# Can I think of both arrival times and service times in a Markov chain as Poisson processes?

According to the Wikipedia article about M/M/1 queues, the rate at which new jobs arrive is a Poisson process with parameter $\lambda$, and the rate at which the jobs are finished is an exponential distribution process with mean service time $\mu$.

To quote directly:

Arrivals occur at rate λ according to a Poisson process and move the process from state i to i + 1. Service times have an exponential distribution with parameter 1/μ in the M/M/1 queue, where μ is the mean service rate.

This seems strange to me. Why not just be consistent and say they are both Poisson processes? After all, exponential distribution is essentially the same thing as a Poisson distribution. Why confuse people and call them by different variations of the same thing?

• From what I remember, a poisson process captures the counting process. So think of jobs arriving as a counting process. An exponential distribution describes the time in between this counting process. – Tyler Hilton Mar 8 '15 at 20:46
• @TylerHilton That would be strange. You would be measuring the rates by different units. From $0$ to $1$ you've got a counting rate, and from $1$ to $0$ you've got an mount of time. How could that work – CodyBugstein Mar 8 '15 at 20:51
• I see your point and I've forgotten most of it so hopefully an answer comes along. But customers entering the system at some rate does not imply they are exiting at the same rate. The exit rate depends on the service node inside the system. But service time is exponentially distributed. And the parameter for this distribution is different than the parameter of the poisson process. – Tyler Hilton Mar 8 '15 at 21:00
• Well ya, I understand that the values for $\lambda$ and $\mu$ are different but under the assumption that they are measuring the same thing. Otherwise it's equivalent of saying the rate to enter State 1 is $5 \ seconds$ and the rate you leave State 1 is $2 \ meters$ – CodyBugstein Mar 8 '15 at 21:19
• No, the "units" are the same. So, you are under the assumption that they are measuring the same thing. They are not. One is measuring how often a customer enters the system - described by the Poisson process. After they enter, they have to go through a service node. This node is a "separate" process as in you spend a certain time at this node - this time is exponentially distributed. Why do you assume that the rate customers entering is the same rate they are leaving at? There is no such assumption. Think of a mcdonalds. Customers enter at a certain rate, but the time spent at the counters... – Tyler Hilton Mar 8 '15 at 21:37

No. Customers arrive according to a Poisson process with rate $\lambda$. It is true that when the server is busy the service rate is $\mu$. But $\lambda < \mu$ if we have steady state. The server is idle some of the time.