Few indices are far above the average #
Let f be a function with natural numbers as values on a nonempty finite set Q. Division is
avoided: "f p is at most k times the average of f" is written
f p * |Q| ≤ k * ∑ p' ∈ Q, f p'.
- Fewer than
|Q|/kindices are more thanktimes as bad as the average (Finset.mul_card_above_average_lt, Markov's inequality). - So for
kfunctions there is one index at which each of them is at mostktimes its average (Finset.exists_forall_mul_card_le_mul_sum, the union bound). - Given a vertex map
v : I → V, nonnegative payoffsW : I → α → R, and vertex-wise sub-probability weightsp : V → α → R(0 ≤ p u aand∑ a, p u a ≤ 1), choosing the best label at each vertex produces a deterministic assignmentf : V → αwhose total payoff is at least the weighted payoff (Finset.exists_choice_sum_le).
For the non-strict, unnormalised finite Markov bound, see
Finset.mul_card_filter_le_sum in Complexity/Foundations/Data/Finset/Markov.lean.
Adapted from Average.lean in
anthropics/formal-math@16a5ce41304fcdaea4e0be7c132e1c361dc26da5.
Main results #
Finset.exists_forall_mul_card_le_mul_sum: simultaneous finite average bound.Finset.exists_choice_sum_le: deterministic vertex-wise maximum choice beats any vertex-wise sub-probability weighting.
theorem Finset.exists_choice_sum_le {I : Type u_1} {V : Type u_2} {α : Type u_3} {R : Type u_4} [Fintype I] [Fintype α] [Nonempty α] [CommSemiring R] [LinearOrder R] [IsStrictOrderedRing R] (v : I → V) (W : I → α → R) (hW : ∀ (i : I) (a : α), 0 ≤ W i a) (p : V → α → R) (hp0 : ∀ (u : V) (a : α), 0 ≤ p u a) (hp1 : ∀ (u : V), ∑ a : α, p u a ≤ 1) :
Deterministic vertex-choice lemma: given a vertex map v : I → V, nonnegative payoffs
W : I → α → R, and vertex-wise sub-probability weights p : V → α → R (0 ≤ p u a and
∑ a, p u a ≤ 1), choosing the best label at each vertex produces a deterministic assignment
f : V → α whose payoff is at least the weighted payoff under p.