Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 353 / 4

For the Poisson process,Differentiating at zero gives

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The scaled cumulant-generating function isBy Cramér theorem, its Legendre transform is stationary at $e^{{\lambda }_{\ast }}=x/\mu$ for $x\ge 0$. Thuswith $I(x)=+\infty$ for $x<0$ and the continuous convention $0\log 0=0$.

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Assume ${\mu }_2>{\mu }_1$, so the upper tail is governed by its boundary. The large deviation principle gives the exponentially equivalent estimatewhere now $I(x)=x\log (x/{\mu }_1)-x+{\mu }_1$. Increasing capacity to ${\mu }_3>{\mu }_2$ reduces this estimate by at least $10^6$ whenEquivalently,

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A Brownian motion has independent stationary Gaussian increments, so

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The increment law immediately givesThe difference quotient has second momentThus increments scale as $\sqrt{h}$, rather than $h$, and no finite derivative is suggested. In fact this heuristic is strengthened by the theorem on nowhere differentiability of Brownian motion.

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Let $Q=(\sigma {\sigma }^T{)}^{-1}$ and use its induced inner product. The autonomous Lagrangian isIts conserved Hamiltonian isOn an infinite-duration fluctuation path $H=0$, so $\Vert \dot{\varphi }\Vert =\Vert a\Vert$. Expanding the action then givesSince metric arclength satisfies $ds=\Vert d\varphi \Vert =\Vert \dot{\varphi }\Vert dt=\Vert a\Vert dt$,This is the geometric minimum action representation and is independent of the speed used to parametrize the path.

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For constant isotropic diffusion and detailed balance, the drift has gradient formfor a positive scalar mobility $M$ after absorbing temperature and diffusion constants into $V$. Deterministic relaxation follows $a$ downhill. The least-action escape trajectory is its time reverse,Its tangent is therefore parallel to $\nabla V$. Since the gradient is normal to every level set $V=\text{constant}$, the escape path from a local minimum crosses all equipotentials orthogonally.

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