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A Gaussian rate-distortion optimizer need not have a density

Statement

Let $X \sim N (0, σ^{2})$ with $σ^{2} > 0$ , use squared-error distortion, and choose $D \geq σ^{2}$ . The constant reconstruction $Y = 0$ attains the Gaussian information rate-distortion value zero, but its law $δ_{0}$ has no density with respect to Lebesgue measure.

Facts & Assumptions

Given: AC, a nondegenerate centered Gaussian source, squared-error distortion, and $D \geq σ^{2}$ .

[A1]

The Axiom of Choice: the Gaussian rate-distortion supplier assumes AC.

[F1]

General single-letter information rate-distortion function: the information rate is the infimum of mutual information over joint laws meeting the expected-distortion constraint.

[F2]

Mutual information for standard-Borel random elements: mutual information is the joint-law KL from the product of the marginals.

[F3]

Gaussian squared-error rate-distortion function: for $D \geq σ^{2}$ , the Gaussian squared-error information rate is zero.

[F4]

The measure with density $f$ relative to $μ$ : having a density with respect to Lebesgue measure requires absolute continuity with respect to that measure.

Proof

1.1A1F1F2F3givenconstruct

The constant choice $Y = 0$ has expected distortion $E X^{2} = σ^{2} \leq D$ . Since $Y$ is constant, $P_{X Y} = P_{X} \otimes δ_{0} = P_{X} \otimes P_{Y}$ , so [F2] gives mutual information zero. By [F3], zero is the minimum rate; hence this reconstruction is optimal.

2.1F4step 1.1givenalgebra∎

The singleton ${0}$ has Lebesgue measure zero but $δ_{0} ({0}) = 1$ . Thus $δ_{0}$ is not absolutely continuous with respect to Lebesgue measure and, by [F4], cannot have a Lebesgue density.

Depends on

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Sources

Robert M. Gray, Source Coding Theory, first edition (standard reference, not scraped)
MIT OpenCourseWare 6.441, Chapter 25: Evaluating R(D) (standard reference, not scraped)