Past exam of the mathematics course of the University of Cambridge 2010 ii Paper 2 27I c Solution Created 2026-09-24 Updated 2026-10-14
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 recursionwhere conditional on , is Poisson with mean , and the descendant busy periods are independent copies of . 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,For , this identity in particular holds in a sufficiently small neighbourhood of zero.