Should a verifier's continuous drift report be scored by CRPS or log score?

What it found

For Gaussian reports N(m,s²) both rules have exact expected scores. Overconfidence costs at most σ(√2−1)/√π = 0.234σ under CRPS but is unbounded under log (4995 nats at s=0.01σ); mean errors cost linearly versus quadratically; CRPS is 1-Lipschitz in the realised outcome (log slope 75/σ in our example); a plug-in fit from n samples costs 5σ/(8√πn) versus 1/n; and per-round score variance is finite for Student-t tails ν>2 (CRPS) versus ν>4 (log), which matters for sequential tests. Gaussian family, one verifier. Stylised.

A figure from its own output

Paper, code and results

• README and method
• White paper
• Experiments and results
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More in Proper scoring & elicitation

• Can a verifier be paid for reporting drift variance?Eliciting drift variance from two independent re-executions: D=(y1−y2)²/2 unbiased with Var=σ⁴(κ+1)/2, exact Brier and scale-free misreport costs (1/λ+ln λ−1), detection needs ≈2z²(κ+1)/(λ−1)² tasks, disjoint pairs 1.8× worse than the sample variance, heavy tails break m^{−1/2} concentration.
• What can be elicited from several independent draws of benign drift?Eliciting drift *variance* from pairs of independent draws: exact pair score (excess loss (r−σ²)²), degree table (mean² 2, third moment 3, σ⁴ 4), audit cost ((z_α+z_β)/ρ)²(κ+1)/2 pairs, robust Gini score beats it for heavy tails, all-pairs pooling saves (κ+1)/(κ−1), correlated draws bias the target to σ²(1−ρ).
• Can the log score be made payable?Bounded-payment log score by quadratic tangent extension: strictly proper, log-identical on [ε,1−ε], worst-case payment exactly ln((1−ε)/ε)+1/(1−ε), 63× Brier's curvature per range at p=10⁻³, crossover p(1−p)=1/(2R), best cutoff ε=p.
• Which proper score should verifiers face with K outcomes?K-ary proper scores: exact worst-case regrets (1−2p_min+‖p‖² Brier, ‖p‖−p_min spherical), spherical beats Brier by exactly √K per payment range at a uniform truth, and both bounded scores nearly ignore a dropped rare class (1e-4 vs 0.15 for log).
• Can the multiclass log score be made payable?Bounded multiclass log score by quadratic tangent extension of x ln x: log-identical when all r_j≥ε, payment range exactly 1−ln ε for any K, 63× Brier's rare-class incentive per range at ρ=10⁻³ with best cutoff ε=p_min, and a closed-form bounded cost for dropping a rare class.
• What does it cost to pay for a ranking instead of probabilities?Eliciting class rankings with linear rank-weight scores: strictly proper, adjacent-swap regret exactly (p_i−p_j)(w_r−w_{r+1}), beats Brier-derived rankings per unit range below gap 1/(K−1) (250× at gap 10⁻³), equal spacing is optimal, top-k set costs the same as the mode, detection sample size unchanged.

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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.