Data processing inequality
LemmaData processing for a single variable#
Let \(X\) be a random variable. Then for any function \(f\) on the range of \(X\), one has \(\mathbb{H}[f(X)] \leq \mathbb{H}[X]\).
Proof
We have
thanks to Entropy and relabeling and Chain rule, giving the claim.
Lean
Formalized as ProbabilityTheory.entropy_comp_le.
Sources
Uses 2 · Used by 1
LemmaOne-sided unconditional data processing inequality#
Let \(X,Y\) be random variables. For any function \(f, g\) on the range of \(X\), we have \(\mathbb{I}[f(X) : Y] \leq \mathbb{I}[X:Y]\).
Proof
By Alternative formulae for mutual information it suffices to show that \(\mathbb{H}[Y|X] \leq \mathbb{H}[Y|f(X)]\). But this follows from Submodularity (and Entropy and relabeling).
Lean
Formalized as ProbabilityTheory.mutual_comp_le.
Sources
Uses 3 · Used by 1
LemmaUnconditional data processing inequality#
Let \(X,Y\) be random variables. For any functions \(f, g\) on the ranges of \(X, Y\) respectively, we have \(\mathbb{I}[f(X) : g(Y )] \leq \mathbb{I}[X : Y]\).
Proof
From One-sided unconditional data processing inequality, Commutativity and associativity of joint entropy we have \(\mathbb{I}[f(X) : Y] \leq \mathbb{I}[X:Y]\) and \(\mathbb{I}[f(X): g(Y)] \leq \mathbb{I}[f(X):Y]\), giving the claim.
Lean
Formalized as ProbabilityTheory.mutual_comp_comp_le.
Sources
Uses 2 · Used by 1
LemmaData processing inequality#
Let \(X,Y,Z\). For any functions \(f, g\) on the ranges of \(X, Y\) respectively, we have \(\mathbb{I}[f(X) : g(Y )|Z] \leq \mathbb{I}[X :Y |Z]\).
Proof
Apply Unconditional data processing inequality to \(X,Y\) conditioned to the event \(Z=z\), multiply by \({\bf P}[Z=z]\), and sum using Conditional mutual information.
Lean
Formalized as ProbabilityTheory.condMutual_comp_comp_le.
Sources
Uses 2 · Used by 1
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