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Uniform samples have logarithmic fill-distance rate on a cube
Statement
Assume the Axiom of Countable Choice. Let , let be a closed cube with , and let be independent random vectors, each distributed according to normalized Lebesgue measure on . For , define the Euclidean fill distance
There are constants and such that, for every ,
Statement references
Facts & Assumptions
Given: Countable Choice, the cube , dimension , and independent uniform sample as in the Statement.
Countable Choice means every countable family of nonempty sets has a choice function (The Axiom of Countable Choice ()).
Lebesgue outer measure agrees with volume on half-open boxes (Assuming countable choice, Lebesgue outer measure is an outer measure that restricts to elementary volume, Lebesgue measurable sets, the family , and the restricted set function ).
The probability of a finite union of events is at most the sum of their probabilities (The finite union bound).
Independence means mutual independence of the whole family of random elements (Identical distribution and IID families).
Under Countable Choice, Lebesgue measure is a complete measure extending volume on half-open boxes, and every coordinate hyperplane is measurable and null (Assuming countable choice, is a sigma-algebra containing every elementary set and is a complete measure extending elementary volume, A box with a degenerate side is Lebesgue null, and so is every coordinate hyperplane in ).
Proof
For large enough that , put and partition each coordinate interval of into equal half-open intervals, closing the last interval at its right endpoint. This gives disjoint cells covering . Each cell differs from a half-open box with the same coordinate bounds only on its boundary faces, which lie in finitely many coordinate hyperplanes. By [F4] those hyperplanes and every subset of their union are measurable and null, so each cell is measurable and has the same measure as that half-open box, namely by [F1]. The same argument for the boundary of gives . Hence normalized Lebesgue measure assigns each cell probability .
For a fixed cell, independence [F3] gives probability that it is empty. By [F2], the probability that some cell is empty is at most , because and hence .
On the complementary event every cell contains a sample point. Every belongs to one cell, whose Euclidean diameter is ; therefore . Since implies , we obtain The event has probability at least by step 2.1.
Depends on
- The Axiom of Countable Choice ($\mathrm{AC}_\omega$)
- Lebesgue measurable sets, the family $\mathcal{L}(\mathbb{R}^n)$, and the restricted set function $\lambda_n$
- Assuming countable choice, Lebesgue outer measure is an outer measure that restricts to elementary volume
- Assuming countable choice, $\mathcal{L}(\mathbb{R}^n)$ is a sigma-algebra containing every elementary set and $\lambda_n$ is a complete measure extending elementary volume
- A box with a degenerate side is Lebesgue null, and so is every coordinate hyperplane in $\mathbb{R}^n$
- The finite union bound
- Identical distribution and IID families
Used by
Dependency tree · two levels
38 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
- Belkin, Hsu, Ma and Mandal, Reconciling modern machine learning practice and the bias-variance trade-off (standard reference, not scraped)