# Poisson distribution puzzle

I asked this on math.stackexchange but have yet to get an answer, so will try here too…

In their monograph “Queues “, Cox and Smith state (paraphrased – this is p5):

In interval (t,t+Δt) the probability of no arrivals in a completely random process is 1−αΔt+o(Δt), for one arrival αΔt+o(Δt) and for more than one arrival o(Δt) where α is the mean rate of arrival.

I cannot follow this… here is my thinking – we take N to be the probability of no arrivals, W to be the probability of one arrival, Z to be the probability of more than one arrival, and A to be the probability of any arrivals.

So 1=N+W+Z=N+A

By my understanding of Cox and Smith:

1=1−αΔt+o(Δt)+αΔt+o(Δt)+o(Δt) =1+3o(Δt) which is surely nonsense.

So, what have I got wrong here?

## 5 thoughts on “Poisson distribution puzzle”

1. Paul Rubin says:

I’m going to use d rather than \delta because I am (blissfully) ignorant of the keycode for a delta.

You’re error lies in assuming that all o(dt) thingies are the same, which they are not (or at least not necessarily). The notation o(dt) just means some term that dies (converges to zero) faster than dt can get to zero. The term can be positive, negative or both (meaning it switches sign as dt changes). So, as an example, 1 = (1 – dt + dt^2) + (dt – dt^3) + (dt^3 – dt^2) = [1 – dt + o(dt)] + [dt + o(dt)] + [o(dt)] whre the first o(dt) is dt^2, the second is -dt^3, the third is dt^3 – dt^2, and they all converge to 0 faster than dt does.

O() and o() notation is in a way the mathematical version of an ellipsis (“I’m leaving this out because it is too scary/boring/lengthy to contemplate”), with an additional restriction that the thing being ignored is unimportant because it vanishes when you start taking limits.

1. Adrian McMenamin says:

Thanks, as always. Incidentally, you can use LaTeX in the comments too, I think – $\delta$ – I cheated in the OP by cutting and pasting from stackechange.

1. dan mackinlay (@howthebodyworks) says:

In an amusing twist I’ve just answered this on stack exchange in ignorance of the lively discourse over here on the blog. I hoe you’ll excuse me not copying and pasting over here!

2. Adrian McMenamin says:

No problem