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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.

For nonnegative independent and identically distributed random variables with unbounded support, this is the displayed two-summand tail asymptotic. The probability that the maximum exceeds $x$ is $1-F(x{)}^2\sim 2\overline{F}(x)$, so the sum and maximum have asymptotically the same tail. This is the basic single-large-arrival effect relevant to queue overflow. The concentration-theory sub-exponential random variable uses a separate light-tail convention.