Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4 / 26J (Applied Probability)

Let $A_{n+1}$ be the number of arrivals during the next customer's service. If $Q_n>0$, that service begins immediately and one of the waiting customers is served; if $Q_n=0$, the next arrival initiates the service, and it is not counted among arrivals during its own service. ThusConditional on a fresh service time $S=t$, $A_{n+1}$ is Poisson with mean $\lambda t$. Independence of service times and the independent-increment/strong Markov property of the Poisson arrival process make these mixed counts identically distributed and independent of the previous departure history. Hence the conditional law of $Q_{n+1}$ depends only on $Q_n$, proving the Markov property of the departure-embedded queue.

PutThe transition probabilities areThe first and second rows coincide because either an arrival to an empty queue or the single existing customer initiates the next service.

The service-period arrival count has probability generating functionwith $g(1)=1$ and $g^{\prime }(1)=\rho$. Let $\Pi (s)={\sum }_{j\ge 0}{\pi }_j{s}^j$. In stationarity the recurrence and independence giveinitially for $0<s<1$. Rearrangement gives $\Pi (s)={\pi }_0(s-1)g(s)/(s-g(s))$. Letting $s\uparrow 1$ and using $\Pi (1)=1$ shows ${\pi }_0=1-\rho$. HenceAnalytic continuation gives the identity inside the unit disc; boundary values follow by convergence of the probability series, with removable limiting values wherever the displayed quotient is indeterminate, particularly at $s=1$.

If $\mathbb{E}{S}^2<\infty$, then $g^{\prime \prime }(1)={\lambda }^2\mathbb{E}{S}^2$. Expand at $s=1$:Substitution into the previous quotient givesThus the stationary mean isOnly a finite first service moment was stated. If the second moment is infinite and $\lambda >0$, the stationary mean is infinite; the formula has that extended-value interpretation. This follows as well by writing the second-order remainder as an integral of $g^{\prime \prime }(s)$ and taking $s\uparrow 1$, where monotone convergence gives $g^{\prime \prime }(s)\uparrow {\lambda }^2\mathbb{E}{S}^2$. A finite stationary distribution need not have a finite mean.