Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 4 / 26J (Applied Probability)

The nonzero off-diagonal rates arewith a transition involving a negative coordinate omitted. These represent arrival, the two metamorphoses, and death. The diagonal generator entry is consequently $-(\lambda +\mu (c+r+b))$, so the holding rate is $\lambda +\mu n$.

Put $\rho =\lambda /\mu$. The proposed stationary probability is the productIt sums to one. Divide the balance equation at $(c,r,b)$ by this probability. Incoming immigration from $(c-1,r,b)$ contributes $\lambda c/\rho =\mu c$; incoming first metamorphosis from $(c+1,r-1,b)$ contributes $\mu (c+1)r/(c+1)=\mu r$; the next metamorphosis contributes $\mu b$; incoming death from $(c,r,b+1)$ contributes $\mu (b+1)\rho /(b+1)=\lambda$. Their sum is precisely the outgoing rate. Missing boundary terms contribute zero, as the corresponding coordinate is zero. Thus $\pi Q=0$.

For $\lambda ,\mu >0$ the chain is irreducible: with positive probability all current insects progress and die before another arrival, and from the empty state specified finite arrivals and stage transitions can reach any state. It is nonexplosive, since only finitely many Poisson arrivals occur in a finite interval and each individual makes at most three stage/death transitions. The existence of this normalized stationary law for an irreducible nonexplosive chain implies positive recurrence and uniqueness of the stationary probability law. “Only solution” here means only probability solution; the homogeneous equation itself also admits zero and scalar multiples.

The three stage counts are independent Poisson variables of mean $\rho$ under the product law. HenceDividing the joint law by this mass givesThus the conditional stage composition is multinomial with equal stage probabilities.

The chain is positive recurrent, by irreducibility, nonexplosion, and the normalized stationary probability law established above. In particular it is neither null recurrent nor transient. The positive arrival and stage rates are understood in the population model.

Treat each arriving caterpillar as one customer in an M-G-infinity queue, with arrival rate $\lambda$ and service time equal to the sum of its three independent exponential stage lifetimes. This is an Erlang distribution of shape three and rate $\mu$, of mean $3/\mu$. Poisson arrivals independently retained according to whether their service has ended give, in stationarity, a Poisson number in service with meanThis is exactly the marginal law of $N$ found above. Each individual remains in service across both metamorphoses; counting a fresh service arrival at each stage would be incorrect.