# Agentic Racing Math Audit This audit checks the main mathematical claims in `agentic_racing.tex`. ## 1. Shifted Reliability and Boundary Value The paper starts with a shifted reliability curve \[ p(q,h)=G(q-h), \qquad g=G'. \] Task surplus on an interval \(H\) is \[ W(q;H)=\int_H a(h)G(q-h)dh. \] Differentiating under the integral sign gives \[ W'(q;H)=\int_H a(h)g(q-h)dh \equiv B(q;H). \] Thus the marginal value of capability is a kernel-weighted value density around the current task-horizon boundary. ## 2. Gompertz Reliability Case The Poisson reliability case is \[ p(q,h)=\exp[-\exp(h-q)]. \] This is \(G(q-h)\) with \[ G(s)=\exp[-\exp(-s)]. \] The derivative is \[ g(s)=G'(s)=\exp[-s-\exp(-s)]. \] It integrates to one: \[ \int_{-\infty}^{\infty}\exp[-s-\exp(-s)]ds=1. \] Set \(z=\exp(-s)\), so \(dz=-\exp(-s)ds\). Then the integral becomes \[ \int_\infty^0 -e^{-z}dz =\int_0^\infty e^{-z}dz=1. \] ## 3. Bounded Menus Saturate If \(H=(-\infty,\bar h]\), \(A_H=\int_H a(h)dh<\infty\), and \(g(s)\to0\) as \(s\to\infty\), then for every fixed \(h\le \bar h\), \[ g(q-h)\to0 \quad\text{as }q\to\infty. \] Since \(g\) is bounded and \(a\) is integrable, dominated convergence gives \[ B(q;H)=\int_H a(h)g(q-h)dh\to0. \] So fixed limited-horizon task menus cannot sustain positive marginal capability value indefinitely. ## 4. Moving Agentic Frontiers Need Not Saturate If \(a(h)\ge a_0>0\) for \(h\in[q-w,q+w]\), then \[ B(q;H) \ge a_0\int_{q-w}^{q+w}g(q-h)dh. \] With \(s=q-h\), this is \[ B(q;H) \ge a_0\int_{-w}^{w}g(s)ds>0. \] Thus an expanding or unbounded task-horizon frontier can sustain positive marginal value. ## 5. Scaling-Law Derivatives The paper uses \[ L(C)=L_\infty+\chi(C+\kappa)^{-\eta}. \] Set \(y=C+\kappa\) and \(b=L_\infty/\chi\). Then \[ L(C)=\chi y^{-\eta}(1+by^\eta) \] and \[ q(C)=-\log L(C) =-\log\chi+\eta\log y-\log(1+by^\eta). \] Differentiating: \[ q'(C)=\frac{\eta}{y}-\frac{b\eta y^{\eta-1}}{1+by^\eta} =\frac{\eta}{y(1+by^\eta)}. \] Differentiating again: \[ q''(C)= -\frac{\eta[1+(1+\eta)by^\eta]}{y^2(1+by^\eta)^2}<0. \] A physical training project \(\varepsilon\) gives \[ \Delta q=q(C+e\varepsilon)-q(C) =eq'(C)\varepsilon+O(\varepsilon^2). \] ## 6. Temporary Lead Values A successful frontier step has task-surplus value \[ S(q,\Delta)=W(q+\Delta)-W(q). \] If the lead duration \(\tau\) is exponential with rate \(\mu\), then \[ \mathbb E\left[\int_0^\tau e^{-rt}dt\right] = \int_0^\infty e^{-rt}\Pr(\tau>t)dt = \int_0^\infty e^{-(r+\mu)t}dt =\frac{1}{r+\mu}. \] Therefore \[ L(q,\Delta)=\frac{\pi S(q,\Delta)}{r+\mu}, \qquad D(q,\Delta)=\frac{\beta S(q,\Delta)}{r+\mu}. \] The private contest prize from shifting the lead from the rival to the lab is \[ P(q,\Delta)=L(q,\Delta)+D(q,\Delta) =\frac{\pi+\beta}{r+\mu}S(q,\Delta). \] Locally, \[ P(C,\varepsilon) \simeq \frac{\pi+\beta}{r+\mu}eq'(C)\varepsilon B(q). \] ## 7. Symmetric Contest Equilibrium Let aggregate race effort be \(X=x_i+x_j\). Completion probability is \(\Lambda(X)\), and conditional on completion lab \(i\)'s contest share is \[ \sigma_i=\frac{x_i}{X}. \] Lab \(i\)'s payoff is \[ U_i = \Lambda(X)\{\sigma_iL-(1-\sigma_i)D\} -x_iR(X). \] Rewrite the payoff as \[ U_i = \Lambda(X)\{\sigma_i(L+D)-D\} -x_iR(X). \] At a symmetric profile \(x_i=x_j=x\), \(X=2x\), \(\sigma_i=1/2\), and \[ \frac{\partial\sigma_i}{\partial x_i} = \frac{x_j}{(x_i+x_j)^2} =\frac{1}{4x} =\frac{1}{2X}. \] The first-order condition is therefore \[ \frac12\Lambda'(X)(L-D) +\frac{\Lambda(X)}{2X}(L+D) = R(X)+\frac{X}{2}R'(X). \] Substituting temporary lead values gives \[ \frac{S(q,\Delta)}{r+\mu} \left[ \frac{\pi-\beta}{2}\Lambda'(X) +\frac{\pi+\beta}{2X}\Lambda(X) \right] = R(X)+\frac{X}{2}R'(X). \] Thus the task and scaling primitives enter through \(S(q,\Delta)\), locally \(eq'(C)\varepsilon B(q)\). ## 8. Joint-Profit Benchmark The two labs' total payoff from completing the step is \(L-D\), because one lab's productive lead value is offset by the other lab's defensive loss. A joint-profit planner solves \[ \max_X\ \Lambda(X)(L-D)-XR(X). \] The interior first-order condition is \[ \Lambda'(X^J)(L-D)=R(X^J)+X^JR'(X^J). \] Evaluate the private marginal payoff at \(X^J\): \[ M^P(X^J) = \frac12\Lambda'(X^J)(L-D) +\frac{\Lambda(X^J)}{2X^J}(L+D) -R(X^J)-\frac{X^J}{2}R'(X^J). \] Using the joint condition, \[ M^P(X^J) = \frac{\Lambda(X^J)}{2X^J}(L+D)-\frac12R(X^J). \] Hence decentralized symmetric effort is locally above the joint-profit benchmark iff \[ \frac{\Lambda(X^J)}{X^J}(L+D)>R(X^J). \] With temporary lead values, this condition becomes \[ \frac{\Lambda(X^J)}{X^J} \frac{\pi+\beta}{r+\mu}S(q,\Delta) > R(X^J). \] ## 9. Social Benchmark A social planner uses total social step value \(S^W(q,\Delta)\), not only private appropriable value. If \(\Psi(X)\) is an external cost of aggregate frontier compute, the social problem is \[ \max_X\ \Lambda(X)S^W(q,\Delta)-XR(X)-\Psi(X). \] The interior condition is \[ \Lambda'(X^S)S^W(q,\Delta) = R(X^S)+X^SR'(X^S)+\Psi'(X^S). \] Therefore racing can be socially excessive or socially insufficient. Defensive displacement and unpriced compute externalities push toward excessive racing. Consumer surplus, spillovers, and low appropriability push toward insufficient racing. ## Result The revised paper's main mathematical claims reduce to: - the shifted-link boundary sufficient statistic; - bounded-menu saturation and moving-frontier nonsaturation; - bounded scaling-law derivatives and local capability steps; - exponential temporary lead values; - the symmetric contest first-order condition; - the joint-profit over-racing condition; - the social planner wedge.