Calls arrive according to Poisson process with parameter $$. Lengths of the calls are iid with cdf $F_x(x)$. What is the probability distribution of the number of calls in progress at any given time?
I am confused, is the answer then just the pdf of Poisson, that is, $P(X=x) = e^{-\lambda}\frac{\lambda^x}{x!}$?
I feel like I am missing something. I assumed the $$ is the number of the calls and I am not sure how to use this with the cdf of the lengths of the calls.
On
Here is my approach, without queuing theory.
Let $T>0$ be any given time, and take $\Big\{[t_{i-1},t_i),s_i\Big\}_{i=1}^n$ as any uniform tagged partition of $[0,T)$ with $\Delta t=\frac{t_i-t_{i-1}}{n}$.
When $n$ is very large, the number of customers that call the call center on $[t_{i-1},t_i)$ may be regarded as a splitting process$-$ those who remain on the line at time $T$ and those who hang up by time $T$. The former distribution is approximately $\text{Poisson}\left(\lambda \Delta t\left(1-F_X(T-s_i)\right)\right)$.
So, the total number of calls in progress at time $T$ is approximately $\text{Poisson}\left(\sum_{j=1}^n\lambda \Delta t\left(1-F_X(T-s_i)\right)\right)$
Taking $n$ to $\infty$ yields a desired distribution of $$\text{Poisson}\left(\int_0^T\lambda \left(1-F_X(T-t)\right)\mathrm{d}t\right)$$
You have an m/g/infinity queue. The distribution of the number of customers in that queue is known to be Poisson
See for instance Kleinrock book