Expected wait and time since the last Poisson bus

Question

Buses pass a stop according to a Poisson process, with $10$ minutes between buses on average. You arrive at a random moment. (a) How long do you expect to wait for the next bus? (b) On average, how long ago did the previous bus leave?

Short answer

(a) $10$ minutes; (b) $10$ minutes. Exponential gaps are memoryless, so the forward wait is exponential with mean $10$ , and by time reversal so is the time back to the last bus. The gap you land in therefore averages $20$ minutes, twice the typical gap: the inspection paradox, since long gaps are more likely to be hit.

Adapted from Zhou, A Practical Guide to Quantitative Finance Interviews, sec. 4.4 Discrete and Continuous Distributions, 'Property of Poisson process', p.89; question and answer rewritten by SelfMadeQuant.com.

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