Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 3 / 26I (Applied Probability)

Assume the first-fork choices are independent Bernoulli routings, as in Poisson thinning. Put $r=2\lambda /3$. Cars assigned to A form a Poisson process of rate $\lambda /3$, also an ordinary renewal process with independent exponential holding times of rate $\lambda /3$. The stream to the second fork is an independent Poisson process of rate $r$.

B receives its odd-numbered arrivals. Its initial delay is exponential of rate $r$, followed by independent Erlang distributions of shape two and rate $r$ between successive B arrivals. Thus B is a delayed renewal process, not Poisson. C receives the even-numbered arrivals: its first and all subsequent holding times are independent Erlang$(2,r)$, so C is an ordinary renewal process, not Poisson. Their holding-time density and survivor are $r^{2}t{e}^{-rt}$ and $e^{-rt}(1+rt)$. Both have asymptotic rate $r/2=\lambda /3$; equality of these mean rates with A does not make the processes Poisson.

For a renewal holding time $T$ with mean $m$, the equilibrium residual-life distribution has survivor $m^{-1}{\int }_t^{\infty }\mathbb{P}(T>s)ds$. For C, and also the common long-run B/C renewal structure, $m=2/r$ and the Erlang survivor giveFor A the exponential law is already its own equilibrium delay.

Let $K$ count second-fork arrivals. Its independent increment over $(s,s+t]$ is Poisson$(rt)$, andobtained by evaluating $\mathbb{E}(-1{)}^{K(s)}=e^{-2rs}$. If $K(s)$ is even, the next second-fork arrival goes to B, so no B arrival requires zero new arrivals. If it is odd, the next goes to C, so up to one new arrival is allowed. Reverse these alternatives for C. ThereforeAt $s=0$ these reduce to the distinct initial-delay survivors; as $s\to \infty$, both non-Poisson probabilities approach the equilibrium-delay survivor from part (b).