Passengers arrive at a bus stop according to a Poisson process with rate $\lambda$ per hour the bus arrives empty at time $t$ and leaves immediately.
If exactly three passengers get on the bus, Find the expected waiting time of the third passenger.
I am a bit confused on finding this. Intuitively it seems it should be $E[t-T_3] = t – E[T_3]$.
Now $E[T_3] = E[E[T_3\mid N(t) = 3 ] ]$
If I understand correctly using that arrivals are uniformly distributed : $E[T_3\mid N(t) = 3 ] = \frac{3}{4}t$
So is the correct answer $t – \frac{3}{4}t = \frac{1}{4}t$?
I would think the the rate $\lambda $ would still appear in the final answer.
Because according to Basic concepts of Poisson processes
It seems $E[T_3] = \frac{3}{\lambda}$
Best Answer
Your answer is correct, so it remains to address the question of Why is the solution independent of $\lambda$?
This in turn is equivalent to asking why conditioning on $N=3$ arrivals in $[0,t]$, the arrival times are uniformly distributed and independent of $\lambda$, as this is the only fact you rely on from the theory of Poisson processes.
You can either prove this for yourself (I suspect you have a proof given that you were aware of the result in your post), but doing so from the arrival process construct of Poisson processes doesn't give much intuition.
Instead I'll walk through the non-arrival time construct of the Poisson process and argue that from this perspective there's no intuition that the distribution of the points in an interval should depend on $\lambda$.
The spatial (non-time dependent) definition of the process is roughly as follows: for any interval $(a,b)$ the number of points that occur in the interval is Poisson distributed with mean $\lambda(b-a)$, and any two disjoint intervals are independent.
Now with this in mind, suppose we consider the intervals $(0,t/2)$ and $(t/2,t)$. From the definition above both are independent, and both will have $\text{Poi}(\lambda t/2)$ points.
If we now condition that there is a single point in $(0,t)$, then is it more likely this point falls into $(0,t/2)$ or $(t/2,t)$? From the preceding paragraph, a priori there's no reason to prefer one set over the other, eg. the point is equally likely to be in either - independently of $\lambda$.
Extrapolating out: we could equally consider the case of conditioning on multiple points $n$, and breaking an interval $(0,t)$ into an arbitrary number of sub-intervals, and we'd still conclude that any given point is no more likely to land in one of the boxes than any other. So informally we've concluded that the distribution of points is independent of $\lambda$, and uniformly distributed.