claimroutine

Noise factorizes into smoothed function payoff

Noisy payoff equals payoff of smoothed T1−λFT_{1-\lambda}F on clean inputs
reduces to the smoothed function for invariance
Claim·§A.2 The finite ordering transfer·inside lemLow-influence functions cannot beat opt_q(J)

Statement (paper)

With H=T1−λFH=T_{1-\lambda}F, independence of the noise between positions and multilinearity of P0mlP_0^{\mathrm{ml}} give Ec,Z,Z~P0ml(F(Z~1),…,F(Z~k))=EcEZ∼μc⊗RP0ml(H(Z1),…,H(Zk)).\mathbb E_{c,Z,\widetilde Z}P_0^{\mathrm{ml}}\bigl(F(\widetilde Z_1),\ldots,F(\widetilde Z_k)\bigr)=\mathbb E_c\mathbb E_{Z\sim\mu_c^{\otimes R}}P_0^{\mathrm{ml}}\bigl(H(Z_1),\ldots,H(Z_k)\bigr).

In plain English

Writing H=T1−λFH=T_{1-\lambda}F, the expected payoff P0mlP_0^{\mathrm{ml}} of the noisy outputs F(Z~1),…,F(Z~k)F(\widetilde Z_1),\ldots,F(\widetilde Z_k) equals the expected payoff of the smoothed outputs H(Z1),…,H(Zk)H(Z_1),\ldots,H(Z_k) evaluated on the unperturbed strings Z∼μc⊗RZ\sim\mu_c^{\otimes R}, averaged over cc.

Why it's here

Each Z~i\widetilde Z_i is resampled independently of the others given ZiZ_i, and P0mlP_0^{\mathrm{ml}} is multilinear in its kk arguments. So the expectation over noise can be taken separately in each slot, replacing F(Z~i)F(\widetilde Z_i) by its conditional mean H(Zi)H(Z_i). This turns the problem into one about the smoothed function HH, whose influences are controlled by hypothesis, which is what the later invariance step needs.
Classified routine:
Standard step in Raghavendra-style dictatorship-test analyses: conditional independence of the per-coordinate noise plus multilinearity lets the noise expectation pass inside each slot, replacing F by its Bonami-Beckner smoothing T_{1-λ}F.

Uses

Nothing else in the paper

Used by

Sources

PaperOpen
Section
§A.2 The finite ordering transfer
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appendices/finite-consequences.tex:545
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