The third-moment inequality #
Let f be a real function on the cube with rows of cost at most K at every point, and let
M ≥ 0 bound all the mean absolute increments L_j = meanAbsIncr f j. Then, for every real s,
E[((f - s)⁺)^3] · Pr[f ≤ s] ≤ (3/2) K² M
(Theorem 7.6 of the paper; thirdMoment_of_integratedAllPairs : IntegratedAllPairsInequality → ThirdMomentInequality; the unnormalised form is
Level.thirdMoment_core).
Proof: in the integrated all-pairs inequality (Proposition 7.4) keep the pairs (x, y) with
f y ≤ s < f x, where (f x - f y)^3 ≥ (f x - s)^3. Dividing the row at x by (f x - s)^3
turns the inequality into the hypothesis of the lemma that removes the weights (Lemma 5.3 of the
paper; abstract_cs in TwoSided/CauchySchwarz.lean), whose conclusion is the claim.
Two consequences of the third-moment inequality:
Level.thirdMoment_lower— the inequality on the other side,E[((s - f)⁺)^3] · Pr[f ≥ s] ≤ (3/2) K² M(the second inequality of Theorem 7.6);Level.separated_sets— iff ≥ s + θonAandf ≤ sonB, then2 θ^3 μ(A) μ(B) ≤ 3 K² M(Corollary 7.7).
The integrated all-pairs inequality multiplied by 4, so that 4^n appears in place of
4^(n-1) (for n = 0 both sides are 0).
The third-moment inequality, unnormalised:
(∑_x ((f x - s)⁺)^3) · #{y : f y ≤ s} ≤ (3/2) 4^n M K².
The third-moment inequality (Theorem 7.6 of the paper, first inequality) from the integrated all-pairs inequality (Proposition 7.4).
The third-moment inequality on the other side (Theorem 7.6 of the paper, second inequality):
E[((s - f)⁺)^3] · Pr[f ≥ s] ≤ (3/2) K² M. It is the third-moment inequality for -f and -s.
Two separated sets (Corollary 7.7 of the paper). If f has rows of cost at most K
everywhere, M ≥ 0 bounds all L_j = meanAbsIncr f j, and f ≥ s + θ on A, f ≤ s on B
with θ ≥ 0, then 2 θ^3 μ(A) μ(B) ≤ 3 K² M (μ the uniform probability on the cube).