Continuity of the stationary distribution of $M/G/1$ queue w.r.t. the input rate

Question

Let $(\lambda_n)_{n\geq0}$ be a sequence of positive numbers such that $\lambda_n\rightarrow \lambda$ as $n\rightarrow +\infty$. These $\lambda_n$ are the parameters of a sequence of Poisson Processes $N(\lambda_n)$. Let $(S_i)_{i\geq0}$ be a sequence of reasonably smooth, positive i.i.d. random variables (say they have finite first and second moment).

To these objects we can associate a sequence of $M/G/1$ queues $(Q_n(t))_{n\geq0}$ by defining the arrival process as $N(\lambda_n)$ and the service times as $(S_i)_{i\geq0}$. If $\rho_n := \lambda_n/\mathbb E [ S_1] < 1$, their stationary distributions exists and are defined as $(Q_n)_{n\geq0}$. Assume also that $\rho := \lambda /\mathbb E[S_1] <1$.

My question is: is it true that $Q_n \stackrel{\text{d}}{\rightarrow} Q$, where $Q$ is the stationary distribution of the $M/G/1$ queue with input rate $\lambda$ and service times $(S_i)_{i\geq0}$?

Answer 1

Yes. To be precise about this you might have to specify more about what space you are working on, etc. For example, the number of customers in the queue is not Markov for a general service time distribution, so you have to be careful what you mean by "stationary distribution". You could look at the total amount of work in the queue, or alternatively at the number of customers in the queue along with the remaining service time of the customer currently in service.

In any case, by coupling the arrival processes for different arrival rates, and using a coupling-from-the-past construction (i.e. Loynes construction) to define a stationary evolution of the queue, you can obtain convergence in distribution (indeed, convergence in total variation). Couple the Poisson arrival processes so that the arrival points in $N(\lambda_i)$ are included in those of $N(\lambda_j)$ for whenever $\lambda_i < \lambda_j$. Couple the service time processes so that a customer arriving at the same time in different systems has the same service time.

Now consider the state of the queue at time 0. Fix some $\delta$ with $\lambda+\delta<\mathbb{E}[S_1]$. Let $\epsilon>0$. For $T$ large enough, with probability at least $1-\epsilon$ the queue with arrival rate $\lambda+\delta$ is empty at some point in $[-T, 0]$. From the coupling, all systems with arrival rate $\lambda_n <\lambda+\delta$ will also be empty whenever the $(\lambda+\delta)$-system is empty. Now, as $n\to\infty$ so that $\lambda_n\to\lambda$, the probability that the arrival process at rate $\lambda_n$ is identical to the arrival process at rate $\lambda$ on the interval $[-T,0]$ tends to 1. In particular, for $n$ large enough this even happens with probability at least $1-\epsilon$.

But if two systems are empty at the same point in $[-T,0]$, and their arrival processes coincide on $[-T,0]$, then they are in the same state at time $0$. So we have that for $n$ large enough, the time-0 state in the $\lambda$ system and the time-$0$ state in the $\lambda_n$-system are the same with probability at least $1-2\epsilon$.

So indeed the total variation distance between the time-0 state in the $\lambda_n$ system and the time-0 state in the $\lambda$ system tends to 0. But everything is stationary, so the time-0 state is just a sample from the stationary distribution.