0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $\mathcal E_i \sim \mathcal E(\mu)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.