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Exponential mechanism privacy and utility

Statement

Let D be the set of finite-support record-count databases, with neighbors defined by adding or removing one record occurrence. Let R be a nonempty finite set independent of the database, and let u:D×R→R be a real-valued utility with finite positive attained global utility sensitivity Δu. For ε>0, use weights wD(r)=exp⁡(εu(D,r)/(2Δu)) and normalize them over R as in the exponential mechanism. Its output law is (ε,0)-differentially private. For a fixed database D, let OPT⁡u(D)=max⁡r∈Ru(D,r),ROPT⁡={r∈R:u(D,r)=OPT⁡u(D)}. For every t>0, if M(D) denotes the random output, then

Pr⁡ ⁣[u(D,M(D))≤OPT⁡u(D)−2Δuε(ln⁡∣R∣∣ROPT⁡∣+t)]≤e−t.

Facts & Assumptions

Given: A nonempty finite database-independent range R, utility u, ε>0, Δu>0, neighboring databases D∼D′, and t>0.

[F1]

Exponential mechanism assigns weight wD(r)=exp⁡(εu(D,r)/(2Δu)), normalizer ZD=∑r∈RwD(r), and output mass wD(r)/ZD.

[F2]

Utility sensitivity gives ∣u(D,r)−u(D′,r)∣≤Δu for every neighboring pair and every r∈R.

[F3]

Finiteness and nonemptiness of R make OPT⁡u(D) attained and 1≤∣ROPT⁡∣≤∣R∣.

[F4]

Differential privacy requires the event inequality for both ordered directions of every neighboring pair.

[F5]

Natural logarithm and its inverse/exponential laws give e−ln⁡x=1/x for x>0.

[F6]

The exponential addition formula, strict monotonicity, and positivity and the reciprocal identity give exp⁡(u+v)=exp⁡(u)exp⁡(v), make exp⁡ increasing, and give exp⁡(−u)=1/exp⁡(u)>0 for real u,v.

Proof

1.1F1F2F4F6givenalgebra

For every r∈R, [F2] and monotonicity and addition for exp⁡ from [F6] give wD(r)≤eε/2wD′(r); summing gives ZD≤eε/2ZD′. Applying the same argument with D,D′ interchanged gives ZD′≤eε/2ZD. Since the weights and normalizers are positive, [F6] then yields (wD(r)/ZD)/(wD′(r)/ZD′)≤eε. Interchanging D,D′ gives the reverse point-mass bound. Summing over any event A⊆R gives both (ε,0) event inequalities, so [F4] proves privacy.

2.1F1F3F5F6givenalgebra∎

By [F3], the maximizing set is nonempty and its elements each have weight eεOPT⁡u(D)/(2Δu). Let B be the event in the Statement. For each r∈B, its defining score inequality, monotonicity and addition for exp⁡ from [F6], and [F5] imply wD(r)≤eεOPT⁡u(D)/(2Δu)(∣ROPT⁡∣/∣R∣)e−t. There are at most ∣R∣ such elements, while the normalizer is at least the sum of the maximizing weights, namely ∣ROPT⁡∣eεOPT⁡u(D)/(2Δu). Therefore the probability of B is at most e−t. If B is empty its probability is zero and the same bound holds.

Depends on

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