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Consider a type of event that has fixed duration $\delta$. Let $\lambda$ be the rate of events starting / ending (assuming steady state). I want to derive the expected number of events observed (partially or fully) in a given period $T$. If $\delta \approx 0$, this should converge on the Poisson distribution.

diagram

Edit: sorry, I'm not just seeking the expected number of events (i.e. mean), but the expected distribution of observed event counts: $p(x|\lambda,\delta,T)$.

I thought to adapt the derivation of the Poisson distribution, but it's a bit over my head.

Bonus: can we derive a more complete expression for $\delta \sim p(\delta)$ ?

jessexknight
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  • I think you have an infinite server system where each new job immediately gets its own server and has fixed service time $\delta$. Jobs arrive at rate $\lambda$ jobs per unit time, meaning $$\lim_{t\rightarrow\infty} \frac{\mbox{Number of arrivals during $[0,t]$}}{t}=\lambda \quad (wp1)$$ Then, Little's theorem $\overline{N}=\lambda \overline{W}$ says the average number of jobs in the system is $\overline{N}=\lambda \delta$. Little's theorem does not require the arrival process to be a Poisson process. The same result holds if service times are random i.i.d. with average $\delta$. – Michael Jul 07 '23 at 19:42
  • Thanks, but I am interested in the expected number of jobs observed over the interval (including 'tail ends', complete jobs (if $\delta < t$), and 'ongoing' jobs), not the expected number of jobs at any instantaneous moment. – jessexknight Jul 07 '23 at 19:48
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    If you fix a particular interval, say, $[1.3, 2.4]$, then nothing can be said without more assumptions since a general arrival process of rate $\lambda$ may very well have (with prob 1)no arrivals until after time $2.4$, or it could have a million arrivals before time $2.4$. The "rate $\lambda$" is describing a limit as time goes to infinity, so transient behavior does not affect the limit. On the other hand, if the arrival process is Poisson of rate $\lambda$, the average number of arrivals over any interval of size $t+\delta$ is exactly $\lambda (t+\delta)$. – Michael Jul 07 '23 at 19:50
  • Sorry, I don't understand how this is different from deriving the Poisson - is it sufficient to assume steady state? I am reading this re. Little's Law in practice... – jessexknight Jul 07 '23 at 19:55
  • You need to specify your assumptions on the arrival process. Is it a Poisson arrival process? If not, what properties does it have? If you say nothing else, I will assume it is a deterministic process that has its first arrival at time 9.8 and periodically has a single arrival every $1/\lambda$ seconds thereafter. – Michael Jul 07 '23 at 19:56
  • Ah, I think I understand... yes, let's assume arrivals are Poisson-distributed with rate $\lambda$. – jessexknight Jul 07 '23 at 20:02
  • So we have a $M/D/\infty$ queue with fixed service times $\delta$. Now consider our interval $[t_0, t_0+T]$ and suppose $t_0\geq \delta$. The number of distinct jobs we observe in the queue over that interval are all the arrivals during $(t_0-\delta, t_0]$ (all such are still in the queue at time $t_0$) and all the arrivals during $(t_0, t_0+T]$. This is the number of arrivals over the interval $[t_0-\delta, t_0+T]$ and has expectation $\lambda (\delta + T)$. This answer changes slightly if $t_0<\delta$, in which case all arrivals during $[0, t_0)$ are still in the system at time $t_0$. – Michael Jul 07 '23 at 20:04
  • Thanks! I think I can confirm this result via simulation. However, I am interested in a parametric distribution $p(x | \lambda, \delta, T)$, which should then have mean $\lambda (\delta+T)$. – jessexknight Jul 07 '23 at 20:09
  • The number of Poisson arrivals during an interval of size $\delta + T$ has a distribution given by the Poisson mass function of parameter $\lambda (\delta + T)$. So if $N$ is the number then $P[N=k]=\frac{\lambda^k(\delta+T)^k}{k!}e^{-\lambda(\delta+T)}$ for $k \in {0, 1, 2, ...}$. – Michael Jul 07 '23 at 20:11

1 Answers1

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Suppose we have a Poisson arrival process of rate $\lambda$. We consider a $M/G/\infty$ queue where all arrivals get their own server and have i.i.d. service times $\{X_i\}_{i=1}^{\infty}$ with some general distribution for the (nonnegative) service times. The service times are independent of the arrival times. For simplicity let $X=X_1$ and let $F(x)=P[X\leq x]$ be the CDF of service time. Assume the queue is initially empty (at time 0).

For an interval of time $(x,y]$ (where $x,y$ are given real numbers and satisfy $0\leq x<y$), let $A(x,y]$ denote the number of arrivals during $(x,y]$. Because the arrival process is Poisson, we know $A(x,y] \sim Poisson(\lambda(y-x))$, so $E[A(x,y]]=\lambda(y-x)$ and $$P[A(x,y]=k] = \frac{\lambda^k (y-x)^k}{k!}e^{-\lambda(y-x)} \quad \forall k \in \{0, 1, 2, ...\}$$

Fix an interval $[a,a+T]$ for some constants $a\geq 0, T>0$. Define $Y$ as the number of distinct jobs we see in the queue during this interval: $$ Y = N(a)+A(a, a+T]$$ where $N(a)$ is the number in the queue at time $a$. Note that $N(a)$ is independent of $A(a,a+T]$.

  • Case 1 (deterministic service times of size $\delta$ and $a\geq \delta$): Then $N(a)=A(a-\delta, a]$ and so $Y=A(a-\delta, a+T] \sim Poisson(\mu)$ where $\mu=\lambda (\delta + T)$. In particular, $E[Y]=\lambda (\delta +T)$ and $P[Y=k]=\frac{\mu^k}{k!}e^{-\mu}$ for $k \in \{0, 1, 2, ...\}$.

  • Case 2 (general service time distribution): It can be shown that $N(a)$ is Poisson with parameter $\mu_a=\lambda\int_0^a P[X>u]du$. Then $Y \sim Poisson(\mu_a+\lambda T)$ and $E[Y]=\mu_a+\lambda T$. This reduces to the same answer of Case 1 when service times are deterministic with size $\delta$ and $a\geq \delta$.

Michael
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  • I'm now considering case 2 and I'm not sure the model $Y \sim \textrm{Pois}(\mu_a+\lambda T)$ is correct, since we can simplify $\mu_a = \lambda E[X]$ from here (for $a \rightarrow \infty$). Since $\mu_a$ only reflects the mean of $X$, I think this model implies the variation in $Y$ will not be influenced by the dispersion of $X$, which feels wrong. I vaguely expect that the Poisson stuff would be inside the integral $\int_0^a$, and then something like $Y \sim \textrm{Pois}(\lambda T) + \int_0^a ...$ – jessexknight Aug 01 '23 at 13:57
  • Well you can try a few examples to verify. The example when $X$ is deterministic of size $\delta$ means $E[X]=\delta$ and so indeed as $a\rightarrow\infty$ we eventually have $a\geq \delta$, so $\lim_{a\rightarrow\infty}\mu_a=\lambda E[X] = \lambda\delta$. – Michael Aug 01 '23 at 15:20
  • Damn. I just explored some different duration distributions $f(x)$ in simulation, and indeed as you state, the distribution of number of observed events $Y$ is identical under any $f(x)$ with the same mean $E[X]$! Thanks very much. – jessexknight Aug 02 '23 at 12:58
  • If the service time $X$ surely satisfies $0\leq X\leq b$ for some $b>0$, then using any $a\geq b$ gives $\mu_a=\lambda E[X]$ exactly (for all service time distributions with support on $[0,b]$ and with the same mean $E[X]$). Similarly, as you note, for general service distributions, if $a$ is "very large" then $\mu_a\approx \lambda E[X]$. However if $X_1\sim U[0,1]$ and $X_2\sim Exp(2)$ then $E[X_1]=E[X_2]=1/2$ but using $a=1/2$ (not "very large") gives $\mu_{a1}=\lambda(3/8)=0.375\lambda$, $\mu_{a2}=\frac{\lambda(e-1)}{2e}\approx 0.316\lambda$. So different distributions do impact things. – Michael Aug 02 '23 at 14:28
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    Right, but in my case $a$ is effectively infinite, as I assume the event process is at steady state, and we just happen to observe it during a random interval, $(a,a+T]$. Thanks again, – jessexknight Aug 02 '23 at 15:40