LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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An absolutely continuous upper quantile has its specified tail mass

Statement

Let $Z$ be a real random variable whose law is absolutely continuous with respect to Lebesgue measure on $R$ . For $0 < u < 1$ , define

Q_{u} (Z) = \sup {t \in R : P (Z \geq t) \geq 1 - u} .

Then $Q_{u} (Z)$ is finite and

P (Z \geq Q_{u} (Z)) = P (Z > Q_{u} (Z)) = 1 - u .

Facts & Assumptions

Given: A real random variable $Z$ with an absolutely continuous law and $0 < u < 1$ .

[A1]

Countable Choice is used by the published singleton-null result, which is the only reason for an axiom assumption here (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Every at most countable subset of $R^{n}$ is Lebesgue null; in particular $λ_{1} (Q) = 0$ ).

[F1]

Absolute continuity means every Lebesgue-null Borel set has probability zero (Absolute continuity of a signed or complex measure with respect to a positive measure).

[F2]

Every singleton in $R$ is Lebesgue null under Countable Choice (Every at most countable subset of $R^{n}$ is Lebesgue null; in particular $λ_{1} (Q) = 0$ ).

[F3]

A probability measure has total mass one and is continuous from below and from above; the latter applies because the first term has mass at most one (Probability measures and probability spaces, Continuity from below for measures, Continuity from above when one set has finite measure).

[F4]

Every nonempty subset of $R$ that is bounded above has a finite least upper bound (The Cauchy-sequence reals have the least-upper-bound property).

Proof

technique · direct

Given: Countable Choice, a real random variable $Z$ with an absolutely continuous law, and $0 < u < 1$ .

1.1A1F1F2F3givenalgebra

For each $t \in R$ , the event ${Z = t}$ is measurable and its singleton is Lebesgue null by [F2]. Absolute continuity [F1] gives $P (Z = t) = 0$ . Therefore, if $t_{n} ↑ t$ , the events ${Z \geq t_{n}}$ decrease to ${Z \geq t}$ and [F3] gives $P (Z \geq t_{n}) ↓ P (Z \geq t)$ . If $t_{n} ↓ t$ with $t_{n} > t$ , these events increase to ${Z > t}$ , so [F3] gives convergence to $P (Z > t) = P (Z \geq t)$ . Thus the nonincreasing tail function $S (t) = P (Z \geq t)$ is continuous at every $t$ .

1.2F3F4givenalgebra

As $n \to \infty$ , the events ${Z \geq n}$ decrease to the empty set and ${Z \geq - n}$ increase to the whole sample space. By [F3], $S (n) \to 0$ and $S (- n) \to 1$ . Since $1 - u \in (0, 1)$ , the set $A = {t : S (t) \geq 1 - u}$ is nonempty and bounded above; [F4] gives its finite supremum. Continuity of $S$ makes $A$ closed, so that supremum $Q_{u} (Z)$ belongs to $A$ .

2.1step 1.1step 1.2algebra∎

If $S (Q_{u} (Z)) > 1 - u$ , continuity at $Q_{u} (Z)$ would give some $t > Q_{u} (Z)$ with $S (t) > 1 - u$ , contradicting that $Q_{u} (Z)$ is an upper bound for $A$ . Hence $S (Q_{u} (Z)) = 1 - u$ . Step 1.1 gives $P (Z = Q_{u} (Z)) = 0$ , so $P (Z > Q_{u} (Z)) = P (Z \geq Q_{u} (Z)) = 1 - u$ .

Depends on

Used by

The claimed univariate clean-quantile count exponent is too strong Counterexample
The excess-tail functional cannot upper-bound the trimmed mean Counterexample
Univariate trimmed-mean bound under sample replacement Theorem

Dependency tree · two levels

36 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

Gábor Lugosi and Shahar Mendelson, Robust Multivariate Mean Estimation: The Optimality of Trimmed Mean, §2.1 (standard reference, not scraped)
Gábor Lugosi, Mean Estimation in High Dimension, §2 (standard reference, not scraped)