Why is waiting time distributed exponentially?

I have admittedly not searched far and wide, but I haven’t seen an intuitive demonstration of why waiting time should be distributed exponentially. The following, which doesn’t rely on simulation, or knowledge about Poisson distributions, seems to work for me.

The problem is as follows: An anticipated event — such as a bus arriving at a bus stop — happens at the rate of λ. I.e. it happens λ times in one time unit. When you show up at the bus stop, what is the probability distribution of the time you have to wait for the bus?

So, we’re given that the probability that the bus will arrive in any small interval dt is λ dt. We are to write the probability that one will have to wait for precisely time t for the bus. I.e., the bus does not arrive in (0, t) and does arrive in (t, t + dt).

Dividing the time line into slivers of dt, the probability of the bus not arriving inside any one sliver is (1 - λ dt). Because there are t/dt slivers in (0, t), the probability of the bus not arriving in any of those slivers is

(1 - λ dt)t/dt
= (1 - λ dt) (1/λ dt)(λ t)
= [(1 - λ dt)1/λ dt]λ t

As dt → 0, this becomes

[e-1]λ t
= e-λ t

We already have the probability that the bus does arrive in (t, t + dt) is λ dt. So probability that one had to wait time t is the product Prob[did not arrive in (0, t)] × Prob[did arrive in (t, t + dt)], which is

e-λ t × λ dt

The pdf of waiting time simply divides by dt:

λ e-λ t

Last modified: Thursday, April 9th, 2009 7:29:35pm US/Eastern