Papers
Manuscripts of the budgeted Brownian race program.
Wavefront Pruning in Budgeted Brownian Races
Defines the budgeted Brownian race: a controller observes a cloud of Brownian paths, pays per unit time for each survivor, may irreversibly prune, and allocates a finite path-time budget to maximize the expected terminal maximum. The planned scope is the mean-field limit — the coupled obstacle/Fokker–Planck system whose discrete counterpart this repository already solves — the wavefront characterization of the optimal policy, and the relation to the finite-n pathwise-budget problem. The numerical certificates will be the committed scripts in this repository.
Posterior-Predictive Pass-at-k
The estimation side of the rollout race: pass@k is the expected-maximum payoff of a best-of-k batch, and the standard extrapolation 1 − (1 − p̂)k evaluates it at a noisy point estimate. Jensen's inequality makes that plug-in biased downward for every difficulty, and a prompt with zero observed successes is rated impossible at every k. Replacing the plug-in with the posterior expectation under a two-parameter difficulty prior, fitted once by marginal maximum likelihood, cuts held-out log loss from 3.37 to 0.48 on nine thousand released rollout prompts and brings extrapolated aggregate pass@8 within two points of truth; beta-binomial and probit-normal priors tie to three decimals.
Clay Shooting: Equilibrium Effort Solving Outstanding Problems
Two mathematicians on the same problem, three years to an attempt, $300k a year to keep one of them fed, and one Millennium problem solved in 159 problem-years. Those four numbers imply that solving one is worth about $96M to whoever is attacking it, against a posted purse of $1M. The inversion runs through a price list: a Brownian attempt buys success probability p at expected cost ζ2p/(1 − p), the identity holds at every abandonment floor, and the observed hazard then fixes p, leaving A = cN/h with the duration of an attempt cancelling out. One event does not pin that level: the same arithmetic gives $18M at one end of the hazard interval and $3.8B at the other. What survives is the surplus, which optimal sizing forces to about twice the cost of one attempt whatever the reward. So the purse is 1% of the value and 55% of the surplus, and that second share moves a single percentage point across the whole interval. A rounding error against what a Clay problem is worth, and a majority stake in what makes attacking one worthwhile.
Research Prizes with Uncertain Compute Costs
The machinery behind Clay Shooting. A one-period entry game with heterogeneous costs, explicit mixed-entry conditions and an exact harmonic potential, hence a pure equilibrium; then a continuous-time extension in which committed attempts become privately observed diffusions with irreversible abandonment and the first solution wins. Against fixed rival strategies their solution hazard acts as additional discounting, the constant-hazard driftless case is solved in closed form, and the general equilibrium is characterised as coupled stopping and forward-distribution equations — a strategic fixed point where the budgeted race has a planner's optimisation.
Working notes
What is a Millennium Prize really worth?
The implied-value argument with no calculus. What money buys when you attack a hard problem is odds, and it buys them at a flat rate, so spending and success rate together reveal the payoff someone must be chasing. Reads the Millennium record backwards to about $96M a problem, explains why that number cannot be trusted and why the 55% figure sitting underneath it can, and ends on the one quantity worth measuring.
The mean-field solver
The Poissonized, expected-budget, discrete-time formulation: the exact layer-cake value of the Poisson terminal maximum, its pivotal-value gradient, the obstacle best response, and the fully-corrective Frank–Wolfe outer loop with dual bisection. The module docstring is the current write-up of record.