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Tom
Tom
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No. Let $Y,Z$ be iid Bernoulli(1/2), and let $X=Y+Z$ mod 2, which induces pairwise independence. TheThe only non-injective deterministic functions $f_Y,f_Z$ are constants, rendering the RHS zero. For the LHS, we can take $f_{Y,Z}(Y,Z)$ equal to the the binary AND of $Y$ and $Z$, which is not injective, and not independent of $X$. Hence, the LHS is something positive, showing there can be strict inequality.

No. Let $Y,Z$ be iid Bernoulli(1/2), and let $X=Y+Z$ mod 2. The only non-injective deterministic functions $f_Y,f_Z$ are constants, rendering the RHS zero. For the LHS, we can take $f_{Y,Z}(Y,Z)$ equal to the the binary AND of $Y$ and $Z$, which is not injective, and not independent of $X$. Hence, the LHS is something positive, showing there can be strict inequality.

No. Let $Y,Z$ be iid Bernoulli(1/2), and let $X=Y+Z$ mod 2, which induces pairwise independence. The only non-injective deterministic functions $f_Y,f_Z$ are constants, rendering the RHS zero. For the LHS, we can take $f_{Y,Z}(Y,Z)$ equal to the the binary AND of $Y$ and $Z$, which is not injective, and not independent of $X$. Hence, the LHS is something positive, showing there can be strict inequality.

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Tom
Tom
  • 716
  • 5
  • 14

No. Let $Y,Z$ be iid Bernoulli(1/2), and let $X=Y+Z$ mod 2. The only non-injective deterministic functions $f_Y,f_Z$ are constants, rendering the RHS zero. For the LHS, we can take $f_{Y,Z}(Y,Z)$ equal to the the binary AND of $Y$ and $Z$, which is not injective, and not independent of $X$. Hence, the LHS is something positive, showing there can be strict inequality.