-H(X|Y,Z)=I(X;Z|Y)\leq H(Z|Y).$$

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math-Student
math-Student
  • 1.1k
  • 7
  • 12

The answer is as follows. We need to show that

$$I(Z; X)-I(X; Y)\leq H(Z|Y).$$ The right side of this is simplified as $H(X|Y)-H(X|Z)$, so we need to show that $H(X|Y)-H(X|Z)\leq H(Z|Y)$. Since conditioning reduces the entropy we have $$H(X|Y)-H(X|Z)\leq H(X|Y)-H(X|Y,Z)=I(X;Z|Y)\leq H(Z|Y).$$