The text I am reading says that the infinitesimal generator of the waiting time process in a M/G/1 is given by $$Gf(x)=\lambda \int_0^\infty [f(y+s)-f(y)]dF(s)-f'(y)1(y>0)$$ where $F$ is the distribution of the service time. Can anyone please explain how this is derived. My intuition is that the first term is due to the arrival process and the second term corresponds to the departures, but I don't know how to derive it. I am particularly interested if this can be extended to G/G/1 queue case. Any reference which discusses infinitesimal generators in the context of queues will be appreciated

  • $\begingroup$ It would be helpful to mention what text you are reading... $\endgroup$
    – Math1000
    Commented Jun 16, 2023 at 9:28
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    $\begingroup$ I need a proper, rigorous definition for the waiting time. As I see this process, the first part $\lambda\int_0^{\infty} [f(y+s)-f(y)]dF(s)$ corresponds to the infinitesimal generator of a compound Poisson process, whose intensity is $\lambda$ and whose jump distribution is given by the cumulative distribution function $F$. On the other hand, the function $f'(y)1_{y>0}$ is the infinitesimal generator of a continuous drift process. I can't say anything without a rigorous definition of W_t, which is absent in the paper. $\endgroup$ Commented Jun 19, 2023 at 11:55
  • $\begingroup$ @SarveshRavichandranIyer Waiting time is the total time the job spends in the system. So, it is the sum of the service times of the jobs that are before it in the queue plus its own service time. $\endgroup$
    – abc
    Commented Jun 19, 2023 at 15:25
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    $\begingroup$ @abc That gives me a better idea. I'd like formulas, proper mathematical formulas though. Given these, I can definitely retrieve the generator. Even if you cannot provide them, any chance you have a good textbook that has some actual formulas for stuff like this? I looked through a couple of books, but they don't give rigorous definitions, instead preferring to skip directly to the stationary distributions. Thank you for your comment. $\endgroup$ Commented Jun 19, 2023 at 15:35
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    $\begingroup$ Maybe I'll look at the references in that paper, which are Kleinrocks' book and Takacs' book. If I have some clarity, I'll let you know. I can confirm that the IG is not very difficult to derive in general. $\endgroup$ Commented Jun 19, 2023 at 15:39

1 Answer 1


As I recall, the arrival process for the M/G/1 queue is a Poisson process of rate $\lambda$, and the service time is a general random variable $S$.

After staring a great deal at the formula for the infinitesimal generator, I must admit that my strategy in solving this question went backwards : I used the generator formula to guess what $W_t$ might be like, and then ensured that the required model fit with known models.

Anyway, the infinitesimal generator of a Markov process $X_t$ is an operator $A$ (whose domain consists of "nice enough" functions $f$) such that $$ Af(x) = \lim_{t \to 0} \frac{\mathbb E_xf(X_t) - f(x)}{t}, $$ where $\mathbb E_x$ denotes the probability distribution on the paths of $X_t$ conditioned on $X_0=x$. Roughly speaking, the idea is the following : even if $X_t$ is not differentiable with respect to $t$ (we can't even define what this means if $X_t$ takes values in some weird space), we can always consider the process $f(X_t)$ for a real valued function $f$ defined on the state space of $X$ and this takes values in the real numbers. We intend to capture the instantaneous rate of change of $X$ by capturing the rates of changes of these functions of $X$ instead.

Our task is to find the infinitesimal generator of the waiting time process $W_t$ of an $M/G/1$ queue. Remember that the waiting time process takes values only in the set of non-negative real numbers.

Let us first make sure that we have the distribution of $W_t$ right. Suppose that $W_0=y \geq 0$. Let $t>0$ be arbitrary. $W_t$ is the waiting time for the queue at time point $t$. We must compare this to the waiting time at time $0$, namely $y$.

To do this, observe that after time $t$, certainly $t$ seconds have elapsed. At this point, we must make two considerations : either $y\geq t$ or $y<t$. If $y>t$ then the person who came in at time $0$ is has still not been serviced at time $t$, and if $y<t$ then the person who came in at time $0$ has been serviced at time $y-t$.

Now, between the time points $0$ and $t$, some customers would have arrived, increasing the waiting time. How many such customers have arrived? Well, it depends on the arrival process, which in this case is Poisson, but we'll use a general arrival process $G(t)$ to denote it. This ensures that we are actually studying the $\mathbf{G/G/1}$ situation for as long as possible, answering your bountied request to the best of my ability.

The number of arrivals at time $t$ is some random variable $G(t)$. Each such arrival will require $S_i$ time to be served, where $S_i$ has the same distribution as $S$. Furthermore, each customer's associated $S_i$ is independent of the others. All in all, one sees that for a $\mathit{G/G/1}$ queue,$$ W_t = \max\{y-t,0\} + \sum_{i=1}^{G(t)} S_i. $$ The second part reflects the service times of arrivals in $[0,t]$ (it's $0$ if $G(t)=0$). The first part says that if $y\geq t$, then at time $t$, there is still $y-t$ amount of time left till this person is served, elsewise $y<t$ implies that at time $t$ the person being served at time point $0$ doesn't need to be counted in the waiting time calculations : they've already been serviced.

That's how the formula for a $G/G/1$ queue looks like. Now, let's see how we can find the infinitesimal generator.

The idea is the following : clearly, for any function $f$, $$ f(W_t) - f(y) = f\left(\max\{y-t,0\} + \sum_{i=1}^{G(t)}S_i\right)-f(y) $$ by merely applying $f$ to both sides of the definition of $W_t$ and then subtracting $f(y)$ from both sides. Applying the expectation (conditioned on $W_0=y$) on both sides of this equation,$$ \mathbb E_y (f(W_t)-f(y)) = \mathbb E_y\left[f\left(\max\{y-t,0\} + \sum_{i=1}^{G(t)}S_i\right)-f(y)\right]. $$

We must obviously work with the right hand side. We do what is expected to be done in this case : condition on $G(t)$, which is a random variable that represents the number of arrivals between time points $0$ and $t$, hence is a non-negative integer. Basically, by the law of iterated expectation, $$ \mathbb E_y\left[f\left(\max\{y-t,0\} + \sum_{i=1}^{G(t)}S_i\right)-f(y)\right] \\ = \sum_{n \in \mathbb N} \mathbb P(G(t)=n) \mathbb E_y \left[f\left(\max\{y-t,0\} + \sum_{i=1}^{n}S_i\right)-f(y)\right]. $$ At this point, recalling the definition of the infinitesimal generator,$$ Af(y) = \lim_{t \to 0} \frac{\mathbb E_y f(W_t) - f(y)}{t} = \lim_{t \to 0} \frac 1t \mathbb E_y(f(W_t) - f(y))\\ = \lim_{t \to 0} \sum_{n \in \mathbb N} \frac{P(G(t)=n)}{t} \mathbb E_y \left[f\left(\max\{y-t,0\} + \sum_{i=1}^{n}S_i\right)-f(y)\right]. $$

This is a general formula for the infinitesimal generator of the waiting time process of the G/G/1 queue. We still haven't used the fact that the arrival process is Poisson : but we will do that shortly.

The first thing we do is assume that we can switch the limit and the sum. Now, this can be justified for $G$ being Poisson : but in general, I am not the most sure regarding why it can be done. To justify this for $G=N$ being Poisson, let me just proceed with the switch.

Switching the two of them, $$ \lim_{t \to 0} \sum_{n \in \mathbb N} \frac{P(N(t)=n)}{t}\mathbb E_y \left[ f\left(\max\{y-t,0\} + \sum_{i=1}^{n}S_i\right)-f(y)\right] \\ = \sum_{n \in \mathbb N} \lim_{t \to 0}\left(\frac{P(N(t)=n)}{t}\right)\mathbb E_y \left[ f\left(\max\{y-t,0\} + \sum_{i=1}^{n}S_i\right)-f(y)\right]. $$

Observe that $\mathbb P(N(t) = n) = \frac{e^{-\lambda t}(\lambda t)^n}{n!}$. In particular, $\lim_{t\to 0}\frac{\mathbb P(N(t) = 0)}{t} = +\infty$, $\lim_{t\to 0}\frac{\mathbb P(N(t) = 1)}{t} = \lambda$ and $\lim_{t\to 0}\frac{\mathbb P(N(t) = k)}{t} = 0$ for all $k>1$. (In the case of a general process $G$, you will have to study the limits $\lim_{t \to 0}\frac{\mathbb P(G(t)=n)}{t}$ and see how quickly these go to zero. Then you will understand which of the limits remain).

Using the above observation, we break the sum over $n \in \mathbb N$ into three pieces : one for $n=0$, one for $n=1$ and one for the the rest. In the first sum, we transfer the $\frac 1t$ from the $N$ term to the term containing the expectation : you'll see why. In the other terms we retain it. $$ \sum_{n \in \mathbb N} \lim_{t \to 0}\left(\frac{P(N(t)=n)}{t}\right)\mathbb E_y \left[ f\left(\max\{y-t,0\} + \sum_{i=1}^{n}S_i\right)-f(y)\right].\\= \lim_{t \to 0}\left(P(N(t)=0)\right)\mathbb E_y \left[\frac{f\left(\max\{y-t,0\}\right)-f(y)}{t}\right] \\ + \lim_{t \to 0}\left(\frac{P(N(t)=1)}{t}\right)\mathbb E_y \left[f\left(\max\{y-t,0\} + S\right)-f(y)\right] \\ + \sum_{n \in \mathbb N} \lim_{t \to 0}\left(\frac{P(N(t)=n)}{t}\right)\mathbb E_y \left[ f\left(\max\{y-t,0\} + \sum_{i=1}^{n}S_i\right)-f(y)\right]. $$

Now, the last term goes to $0$. The first term, the $P(N(t)=0)$ goes to $1$, and $$\lim_{t \to 0}\mathbb E_y \left[\frac{f\left(\max\{y-t,0\}\right)-f(y)}{t}\right] = \mathbb E_y \left[\lim_{t \to 0} \frac{f\left(\max\{y-t,0\}\right)-f(y)}{t}\right]$$ (exchange of limit and expectation justified by e.g. assuming that $f'$ is uniformly bounded). If $y=0$ then this quantity is $0$ : otherwise it's equal to $-f'(y)$ as you can easily see. Therefore, the limit of the first term is $-f'(y) 1_{y>0}$.

How about the second term? Here, $\lim_{t \to 0}\left(\frac{P(N(t)=1)}{t}\right) = \lambda$. On the other hand, $$ \mathbb E_y \left[ f\left(\max\{y-t,0\} + S\right)-f(y)\right] = \int_{0}^{\infty} (f\left(\max\{y-t,0\} + s\right) - f(y))dF(s) $$ as $t \to 0$, using the dominated convergence theorem (noting that $f$ is uniformly bounded), this converges to $$\int_{0}^{\infty} (f\left(\max\{y,0\} + s\right) - f(y))dF(s)=\int_{0}^{\infty} (f\left(y + s\right) - f(y))dF(s).$$ Thus, the first term converges to $\lambda\int_{0}^{\infty} (f\left(y + s\right) - f(y))dF(s)$.

All in all, $$ \bbox[border:2px solid red] {Af(y) = \lambda\int_{0}^{\infty} (f\left(y + s\right) - f(y))dF(s) - f'(y)1_{y>0}}, $$ as desired.

Recall that I mentioned earlier that the only things that matter earlier were the quantities $\lim_{t \to 0} \frac{G(t)=n}{t}$. For $n=0$, you can transfer the first of these to the derivative part like we did with $G=N$. For $n\geq 1$, though, (to me at least) it is not obvious that the limits will all have some pattern, so once you find those limits, then you get the infinitesimal generator for the waiting time of the G/G/1 process, because for $n \geq 1$, as $t \to 0$, $$ \mathbb E_y \left[ f\left(\max\{y-t,0\} + \sum_{i=1}^n S\right)-f(y)\right] \to \int_{0}^{\infty} (f\left(y + s\right) - f(y))dF^n(s), $$ where $F^n$ is the distribution function of the random variable $\sum_{i=1}^n S_i$ and is related to the CDF of $S$ by some convolution identity. Hence, the limits of these quantities always exist, and you must merely investigate $\lim_{t \to 0} \frac{P(G(t)=n)}{t}$ for each $n \geq 1$ to find the infinitesimal generator of the $G/G/1$ queue with arrival process $G$ and service time $S$.

  • $\begingroup$ Should the equation for the waiting time be $$ W_t = \max\{y-t,0\} + \sum_{i=1}^{G(t)+1} S_i. $$Because for example if $G(t)=0$ the amount of time a customer has to wait is $max(y-t,0)$ plus the service time of the customer that came in at time 0? $\endgroup$
    – stochs
    Commented Sep 5, 2023 at 16:06

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