I am presented with the following homework problem:

Let $X(t)$, $t > 0$, be the infinite server queue and suppose that initially there are $x$ customers present. Compute the mean and variance of $X(t)$.

The solution for the mean is $$E[X(T)]=\frac{\lambda}{\mu}\left(1-e^{-\mu t}\right)+xe^{-\mu t}$$ and the variance is $$\operatorname{Var}(X(T))=\left(\frac{\lambda}{\mu}+xe^{-\mu t}\right)\left(1-e^{-\mu t}\right).$$ However, we just started learning about this a few days ago and I'm totally lost on where to even begin. Could someone provide a proof of this?

  • $\begingroup$ what are $\lambda, \mu$? $\endgroup$ – gunes Apr 9 at 19:21
  • $\begingroup$ @gunes 𝜆 is the arrival rate while 𝜇 is the service rate. Since this is a birth-death process, you can also think of them as the birth and death rates, respectively. $\endgroup$ – Jin Yu Li Apr 9 at 19:35
  • $\begingroup$ Would it be fair to assume you are asking about the $M/M/\infty$ queue as defined at en.wikipedia.org/wiki/M/M/%E2%88%9E_queue? $\endgroup$ – whuber Apr 9 at 19:48
  • $\begingroup$ @whuber yes, it seems like that's the queue in question. $\endgroup$ – Jin Yu Li Apr 9 at 20:18

The question concerns a queue of infinite capacity in which people are served as soon they arrive, the so-called $M/M/\infty$ queue. All servers work independently.

We are interested in keeping track of how many people are in the queue at any time $t.$ Let this be $X(t),$ a random variable that varies with $t.$ In order to appreciate what's going on, it will be important to simplify the notation, so let's begin there.


The parameter $\lambda$ is the rate at which people arrive in the queue, according to a homogeneous Poisson process. This implies that $\lambda\cdot(t-s)$ people are expected to arrive in any time interval from $s$ to $t.$ More generally, they might arrive at rates varying with time, $\lambda=\lambda(t),$ in which case the number of people arriving in the interval $[s,t]$ has a Poisson distribution with parameter $\int_s^t \lambda(u)\mathrm{d}u.$

Let's temporarily ignore the parameter $\mu$ in order to focus on what it represents: namely, how long it takes to serve people after they arrive. Quite generally, suppose that anybody who arrives at time $s$ will be served by time $t\ge s$ with probability $F(t,s)$ and therefore will still be in the queue at time $t$ with probability $S(t,s)=1-F(t,s).$ This is the survival function for service beginning at time $s.$

There are two kinds of people in this queue: the customers who are initially present and those who arrive later. We might as well let $S(t,0)$ be the common survival function for all those initially present.

We're ready to analyze the queue.

  1. An individual who was initially present is still in the queue at time $t$ with probability $S(t,0).$ Because everyone is served independently, the number of these individuals $Z(t)$ still in the queue therefore has a Binomial$(x, S(t,0))$ distribution..

  2. Any individual arriving between nearby times $s$ and $s+\mathrm{d}s$ (which occurs with probability $\lambda(s)\mathrm{d}s$) has a chance $S(t,s)\lambda(s)\mathrm{d}s$ of still being in the queue at time $t.$ This makes the number of people in the queue at time $t$ (say $Y(t)$) equivalent to the number in the interval $[0,t]$ under the operation of an inhomogeneous Poisson process with rate $$\Lambda(t)=\int_0^t S(t,s)\lambda(s)\mathrm{d}s.\tag{1}$$ $Y(t),$ as previously noted, has a Poisson distribution. (The thread on distribution of successes of a poisson process followed by a binomial distribution presents a similar analysis of this "thinning" operation.)

We have derived the entire distribution of the number of people at any time $t,$ not just its moments: it is the sum of the Binomial distribution with parameters $(n, p(t))=(x, S(t,0))$ and the Poisson distribution with parameter $\Lambda(t).$

Solution details

The rest is just algebra (and an easy integration): the mean and variance of the Binomial distribution are $np(t)$ and $np(t)(1-p(t)),$ respectively, while the mean and variance of the Poisson distribution are both $\Lambda(t).$ Taking expectations gives

$$E[X(t)] = E[Y(t)+Z(t)] = E[Y(t)] + E[Z(t)] = \Lambda(t) + np(t)\tag{2}$$

and the independence of the servers implies the variances add,

$$\operatorname{Var}(X(t)) = \operatorname{Var}(Y(t)) + \operatorname{Var}(Z(t)) = \Lambda(t) + np(t)(1-p(t)).\tag{3}$$

In the queue in question, the survival function is assumed to be exponential with rate $\mu:$

$$S(s,t) = \exp(-\mu(t-s)).$$

Plugging this into formula $(1)$ and recalling $\lambda$ is constant gives

$$\Lambda(t)=\int_0^t \exp(-\mu(t-s))\lambda\,\mathrm{d}s = \lambda\, \frac{1-\exp(-\mu t)}{\mu}$$

(provided $\mu \ne 0;$ when $\mu=0$ this reduces to $\lambda t$).

Insert this result along with $p(t) = S(t,0) = \exp(-\mu t)$ into $(2)$ and $(3)$ to obtain the answers.

  • $\begingroup$ Thanks for the detailed explanation, whuber. This stochastic processes course is so different from other stats and math courses that everyone in the class is struggling with it, especially since the topic isn't very intuitive to begin with. $\endgroup$ – Jin Yu Li Apr 10 at 16:51

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