Suppose we have a 1-Lipschitz function $f$ such that 1-Lipschitzness is preserved, with $D_A(f(X), f(Y)) \leq D_B(X, Y)$ for some metric spaces $A$ and $B$.

Does this also imply that $I(f(X); f(Y)) = I(X;Y)$? It seems intuitive that this would be the case if the 1-lipschitz function is injective, but I don't know if this is true, so that the mutual information is preserved under such invertible transformation. Is there a way to prove this?


1 Answer 1


No, it does not. Let $X$ and $Y$ be non-independent $A$-valued random variables, and let $f:A\to B$ be a constant function. Then $I(f(X);f(Y))=0<I(X;Y)$.

  • 3
    $\begingroup$ Note the OP assumed $f$ is injective $\endgroup$ Oct 31, 2019 at 3:17
  • $\begingroup$ Whoops! Thank you and sorry. $\endgroup$ Oct 31, 2019 at 10:39
  • $\begingroup$ Sorry I should have made it clearer: I don't know if $f$ is injective or not, but from @e.lipnowski comments it seems that $f$ isn't injective even if it's 1-Lipschitz, in which case I should suppose the MI is not preserved? $\endgroup$
    – minimore99
    Oct 31, 2019 at 11:08

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