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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.

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.