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Laplace noise has its stated Lebesgue density

Statement

Assume Countable Choice. For every $b > 0$ , the probability measure $Lap (b)$ is absolutely continuous with respect to Lebesgue measure on $R$ and has density

z ⟼ \frac{1}{2 b} e^{- ∣ z ∣ / b} .

Statement references

The mixture definition of Laplace distribution with scale b

Dwork and Roth, The Algorithmic Foundations of Differential Privacy, Definition 3.2, printed p. 31.

Vadhan, The Complexity of Differential Privacy, §1.5, PDF pp. 7–8.

Facts & Assumptions

Given: $b > 0$ and a Borel set $A \subseteq R$ .

[A1]

Countable Choice is the axiom assumed by the cited exponential-law and Lebesgue change-of-variables suppliers; no full Axiom of Choice is invoked.

[F1]

Laplace distribution: $Lap (b) (A) = \frac{1}{2} μ_{b} (A \cap (0, + \infty)) + \frac{1}{2} μ_{b} ({x > 0 : - x \in A})$ , where $μ_{b}$ is the exponential probability law of rate $1 / b$ .

[F2]

Exponential-law density: for $λ > 0$ , the density is $f_{λ} (t) = λ e^{- λ t}$ when $t > 0$ and $0$ otherwise, and it defines a probability law.

[F3]

Lebesgue change of variables: a $C^{1}$ diffeomorphism $T$ transforms nonnegative Lebesgue integrals by the absolute Jacobian; for $T (x) = - x$ the absolute Jacobian is $1$ .

[F4]

Measure of a degenerate interval: a one-dimensional box with equal endpoints has Lebesgue measure $0$ .

Proof

1.1A1F1F2F3F4givenconstructalgebra

Put $A_{+} = A \cap (0, + \infty)$ and $A_{-} = A \cap (- \infty, 0)$ . By [F1] and [F2], the positive mixture half contributes $\frac{1}{2} \int_{A_{+}} b^{- 1} e^{- x / b} d λ (x)$ . Applying [F3] to reflection $T (x) = - x$ and $y \mapsto 1_{A_{-}} (y) b^{- 1} e^{y / b}$ shows that the reflected half contributes $\frac{1}{2} \int_{A_{-}} b^{- 1} e^{y / b} d λ (y)$ . On $A_{+}$ , $e^{- x / b} = e^{- ∣ x ∣ / b}$ ; on $A_{-}$ , $e^{y / b} = e^{- ∣ y ∣ / b}$ . The singleton at zero has Lebesgue measure zero by [F4]. Hence $Lap (b) (A) = \int_{A} (2 b)^{- 1} e^{- ∣ z ∣ / b} d λ (z)$ .

2.1step 1.1givenalgebra∎

The identity in step 1.1 holds for every Borel set $A$ . Its integrand is nonnegative and measurable; thus every Lebesgue-null Borel set has zero $Lap (b)$ measure, and the displayed function is the claimed Radon–Nikodym density.

Depends on

Used by

Finite independent Laplace noise has product density Lemma

Dependency tree · two levels

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Sources

Cynthia Dwork and Aaron Roth, The Algorithmic Foundations of Differential Privacy (standard reference, not scraped)
Salil Vadhan, The Complexity of Differential Privacy, 2017 manuscript (standard reference, not scraped)