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Use the large-deviation speed $a_N=N^{1+\epsilon }$ from part (a). For every $\eta >0$, the Chernoff bound for a normal distribution gives, for all sufficiently large $N$,Consequentlybecause the exponent has order $N^{1+2\epsilon }$, exceeding $a_N$ by a factor $N^{\epsilon }$. Thus $(A+B_N)/a_N$ and $A/a_N$ satisfy exponential equivalence. This conclusion does not require independence between $A$ and $B_N$, since their difference is exactly $B_N/a_N$.

The exponential equivalence theorem transfers the large deviation principle and its good rate function, givingLarge positive excursions are cheapest through the single exponential random variable; a negative excursion would require the Gaussian random variable sum to move on its faster scale. If $\sigma =0$, the deterministic shift $N\mu /a_N\to 0$ gives the same conclusion directly.

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.

Let $\kappa =2(1-H)$ and fix $a>0$. Starting with part (b), the contraction principle for large deviations under the map $q\mapsto q/a$ gives an LDP for $r(X)/(aN)$ with speed $N^{\kappa }$ and rate function $q\mapsto J(aq)$.

Alternatively, replace $N$ by $aN$ in parts (a) and (b), using the real-scale justification in part (a). The same family has large-deviation speed $(aN{)}^{\kappa }$ and rate function $J(q)$. When expressed at speed $N^{\kappa }$, that rate function is $a^{\kappa }J(q)$. The two descriptions of the same large deviation principle must agree:For clarity, uniqueness here is a general property of lower semicontinuous rate functions: at a point $x$, the shrinking-neighborhood logarithmic bounds recover $I(x)$, since ${\inf }_{B(x,\eta )}I\to I(x)$ as $\eta \downarrow 0$. Thus two rate functions for the same family and speed cannot differ.

Taking the base argument $q=1$ and replacing $a$ by any positive queue size givesFor $q<0$ the rate function is infinite. In the usual nondegenerate case $J(1)$ is finite, so the formula also holds at $q=0$. This is the self-similar Gaussian workload rate; the exponent comes from the Hurst parameter, while the constant comes from the path rate function and the drift and noise scales.

For a fixed finite number of nonnegative terms and $a_L\to \infty$, use ${\max }_j{p}_{j,L}\le {\sum }_j{p}_{j,L}\le r{\max }_j{p}_{j,L}$. The logarithm of $r$, divided by the large-deviation speed, tends to zero, and a finite maximum commutes with the limit superior. If all individual logarithmic limits exist, the limit of the sum is their maximum.

For independent and identically distributed exponential random variables of rate $\lambda >0$, the sample mean has large-deviation speed $n$ and the displayed good rate function, infinite for $x\le 0$. The cumulant-generating function is $\log [\lambda /(\lambda -\theta )]$ for $\theta <\lambda$. Its Legendre-Fenchel transform has maximizing parameter $\theta =\lambda -1/x$. For $x\le 0$, sending $\theta \to -\infty$ gives infinite cost. The cost has its unique zero at $1/\lambda$ and diverges at both ends of its positive domain.

A process $Z$ is self-similar with exponent $H$ if $Z(a\cdot )$ and $a^{H}Z$ have the same finite-dimensional distributions for every $a>0$, or the same path law when the process is considered as a random path. Fractional Brownian motion has this property with its Hurst exponent. It converts dilation of time to deterministic scaling of amplitude: $Z(N\cdot )/N\stackrel{d}{=}{N}^{H-1}Z$. Consequently a small-noise large deviation principle with inverse-variance speed produces the large-deviation speed $N^{2(1-H)}$ for those rescaled paths.

Let $S$ be a fixed sum of independent exponential random variables with smallest rate $\lambda >0$, and let $Z_N$ be the sum of $N-k$ independent rate-$r$ exponential random variables, independent of $S$, with fixed $k$ and $r>\lambda$. At large-deviation speed $N$, $S/N$ has rate function $\lambda y$ for $y\ge 0$; the sample mean has rate function $K_r(z)=rz-1-\log (rz)$ for $z>0$. The product large-deviation principle and contraction principle for large deviations giveIndeed the derivative of the quantity minimized with respect to $z$ is $r-\lambda -1/z$. Its minimizer is $\min (x,(r-\lambda {)}^{-1})$. Beyond that transition, a single slow exponential random variable carries the extra excursion; the rate function becomes affine.