Finding the mean and variance of an infinite server queue 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?
 A: 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.
Notation
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.


*

*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..


*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.
