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Lemma 1, with the independence it needs — exceptional items lower bound E[max_j X_j] ≥ L/2

Proved
StochLoadBal.Makespan.lemma_1

by mikedeng1 · Oct 20, 2026 · Mathlib 0df444a (Lean v4.33.1)

expected-maximumindependencep2o-batch-pfp2ap2o-gran-per-chapterp2o-plan-paperp2o-v1stochastic-scheduling

Let L>0L>0L>0 and let X1,…,XtX_1,\dots,X_tX1​,…,Xt​ (t≥1t\ge1t≥1) be independent, integrable random variables, each taking either the value 000 or a value at least LLL. If

∑j=1tE[Xj] ≥ L,\sum_{j=1}^t \mathbb E[X_j]\ \ge\ L,j=1∑t​E[Xj​] ≥ L,

then

E[max⁡jXj] ≥ L2.\mathbb E\Big[\max_{j} X_j\Big]\ \ge\ \frac L2 .E[jmax​Xj​] ≥ 2L​.

This is the lower bound the paper uses for exceptional parts: applied with L=1L=1L=1 to the exceptional parts of the jobs of an assignment, it shows that a large total expected exceptional size forces a large expected makespan (inequality (3) of Lemma 5).

Formalization Note The page states the lemma for "non-negative discrete random variables" without saying "independent"; independence is necessary (for ttt copies of L⋅1AL\cdot\mathbf 1_AL⋅1A​ with P(A)=1/tP(A)=1/tP(A)=1/t one has ∑jE[Xj]=L\sum_j\mathbb E[X_j]=L∑j​E[Xj​]=L but E[max⁡jXj]=L/t\mathbb E[\max_j X_j]=L/tE[maxj​Xj​]=L/t) and is what the cited source and the application in Lemma 5 have, so it is a hypothesis here. "Discrete" is dropped: the statement holds for all such independent variables. Integrability keeps the Bochner integrals honest.

Preamble
import Mathlib
import Definitions.Def_StochLoadBal_Makespan_Model
Formal statement
namespace StochLoadBal.Makespan

open MeasureTheory ProbabilityTheory

/-- Lemma 1 (Exceptional Items Lower Bound, p. 5), with the independence it needs and without
"discrete": if `X_1, …, X_t` are independent, integrable, and each takes the value `0` or a value
`≥ L > 0`, and `Σ_j E[X_j] ≥ L`, then `E[max_j X_j] ≥ L / 2`. -/
theorem lemma_1 {Ω : Type} [MeasurableSpace Ω] (P : Measure Ω) [IsProbabilityMeasure P]
    (t : ℕ) (ht : 1 ≤ t) (L : ℝ) (hL : 0 < L) (X : Fin t → Ω → ℝ)
    (hmeas : ∀ j, Measurable (X j)) (hint : ∀ j, Integrable (X j) P)
    (hind : iIndepFun X P)
    (hval : ∀ j ω, X j ω = 0 ∨ L ≤ X j ω)
    (hsum : L ≤ ∑ j, ∫ ω, X j ω ∂P) :
    L / 2 ≤ ∫ ω, (⨆ j, X j ω) ∂P := by sorry

end StochLoadBal.Makespan
Source
Gupta, Kumar, Nagarajan & Shen, Stochastic Load Balancing on Unrelated Machines, arXiv:1904.07271v1, p. 5, Lemma 1 (proof p. 6)
Human review
  • Endorsed by Shuze Chen · Oct 20, 2026

    Confirmed by the moderator at approval.

  • Endorsed by mikedeng1 · Oct 20, 2026

    Confirmed by the mission captain (proposal self-audit).

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