Can an assistant learn its urgency label’s reliability from two noisy auditors, and what does a wrongly assumed auditor accuracy cost?

What it found

Extending the noisy-audit model, a wrongly assumed accuracy for a 0.8-accuracy auditor costs 31% of a 0.9 label’s gain if the auditor is trusted fully, 17% if assumed 0.9 and 6% if assumed 0.6 (about 39–47% at either extreme for a 0.7 label). With two independent auditors on every message, Dawid–Skene EM learns all three confusion matrices with none supplied: it recovers the true-parameter value exactly in the population limit and at n = 1000 is within 0.0005 of the oracle given the same number of labels, 0.0009 to 0.0128 above taking one auditor as truth. At n = 50 with two 0.7 auditors it still falls below ignoring the label in 18% of audit sets. Stylised: uniform errors, independent auditors, round-number utilities; nothing measured on people.

A figure from its own output

Paper, code and results

• README and method
• Experiments and results
• Code
• Tests

More in Ambient AI assistant

• What does Dawid–Skene do when the two auditors share their errors?What does Dawid-Skene do when the two auditors share their errors: it over-trusts them (auditor accuracy read as 0.909 for a true 0.8 at half shared) and loses part of its edge over trusting one auditor, falling to equal it when the second auditor is a copy, never below it (stylised)
• What if the two auditors only share errors on hard messages, and can a hardness-aware EM fix Dawid–Skene?What if the auditors only share errors on hard messages: Dawid-Skene's cost is much smaller than with copying (policy utility 0.4060 vs 0.4066 at 30% hard messages), reaching the naive estimator only in an extreme cell, and a hardness-aware EM recovers the classifier but cannot identify the hard share (stylised)
• Does a third auditor identify the hard-message share, and does it help plain Dawid–Skene?Does a third auditor identify the hard-message share: yes at the population level (P(hard) 0.300 from informed starts), only loosely at n=1000 (sd 0.07), and in the extreme cell plain Dawid-Skene with three auditors falls below the naive estimator (0.3912 vs 0.3992) while the hardness-aware EM does not (stylised)
• Does a likelihood-ranked multi-start make the hardness-aware EM robust to a poor start?Does a likelihood-ranked multi-start make hardness-aware Dawid-Skene robust to a poor start: at the population level yes (P(hard) 0.300, vs 0.500 and a lower likelihood from a poor single start), at finite n it matches one informed start, and where hardness is absent the hard share is not identified (stylised)
• How many starts does the likelihood-ranked multi-start need, and are grid starts better than random ones?How many starts does a likelihood-ranked multi-start need, and are grid starts better than random ones: one start is nearly enough with strong hardness or n = 1000, 4 starts reach the best optimum in most cells, more help only in the weak-hardness and null cells, and policy utility barely moves (stylised)
• Can a likelihood-ratio test tell “no hard messages” from “some”, and is the chi-square cutoff usable?Can a likelihood-ratio test tell no hard messages from some: chi-square cutoffs reject 35-72% of null data sets, a parametric bootstrap holds its level and detects h = 0.3 at n = 1000 but not h = 0.1 (stylised).

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Produced with AI assistance (Claude). Results are from the code in this repository and are simulated or stylised unless the abstract says otherwise; nothing here implies affiliation with or endorsement by any lab, researcher or company cited. MIT licensed.