Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 26 / 3

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.

The downstream service rate of a queue now exceeds the upstream cap for all sufficiently large $L$, sinceBut the bufferless queue output is always at most $u_L$. Therefore the downstream queue overflow probability is eventually exactly zero, andThis is a deterministic cap argument, stronger than merely proving superexponentially small probability.

The input admits positive moderate deviations on every scale $L^{(1+\delta )/2}$ with $0<\delta <1$. The bufferless queue output retains exactly the same overflow events at thresholds below the cap, so clipping does not improve the exponents on the smaller $\alpha$ scale. It removes every positive burst at thresholds on the larger $\gamma$ scale.

At the cap's own scale $\beta$, apply the contraction principle for large deviations to the map $x\mapsto \min (x,C)$. The output has good rate functionAt $C$ the cost is ${\inf }_{x\ge C}{x}^2/(2\lambda )=C^2/(2\lambda )$. Thus the burstiness of queue input has been reduced in its sufficiently large positive tail, while its smaller-scale rare fluctuations persist. These are the moderate deviation scales of a clipped Poisson aggregate.