# Poisson time arrival process

Assume that calls arrive at a call center according to a Poisson arrival process with a rate of λ calls per hour. For $0 <= s < t$, what is the probability of $N((0,s]) = m$, when conditioned on the event $N((0,t]) = n$ (assume $n >= m$)?

I know the formula for a Poisson distribution. And I think that $E(N((0,s])) = s\cdot λ$.

I'm not sure how to work with this information though.

Any hints are greatly appreciated!

## 2 Answers

Given the number of calls, the resulting process is equivalent to sampling $n$ independent uniform variables in the interval $(0,t]$. For each of these variables, the probability of falling inside $(0,s]$ is $p = s/t$. This gives a binomial process.

• So I would take the following: P(N((0,s]) = m, and taking p = s/t, plug it into P(N() = m) = m!/m!(n-m)! * (s/t)^m(1-s/t)^(n-m) ? – starter1011 Mar 24 '15 at 20:27

Hint: The number of calls in the interval (0,s] conditioned on number of calls in (0,t] being n follows a Binomial(n,s/t) distribution.

• Same comment from filipos' post: So I would take the following: P(N((0,s]) = m, and taking p = s/t, plug it into P(N() = m) = m!/m!(n-m)! * (s/t)^m(1-s/t)^(n-m) ? – starter1011 Mar 24 '15 at 20:29