Expected Wait for a Poisson Bus
Buses arrive at a stop according to a Poisson process, averaging one bus every 10 minutes.
(a) You walk up to the stop at the exact instant the previous bus pulls away. How long should you expect to wait for the next bus?
Now suppose instead that you show up at an arbitrary moment, unrelated to the bus schedule.
(b) What is your expected wait until the next bus?
(c) On average, how long ago did the previous bus leave?
(d) What is the expected length of the gap between buses that contains your arrival time? Reconcile your answers with the fact that the average gap between buses is 10 minutes.
Hints
- The word "Poisson" is doing all the work here -- what property of the interarrival times does it hand you for free?
- Interarrival times of a Poisson$(\lambda)$ process are exponential and memoryless; the mean of an $\text{Exp}(\lambda)$ is $1/\lambda$.
- In (a) you arrive exactly at a departure, so you wait one full fresh gap $T \sim \text{Exp}(1/10)$; compute $E[T]$.
- For (b)-(d): memorylessness gives the forward wait from any instant, and a Poisson process run backward in time is still Poisson with the same rate. The gap containing you is age plus wait, and long gaps are more likely to contain a random instant.
Worked Solution
How to Think About It: The magic word is "Poisson." For a Poisson process the gaps between arrivals are i.i.d. exponential, and the exponential is memoryless -- the process has no memory of how long it has been since the last event. In (a) you are at the stop at a departure instant, so you are simply waiting for one fresh interarrival gap. Mean gap = mean wait. When you arrive at a random moment instead, memorylessness still settles the forward wait, time reversal settles the time since the last bus, and length-biased sampling (the inspection paradox) explains why those two add up to more than one average gap.
Quick Estimate: Rate is $\lambda = 1/10$ bus per minute, so the mean gap between buses is $1/\lambda = 10$ minutes. You wait one full gap: about 10 minutes. For a random arrival the forward wait is still 10 (memorylessness), the time back to the last bus is also about 10, so the gap you land in is about 20, twice a typical gap.
Formal Solution:
Part (a): Arriving at a departure. Let $T$ be the time from the previous departure to the next arrival. In a Poisson$(\lambda)$ process, interarrival times are $T \sim \text{Exp}(\lambda)$ with $\lambda = 1/10$ per minute, so
$$E[T] = \frac{1}{\lambda} = 10 \text{ minutes.}$$
A quick simulation of exponential gaps with mean 10 confirms an average wait of $\approx 10.0$ minutes.
Part (b): Forward wait from a random arrival. Suppose $s$ minutes have passed since the last bus when you arrive. Memorylessness says $P(T > s + t \mid T > s) = P(T > t)$, so the remaining time is again $\text{Exp}(\lambda)$, whatever $s$ is. Hence $E[\text{wait}] = 1/\lambda = 10$ minutes. The tempting answer of 5 ("I land in the middle of a 10-minute gap") would be right for buses on a fixed 10-minute schedule, not for exponential gaps.
Part (c): Time since the last bus. Run the process backward in time from your arrival instant. A Poisson process reversed in time is again a Poisson process with the same rate, so the time back to the most recent bus (the age $A$) is the first arrival of the reversed process: $A \sim \text{Exp}(\lambda)$ and $E[A] = 10$ minutes. (Strictly, $A = \min(\text{Exp}(\lambda), t_0)$ if the process started $t_0$ minutes ago; for a process that has been running a long time this is 10.)
Part (d): The gap you land in (inspection paradox). The gap containing your arrival is $A + W$, with age $A$ and forward wait $W$ independent $\text{Exp}(\lambda)$. So it is $\text{Gamma}(2, \lambda)$ with mean $2/\lambda = 20$ minutes. This does not contradict the 10-minute average gap. A random instant is more likely to fall in a long gap than a short one: the chance of landing in a gap of length $x$ is proportional to $x$, so the gap you sample is length-biased, with density $$\frac{x f(x)}{E[X]} = \lambda^2 x e^{-\lambda x},$$ exactly $\text{Gamma}(2, \lambda)$. You land, on average, in the middle of that longer gap, so the age and the wait are $20/2 = 10$ each. A simulation of a long run of exponential gaps gives forward wait $\approx 10.0$, age $\approx 10.0$ and containing gap $\approx 19.9$ minutes.
Answer: (a) $10$ minutes. (b) $10$ minutes: memorylessness makes the arrival time irrelevant. (c) $10$ minutes: the time-reversed process is Poisson with the same rate. (d) $20$ minutes: the gap containing a random instant is length-biased, $\text{Gamma}(2, \lambda)$, which is twice a typical 10-minute gap and is why $10 + 10 = 20$.
Intuition
This is the cleanest illustration of memorylessness in the exponential/Poisson world: the expected time to the next event never depends on the past. In practice this is exactly why Poisson models are the default for order arrivals, trade prints, and default events -- the hazard rate is constant, so "time already waited" carries no information about "time still to wait." The classic trap is the bus-waiting (inspection) paradox: sample a random instant and you are more likely to land in a long gap, so the gap you observe averages twice the naive value. Part (a) defuses that trap because you start the clock at a departure. For a random arrival the wait and the time since the last bus are both 10, and the gap containing you is 20: knowing both stories, and how they reconcile, is what separates a careful answer from a lucky one.