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A finite kernel criterion for perfect privacy

Statement

Let $(S, X)$ have a finite joint law and let $K (z ∣ x)$ be a finite release kernel, giving $P_{S X Z} = P_{S X} K_{Z ∣ X}$ as in Finite-source privacy–utility region. Then $I (S; Z) = 0$ if and only if, for each $z$ with $P_{Z} (z) > 0$ , the conditional mixture

$\sum_{x} P_{X ∣ S = s} (x) K (z ∣ x)$

is independent of $s$ over $supp P_{S}$ . No full-support assumption is imposed; zero-mass values of $s$ are not constrained.

Facts & Assumptions

Given: A finite joint law $P_{S X}$ and a finite stochastic release kernel $K (z ∣ x)$ .

[F1]

Finite-source privacy–utility region: $P_{S X Z} (s, x, z) = P_{S X} (s, x) K (z ∣ x)$ , so $P_{Z ∣ S = s} (z) = \sum_{x} P_{X ∣ S = s} (x) K (z ∣ x)$ for $P_{S} (s) > 0$ .

[F2]

Finite mutual information: $I (S; Z) = D (P_{S Z} ∥ P_{S} \otimes P_{Z})$ .

[F3]

Finite KL divergence is nonnegative with equality exactly at equal laws: $D (P ∥ Q) = 0$ if and only if the finite probability vectors $P, Q$ are equal; a positive numerator above a zero denominator gives $+ \infty$ .

Proof

1.1F1F2F3givenalgebra

If $I (S; Z) = 0$ , [F2] and [F3] give $P_{S Z} = P_{S} \otimes P_{Z}$ . For every $s$ with $P_{S} (s) > 0$ , divide the equality $P_{S Z} (s, z) = P_{S} (s) P_{Z} (z)$ by $P_{S} (s)$ and use [F1]. The conditional mixture in the Statement is therefore $P_{Z} (z)$ , independent of $s$ for every $z$ , in particular for every positive-mass $z$ .

1.2F1givenalgebra

Conversely, assume the stated mixture is a constant $c_{z}$ in $s$ on $supp P_{S}$ for each $z$ with $P_{Z} (z) > 0$ . Averaging it over $P_{S}$ shows $c_{z} = P_{Z} (z)$ . If $P_{Z} (z) = 0$ , [F1] gives $0 = P_{Z} (z) = \sum_{s : P_{S} (s) > 0} P_{S} (s) P_{Z ∣ S = s} (z)$ , a finite sum of nonnegative terms with positive weights; hence each $P_{Z ∣ S = s} (z) = 0 = P_{Z} (z)$ . Thus for every $s$ in the support and every $z$ , $P_{Z ∣ S = s} (z) = P_{Z} (z)$ , so [F1] yields $P_{S Z} = P_{S} \otimes P_{Z}$ .

2.1F2F3step 1.2∎

Applying [F2] and [F3] to this equality gives $I (S; Z) = 0$ . This proves both directions, including zero-probability release symbols.

Depends on

Used by

A finite information-bottleneck privacy model Example

Dependency tree · two levels

10 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Ali Makhdoumi, Salman Salamatian, Nadia Fawaz, and Muriel Médard, From the Information Bottleneck to the Privacy Funnel, ITW 2014 (standard reference, not scraped)
Abbas El Gamal and Young-Han Kim, Network Information Theory, arXiv:1001.3404v3 (standard reference, not scraped)