 Solution

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.