LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Finite type-S and type-H prior draws are L2 probability densities

Statement

Every finite type-S draw and every type-H draw in Type-S log-density and type-H histogram prior families is a well-defined nonnegative $L^{2} ([0, 1])$ probability density.

Facts & Assumptions

Given: A finite type-S coefficient array or a type-H Dirichlet vector, and the bounded finite-level wavelet basis and normalized prior constructions of Type-S log-density and type-H histogram prior families.

[F1]

Type-S log-density and type-H histogram prior families: type S sets $f = e^{T - c (T)}$ for a finite bounded wavelet sum $T$ , with $c (T) = \log \int_{0}^{1} e^{T}$ ; type H sets $f = 2^{L} ω_{k}$ on dyadic cells of length $2^{- L}$ , where $ω_{k} \geq 0$ and $\sum_{k} ω_{k} = 1$ .

Proof

1.1F1givenalgebra

For a type-S draw, boundedness of the finite wavelet sum gives $M = ∥ T ∥_{\infty} < \infty$ , hence $e^{- M} \leq e^{T (x)} \leq e^{M}$ on $[0, 1]$ . Therefore $0 < e^{- M} \leq \int_{0}^{1} e^{T (x)} d x \leq e^{M} < \infty$ , so $c (T)$ is finite and $f = e^{T - c (T)}$ is positive and integrates to one. The same bounds make $f$ bounded, so $f \in L^{2} ([0, 1])$ .

2.1F1givenalgebra∎

For type H, each dyadic cell has length $2^{- L}$ , so $\int_{0}^{1} f (x) d x = \sum_{k < 2^{L}} 2^{L} ω_{k} 2^{- L} = \sum_{k} ω_{k} = 1$ . Also $∥ f ∥_{2}^{2} = 2^{L} \sum_{k} ω_{k}^{2} \leq 2^{L} (\sum_{k} ω_{k})^{2} = 2^{L} < \infty$ , since all weights are nonnegative. Thus every type-H draw is a nonnegative $L^{2}$ probability density.

Depends on

Type-S log-density and type-H histogram prior families

Used by

Cited to discharge well-definedness by Type-S log-density and type-H histogram prior families.

Dependency tree · two levels

2 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

Castillo and Nickl, On the Bernstein–von Mises phenomenon for nonparametric Bayes procedures, §3.2 (standard reference, not scraped)
Giné and Nickl, Mathematical Foundations of Infinite-Dimensional Statistical Models, §7.4 Notes (standard reference, not scraped)