Help with the poisson process and jackson networks.

Question

I'm posting the following question from my notes as an image because it has a diagram within it. It's from my lecture notes.

I start to get confused when $\delta $ is choses to be much smaller than $1/\lambda$. I also don't understand when he computes the probability of there being at least one customer in the queue, and I'm pretty sure that's the crux of the problem.

Any help would really be appreciated! Thanks.

Answer 1

These notes are using the infinitesimal definition of the Poisson process, which requires the following properties of the arrival process $A:=\{A(t):t\geqslant 0\}$:

• $A$ has independent increments; if $0\leqslant t_1<t_2<\ldots<t_n<\infty$, then $A(t_2)-A(t_1),\ldots,A(t_n)-A(t_{n-1})$ are independent.
• $A$ has stationary increments; the distribution of $A(t+s)-A(s)$ is independent of $s$.
• $\mathbb P(A(\delta)=1)=\lambda\delta + o(\delta)$
• $\mathbb P(A(\delta)\geqslant 2) = o(\delta).$ Here $o(\delta)$ denotes a function $g$ that satisfies $$\lim_{\delta\to 0}\frac{g(\delta)}\delta=0. $$ In other words, $\lim_{\delta\to 0}g(\delta)=0$, and for small $\delta$, $g(\delta)$ is so small compared to $\delta$ that it is negligible.

Recall that a queueing system must have $\lambda<\mu$ in order to be stable. Equivalently, the mean service time $\frac1\mu$ is less than the mean interarrival time $\frac1\lambda$. Under the assumptions that $p\to1$ and $\mu\to\infty$ as time tends to infinity, $\frac1\mu$ becomes negligibly small compared to $\frac1\lambda$ and the probability that a customer leaves the system after completing service becomes negligibly small as well. The choice of $\delta$ such that $$\frac1{\lambda'}\gg \delta\gg \frac1\mu $$ means that $\delta$ is negligible compared to $\frac1{\lambda'}$, and also $\frac1\mu$ is negligibly small compared to $\delta$. The probability of having at least one arrival in $(t,t+\delta]$, conditioned on there being at least one customer in the system at time $t$, is essentially $1$, because the amount of time it takes a customer to complete service and return to the queue is very small in comparison to $\delta$.

However, conditioned on there being no customers in the system at time $t$, from $\delta\ll\frac1{\lambda'}$ we have that $\lambda'$ and $\delta$ are of similar magnitude. The probability of having at least one arrival in $(t,t+\delta]$ is approximately $\lambda'\delta$, since it is proportional to both the arrival rate and the length of the time interval.