Poisson-like distribution for partially or fully observed events with "duration"

Question

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.

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 $\delta \sim p(\delta)$ ?

Answer 1

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 $a\geq \delta$.