I've been trying to solve the following exercise and I was hoping for your input.
If $Q$ is a queueing process with arrival rate $\lambda$ and service rate $\mu$, and a customer arrives to find exactly $k$ customers waiting ahead (including the person being served), show that this customer leaves the queueing system after a length of time which has the gamma distribution with parameters $k+1$ and $\mu$.
source: Probability: An Introduction - Grimmet & Welsh
If we call $Z$ the random variable that is equal to the time that $k+1$ customers have been dealt with, we know that $Z = \sum_{i=1}^{k+1}X_i$ with $X_i$ being the service time for customer $i$. We know that $X_i$ for all $i$ is distributed exponentially with parameter $\mu$. That is: $\forall i\in \{ 1,2,\ldots ,k+1\}: X_i \sim Exp(\mu)$. Thus I assume that we need to find that a sum of $b$ exponentially distributed random variables, all with parameter $a$ is Gamma distributed with parameters $a$ and $b$ respectively.
How could I show this in a neat manner? I'd assume using convolution integrals combined with induction, but I feel as if there should be an easier/smarter way of showing this fact.
Thanks for your time
K. Kamal
I would suggest the easiest way is with moment generating functions. These are often useful when dealing with sums of iid random variables.
First, note that $\Gamma(1,\mu) = \mathcal E(\mu)$, ie an exponential with rate $\mu$. Now our claim is that $\Gamma(\ell,\mu) + \Gamma(k,\mu) = \Gamma(\ell+k,\mu)$, in distribution, where the two random variables on the left-hand side are independent -- write these two as $X_\ell$ and $X_k$. Using the pdf for the $\Gamma$ distribution, one can calculate that $$ E(e^{\theta X_k}) = (1 - \theta/\mu)^{-k} \quad\text{for}\quad \theta < \mu. $$ Hence we see that $$ E(e^{\theta(X_k + X_\ell)}) = E(e^{\theta X_k}) E(e^{\theta X_\ell}) = (1 - \theta/\mu)^{-(k+\ell)}, $$ which is the mgf of $\Gamma(k+\ell,\mu)$.
A simply application of induction now implies that $\sum_{i=1}^n \mathcal E_i \sim \Gamma(n,\mu)$ if $\mathcal E_i \sim \mathcal E(\mu)$.