Bus Waiting Time

Buses arrive according to a Poisson process averaging one bus every 10 minutes. You walk up to the stop at a random time, knowing nothing about when the last bus came. How long do you expect to wait for the next bus, and how long ago did the last one leave?

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#Forward wait, by memorylessness

A Poisson process has exponential gaps, here with mean $10$ minutes. From any fixed instant, the time to the next arrival is again exponential with the same mean, no matter how long it has already been since the last bus. The exponential simply forgets its past, so

\E[\text{wait for next bus}] = 10 \text{ minutes}. \tag{1}

#Backward age, by symmetry

Run the clock backwards. A Poisson process reversed in time is another Poisson process with the same rate, so the time back to the previous bus is exponential with mean $10$ as well,

\E[\text{time since last bus}] = 10 \text{ minutes}. \tag{2}

Arriving at a random instant lands you more often inside a long gap. Forward memorylessness and backward time-reversal each give an exponential wait averaging ten minutes, so the gap you hit averages twenty, twice the mean spacing.

#The inspection paradox

The gap you happen to land in therefore has expected length $\E[\text{age}] + \E[\text{wait}] = 20$ minutes, twice the $10$ -minute mean spacing. This is not a contradiction. Arriving at a uniform instant is length-biased sampling, since a gap of length $\ell$ is hit with probability proportional to $\ell$ , so you fall into long gaps far more often than short ones, and the gap you observe averages double the typical gap.