Waiting for the Bus: Memorylessness Forward and Backward
Buses arrive at a stop according to a Poisson process with an average time of 10 minutes between consecutive buses. You show up at the stop at an arbitrary moment, unrelated to the bus schedule.
(a) What is the expected time you will wait until the next bus arrives?
(b) On average, how long ago did the previous bus leave the stop?
Reconcile your two answers with the fact that the average gap between buses is 10 minutes.
Hints
- Interarrival times of a Poisson process are exponential. What special property does the exponential distribution have regarding the time already elapsed?
- The same memorylessness argument works looking backward in time: a Poisson process reversed in time is still a Poisson process with the same rate.
- Both answers are 10 minutes. The gap you land in is not a typical gap: long gaps are more likely to contain your arrival time (length-biased sampling), and its expected length is $20$ minutes, not $10$.
Worked Solution
How to Think About It: The intuitive answer "5 minutes each, since I land in the middle of a 10-minute gap on average" fails because interarrival times are random and you are more likely to arrive during a long gap than a short one. Memorylessness settles the forward wait; time-reversal settles the backward one; and the inspection paradox explains why the sum is 20.
Quick Estimate: If buses came exactly every 10 minutes, your wait would be uniform on $[0, 10]$ with mean 5. Random (exponential) gaps are worse: you tend to land in long gaps, and the mean wait doubles to 10.
Formal Solution:
Part (a): Expected wait for the next bus
*Step 1 -- Exponential gaps.* Interarrival times of a Poisson process with rate $\lambda = 1/10$ per minute are i.i.d. $\text{Exp}(\lambda)$ with mean $1/\lambda = 10$ minutes.
*Step 2 -- Memorylessness.* For an exponential $T$, $P(T > s + t \mid T > s) = P(T > t)$. Conditional on the time $s$ already elapsed since the last bus, the remaining time until the next bus is still $\text{Exp}(\lambda)$. So regardless of $s$,
$$E[\text{wait}] = \frac{1}{\lambda} = 10 \text{ minutes}.$$
Part (b): Expected time since the last bus
*Step 1 -- Time reversal.* A Poisson process viewed backward in time from a fixed instant is again a Poisson process with the same rate: the time back to the most recent bus is the first arrival of the reversed process, hence $\text{Exp}(\lambda)$.
*Step 2 -- Conclude.* $E[\text{time since last bus}] = 1/\lambda = 10$ minutes. (Strictly, it is $\min(\text{Exp}(\lambda), t_0)$ if the process started $t_0$ ago, but for a process that has been running a long time this is 10 minutes.)
Reconciliation: the inspection paradox
The gap that contains your arrival time is $A + W$ where $A$ is the age (time since last bus) and $W$ the wait, both $\text{Exp}(\lambda)$ and independent, so it has a $\text{Gamma}(2, \lambda)$ distribution with mean $2/\lambda = 20$ minutes. This does not contradict the 10-minute average gap: a random point in time is more likely to fall into 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. Its density is $x f(x)/E[X] = \lambda^2 x e^{-\lambda x}$, exactly $\text{Gamma}(2,\lambda)$.
Answer: (a) Expected wait $= 10$ minutes. (b) The last bus left, on average, $10$ minutes ago. The gap you happen to land in averages 20 minutes because of length-biased (inspection-paradox) sampling, not 10.
Intuition
Because exponential waiting times are memoryless, the bus does not "know" how long it has been since the last one, so your expected wait is the full 10 minutes no matter when you arrive; and since a Poisson process looks the same run backward, the time since the last bus is also 10 minutes on average. The apparent contradiction ($10 + 10 = 20$, not 10) is the inspection paradox: by arriving at a random time you are more likely to land inside a long gap, so the gap containing you averages 20 minutes. This length-biased sampling shows up everywhere in trading data, from the average duration of the trade interval you happen to observe to the size of the bucket a random order lands in.