Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 1 / 24K (Applied Probability)

A birth-death chain on ${\mathbb{N}}_0$ has generator entries $q_{n,n+1}={\lambda }_n$, $q_{n,n-1}={\mu }_n$ for $n\ge 1$, ${\mu }_0=0$, $q_{nn}=-({\lambda }_n+{\mu }_n)$, and zero elsewhere. Assume the usual conservative nonexplosive chain. The detailed balance for a birth-death process are ${\pi }_n{\lambda }_n={\pi }_{n+1}{\mu }_{n+1}$.

If they hold, each incoming term cancels an outgoing term, giving $\pi Q=0$ and thus invariance. Conversely the invariance equation at zero gives ${\pi }_0{\lambda }_0={\pi }_1{\mu }_1$. At state $n$, write the stationarity equation as equality of the neighbouring probability currentsInduction from the zero boundary current gives detailed balance for a birth-death process for every edge. Thus invariant measures are exactly the detailed-balance measures in this standard birth–death setting. The argument is the generator characterization; a stationary probability law additionally needs finite normalization. For potentially explosive processes, formal generator balance alone would require additional boundary qualifications.

The number present is an M-M-s queue: births occur at rate $\lambda$, deaths at rate $\mu \min (n,s)$. Both rates are bounded, so the chain is nonexplosive. For $\lambda ,\mu >0$ and an integer $s\ge 1$, the detailed-balance weights with ${\pi }_0=1$ areTheir sum is finite precisely when $\lambda <s\mu$, giving positive recurrence and a stationary probability distribution after normalization.

For completeness, the birth–death recurrence criterion is ${\sum }_{n\ge 0}({\lambda }_n{\pi }_n{)}^{-1}=\infty$. It follows by solving the harmonic hitting-probability recurrence: successive differences are proportional to $({\lambda }_n{\pi }_n{)}^{-1}$, and divergence makes the probability of escaping to infinity before reaching zero vanish. The geometric tail of the weights makes this sum diverge when $\lambda \le s\mu$ and converge when $\lambda >s\mu$. Thus $\lambda =s\mu$ gives null recurrence, since the chain is recurrent but has no finite invariant normalization, and $\lambda >s\mu$ gives transience. The critical case is not positive recurrent merely because arrival and maximum service rates balance.