= Solution
For the <busy-period branching equation>, during the first service of a <busy period>, every arriving customer initiates a subtree of further work. The independence and stationary increments of the <Poisson process>, together with independent services, give the distributional recursion
$$
B_k\overset d=S_k+\sum_{j=1}^{N(S_k)}B_{k,j},
$$
where conditional on $S_k=s$, $N(S_k)$ is Poisson with mean $\lambda s$, and the descendant busy periods are independent copies of $B_k$. One may explore these subtrees in order; the server's total workload is unchanged by that ordering. Consequently, wherever the <moment-generating functions> are finite,
$$
\phi_{B_k}(\theta)=\mathbb E\!\left[e^{\theta S_k}
\exp\{\lambda S_k(\phi_{B_k}(\theta)-1)\}\right]
=\boxed{\phi_{S_k}(\theta+\lambda(\phi_{B_k}(\theta)-1))}.
$$
For $\lambda<\mu$, this identity in particular holds in a sufficiently small neighbourhood of zero.
Back to article page
