claimroutineinferred_dep

Wedge square identity bounds center discrepancy

Claim·§3.1 The probability laws and the entropy comparison·inside lemPrefix history predicts test discrepancy

Statement (paper)

For each j∈[K]j\in[K], E(p,r)∼π⟨Uj(p),Uj(r)⟩=Eq∼κ∥Ep∼ω(⋅∣q)Uj(p)∥2≥0\E_{(p,r)\sim\pi}\ip{U_j(p)}{U_j(r)}=\E_{q\sim\kappa}\left\|\E_{p\sim\omega(\cdot\mid q)}U_j(p)\right\|^2\ge0. Consequently, by Cauchy--Schwarz first over qq and then over the tests and step, the average absolute discrepancy is at most B′(8B2D2K)1/2≤4BB′(D/K)1/4B'\left(8B^2\sqrt{\frac{D}{2K}}\right)^{1/2}\le4BB'(D/K)^{1/4}.

In plain English

For each step j∈[K]j\in[K], the averaged inner product E(p,r)∼π⟨Uj(p),Uj(r)⟩\E_{(p,r)\sim\pi}\ip{U_j(p)}{U_j(r)} equals Eq∼κ∥Ep∼ω(⋅∣q)Uj(p)∥2\E_{q\sim\kappa}\|\E_{p\sim\omega(\cdot\mid q)}U_j(p)\|^2, which is nonnegative. Hence, by Cauchy–Schwarz over qq and then over tests and step, the average absolute discrepancy is at most B′(8B2D/(2K))1/2≤4BB′(D/K)1/4B'(8B^2\sqrt{D/(2K)})^{1/2}\le 4BB'(D/K)^{1/4}.

Why it's here

This turns the small averaged cross inner products from the previous claim into a bound on the discrepancy itself. Because pp and rr are conditionally independent given qq under π\pi, the cross term is a squared norm of a conditional mean, so it is a nonnegative quantity that dominates the square of the discrepancy after Cauchy–Schwarz. This gives the (D/K)1/4(D/K)^{1/4} rate in the parent lemma.
Classified routine:
Standard step: conditional independence of p and r given q turns the cross term into the squared norm of a conditional mean. Two Cauchy–Schwarz applications then convert the averaged inner-product bound from claim-pred-device-2 into the (D/K)^{1/4} discrepancy rate.

Sources

PaperOpen
Section
§3.1 The probability laws and the entropy comparison
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sections/02-prediction.tex:154
Leannot aligned

Phase 2 aligns this node to a Lean declaration.

Friction

  • info
    inferred_dep
    outline: "Wedge square identity bounds center discrepancy" rests on claim-pred-device-2