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For a bufferless queue with Poisson mean input $L\lambda$, place the service margin at $C{L}^{(1+\beta )/2}$, $0<\beta <1$. It exceeds the ordinary square-root fluctuation scale but is sublinear in total input. The Poisson moderate deviation principle gives the displayed overflow exponent. This interpolates between a central-limit safety margin and a fixed per-source excess capacity.

Clip Poisson input of mean $L\lambda$ at $L\lambda +C{L}^{(1+\beta )/2}$. The Poisson moderate deviation principle and contraction principle for large deviations give the displayed output rate at the clipping scale. At smaller exponents the output retains the original rate by moderate deviations below a bufferless queue cap. At larger positive-deviation scales every overflow event is eventually impossible because the requested threshold exceeds the cap.

Write $S_L=X^{\oplus L}$. Its Poisson distribution has mean $L\lambda$, since the probability-generating functions of the independent summands multiply. DefineThe scaled cumulant-generating function isA Taylor expansion of the exponential, valid for each fixed $\theta$ because $\beta <1$, givesThe remainder is $O(L^{(\beta -1)/2})$. The limit is finite and differentiable on all of $\mathbb{R}$, hence satisfies the essential-smoothness hypotheses of the Gärtner–Ellis theorem. Its Legendre-Fenchel transform isThis proves the Poisson moderate deviation principle with large-deviation speed $L^{\beta }$ and a good rate function.

For an arbitrary open set $B$, the lower bound uses ${\inf }_B{I}_{\beta }$ and the upper bound uses ${\inf }_{\overline{B}}{I}_{\beta }$. These infima agree: every point of the closure is a limit of points of $B$ and $I_{\beta }$ is finite and continuous. Thus the logarithmic probabilities of open sets for a continuous rate function give the stated logarithmic limit for every open $B$, not just intervals. The empty-set case follows from the usual infinite-value conventions.

In this bufferless queue, queue overflow is the eventThe Poisson moderate deviation principle therefore givesThe same infimum is obtained for $x\ge C$, so replacing $Q_L>C$ by $Q_L\ge C$ gives the same logarithmic exponent. Any integer rounding of the service rate of a queue changes the scaled threshold by a quantity tending to zero.

The bufferless queue output isPut $r_L=L\lambda +B{L}^{(1+\alpha )/2}$. Since $\alpha <\beta$, we have $r_L<u_L$ for all sufficiently large $L$. Consequently the two queue overflow events are exactly equal eventually:Apply the Poisson moderate deviation principle with exponent $\alpha$ to obtainThere is also an exponential equivalence proof, which describes the whole output law at this smaller scale. For each $\epsilon >0$,The right side has strictly negative logarithmic exponent at speed $L^{\beta }$. Dividing instead by $L^{\alpha }$ multiplies that negative exponent by $L^{\beta -\alpha }\to \infty$, giving $-\infty$. Thus the centered, scaled input and output are exponentially equivalent and have the same good rate function. This is the moderate deviations below a bufferless queue cap.

For sample means of Poisson random variables of mean $\lambda >0$, the cumulant-generating function is $\lambda (e^{\theta }-1)$. Its Legendre-Fenchel transform is the displayed good rate function, with $0\log 0=0$, and is infinite for $x<0$. Near its zero, $I_P(\lambda +h)=h^2/(2\lambda )+O(h^3)$. This local quadratic explains the Poisson moderate deviation principle.