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Detecting a Shift Is Not Enough: Exact Minimax Limits of Linear Representation Repair

arXiv cs.AI / cs.LG / cs.CLby Anuar Aimoldin ()1 min read2 · 1 read
AI summary · grok-4.7Summaries by Vani AI

Exact minimax limits show linear shift repair needs far more calibration signal than detection.

The paper treats removal of a mean shift between two sources as learning one linear map under a hard distortion budget from noisy paired calibration measurements. It derives the exact finite-sample minimax risk, (d-k) times the expectation of 1/(d+2J) with J Poisson(kappa/2), and shows that projecting out the mean calibration difference attains it. Detecting the shift needs only kappa much larger than sqrt(d), while removing a fixed fraction at constant distortion needs kappa on the order of d. On paired clinical and wearable sleep EEG, the formula predicts residual device shift in new participants.

Key facts
  • Exact minimax risk is derived for budgeted linear shift repair.
  • Detection needs kappa much larger than sqrt(d); fixed-fraction repair needs kappa on the order of d.
  • Projecting out the mean calibration difference attains the bound without knowing kappa.
  • The formula predicts residual shift left by MP, SAL, and LEACE.
  • On paired sleep EEG, extra recordings per person soon stop helping.
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From the original article

1 min read at arxiv.org

A mean shift between two data sources can be easy to detect but hard to remove without substantially changing their representations. We cast its removal as a statistical decision problem: from noisy differences between paired calibration measurements in $\mathbb{R}^d$, learn one linear map, applied to both sources under a hard distortion budget, that leaves as little of the shift as possible on fresh data.

arXiv cs.AI / cs.LG / cs.CL

By Anuar Aimoldin · arxiv.org

Read the full article at arXiv cs.AI / cs.LG / cs.CL

A short excerpt; the article and its images belong to the publisher. Summary, key facts and assessment above are AI-written from the article and can be wrong.