= Solution
The <bufferless queue output> is
$$
Y^L=\min(S_L,u_L),\qquad u_L=L\lambda+C L^{(1+\beta)/2}.
$$
Put $r_L=L\lambda+B L^{(1+\alpha)/2}$. Since $\alpha<\beta$, we have $r_L<u_L$ for all sufficiently large $L$. Consequently the two <queue overflow> events are exactly equal eventually:
$$
\{Y^L>r_L\}=\{S_L>r_L\}.
$$
Apply the <Poisson moderate deviation principle> with exponent $\alpha$ to obtain
$$
\boxed{\lim_L\frac1{L^\alpha}\log\mathbb P(Y^L>r_L)=-\frac{B^2}{2\lambda}.}
$$
There is also an <exponential equivalence> proof, which describes the whole output law at this smaller scale. For each $\varepsilon>0$,
$$
\mathbb P\!\left(\frac{|S_L-Y^L|}{L^{(1+\alpha)/2}}>\varepsilon\right)
\le\mathbb P(S_L>u_L).
$$
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>.
Back to article page
