Lesson 1

A call center receives a certain number of calls from noon to 1 PM. How many operators should be available during this hour to handle these calls? Of course, it is impossible to answer such an abstract question, so we will refine it to construct a well-defined mathematical model. Suppose observations show that, on average, \(\lambda\) calls are received during this hour. Then we can consider a discrete model: divide our time interval (one hour) into \(n\) parts and assume that in each such interval, a call arrives with probability \(p=\frac{\lambda}{n}\) (thus, we assume that in each interval, a call arrives independently of what happened in other intervals, with the same probability). We obtain a binomial distribution with parameters \(n,p\). Let the random variable \(\alpha\) represent the number of calls received in an hour. Then \[\Pr[\alpha=k]=\binom{n}{k}p^{k} (1-p)^{n-k}\ .\] Naturally, we expect that our estimates will be more accurate as \(n\) increases, so we let \(n\) tend to infinity and evaluate our expression (where \(\lambda\) and \(k\) are considered fixed, i.e., constants). \[\begin{align*}\Pr[\alpha=k]&=\binom nk p^{k} (1-p)^{n-k}=\\&=\frac{n!}{k!(n-k)!}\left(\frac{\lambda}{n}\right)^{k}\left(1-\frac{\lambda}{n}\right)^{n-k}=\\&=\frac{\lambda^k}{k!}\cdot \frac{n\dotsb(n-k+1)}{n^k}\left(1-\frac{\lambda}{n}\right)^{-k}\left(1-\frac{\lambda}{n}\right)^{n} \ .\end{align*}\] Substituting the limiting relations as \(n \to \infty\) \[\begin{align*}\frac{n(n-1)\dotsb(n-k+1)}{n^k}&\to 1\ ,\\ \left(1-\frac{\lambda}{n}\right)^{-k}&\to 1\ ,\\ \left(1-\frac{\lambda}{n}\right)^{n}&\to e^{-\lambda}\ ,\end{align*}\] we obtain \[\Pr[\alpha=k] \to \frac{\lambda^ke^{-\lambda}}{k!}\ .\]

This is the Poisson distribution. It plays an important role in queuing theory (the call center problem is a standard example from this field).