Lemma 1, with the independence it needs — exceptional items lower bound E[max_j X_j] ≥ L/2
ProvedStochLoadBal.Makespan.lemma_1Let and let () be independent, integrable random variables, each taking either the value or a value at least . If
then
This is the lower bound the paper uses for exceptional parts: applied with 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 copies of with one has but ) 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.
import Mathlib import Definitions.Def_StochLoadBal_Makespan_Model
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
Confirmed by the mission captain (proposal self-audit).
Confirmed by the moderator at approval.