Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 36 / 3 / b / Solution

Let $\delta =C-\mu >0$, and first suppose $\sigma >0$. For the infinite-horizon contraction, use the usual sublinear path space for fluid queues,We interpret the supplied path-space large deviation principle in this norm, or in a topology with the same workload-continuity property. The paper does not specify its norm explicitly; mere convergence on compact time intervals would not justify an infinite-horizon contraction.

Define the queue workload functionalIt is finite and nonnegative on the sublinear path space for fluid queues. Here is the required continuity argument. Fix $f$ and choose $T\ge 1$ so that $|\sigma f(t)|\le \delta t/4$ for $t\ge T$. If $\Vert g-f{\Vert }_{\mathrm{sl}}<\delta /(4\sigma )$, then for $t\ge T$,The corresponding expression for $f$ is also negative there, whereas at zero both are zero. Each supremum can therefore be restricted to $[0,T]$, givingThus $R$ is a continuous map; this proof also shows why the negative drift is essential.

Substituting $t=Ns$ into the queue workload givesThe contraction principle for large deviations says that a continuous image of a family with good rate function $I$ has the same large-deviation speed and good rate function obtained by minimizing $I$ over each fibre. ConsequentlyIn particular $J(0)=0$: $Z/\sqrt{L}\to 0$ in the path norm almost surely, so the given large deviation principle and lower semicontinuity give $I(0)=0$, and $R(0)=0$. If $\sigma =0$, the queue workload is identically zero and the rate function is zero at zero and infinite elsewhere. No unspecified variance normalization of the fractional Brownian motion is needed for this variational answer.